Category: 1st PUC

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2018 (North), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2018 (North)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the imaginary part of the complex number -3i + 7
Answer:
-3.

Question 2.
If (2x + 4, 0) = (8,0) find x.
Answer:
2x + 4 = 8
2x = 4
x = 2

Question 3.
Simplify: \(\left(\frac{9}{4}\right)^{-3 / 2}\)
Answer:
\(\left[\left(\frac{3}{2}\right)^{2}\right]^{-3 / 2}=\left(\frac{3}{2}\right)^{-3}=\left(\frac{2}{3}\right)^{3}=\frac{8}{27}\)

Question 4.
Express 3⁴ = 81 in logarithmic form.
Answer:
log₃81 = 4

Question 5.
Find the tenth term of HP \(\frac{1}{2}, \frac{1}{4}, \frac{1}{6}\)………..
Answer:
AP → 2, 4, 6,…. T₁₀ = 2 + (n – 1)2 = 2 + 2n – 2 = 2n = 2(10) = 20
∴ The tenth term of HP = \(\frac{1}{20}\)

Question 6.
Form the cubic equation whose roots are 3, 5, 7.
Answer:
(x – 3)(x – 5) (x – 7) = 0
(x – 3)(x² – 5x – 7x + 35) = 0
(x – 3)(x² -12x + 35) = 0
x³ – 12x² + 35x – 3x² + 36x – 105 = 0
x³ – 15x² + 71x – 105 = 0

Question 7.
What is the simple interest for 245 days for ₹ 6000 at 8% p.a simple interest?
Answer:
SI = 6000 × \(\frac{245}{365} \times \frac{8}{100}\) = 402.39

Question 8.
What is the present value of a perpetuity of 5000 at 8% p.a
Answer:
A = \(\frac{A}{r}=\frac{5000}{0.08}\) = 62,500

Question 9.
If the cost price of an item is 25000 and the profit is 10,000 find the profit percentage.
Answer:
CP = 25,000 profit = ₹ 10,000
Profit% = \(\frac{\text { Profit }}{\mathrm{CP}}\) × 100 = 40%

Question 10.
Express 144° ¡n radian measure.
Answer:
\(\frac{144 \times \pi}{180}=\frac{4 \pi}{5}\)

Question 11.
Find the value of cos 1500
Answer:
cos150 = cos(180 – 30) = -cos 30 = \(\frac{\sqrt{3}}{2}\)

Question 12.
Find k if the slope of the line joining the points (2,5) and (k, 13) is 20.
Answer:
\(\frac{13-5}{k-2}\) = 20 ⇒ \(\frac{8}{k-2}\) = 20 ⇒ 8 = 20k – 40 ⇒ 20k = 48 ⇒ k = \(\frac{48}{20}=\frac{12}{5}\)

Part-B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.
Find the sum of all positive divisors of 768.
Answer:

768 = 2⁸ × 3¹
= p₁^α₁ × p₂^α₂
p₁ = 2 p₂ = 3
α₁ = 8, α₂ = 1

Question 14.
Find the least integer which when divided by 36,40 and 48 leaves the same remainder 5.
Answer:
LCMof 36, 40, 48

LCM = 2⁴ × 3² × 5¹ = 16 × 9 × 5 = 720
∴ Required member = 720 + 5 = 725

Question 15.
Find the domain and range of relation R = {(1, 2) (2, 5) (1, 6) (3, 6)}
Answer:
Domain = {1, 2,3)
Range = {2, 5, 6}

Question 16.
If a^x = b^y = c^z and b² = ac show that \(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)
Answer:
Let a^x = b^y = c^z = k (say)
∴ a^x = k ⇒ a = k^1/x, b^y = k ⇒ b = k^1/y, c^z = k ⇒ c = k^1/z
Now, b² = ac
∴ (k^1/y)² = k^1/x. k^1/z
∴ k^2/y = k^1/x+1/z
Bases are the same ∴ Equating powers on both sides, we get.
\(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)

Question 17.
If the second term of the GP is 6 and fifth term is 162 then find the GP.
Answer:
T₂ = 6, T₅ = ar⁴ = 162 ar = 6 ar⁴ = 162
ar.r³ = 162
6r³ = 162
r³ = \(\frac{162}{6}\) = 27
r = 3

Question 18.
The sum of three consecutive numbers is 243. Find the three numbers.
Answer:
x + x + 1 + x + 2 = 243
3x + 3 = 2433x = 240
x = 80
80, 81, 82

Question 19.
If a and P are the roots of the equation 2x² + 4x + 1 = 0 find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)
Answer:
α + β = –\(\frac{4}{2}\) = -2
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}\)
αβ = \(\frac{c}{a}=\frac{1}{2}=\frac{-2}{\frac{1}{2}}\) = -4

Question 20.
Solve 7x + 3 < 5x + 9, x ∈ R and represent on the number line.
Answer:

Question 21.
Find the future value of an annuity of ₹ 5000 at 12% p.a for 6 years.
Answer:
A = Rs.5,000 r = 0.12 n = 6
F = \(\frac{A\left[(1+r)^{n}-1\right]}{r}=\frac{5000\left[(1+0.12)^{6}-1\right]}{0.12}=\frac{4869.11}{0.12}\) = 40,575.9

Question 22.
Prove that (1 + cotA)² + (1 – cotA)² = 2 cosec²A
Answer:

Question 23.
Find the value of cos 60° -sin 30° – cot³ 45°
Answer:
\(\frac{1}{2}-\frac{1}{2}\) – (1)³ = -1

Question 24.
Find the distance of the centroid of the triangle formed by the points (7, 1) (1, 5) and (1, 6) from origin.
Answer:
G = \(\left[\frac{7+1+1}{3}, \frac{1+5+6}{3}\right]\) = (3, 4)
O = (0, 0)
OG = \(\sqrt{(3)^{2}+(4)^{2}}=\sqrt{9+16}=\sqrt{25}\) = 5

Question 25.
Find the equation of the locus of the point which moves such that it is equidistant from (4,2) and the X – axis.
Answer:
P = (x, y) distance from x – axis = y A = (4,2)
PA = y
\(\sqrt{(x-4)^{2}+(y-2)^{2}}\) =y
(x-4)² + (y-2)² = y²
x² – 8x + 16 – 4y + 4 = 0

Part-C

III. Answer any TEN questions. (10 × 3 = 30)

Question 26.
Prove that \(\sqrt{2}\) is an irrational number.
Answer:
We shall prove it by the method ofcontradiction.
If possible Let \(\sqrt{2}\) be a rational number
Let \(\sqrt{2}\) = \(\frac{p}{q}\) where p and q are Integers and q ≠ 0.
Further let p and q are coprime i.e. H.C.F. of p and q = 1.
\(\sqrt{2}\) = \(\frac{p}{q}\) ⇒ \(\sqrt{2}\)q = p
⇒ 2q² = p²
⇒ 2 divides p² ⇒ 2 divides p
⇒ p is even
Let p = 2k where k is an integer p² = 4k²
2q² = 4k²
q² = 2k² ⇒ q² is even
⇒ q is even.
Now p is even and q is even which implies p and q have a common factor 2. which is a
contradiction of the fact that p and q are co-prime.
∴ our assumption that \(\sqrt{2}\) is rational is wrong and hence \(\sqrt{2}\) is irrational.

Question 27.
A relation R is defined on the set of integers by
R = {(X, Y): X-Y is a multiple of 5} Show that R is an equivalence relation on Z.
Answer:
x – x = 0 is a multiple of 5
∴ (x, x) ∈ R ∀ x
R is reflexive

Let (x, y) ∈ R
x – y is a multiple of 5
y – x is a multiple of 5 (y, x) ∈ R
∴ R is symmetric

Let (x, y) ∈ R
(y, z) ∈ R
x -y is a multiple of 5 and y – z is a multiple of 5
(x – y) +(y – z) is a multiple of 5
x-z is a multiple of 5 (x, z) ∈ R

∴ R is transitive
R is an equivalence relative

Question 28.
Solve : 3^2x – 10.3^x + 9 = 0
Answer:
Let 3^x = a ⇒(3^x)² = a²
⇒ 3^2x = a²
∴ a² – 10a + 9 = 0
a² – 9a – a + 9 = 0
a(a – 9) -1 (a – 9) = 0
(a – 9)(a – 1) = 0
a – = 0 a – 9 = 0
a = 1 a = 9
3^x = 3⁰ 3^x = 3²
x = 0 x = 2

Question 29.
If x² + y² = 12xy, show that 2 log (x – y) = log2 + log 5 + log x + log y.
Answer:
x² + y² = 12xy
x² + y² – 2zy = 12xy – 2xy ⇒ (x – y)2 =10xy
log(x-y)² = log10xy
2 log (x – y) = log 10 + log x + log y
2 log (x – y) = log (2 x 5)+log x + log y
2 log (x – y) = log 2 + log 5 + log x + logy

Question 30.
Insert 4 Arithmetic means between 14 and 34.
Answer:
14, a₁, a₂, a₃, 34
a = 14
T =34
n = 5
T_n = a + (n-1)d
3 = 14 + (4 )d
20 = 4 d
d = 5
14, 19, 24, 29, 34

Question 31.
A number consists of 2 digits whose sum is 4, if 18 is added to the number, the digits get interchanged. Find the number.
Answer:
Let the digit in the tens place be x and the digit in the unit place be y
The nunter is 10x + y
Given x + y = 4 …………(1)
Also given 10x + y + 18 = 10y + x
10x – x + y – 10y + 18 = 0
9x – 9y + 18 = 0
x – y + 2 = 0
x – y = -2……………..(2)

x + y = 4
y = 4 – x = 4 – 1 = 3
The numbcr is 10x + y
10(1) + 3 = 10 + 3 = 13
The number is 13.

Question 32.
Solve graphically: 2x + y ≥ 8 ; and x + y ≥ 110
Answer:

Question 33.
Preitha bought a car for ₹ 4,00,000. If it depreciates at the rate of 12% per year, how much will it be worth after 10 years?
Answer:
C = Rs.4,00,000 r = 12% = 0.12 n = 10
B = C(1 – r)ⁿ
= 4,00,000 (1 – 0.12)¹⁰ = 4,00,000 (0.88)10
B = Rs. 1,11,400.39

Question 34.
The average age of 10 students in a class increases by 4.8 months, when a boy of 6 years is replaced by a new boy. What is the age of the new boy?
Answer:
Let the average age of 10 students x
∴ Sum of ages of 10 students = 10x
A boy of 6 years is replaced by a new boy whose age is y
∴ Total age of 10 students 10x + y – 6
∴ Average age of 10 students = \(\frac{10 x+y-6}{10}\)
∴ \(\frac{10 x+y-6}{10}\) = x + \(\frac{4.8}{12}\) 4.8 months = \(\frac{4.8}{12}\) years
x by 10
∴ 10x + y – 6 = 10x + \(\frac{48}{12}\)

y = 6 + 4 = 10
∴ Age of the new boy = 10 years

Question 35.
In a dance competition, 70% of the participants were girls. 35% of the boys and 65% of the girls are selected for the next round. If 49 girls were eliminated, And the number of boys who got selected.
Answer:
Let the total number of participants – x
If 65% of girls got qualified then 35% of girls got eliminated
Total number of girls = \(\frac{70}{100}\)x = 0.7x
Total number of boys = \(\frac{30}{100}\) × x = 0.3x
∴ total girls eliminated = \(\frac{35}{100}\) × 0.7x = 0.245x
Hence 0.245x = 49
x = \(\frac{49}{0.245}\)= 200
Hence total number of boys eliminated = \(\frac{35}{100}\) × 0.3 × 200 = 21

Question 36.
If tan A = \(\frac{-12}{5}\) and 270% < A < 360° find the value of \(\frac{3 \sin A-2 \cos A}{9 \cos A+4 \sin A}\)
Answer:
\(\frac{3 \sin A-2 \cos A}{9 \cos A+4 \sin A}=\frac{3\left(-\frac{12}{13}\right)-2\left(\frac{5}{13}\right)}{9\left(\frac{5}{13}\right)+4\left(\frac{-12}{13}\right)}=\frac{-36-10}{45-48}=\frac{-46}{-3}=\frac{46}{3}\)

Question 37.
Find area of the triangle whose vertices are A(6, 3) B(-3, 5) C(4, -2)
Answer:
A = (6,3) B = (-3,5) C = (4, -2)
∆ABC = \(\frac{1}{2}\)[x1(y2 – y3) + x3 (y3 – y) + X3 (y1 – y2)]
= \(\frac{1}{2}\) [6 (5 + 2) – 3 (-2 – 3) + 4 (3 – 5)] = \(\frac{1}{2}\) [6(7) – 3(-5) + 4(-2)]
= \(\frac{1}{2}\) [42 + 15 – 8] = \(\frac{1}{2}\) [49] squarts

Question 38.
Find the equation of the line passing through (5,2) and cutting off intercepts that are equal in magnitude but opposite in sign.
Answer:
b = -a \(\frac{x}{a}+\frac{y}{-b}\) = 1 put x = 5, y = 2\(\frac{5}{a}-\frac{2}{a}\) = 1
3 = a, b = -3
∴ \(\frac{x}{3}+\frac{y}{-3}\) = 1 ⇒ \(\frac{x-y}{3}\) = 1 ⇒ x – y – 3 = 0

Part – D

Answer any SIX questions. (6 × 5 = 30)

Question 39.
In a class of 250 students if 75 students have taken statistics and 25 have taken both statistics and economics and 100 have taken neither statistics nor economics. Find how many have taken
(a) economics (b) economics only Also represent in Venn diagram.
Answer:

n(S∪E) = 250 – 100 = 150
n(S) = 75, n(S ∩ E) = 25
n(S ∪ E) = n(S) + n(E) – n(S ∩ E)
150 = 75 + n(E) – 25
150 = n(E)
n(E)= 100 → Economics
n(E – S) = n(E) – n[E ∩ 5) = 100 – 25 = 75 →Economics only

Question 40.
Evaluate using log \(\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\)
Answer:
Let x = \(\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\)
log x = log[latex]\frac{0.5634 \times 0.0635}{2.563 \times 12.5}[/latex]
= log 0.5634 + log 0.0635 – log 2.563 – log 12.5
=1.7508 + 2.8028-0.4087-1.0969
= 2.5536- 1.5056 = 2 + 0.5536-1.5056
= -2 – 0.952 = -2.952
= -3 + 3 -2.952 = – 3 + 0.048 = 3.048

Question 41.
Find the sum of all integers between 200 and 500 which are divisible by 7.
Answer:
S_n = 203 + 210 + 217 +………….+ 497 a = 203, d = 7, T = 497 , n = ?
T_n = a + (n – 1)d
497 = 203 + (n- 1)7
497 = 203 + 7n-7
497= 196 +7n
7n = 301
n = 43
Sn = \(\frac{n}{2}\)(a + T_n) = \(\frac{43}{2}\)(203 + 497) = 15,050

Question 42.
Find an integral root between -3 and 3 by inspection and they using synthetic division solve the equation x³ – 10x² + 29x – 20 = 0
Answer:
Put x = 1
1 – 10(1)² + 29(1) – 20
1 – 10 + 29 – 20
30 – 30 = 0

Quotient = x² – 9x + 20
Remainder = 0
x³ – 10x² + 29x – 20 = 0
(x + 1)(x² – 9x + 20) = 0
x – 1, x² – 9x + 200
x = 4, x = 5

Question 43.
A sum of money amounts to Rs. 19500 in 5 years and 22,200 after 8 years at the same rate of simple interest. Find the principal and rate of interest.
Answer:
P = ?, A = Rs.19,500, T = 5years
A = Rs.22200, T = 8years R = ?
A = P + SI
A = P + \(\frac{\mathrm{PTR}}{100}\)
19,500= P(1 + \(\frac{5 \mathrm{R}}{100}\)) ………(1)
22,220= P(i + \(\frac{8 \mathrm{R}}{100}\)) ………………(2)

(100 + 5R)222 = (100 + 8R) 195,
22200 + 1110R = 19500 + 1560 R
22200 – 19500 = 1560 R – 1110
R = \(\frac{2700}{450}\) = 6%p.a. ∴ R = 6%p.a.
From (1)
19,500 = P(1 + \(\frac{5 \mathrm{R}}{100}\))
19500 = P(1 + \(\frac{5 \cdot 6}{100}\)) = P(1 + \(\frac{30}{100}\)) = P(1 + \(\frac{3}{10}\)) = P(\(\frac{10+3}{10}\)) = \(\frac{13}{10}\)P

P = \(\frac{19500 \times 10}{13}\) = 15,000
∴ P = Rs. 15,000

Question 44.
Find the present value of an annuity of 2000 payable at the beginning of each quarter for the next 3 years if the rate of interest is 4% p.a. compounded quarterly.
Answer:
A = Rs.2000 n = 3 x 4 = 12 r = = 0.01
P = \(\frac{A\left[(1+r)^{n}-1\right]}{r(1+r)^{n}}\)(1 + r) = \(\frac{2000\left[(1+0.01)^{12}-1\right]}{0.01(1+0.01)^{12}}\)(1 + 0.01)
= \(\frac{2000\left[(0.01)^{12}-1\right]}{0.01(1.01)^{12}}\)(1.01) = \(\frac{256.18}{0.011}\) = 23,289.09

Question 45.
A radio is sold at a profit of 25%. Cost price and selling price both are increased by 100. If the new profit is 20% find the original cost of the ratio.
Answer:
Let x be the C.P. of the radio.
then SP = \(\frac{125 \mathrm{x}}{100}\)
When CP changes to x + 100SP = \(\frac{125 \mathrm{x}}{100}\) + 100
i.e., x + 1.00 +20% of(x + 100) = \(\frac{125 \mathrm{x}}{100}\) + 100
x + 100 + \(\frac{20}{100}\)(x + 100) = \(\frac{125 x+10000}{100}\)
100x + 10000 + 20x + 2000 = 125x + 10000
5x = 2000
x = Rs.400
The original cost of the radio = Rs.400

Question 46.
Prove that \(\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}\) = 1 + sec A cosec A
Answer:

= secA.cosecA + 1 = RHS

Question 47.
Show that (-3 1), (-6, -7) (3, -9) (6, -1) taken in order are vertices of a parallelogram.
Answer:

Question 48.
Find the image of the point (2, 4) on the line x + y – 10 = 10
Answer:
Let P be the point (2,4) and Q(x,y) be its reflection on the line x + y – 10 = 0. Let PQ cut the line at R. Then R is the midpoint of PQ and PQ is perpendicular to the given line.
Now R = \(\left(\frac{x_{1}+2}{2}, \frac{y_{1}+4}{2}\right)\) Since R lies on x + y – 100
We have \(\left(\frac{x_{1}+2}{2}+\frac{y_{1}+4}{2}\right)\) – 10 = 0 i.e. x1 + y1 – 14 = 0 ………………(1)

Slope ofPQ = \(\frac{y_{1}-4}{x_{1}-2}\) Slope ofthegiven line x + y – 100 is -1.
Since PQ is perpendicular to the line x + y – 10 = 0 m1m2 = -1
Solving (1) and (2) we get the point of intersection as JC, = 6 and y] = 8 ∴Image = Q (6,8)

Part – E

V. Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) If A = {1, 2, 3} B = {2, 3, 4}, C = {3, 4, 5} verify (A × B) (A × C) = A × (B ∩ C)
Answer:
A × B = {(1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}
A × C = {(1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,3), (3,4), (3,5)}
(A × B)∩(A × C) = {(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)} B∩C = {3,4}
A × (B∩C) = {(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)} ∴ (A × B)∩(A × C) = A × (B∩C)

(b) The daily cost of production C for x units is given by C(x) = 17.5 x + 7000.
(i) If each unit is sold for 30, then determine the minimum units that should be produced and sold to ensure no loss.
(ii) If it is known that 500 unit can be sold daily what price / unit should be charged to guarantee no loss.
Answer:
(i) C(x) = 17.5x + 7000
R(x) = 30x, R(x) = C(x)
30x = 175x + 7000,
12.5x = 7000
x = \(\frac{7000}{12.5}\) = 560 BEP

(ii) If the S.P. is reduced by Rs.3/unit then R(x) = 27x
For BEP R(x) = C(x)
27x = 17.5x + 7000,
27x – 17.5x = 7000
9.5x = 7000,
x = \(\frac{7000}{9.5}\) = 73684 = 737 units

(c) The average score of 20 boys is 60% and average score of 30 girls is 70% Find the combined averaged score.
Answer:
x₁ = 60%
x₂ = 70%
N₁ = 20
N₂ = 30

X̄ = \(\frac{\mathrm{X}_{1} \mathrm{~N}_{1}+\mathrm{X}_{2} \mathrm{~N}_{2}}{\mathrm{~N}_{1}+\mathrm{N}_{2}}=\frac{(60)(20)+(70)(30)}{(20)+(30)}=\frac{1200+2100}{50}=\frac{3300}{50}\) = 66%

Question 50.
(a) Find the sum to n terms of the series 9 + 99 + 999 + ………………….
Answer:
S_n = 9 + 99 + 999 +…………. = (10 – 1) + (102 – 1) + (103 – 1) + ………………
= (10 + 102 + 103 +………………. ) – (1 + 1 + 1 +……………. ) = \(\frac{10\left(10^{n}-1\right)}{10-1}\) – n
S_n = \(\frac{10\left(10^{n}-1\right)}{9}\) – n
n ∈ N

(b) Find k if line x – 2y + 1 = 0; 2x – 5y + 3 = 0; 5x – 9y + k = 0 are concurrent.
Answer:

(c) If the product of two numbers ¡s 216 and their LCM ¡s 36 find their HCF.
Answer:
ab = 216 LCM = 36 HCF = ?
HCF × LCM = ab = 216
HCF = \(\frac{216}{\mathrm{LCM}}\) HCF = \(\frac{216}{36}\) = 6

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2018 (South), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2018 (South)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the conjugate of 2 – i3.
Answer:
2 + i3

Question 2.
If A = {5,6}, Find power set of A.
Answer:
P(A) ={Φ, {5}, {5,6}, {6}}

Question 3.
Simplify \(\left[\left\{\sqrt[3]{x^{2}}\right\}^{3}\right]^{\frac{1}{2}}\)
Answer:
x

Question 4.
Express log₅ 0.2 = -1 in exponential form
Answer:
5^-1 = 0.2

Question 5.
Find the 8lh term of an AP -2, -4, -6 ………….
Answer:
T₈ = -2 + 7 (-2) = -2 – 14 = -16

Question 6.
Solve for x : 2(1+x) – 10 = 16 – 2 (x – 24).
Answer:
14 + 2x- 10 = 16 – 2x + 48 ⇒ x = 15

Question 7.
Find the simple interest on ₹ 1500 at 4% p.a for 145 days.
Answer:
SI = \(\frac{\mathrm{P} t r}{100}\) = 23.84

Question 8.
Define Annuity.
Answer:
An annuity is a fixed sum of money paid at regular intervals of time under certain conditions.

Question 9.
Convert 42% to a decimal.
Answer:
\(\frac{42}{100}\) = 0.42

Question 10.
Convert \(\frac{3 \pi^{c}}{2}\) into degree measure.
Answer:
\(\frac{3 \pi}{2} \times \frac{180}{\pi}\) = 270

Question 11.
Prove that Sin 30°. Cos 60° + cos 30°. Sin 60° = 1
Answer:
\(\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}\) = 1

Question 12.
Find the slope of the line with the inclination \(\frac{\pi}{4}\) with respect to x – axis
Answer:
θ = π/4 ⇒ m = tanπ/4 = 1

Part – B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.
Find the LCM of 36, 40, and 48 by the factorization method.
Answer:
36 = 2² × 3², 40 = 2³ × 5¹, 48 = 2³ × 3¹
LCM = 2⁴ × 3² × 5¹ = 720

Question 14.
Find the number of positive divisors of 768.
Answer:

768 = 2⁸ × 3¹

Question 15.
If A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6, 7} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9} verify (AuB)’ = A’∩B’
Answer:
(A∪B)1 = U – (A∪B) = {8,9}
A’ = {6, 7, 8, 9} B’ = {1,2, 8,9}
∴ A’∩B1 = {8, 9} (A∪B)’ = A’∩B’

Question 16.
Prove that (x^b-c)^a .(x^c-a)^b (x^a-b)^c = 1
Answer:
(x^b-c)^a .(x^c-a)^b (x^a-b)^c ⇒ x^{ab-ac+bc-ca-cb} ⇒x° = 1

Question 17.
The third term of a HP is \(\frac{1}{7}\) and Fifth term is \(\frac{1}{11}\) then find the seventh term.
Answer:
3rdterm of an A.P = 7 a + 2d = 7
5th term of A.P = 11 a + 4d = 11 ∴ a = 3 d = 2
seventh term of HP = \(\frac{1}{a+(n-1) d}\) = \(\frac{1}{3+6(2)}\) = 15

Question 18.
The sum of four consecutive numbers is 366, find them.
Answer:
x + x + 1 + x + 2 + x + 3 = 366
⇒ x = 90 numbers are 90, 91, 92, 93

Question 19.
Solve x² + 3x – 28 = 0 by formula method.
Answer:
x = \(\frac{3 \pm \sqrt{9-4(1)(-28)}}{2}\) ⇒ x = -4 or x = 7

Question 20.
Solve : 5x – 3 > 3x + 1 ; xeR and represent on the number line.
Answer:
5x – 3 < 3x + 1 ⇒ 2x < 4 ⇒ x < 2

Question 21.
Find the present value of an annuity of ₹400 for 3 years at 16% p.a compound interest
Answer:
P = \(\frac{400\left[(1+0.16)^{3}-1\right]}{0.16(1+0.16)^{3}}\) = 898.35

Question 22.
The angles of a triangle are in the ratio 3:4:5. Find them in degrees.
Answer:
3A + 4A + 5A = 180 ⇒ A= 15 45°, 60°, 75°

Question 23.
Prove that Sin (480°).cos (690°) + cos (780°). Sin (1050°) = 1/2
Answer:
sin (360 + 120) = sin 120 = sin (180 – 60 = sin 60 = \(\frac{\sqrt{3}}{2}\)
cos (720 – 30) = cos 30 = \(\frac{\sqrt{3}}{2}\) cos (720 + 60) = cos 60 = \(\frac{1}{2}\)
sin (1080 – 30) = sin (-30) = -1/2 .
\((\sqrt{3} / 2)(\sqrt{3} / 2)-(1 / 2)(1 / 2)=3 / 4-1 / 4=2 / 4=1 / 2\)

Question 24.
Show that the points (1,1), (5,2) and (9,5) are collinear.
Answer:
AB = 5 BC = 5 AC =10
∴ AB + BC = AC
5 + 5 = 10

Question 25.
Find the equation of the locus of a point that moves such that the square of its distance from (2,3) is 3.
Answer:
Let p (x, y) be any point on the locus PA² = 3
(x – 2)² + (y – 3)² = 3 ⇒ x² + y² – 4x – by +10 = 0

Part – C

III. Answer any TEN questions. (10 × 3 = 30)

Question 26.
Prove that \(\sqrt{2}\) is an irrational number.
Answer:
We shall prove it by the method ofcontradiction.
If possible Let \(\sqrt{2}\) be a rational number
Let \(\sqrt{2}\) = \(\frac{p}{q}\) where p and q are Integers and q ≠ 0.
Further let p and q are coprime i.e. H.C.F. of p and q = 1.
\(\sqrt{2}\) = \(\frac{p}{q}\) ⇒ \(\sqrt{2}\)q = p
⇒ 2q² = p²
⇒ 2 divides p² ⇒ 2 divides p
⇒ p is even
Let p = 2k where k is an integer p² = 4k²
2q² = 4k²
q² = 2k² ⇒ q² is even
⇒ q is even.
Now p is even and q is even which implies p and q have a common factor 2. which is a
contradiction of the fact that p and q are co-prime.
∴ our assumption that \(\sqrt{2}\) is rational is wrong and hence \(\sqrt{2}\) is irrational.

Question 27.
Let f = {(1, 1), (2, 3), (0, -1)} be a function from z to z defined by f(x) = ax + b some integers a and b. Determine a and b.
Answer:
f(x) = ax + b
when x = 1 f(x) = 1
∴ a + b = 1
x = 0 f(x) = -1
a(0) + b – 1 ⇒ b = -1
a + b = 1
a – 1 = 1 ⇒ a = 1 + 1 =2
∴ a = 2
∴ a = 2, b = 1

Question 28.
If ax = by = cz and b2 = ac. Show that \(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)
Answer:
Let a^x = b^y = c^z = k (say)
∴ a^x = k ⇒ a = k^1/x, b^y = k ⇒ b = k^1/y, c^z = k ⇒ c = k^1/z
Now, b² = ac
∴ (k^1/y)² = k^1/x. k^1/z
∴ k^2/y = k^1/x+1/z
Bases are the same ∴ Equating powers on both sides, we get.
\(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)

Question 29.
Prove that \(\frac{1}{\log _{2} 4}+\frac{1}{\log _{8} 4}+\frac{1}{\log _{16} 4}\) = 4
Answer:
LHS = \(\frac{1}{\log _{2} 4}+\frac{1}{\log _{8} 4}+\frac{1}{\log _{16} 4}\) = log₄2 + log₄8 + log₄ 16
= log₄2.8. 16 = log₄ 256 = log₄ 4⁴ = 4log₄4 = 4(1) = 4 = RHS

Question 30.
Find the three numbers in GP whose sum is \(\frac{31}{5}\) and their product is 1.
Answer:
Let the number \(\frac{a}{r}\), a, ar
Product of extremes = 1
\(\frac{a}{r}\) × ar =1 ⇒ a² = 1 ⇒ a= 1
Sum \(\frac{13}{3} \Rightarrow \frac{a}{r}\) + a + ar = \(\frac{13}{3} \Rightarrow \frac{1}{r}\) + 1 + (1)r = \(\frac{13}{3}\)
\(\frac{1}{r}\) + r = \(\frac{13}{3}\) – 1
\(\frac{1+r^{2}}{r}=\frac{10}{3}\)
⇒ 3 + 3r² = 10r³⇒ 3r² – 10r + 30 = 3r² – 9r – r + 3O
⇒ 3r(r – 3) – 1 (r – 3) = 0
(r – 3)(3r – 1) = 0 ⇒ r = 3 or r = \(\frac{1}{3}\)
The numbers are
\(\frac{a}{r}\), a, ar
\(\frac{1}{3}\), 1, 1.3
⇒ \(\frac{1}{3}\), 1, 3

Question 31.
If a and P are the roots of the equation 2x² + 5x + 5 = 0, find the value of \(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}\)
Answer:

Question 32.
Solve the linear inequalities x + 2y ≤ 8, 2x + y ≤ 8, u ≥ 0, x ≥ 0 graphically.
Answer:

Question 33.
In how many years a sum of ₹2000 becomes ₹2205 at the rate of 5% p.a compound interest?
Answer:
A = (1 + r)ⁿ
n = \(\frac{\log A-\log p}{\log (1+r)}=\frac{3.3434-3.3010}{0.0211}\) = 2 years.
Future value = 4,575.9

Question 34.
The average weight of a group containing 25 persons is 70 kg. 5 persons with an average weight of 63kg leave the group and 4 persons with a weight of 72 kg, 78 kg, 70 kg, and 73 kg joins the group. Find the average weight of the new group.
Answer:
Average weight of 25 persons 70 kg
Total weight of 25 persons = 25 × 70 = 1750kg
Average weight of 5 persons who leave the group = 63 kg
∴ Total weight of 5 persons who leave the group 63 x 5 = 315kg

Total weight of4 persons who join the group = 72 + 78 + 70 + 73 = 293kg
∴ Total weight of the group now containing 24 persons = 1750 – 315 + 293 = 1728kg
∴ Average weight of the group now = \(\frac{1728}{24}\) = 72 kg

Question 35.
Savitha sold her bag at a loss of 7%. Had she been able to sell it at a gain of 9%, it would have fetched ₹ 64 more than it did. What was the cost price of the bag?
Answer:
Let CP = 100
loss at 7% ⇒ SP = 93 Gain at 9% ⇒ SP = 109 Difference 109 – 93 = 16
C.P of the bag = \(\frac{100 \times 64}{16}\) = ₹400

Question 36.
If sinθ = \(\frac{-8}{17}\) and π < θ < \(\frac{3 \pi}{2}\). Find the value of \(\frac{\tan \theta-\cot \theta}{\sec \theta+\operatorname{cosec} \theta}\)
Answer:
tan θ = 8/15, cot θ = 15/8
secθ = -17/5, cosec θ = 17/8
\(\frac{\tan \theta-\cot \theta}{\sec \theta-\cos e c \theta}=\frac{8 / 15-15 / 8}{-17 / 5-17 / 8}=\frac{171}{391}=\frac{7}{17}\)

Question 37.
Find the third vertex of a triangle if two of its vertices are at (-2, 4) and (7, -3) and the centroid at (3, -2).
Answer:
A (-2, 4) B = (7, -3) C = (x,, y) G = (3, -2) (3, – 2) = \(\left(\frac{-2+7+x}{3}, \frac{4-3+y}{3}\right)\)
⇒ x = 4, y = -7 ∴ C = (4,-7)

Question 38.
If the lines 2x – y = 5, Kx – y = 6 and 4x – y = 7 are concurrent, find K.
Answer:
2x – y – 5 = 0
4x – y – 7 = 0
Solving we get x = 1, y = -3
kx – y – 6 = 0 k( 1 ) – (-3) -6 = 0 ⇒ K = 3

Part – D

IV. Answer any SIX questions. (6 × 5 = 30)

Question 39.
Out of 250 people, 160 drink coffee, 90 drink tea, 85 drink milk, 45 drink coffee and tea, 35 drink tea and milk, 20 drink all three, i) How many will drink coffee and milk? ii) only milk iii) only coffee. Show the result through the Venn diagram.
Answer:
n(C∪T∪M) = 250
n(C) = 160,
n(T) = 90
n(M) = 85
n(C∩T)=45
n(T∩M) = 35
T(C∩T∩M)=20
n(C∩M)=?
n(C∪T∪M) = n(c) + n(T) + n(M) – n(C∩T) – n(T∩M) – n(M∩C)+ n(C∩T∩M)
250 = 160 + 90 + 85 – 45 – 35 – n(C∩M) + 20
∴ n(C∩M) = 160 + 90 + 85 – 45 – 35 + 20 – 250
∴ n(C∩M) = 25

Question 40.
Using logarthamic tables, find the value of \(\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\)
Answer:
x = \(\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\)
logx = log 0.5634 + log 0.0635 – log 2.563
-log 12.5 =1.7508 + 2.8028-0.4087-1.0969
= -2.952 = -2-1+1-0.952 = -3 + 0.048 = 3.0.048
x = AL[3.048] = 0.001117

Question 41.
Find the sum of all numbers between 50 and 200 which are divisible by 11.
Answer:
S_n = 55 + 66 + 77 +………….+ 198
a = 55
d = 11
T = 198
S_n = \(\frac{n}{2}\)[2 a + (n-1 )d] = 7(253) = 1771

Question 42.
Find an integral root between -3 and 3 by inspection and then using synthetic division solve the equation x3 – 10x² + 29x – 20 = 0
Answer:
Put x = I
1 – 10(1)² + 29(1) – 20
1 – 10 + 29 – 20
30 – 30 = 0

Quotient x² – 9x + 20
Remainder = 0
x³ – 10x² + 29x – 200
(x + 1)(x² – 9x + 20) = 0
x = -1, x² – 9x + 20 = 0
x = 4, x = 5

Question 43.
A person borrows a certain sum of money at 3% p.a Simple interest and invests the same at 5% p.a compound interest compounded annually. After 3 years he makes a profit of 1,082. Find the amount he borrowed.
Answer:
Let the amount invested be x
r =4% p.a.
A= 1352
p = x
n = 2
A = P (1 + r)ⁿ
1352 = x(1 + 0.04)² = x (1,04)² = x (1.0816)
∴ x = \(\frac{1352}{1.0816}\) = 1,250
∴ Amount invested Principal = 1,250
P = 1250 T = 2 r = 4%
ST = \(\frac{\mathrm{P} \times \mathrm{T} \times \mathrm{R}}{100}=\frac{1250 \times 2 \times 4}{100}\) = 100
A = P + SI = 1250 + 100 = 1350
If the same amount at the same rate of simple interest then
1352 – 1350 = Rs. 2 less will be received.

Question 44.
If Poornima deposits ‘600 at the beginning of every year for the next 15 years. Then how much will be accumulated at the end of 15 years, if the interest rate is 7% p.a?
Answer:
a = 600
n = 15
i = 0.07
F =a\(\frac{\left[(1+i)^{n}-1\right]}{i}\)(1 + i) = \(\frac{600\left[(1.07)^{15}-1\right]}{0.07}\)(1 + 0.07) = ₹ 161325.428.

Question 45.
A businessman sells an article for ₹ 720 and earns a profit of 20% Find the a) cost price b) profit percentage at the selling price.
Answer:
SP = Rs.720 profit = 20%
CP = SP × \(\frac{100}{(100+\text { Profit } \%)}\) = 720 × \(\frac{100}{(100+20)}\) = 720 × \(\frac{100}{120}\) = Rs. 600
Profit = SP – CP = 720 – 600 = Rs. 120
Profit percentage at selling price = \(\frac{\text { Profit }}{\mathrm{SP}}\) × 100 = \(\frac{120}{720}\) × 100 = 16.6%

Question 46.
If x = r cos A.cos B, y = r cos A. sin B and z = r sin A. then prove that x² + y² + z² = r²
Answer:
LHS = x² + y² + z² = r²cos²Acos²B + r²cos²Asin²B + r²sin²A = r²cos²A(cos²B + sin²B) + rsin²A = r²cos²A(1) + rsin²A = r²(cos²A + sin²A) = r²(1) = r² = RHS

Question 47.
Find the coordinates of the vertices of the triangle given the mid points of the sides as (4, -1), (7, 9) and (4, 11)
Answer:
Let A = (x₁, y₁) B = (x₂, y₂) and C = (x₃, y₃)

Now D = midpoint of BC .
= (4, -1) = \(\left[\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right]\)
x₂ + x₃ = 8 …………….(1)
y₂ + y₃ = -2 …………(2)

E = mid point of CA
(7, 9) = \(\left(\frac{x_{3}+x_{1}}{2}, \frac{y_{3}+y_{1}}{2}\right)\)
x₃ + x₁ = 14 …………(3)
y₃ + y₁ = 18 ………….(4)

F = mid point of AB
(4, 11) = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)
x₁ + x₂ = 8 …………. (5)
y₁ + y₂ = 22 ………….(6)

Solving (1), (3), (5) we get x₁, x₂ and x₃
Consider (1) + (3) + (5) we get
2(x₁ + x₂ + x₃) = 30
(x₁ + x₂ + x₃) = 15
x₁ + x₂ = 8 and x₁ + x₂ + x₃ = 15 ⇒ x₃ = 7
x₂ + x₃ = 8 and x₁ + x₂ + x₃ = 15 ⇒ x₁ = 7
x₃ + x₁ = 14 and x₁ + x₂ +x₃ = 15 ⇒ x₂ = 1
Consider (2) + (4) + (6) we get
2 (y₁ + y₂ + y₃) = 38
(y₁ + y₂ + y₃) = 19

Now y₂ + y₃ = – 2 and(y₁ + y₂ + y₃) = 19 ⇒ y1 =21
y₃ + y₁ = 18 and (y₁ + y₂ + y₃) = 19 ⇒ y₂ = 1
y₁ + y₂ = 22 and (y₁ + y₂ + y₃) = 19 ⇒ y₃ = 1
Thus A = (7, 21) B = (1, 1) and C = (7, -3)

Question 48.
Find the coordinate of the foot of the perpendicular from (-6, 2) on the line 3x – 4y + 1 = 0.
Answer:

Part-E

V. Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) Find tue domain and Range of the function f(x) = \(\frac{x^{2}-2 x+1}{x^{2}-9 x+13}\) where x ∈ N
Answer:
Domain of F(x) = N{1, 2, 3…………… )
n = 1 ⇒ F(x) = \(\frac{1-2+1}{1-9+13}=\frac{0}{5}\) =0
n = 2 ⇒F(x) = \(\frac{4-4+1}{4-18+13}=\frac{1}{-1}\) = -1
n = 3 ⇒ F(x) = \(\frac{9-6+1}{9-27+13}=\frac{4}{-5}\)…………
Range of F(x) = {0, -1, \(\frac{4}{-5}\),………………}

(b) Find the distance between the parallel lines 5x + 12y + 7 0 and 5x + 12y – 19= 0
Answer:
d = \(\left|\frac{C_{1}-C_{2}}{\sqrt{a^{2}+b^{2}}}\right|=\left|\frac{26}{\sqrt{25+144}}\right|=\left|\frac{26}{\sqrt{169}}\right|=\frac{26}{13}\) = 2 units

(c) What is the present value of an income of 3000 to be received forever if the interest rate is 14% p.a.
Answer:
P_∞ = \(\frac{a}{i}=\frac{3000}{0.14}\) = 21428.5

Question 50.
(a) Find the sum of n terms of the series 7 + 77 + 777 + …… n terms.
Answer:

(b) A confectioner makes and sells biscuits. He sells one pack of biscuits at ₹ 80 His cost of manufacturings is ₹40 pack as variable cost and ₹ 3000 a fixed cost find i) Revenue function ii) Cost function iii) Profit function iv) If he limits his production to 100 packets can. he makes a profit.
Answer:
Revenue function (i) R (X) = 80.x TC = TVC + TFC
(ii) C(x) = ax + b = 40.x + 3000
(iii) P(x) = R(.x) – C(x) = 80 x – (40 x + 3000) = 80x – 40x – 3000 = 40.x – 3000
(iv) x= 100
P(x) = 40x-3000 P(100) = 4(100) -3000 = ₹1000

(c) The profit of a business firm for the 5 years are ₹17,598, ₹20,703, ₹10,085, ₹25,375, and ₹16,315. Find the average profit?
Answer:
Average profit = \(\frac{17598+20703+15085+25375+16315}{5}\) = ₹19015.20

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2019 (North), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2019 (North)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the canonical representation of 360.
Answer:
360 = 2³ × 3² × 5¹.

Question 2.
If A = {1, 2, 3, 4} and B = {2, 4, 6, 8} find A ∪ B
Answer:
A ∪ B = {1, 2, 3, 4, 6, 8}.

Question 3.
Simplify: \(\left(\frac{8}{27}\right)^{3 \times 1 / 3}\)
Answer:
\(\left(\frac{8}{27}\right)^{3 \times 1 / 3}=\left(\frac{2^{3}}{3^{3}}\right)^{1 / 3}=\frac{2}{3}\)

Question 4.
Define perpetuity.
Answer:
If annuity payments are made for an infinite period it is called a perpetuity.

Question 5.
Find the value of x, if log₇ x = 2.
Answer:
x = 7² ⇒ x = 49

Question 6.
Find the 12th term of A.P. 1, 4, 7, 10
Answer:
T_n = a + 11d = 1 +(11 × 3) = 1 +33 =34.
∴ 12th term is 34.

Question 7.
Solve \(\frac{x+2}{x-1}=\frac{5}{2}\)
Answer:
2(x + 2) = 5 (x – 1)
2x + 4 = 5x – 5
4 + 5 = 5x – 2x
3x = 9 ⇒ x = 3

Question 8.
Write the formula to find the present value of an annuity due.
Answer:
Present value of annuity due = \(\frac{\mathrm{A}\left[(1+r)^{n}-1\right](1+r)}{r(1+r)^{n}}\)

Question 9.
Convert 0.20 into a percentage.
Answer:
0.20 × 100 = 20%.

Question 10.
Convert \(\frac{7 \pi}{12}\) into degrees.
Answer:
\(\frac{7 \pi}{12}=\frac{7 \times 180}{12}\) = 105°

Question 11.
Find the value of sinA sec A.
Answer:
sinAsecA = \(\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\) = tan A

Question 12.
If the slope of the line is \(\frac{2}{3}\), find the slope of its perpendicular line.
Answer:
Slope of perpendicular line = –\(\frac{3}{2}\)

Part – B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.
Find the L.C.M. of \(\frac{1}{3}, \frac{5}{6}, \frac{5}{7}\)
Answer:
LCM of Numerators 1, 5, 2 = 10
HCF of denominators 3, 6, 9 = 3
∴ LCM of fractions = \(\frac{\text { LCM of Numerators }}{\text { HCF of denominators }}=\frac{10}{3}\)

Question 14.
Find the product of two numbers is 216 and their L.C.M. is 36. Find the H.C.F.
Answer:

Question 15.
If A = {a,b,c, d}, B = (d, e, f, g). Find A – B and B – A.
Answer:
A – B = {a, b, c} B – A = {e, f, g, h}

Question 16.
If the first term of an A.P. is 3 and the common difference is -2. Find the 11th term.
Answer:
Given a = 3, d = -2, T_n = ?
T_n =a +10d = 3 + 10(-2) = 3 – 20 = —17
∴ 11th term = – 17.

Question 17.
Evaluate: sin²\(\frac{\pi}{6}\) + cos²\(\frac{\pi}{3}\)-tan2\(\frac{\pi}{4}\) + cot²\(\frac{\pi}{4}\)
Answer:
LHS = sin² 30 . cos² 45 – cos² 90 – sin² 60.
= \(\left(\frac{1}{2}\right)^{2} \cdot\left(\frac{1}{\sqrt{2}}\right)^{2}\) – (0)2\(\left(\frac{\sqrt{3}}{2}\right)^{2}\) = \(\frac{1}{4} \cdot \frac{1}{2}=\frac{1}{8}\)

Question 18.
Solve by method of elimination :
x + 2y = 4
3x + y = 7
Answer:
x + 2y = 4 ………..(i)
3x + y = 7 …………..(ii)
Multiply Eqn. 1 by 3 we get

Question 19.
Solve: 3(2 – x) ≥ 2(1 – x), x GR .
Answer:
3(2-x) ≥ 2(1 – x), x ∈ R
6-3x ≥ 2-2x ⇒ 6-2 = 3x-2x.
4 ≥ x ⇒ ∴ x ≤ 4

Question 20.
Simplify : \(\frac{a^{2 m+n} a^{3 m+n}}{a^{4 m+2 n}}\)
Answer:
\(\frac{a^{2 m+n} a^{3 m+n}}{a^{4 m+2 n}}\) = a^{2m+n+3m+n-4m-2n} = aⁿ

Question 21.
Prove that tanA + cotA = secA cosecA
Answer:
LHS = tanA + cotA = \(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\)
⇒ \(\frac{\sin ^{2} A+\cos ^{2} A}{\cos A \sin A}=\frac{1}{\cos A \sin A}\) = secAcosecA = RHS

Question 22.
The average age of 7 members of a family is 18 years. If the head of the family is excluded, the average age of the rest of the members would fall to 13 years. What is the age of the head of the family?
Answer:
Total age of 7 members = 18 × 7 = 126 years
Total age of 6 members = 13 × 6 = 78
∴ the age of head of the family = 126 – 78 = 48 years

Question 23.
Insert 3 geometric means between \(\frac{1}{4}\) and \(\frac{1}{64}\).
Answer:
Insert 3 Gms G₁, G₂, G₃ between \(\frac{1}{4}\) & \(\frac{1}{64}\)
∴ \(\), G₁, G₂, G₃, are in GP
a = \(\frac{1}{4}\), ar⁴ = \(\frac{1}{64} \Rightarrow \frac{1}{4}\)r⁴ =\(\frac{1}{64}\) ⇒ r⁴ = \(\frac{4}{64}\) ⇒ r⁴ = \(\frac{1}{16}=\left(\frac{1}{2}\right)^{4}\) ⇒ r = \(\frac{1}{2}\)

Question 24.
Prove that (1 +tan² θ)(1 – sin² θ) = 1
Answer:
LHS = (1 + tan²θ)(1 – sin²θ) sec2θ.cos2θ = \(\frac{1}{\cos ^{2} \theta}\).cos² θ = 1 = RHS

Question 25.
Find the value of k, if the distance between (2k, 5) and ( -k, -4). If the distance between the points (3, -2) and (-1, a) is 5 units find the values of a.
Answer:
AB = \(\sqrt{(-1-3)^{2}+(a+2)^{2}}\) = 5
Squaring and simplifying we get
a = 1 and a = – 5

Part – C

III. Answer any TEN questions. (10 × 3 = 30)

Question 26.
Prove that \(\sqrt{2}\) is an irrational number.
Answer:
We shall prove it by the method ofcontradiction.
If possible Let \(\sqrt{2}\) be a rational number
Let \(\sqrt{2}\) = \(\frac{p}{q}\) where p and q are Integers and q ≠ 0.
Further let p and q are coprime i.e. H.C.F. of p and q = 1.
\(\sqrt{2}\) = \(\frac{p}{q}\) ⇒ \(\sqrt{2}\)q = p
⇒ 2q² = p²
⇒ 2 divides p² ⇒ 2 divides p
⇒ p is even
Let p = 2k where k is an integer p² = 4k²
2q² = 4k²
q² = 2k² ⇒ q² is even
⇒ q is even.
Now p is even and q is even which implies p and q have a common factor 2. which is a
contradiction of the fact that p and q are co-prime.
∴ our assumption that \(\sqrt{2}\) is rational is wrong and hence \(\sqrt{2}\) is irrational.

Question 27.
If U = {1, 2, 3, 4, 6, 8, 9} A = (2, 3, 6, 8} and B = {1, 3, 6, 9} verify (A∪B) – A’ ∪ B’.
Answer:
Let U = {1, 2, 3, 4, 6, 8, 9}
A = {2, 3, 6, 8}
B = {1, 3, 6, 9}
∴ A ∪ B = {1, 2, 3, 6 ,8, 9}

A’ = U – A = {1, 2, 3, 4, 6, 8, 9} – {2, 3, 6, 8}
= {1, 4, 9}

B’ = U – B = {1, 2, 3, 4, 6, 8, 9} – {1, 3, 6, 9}
= {2, 4, 8}

∴ A’ ∪ B’ = {1, 4, 9} ∪ {2, 4, 8}
= {1, 2, 4, 8, 9}

∴ (A∪B) – A’ ∪ B’ = {1, 2, 3, 6 ,8, 9} – {1, 2, 4, 8, 9}
= {3, 6, 4, 9}

Question 28.
Solve log₂ x + log₄ x = 3.
Answer:
\(\frac{\log x}{\log 2}+\frac{\log x}{\log 4}\) = 3
\(\frac{\log x}{\log 2}+\frac{\log x}{2 \log 2}\) = 3
\(\frac{2 \log x+\log x}{2 \log 2}\) = 3

log x =2 log2 = log 2²= log 4
log x = log 4
∴ x = 4

Question 29.
If a^x = b^y = c^x and b² = ac show that \(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)
Answer:
Let a^x = b^y = c^z = k (say)
∴ a^x = k ⇒ a = k^1/x, b^y = k ⇒ b = k^1/y, c^z = k ⇒ c = k^1/z
Now, b² = ac
∴ (k^1/y)² = k^1/x. k^1/z
∴ k^2/y = k^1/x+1/z
Bases are the same ∴ Equating powers on both sides, we get.
\(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)

Question 30.
The 3rd and 5th element of a GP are 3 and 27 respectively. Find the 8th element.
Answer:
Given T₃ = 3 ⇒ ar² = 3 ………………(1)
T₅ = 27 ⇒ or = 27 …………………(2)
Divide \(\frac{(2)}{(1)}\) gives r² = 9 ⇒ r = 3
ar² = 3 ⇒ a = \(\frac{3}{r^{2}}=\frac{3}{9}=\frac{1}{3}\)
∴ T₈ = ar⁷ = \(\frac{1}{3}\).37 = 36

Question 31.
If a and b are the roots of the equation 3×2 – 6x + 4 = 0, then find the value of \(\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+\left[2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3(\alpha+\beta)\right]\)
Answer:
α + β = \(\frac{-b}{a}=\frac{6}{3}\) = 2
αβ = \(\frac{c}{a}=\frac{4}{3}\)
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\) + 2\(\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\) + 3(α + β)

= \(\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}\) + 2\(\left(\frac{\alpha+\beta}{\alpha \beta}\right)\) + 3(α + β) = \(\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}\) + 2\(\left(\frac{\alpha+\beta}{\alpha \beta}\right)\) + 3(α + β)

= \(\frac{(2)^{2}-2(4 / 3)}{4 / 3}\) + 2\(\left(\frac{2}{4 / 3}\right)\) + 3(2) = \(\frac{4-8 / 3}{4 / 3}+\frac{4}{4 / 3}\) + 6 = \(\frac{\frac{12-8}{3}}{\frac{4}{3}}\) + 3 + 6 = 1 + 3 + 6 = 10

Question 32.
Solve the system of inequalities 2x – y < 1 and x – 2y < -1 graphically.
Answer:
Consider (i) 2x – y = 1
x = 0 ⇒ y = -1 ∴ A(0, 1)

(ii)x – 2y = -1
x = 0 ⇒ y = \(\frac{1}{2}\) ∴ C(0, \(\frac{1}{2}\))
y = 0 ⇒ x = -1 ∴ D(-1, 0)

Question 33.
Find the compound interest on ₹ 22,000 for 21/2 years at 6% p.a.
Answer:
Given P = 22,000, i = \(\frac{6}{100}\) = 0.06
A = P(1 + r)ⁿ = 22000(1 + 6.06)² (1 +0.06 × \(\frac{1}{2}\)) = ₹ 25460.78
C.I = A – P = 3460.78

Question 34.
A bookseller bought 228 notebooks at an average price of ₹ 8.50 of which 80 books he bought at ₹ 7.50 each and 84 at ₹ 10.50 each. Find the price of the remaining books per unit.
Answer:
Total price of 228 books = 228 × 8.50 = 1938
Total price of 164 books = 1482
∴ Total price of 64 books = 1938 – 1482 = 456
∴ Average price of 64 books = \(\frac{456}{64}\) = 7.12

Question 35.
A watch is sold for ₹ 150 at a profit of 25%. At what price should it be sold in order to have a 50% profit?
Answer:
Let C.P be 100, then S.P is 125
If SP is 150, then CP =?
\(\frac{100 \times 150}{125}\) = 120 ⇒ S.P. to get a profit of 50% is \(\frac{120 \times 150}{100}\) = ₹ 180

Question 36.
Find the value of 3tan² 30 +4cos² 30 – \(\frac{1}{2}\)cot² 45 – \(\frac{2}{3}\)sin² 60 + \(\frac{1}{8}\)sec⁴ 60
Answer:
\(3\left(\frac{1}{\sqrt{3}}\right)^{2}+4\left(\frac{\sqrt{3}}{2}\right)^{2}-\frac{1}{2}(1)^{2}-\frac{2}{3}\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{8}\left(2^{4}\right)\)
= 1 + 3 – \(\frac{1}{2}\) – \(\frac{1}{2}\) + 2 = 6 – 1 = 5

Question 37.
Find the points of trisection of the line joining (3, 4) and (5, -2).
Answer:
Let A = (3, 4) B = (5, -2)
Let P and Q be the points of trisection of AB
then P divides AB internally in the ratio 1 : 2 & Q is the midpoint of PB.
P = \(\left(\frac{1 \times 5+2 \times 3}{1+2}, \frac{1 \times-2+2 \times 4}{1+2}\right)\) = (\(\frac{11}{3}\), 2)
Q = \(\left(\frac{\frac{11}{3}+5}{2}, \frac{2+(-2)}{2}\right)\) = (\(\frac{13}{3}\), 0)
∴ the points of trisection = (\(\frac{11}{3}\), 2) (\(\frac{13}{3}\), 0)

Question 38.
Find the point of intersection of the lines 3x + 2y – 5 = 0 and 4x -y – 3 = 0
Answer:
3x + 2y – 5 = 0 ……….. (1)
4x – y – 3 = 0 ………………(2)
Multiply equation (2) by 2 and we get
Add 11x – 11 – 0 ⇒ x = 1
Put x = 1 in 3x + 2y – 5 = 0 ⇒ y = 1
∴ the point of intersection is (1, 1)

Part – D

IV. Answer any SIX questions. (6 × 5 = 30)

Question 39.
In a survey of 100 persons, it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 read magazine A and B, 10 read magazine A and C, 5 read magazine B and C while 3 read all the 3 magazines?
Find (i) How many read none of the 3 magazines, (ii) How many read only magazine C?
Answer:

Question 40.
Evaluate \(\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\) using log tables.
Answer:
Let x = \(\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\)
log x = log\(\left[\frac{0.5634 \times 0.0635}{2.563 \times 12.5}\right]\)
= log 0.5634 + log 0.0635 – log 2.563 – log 12.5
= 1.7508 + 2.8028 – 0.4087 – 1.0969
= 2.5536 – 1.5056 = 2+0.5536 – 1.5056 = -2 – 0.952 = -2.952
= -3 + 3 – 2.952 = -3 + 0.048 = 1048

Question 41.
The sum of four numbers which are in A.P. is 28 and 10 times the least number is 4 times the greatest number. Find the numbers.
Answer:
Let the four numbers are (a – 3d), (a – d)(a + d)(a + 3d).
Given sum = -28

Also given 10(a – 3d) = 4(a + 3d) ⇒ 10a – 30d = 4a + 12
6a = 42d
∴ 42d = 6 × 7 = 42 ⇒ d = 1
∴ the four numbers are 4, 6, 8, 10

Question 42.
Find an integral root between -3 and 3 by inspection and they using synthetic division solve the equation x³ – 3x² – 5x + 6 = 0
Answer:
Letf(x) = x³ – 2x² – 5x + 6
f(1)= 1 – 2 – 5 + 6 = 7 – 7 = 0
∴ x = 1 is a root of the given equation by using Synthetic division.

∴ The resulting equation is x2 – x – 6 = 0 is the quotient and remainder = 0
– x – 6 = 0
(x – 3)(x + 2) = 0
x = 3 or -2
Thus x = 1, -2, 3 are the roots of the given equation.

Question 43.
Find the difference between simple interest and compound interest and compound interest on ₹ 6400 at 9% p.a. for 5 years.
Answer:
Simple Interest
P = 18,000 T = 4 R = 4
SI = \(\frac{\mathrm{PTR}}{100}\)
= \(\frac{18,000 \times 4 \times 8}{100}\)
= ₹5,760
Compound Interest
P = 18,000 n = 4 I = 008
A = P(1 + i)ⁿ
= 18,000 (1 + 0.08)⁴
= 24,488.8
CI = A – P
= 24,488.8 – 18,000
CI = 6488.8
Difference between CI and SI is
CI – SI = 6488.8, – 5760
= 728.8.

Question 44.
Preritha wants to buy a house after 5 years when it is expected to cost ₹ 50 lakhs. How much should she save annually if her savings earn a compound interest of 12%?
Answer:
Given: F = 50,00,000; n =5, i = 0.12
F = \(\frac{\mathrm{A}\left[(1+\mathrm{i})^{\mathrm{n}}-1\right]}{\mathrm{i}}\)
50,00,000 = A(6.3 528)
A = \(\frac{50,00,000}{6.3528}\)
A = ₹ 787054.5

Question 45.
A person gets ₹ 1216 more when selling a product for a profit of 15% instead of a loss of 4%. What would be the percentage profit or loss if it is sold for ₹7552?
Answer:
Let the C.P be ₹ 100
SP making profit of 15% = 115
SP while making a loss of 4% = 96
Difference = 19
If the CP is 100, the difference is 19.
If the CP is ?, difference is 1216.
∴ C.P = \(\frac{100 \times 1216}{19}\) = 6400
If SP = 7552 ⇒ Profit = 7552 – 6400 = 1152
Profit% = \(\frac{\text { Profit }}{\mathrm{CP}}\) × 100 ⇒ Proflt% = \(\frac{1152}{6400}\) × 100 = 18%

Question 46.
If cosθ = \(\frac{4}{5}\) and θ is acute, show that \(\frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{24}{7}\)
Answer:
LHS = \(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\)
If cosθ = \(\frac{4}{5}\) then tanθ = \(\frac{3}{4}\)
L.H.S = \(\frac{2 \cdot \frac{3}{4}}{1-\frac{9}{16}}=\frac{3}{2} \times \frac{16}{16-9}=\frac{3}{2} \times \frac{16}{7}=\frac{24}{7}\) = RHS

Question 47.
Show that the points (1,1)(4,1) and (1,4) are the vertices of a square. Also And the area.
Answer:
AB = \(\sqrt{(4-1)^{2}+(1-1)^{2}}=\sqrt{9+0}\) = 3
BC = \(\sqrt{(4-4)^{2}+(4-1)^{2}}=\sqrt{0+9}\) = 3
CD = \(\sqrt{(1-4)^{2}+(4-4)^{2}}=\sqrt{9+0}\) = 3
DA = \(\sqrt{(1-1)^{2}+(4-1)^{2}}=\sqrt{0+9}\) = 3
AB = BC = CD + AD
AC = \(\sqrt{(4-1)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}\)
BD = \(\sqrt{(1-4)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}\)
∴ AC = BD
∴ ABCD Δ is a Square.

Question 48.
Derive the equation of a straight line in the form x cosα + y sinα = p
Answer:
Let the line cut the x-axis at A and y axis at B.
Let ON be the perpendicular to AB and ON = P & ∠AON = a
∴ ∠NOB = 90 – α

From Δ ONA
cos α = \(\frac{\mathrm{ON}}{\mathrm{OA}}\) ⇒ OA = \(\frac{p}{\cos \alpha}\) = a(say)
cos(90 – α) = \(\frac{p}{\mathrm{OB}}\) ⇒ sin α = \(\frac{p}{\mathrm{OB}}\)
⇒ OB = \(\frac{p}{\sin \alpha}\) = b (say)
Using equation of the line
\(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ \(\frac{x}{p / \cos \alpha}+\frac{y}{p / \sin \alpha}\) = 1 ⇒ \(\frac{x \cos \alpha}{P}+\frac{y \sin \alpha}{P}\) = 1
∴ xcosα + ysinα = p
This is called normal form of the equation.

Part – E

V. Answer any ONE question (1 × 10 = 10)

Question 49.
(a) If f(x) = x and g(x) = x1 + 1, find (i) fog(1) (ii) fog(2) (iii) gof (1), (iv) gof(2)
Answer:
(i) fog(1) = f[g(1)] = f[1³ + 1] = f(2) = 2
(ii) fog (2)= f[g(2)] = f[2³ + 1] = f(9) = 9
(iii) gof(1) = g[f(1)] = g [1] = 1 + 1 = 2
(iv) gof(-1) = g[f(-1)] = g[-1] = (-1)³ + 1 = -1 + 1 = 0

(b) Find the sum to n terms of a G.P 4 + 44 + 444 +………….
Answer:
S = 4 + 44 + 444 + ………………. to n terms
S_n = 4(1 + 11 + 111 + ……………to n terms)
\(\frac{9 \mathrm{~S}_{n}}{4}\) = 9 + 99 + 999 + ……………… to n terms
\(\frac{9 \mathrm{~S}_{n}}{4}\) = (10 – 1) + (10² – 1) + (10³ – 1) +………………
= (10 + 10² + 10³+ ……………. ) – (1 + 1 + 1 + …………..)
\(\frac{9 \mathrm{~S}_{n}}{4}=\frac{10\left(10^{n}-1\right)}{10-1}\) – n
Sn = \(\frac{4}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(c) The average age of 12 boys is 8 years. Another boy of 21 years joins this group. Find the average age of the new group.
Answer:
The average age of 12 boys 8 years
∴ Total age of boys = 8 × 12 = 96 years
when a boy of 21 yrs joins the group,
the total age of 13 boys 96 + 21 = 117years
∴ Average age of the new group = \(\frac{117}{13}\) = 9 years.

Question 50.
(a) A manufacturer produces and sells balloons at ₹8 per unit. His fixed cost is 6500 and the variable cost/balloon is ₹3.50.
Calculate (i) Revenue function (ii) Cost function (iii) Profit function (iv) Break Even point.
Answer:
i.e. (i) R(x) = 8x

(ii) C(x) = 3.5x + 6500

(iii) P(x) = 4.5x – 6500
i.e.,P(x) = 6500 = 0

(iv) At BEP P(A) = 0
4.5x – 6500 = 0
4.5x = 6500
x = 144.4 units

(b) For what values of k are the 3 lines x – 2y + 1 =0, 2x + 5y + 3 = 0 and 5x – 9y + k 0 are concurrent.
Answer:

(c) Find the H.C.F. of 16, 24, and 48.
Answer:
16 = 2 × 2 × 2 × 2
24 = 2 × 2 × 2 × 3
48 = 2 × 2 × 2 × 2 × 3
H.C.F of (16, 24, 48) = 2 × 2 × 2 = 10

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2019 (South), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2019 (South)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Define a prime number
Answer:
Numbers which are divisible by one and the number itself is called prime number. {2,3,5,7,11,…………}

Question 2.
If A = {a, b, c, d}, B = {d, e, f, g}, find A-B.
Answer:
A – B = {a, b, c}

Question 3.
Simplify \(\left\{\frac{5 x^{3}}{y}\right\}^{2}\)
Answer:
\(\frac{25 x^{6}}{y^{2}}\)

Question 4.
Find x if log_x625 = 4.
Answer:
x⁴ = 625 ⇒ x⁴ = 5⁴ ⇒ x = 5.

Question 5.
Find the 11“’ term of the AP 3, 5, 7, 9
Answer:
T₁₁ = 3 + (11 – 1)2 ⇒ T_n = 3 + 20 = 23.

Question 6.
Solve for x if (x + 2) (x + 3) = (x – 2) (x – 4).
Answer:
x² + 6 + 5x = x² + 8 – 6x
11x = 2 ⇒ x = \(\frac{2}{11}\)

Question 7.
Find the simple interest on ₹ 600 for 3 years at 4% p.a.
Answer:
SI = \(\frac{600 \times 3 \times 4}{100}\) = ₹ 72

Question 8.
Find the present value of a perpetuity of ₹ 3000 to be received forever at 4% p.a.
Answer:
A = Rs, 3000
r = 0.04
P∞ = \(\frac{\mathrm{A}}{\mathrm{r}}=\frac{3000}{0.04}\) = 75000

Question 9.
Convert \(\frac{1}{4}\) into percentage.
Answer:
\(\frac{1}{4}\) × 100 = 25%

Question 10.
Express \(\frac{3 \pi}{2}\) in degrees.
Answer:
\(\frac{3 \pi}{5}=\frac{3 \times 180}{5}\) = 108°

Question 11.
Find the value of sin²120 + cos²120.
Answer:
sin²120 + cos²120 = 1

Question 12.
Find the slope of line 2x + 3y – 11 = 0.
Answer:
5y = -2x + 11
y = \(\frac{-2}{5}\)x + \(\frac{11}{5}\) compare with y = mx + c
∴ m = slope = \(\frac{-2}{5}\)

Part – B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.
Find the number of divisors of 825.
Answer:
825 = 3¹ x 5² x 11¹
P₁ = 3, α₁ = 1,P₂ = 5, α₂ = 2, P₃ = 11, α₃ = 11
T (n) = (1 + α₁) (1 + α₂) (1 + α₃)
= (1+ 1)(1 +2)(1 + 1)
= (2)(3)(2)
= 12

S(n) = \(\left(\frac{p_{1}^{\alpha_{1}+1}-1}{p_{1}-1}\right)\left(\frac{p_{2}^{a_{i}+1}-1}{p_{2}-1}\right)\left(\frac{p_{1}^{a_{1}+1}-1}{p_{3}-1}\right)\)
S(825) = \(\frac{3^{2}-1}{3-1} \times \frac{5^{3}-1}{5-1} \times \frac{11^{2}-1}{11-1}\)
= \(\frac{8}{2} \times \frac{124}{4} \times \frac{120}{10}\)
= 4 × 31 × 12 = 1488

Question 14.
Find the number which, when divided by 16, 20 and 40 leaves the same remainder 4.
Answer:
16 = 2⁴, 20 = 2² × 5¹, 40 = 2³ × 5¹
LCM = 2⁴ × 5¹ = 16 × 5 = 80
∴ Required number = 80 + 4 = 84.

Question 15.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6} and U = {1, 2-, 3, 4, 5, 6, 7, 8} verify (A∪B)’ = A’∩B’
Answer:

1200 = 2⁴ × 5² × 3¹ = p₁^α₁× p₂^α₂ × p₃^α₃
P₁ = 2 P₂ = 5; P₃ = 3 α₁ = 4; α₂ = 2; α₃ = 1
T(n) = (1 + α₁)(1 + α₂)(1 + α₃) = (1 + 4)(1 + 2)(1 + 1) = (5)(3)(2) = 30
S(n) = \(\left(\frac{\mathrm{P}_{1}^{\alpha_{1}+1}-1}{\mathrm{P}_{1}-1}\right)\left(\frac{\mathrm{P}_{2}^{\alpha_{2}+1}-1}{\mathrm{P}_{2}-1}\right)\left(\frac{\mathrm{P}_{3}^{\alpha_{3}+1}-1}{\mathrm{P}_{3}-1}\right)\)
= \(\left(\frac{2^{5}-1^{\circ}}{2-1}\right)\left(\frac{5^{3}-1}{5-1}\right)\left(\frac{3^{2}-1}{3-1}\right)=\left(\frac{31}{1}\right)\left(\frac{124}{4}\right)\left(\frac{8}{2}\right)\)
= 31 × 124 = 3844

Question 16.
Simplify \(\left[\frac{x^{a}}{x^{b}}\right]^{a+b}\left[\frac{x^{b}}{x^{c}}\right]^{b+c}\left[\frac{x^{c}}{x^{a}}\right]^{c+a}\)
Answer:

Question 17.
Which clement of the GP 5, 10, 20………. is 80.
Answer:
T_n = ar^n-1
80 = 5.2^x-1
2^n-1 = 16 ⇒ 2^n-1 = 2⁴ ⇒ n-1 = 4⇒ n = 5

Question 18.
The sum of the two numbers is 107 and their difference is 17. Find the numbers.
Answer:

∴ a = 62
a + b = 107
b = 107 – a = 107 – 62 = 45
∴ b = 45

Question 19.
Solve by formula method: 5x² – 7x – 12 = 0.
Answer:
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{-(-7) \pm \sqrt{(-7)^{2}-4(5)(-12)}}{(2)(5)}\) ⇒ x = \(\frac{24}{10}\) and x = -1

Question 20.
Solve : 3x – 2 < 2x + 1 ; x∈ R and represent on the number line.
Answer:
3x – 2 < 2x + 1
3x – 2x < 2+ 1
x < 3

Question 21.
Find the future value of an annuity due of ₹ 800 for 3 years at 5% p.a.
Answer:
F = \(\frac{a\left[(1+i)^{n}-1\right]}{i}\)(1 + i) = \(\frac{800\left[(1+0.05)^{3}-1\right]}{(0.05)}\)(1 + 0.05)
F = ₹ 2287.14

Question 22.
Prove that tan²A(1 – sin2A) = sin2A.
Answer:
LHS = tan² A cos² A = \(\frac{\sin ^{2} A}{\cos ^{2} A}\)cos² A- = sin² A = RHS

Question 23.
Find the value of cot²60° + sin²45° + sin²30° + cos²90°.
Answer:
\(\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}\) + 0 = \(\frac{1}{3}+\frac{1}{2}+\frac{1}{4}=\frac{13}{12}\)

Question 24.
A point P moves such that PA² = 3PB². If A = (5, 0), B = (-5, 0). Find the equation of the locus of P.
Answer:
Let P(x, y) be the point on the locus
Given PA² = 3PB² and A = (5, 0), B (—5, 0)
(x – 5)² + (y – 0)² = 3[(x+ 5)² + (y – 0)²]
x² + 25 – 10x + y² = 3[x² + 25 + 10x + y²]
3x² + 75 + 30x + 3y² – x² + y² – 10x + 25
2x² + 2y² + 40x + 50 = 0 is the required equation.

Question 25.
If the distance between the points (3, -2) and (-1, a) is 5 units find the values of a.
Answer:
AB = \(\sqrt{(-1-3)^{2}+(a+2)^{2}}\) =5
Squaring and simplifying we get
a = 1 and a = -5

Part – C

III. Answer any TEN questions. (10 × 3 = 30)

Question 26.
Prove that \(\sqrt{2}\) is an irrational number.
Answer:
We shall prove it by the method ofcontradiction.
If possible Let \(\sqrt{2}\) be a rational number
Let \(\sqrt{2}\) = \(\frac{p}{q}\) where p and q are Integers and q ≠ 0.
Further let p and q are coprime i.e. H.C.F. of p and q = 1.
\(\sqrt{2}\) = \(\frac{p}{q}\) ⇒ \(\sqrt{2}\)q = p
⇒ 2q² = p²
⇒ 2 divides p² ⇒ 2 divides p
⇒ p is even
Let p = 2k where k is an integer p² = 4k²
2q² = 4k²
q² = 2k² ⇒ q² is even
⇒ q is even.
Now p is even and q is even which implies p and q have a common factor 2. which is a
contradiction of the fact that p and q are co-prime.
∴ our assumption that \(\sqrt{2}\) is rational is wrong and hence \(\sqrt{2}\) is irrational.

Question 27.
If f(x) = x + 1 and g(x) = x² + 1. Find (1) fog(1) (2) fog(2) (3) gof(2)
Answer:
(i) fog(1) =f(g(1))
= f(2)
= 2 + 1 = 3

(ii) fog (2) = f(g (2))
= f(5)
= 5 + 1 = 6

(iii) gof (1)
= gof (1)
= g(f(1))
= g (2) = 4 + 1 = 5

Question 28.
If a^x = b^y = c^z and b² = ac. Show that \(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)
Answer:
Let a^x = b^y = c^z = k(say)
∴ a^x = k ⇒ a = k^1/x, b^y = k ⇒ b = k^1/y, c^z = k ⇒ c = k^1/z
Now, b² = ac
∴ (k^1/y)² = k^1/x. k^1/z
∴ k^2/y = k^1/x+1/z
Bases are the same ∴ Equatlig powers on both sides, we get.
\(\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\)

Question 29.
Solve log x + log (x – 4) – log (x – 6) = 0
Answer:
logx(x-4) – log(x-6) = 0
⇒ log\(\) = 0 ⇒ \(\) = 1 ⇒ x² – 5x + 6 = 0
x = 2 or 3

Question 30.
Find the three numbers in GP whose sum is 39 and their product is 729.
Answer:
Let the three numbers be \(\frac{a}{r}\),a,ar
Product = 729
⇒ a³ = 729 ⇒ a³ = 9³ ⇒a = 9
Sum = 39
\(\frac{a}{r}\) + a + ar = 39
\(\frac{9}{r}\) + 9 + 9r = 39
\(\frac{9}{r}\) + 9r = 39 – 9
\(\frac{9}{r}\) + 9r = 30
⇒ 9 + 9r² = 30
r ⇒ 9r² – 30r + 90
⇒ 9r² – 27r – 3r + 90
9r(r – 3)-3 (r-3) = 0
(r – 3)(9r – 3) = 0
r = 3 or 9r = 3
r = 3 or r = \(\frac{3}{9}=\frac{1}{3}\)
r = 3 or r = \(\frac{1}{3}\)

The numbers are \(\frac{a}{3}\),a,ar
\(\frac{9}{3}\), 9, 9(3)
3,9,27.

Question 31.
Find the quotient and remainder when x⁴ + 10x³ + 39x² + 76x + 65 is divided by x +4.
Answer:

Quotient = x³ + 6x² + 15x + 16
Remainder = 1

Question 32.
Solve graphically : x + 2y < 8, 2x + y < 8
Answer:

Question 33.
Find the compound interest on ₹ 7000 at 5% p.a for 4 years?
Answer:
A = P(l + i)ⁿ ⇒ A = 7000(1 + 0.05)⁴ = 8508.54
CI – A – P = 8508.54 – 7000 ⇒ Cl = ₹ 1508.54

Question 34.
Ramya purchased 3 varieties of cooking oil 4 kg of oil at ₹ 50/kg. 5 kg of oil at ₹ 60/ kg and 9 kg of oil at ₹ 70/kg. What is the average price of oil/kg?
Answer:
X̄w = \(\frac{x_{1} w_{1}+x_{2} w_{2}+x_{3} w_{3}}{w_{1}+w_{2}+w_{3}}=\frac{50 \times 4+60 \times 5+70 \times 9}{4+5+9}=\frac{1130}{18}\) = ₹ 63.2 /kg

Question 35.
The price of a pair of trousers was decreased by 22% of ₹ 390. What was the original price of the trousers.
Answer:
Let the original price be x. .
Then (x – \(\frac{22 x}{100}\)) = 390 ⇒ x(1 – \(\frac{22}{100}\)) = 390 ⇒ x\(\frac{78}{100}\) = 390
x = \(\frac{390 \times 100}{78}\) = ₹ 500

Question 36.
Find the value of \(\frac{\sin \frac{\pi}{2} \cos ^{2} \frac{\pi}{6} \sec ^{2} \frac{\pi}{4}}{\tan \frac{\pi}{3}+\cot \frac{\pi}{3}}\)
Answer:
\(\frac{\sin \frac{\pi}{2} \cos ^{2} \frac{\pi}{6} \sec ^{2} \frac{\pi}{4}}{\tan \frac{\pi}{3}+\cot \frac{\pi}{3}}=\frac{1 \times\left(\frac{\sqrt{3}}{2}\right)^{2} \cdot(\sqrt{2})^{2}}{\sqrt{3}+\frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}}{4} \times 2}{\frac{(\sqrt{3})^{2}+1}{\sqrt{3}}}=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{4}=\frac{3}{8}\)

Question 37.
Find the third vertex of a triangle if two of its vertices are at (-2, 4) and (7, -3) and the centroid at (3, -2).
Answer:
Centroid G(x, y) = \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)
(3, 2) = \(\left(\frac{-2+7+x_{3}}{3}, \frac{4-3+y_{3}}{3}\right)\) ⇒ x₃ = 4, y₃ = -7
∴ Third vertex (x₃, y₃) = (4, -7)

Question 38.
Find the distance between the two lines 2x – 3y + 4 = 0 and 4x – 6y – 5 = 0.
Answer:
Distance between parallel lines = \(\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|=\left|\frac{8-(-5)}{\sqrt{4^{2}+(-6)^{2}}}\right|=\frac{13}{\sqrt{52}}=\frac{\sqrt{13}}{2}\) units

Part – D

IV. Answer any SIX questions. (6 × 5 = 30)

Question 39.
In a group of 50 people, 35 speak Kannada and 25 speak both English and Kannada and all the people speak at least one of the two languages. How many speak English? How many speak only English and not Kannada? How many speak only Kannada?
Answer:
n(K ∪ E) = n[K) + n(E) – n(K ∩ E)
50 = 35 + n(E) – 25 ⇒ K(E) = 40
(i) n(E-K) = n(E)-n(E∩K) = 40-25 = 15
(ii) n(K-E) = n(K) – n[K ∩ E) = 35 -25 = 10

Question 40.
Evaluate \(\frac{213.78 \times 7.434}{6.321}\) using logarithmic tables.
Answer:
Let x = \(\frac{213.781 \times 7.434}{6.321}\)
logx = \(\left[\frac{213.781 \times 7.434}{6.321}\right]\)
= log213.781 + logl.434 – log6.321
= 2.3298 + 0.87 12 –0.8008.
= 2.4002
x = Antilog 2.4002
x = 25 1.3.

Question 41.
Find the sum of all numbers between 60 and 400 which are divisible by 13.
Answer:
S_n = 65 + 78 + 91 +…………..+390
a = 5 d = 13 n =? T_n = 390
T_n = a + (n – 1)d
390 = 65 + (n – 1)13
390 = 65 + 13n – 13
13n = 338
∴ n=26
S_n = \(\frac{n}{2}\)[a + l] = (65+390) = 13 (455)= 5,915
∴ S = 5,915

Question 42.
A mother is 32 years older than her son. After 4 years the mother’s age will be 8 years more than twice that of her son. Find their present ages.
Answer:
Let the present age of the son = x yrs.
and the present age of the mother = x + 32 years
After 4 years age of the mother = x + 36
Given that x + 36 = 8 + 2 (x + 4) = 8 + 2x + 8
36 – 16 = 2x – x
x = 20 years
∴ son’s present age is x = 20 yrs
mother’s present age is 20 + 32 = 52 years.

Question 43.
Find the difference between simple interest and compound interest on ₹ 18,000 invested for 4 years at 8% p.a. where compound interest is compounded annually.
Answer:
Simple Interest
P = 18,000 T = 4 R = 4
SI = \(\frac{\mathrm{PTR}}{100}\)
= \(\frac{18,000 \times 4 \times 8}{100}\)
= ₹5,760

Compound Interest
P = 18,000 n = 4 I = 008
A = P(1 + i)n
= 18,000 (1 + 0.08)4
= 24,488.8
CI = A – P
= 24,488.8 – 18,000
CI = 6488.8

Difference between CI and Si is
CI – SI = 6488.8, – 5760
= 728.8.

Question 44.
If you want to have ₹ 80,000 after 5 years, how much should you deposit every year if the bank offers 12% p.a. interest compounded quarterly.
Answer:
F = 80,000, n = 5yrs, r 12% 0.12, i = \(\left(1+\frac{0.12}{4}\right)^{4}\) – 1 = 0.125
F = \(\frac{a\left[(1+i)^{n}-1\right]}{i}\)
a = \(\frac{\mathrm{F} \times i}{(1+i)^{n}-1}=\frac{80,000 \times 10.125}{(1.125)^{5}-1}=\frac{80,000}{6.416}\) = ₹12,468.8

Question 45.
A businessman sells an article for ₹ 720 and earns a profit of 20% Find the (a) cost price (b) profit percentage at the selling price.
Answer:
S.P = 720 = \(\frac{120}{100}\)x ⇒ x = \(\frac{720 \times 100}{120}\) = 600
∴ Profit = 720 – 600 = 120 ⇒ profit% = \(\frac{\text { profit in Rs. }}{\text { S.P. }}\) × 100 = \(\frac{120}{720}\) × 100 = 16.66%

Question 46.
If cot \(\frac{\theta}{2}=\frac{5}{2}\) q is acute. Show that \(\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}=\frac{19}{7}\)
Answer:
cotθ = \(\frac{5}{2}\) ∴ Hyp = \(\sqrt{29}\)
cosθ = \(\frac{5}{\sqrt{29}}\), sinθ = \(\frac{2}{\sqrt{29}}\)

Substituting and simplifying we get
LHS = \(\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}=\frac{3 \cdot \frac{5}{\sqrt{29}}+2 \cdot \frac{2}{\sqrt{29}}}{3 \cdot \frac{5}{\sqrt{29}}-4 \cdot \frac{2}{\sqrt{29}}}=\frac{15+4}{15-8}=\frac{19}{7}\)

Question 47.
Find the ratio in which the line joining the points (3, 5) and (-7, 9) ¡s divided by the point (\(\frac{1}{2}\),6).
Answer:

Question 48.
Find the equation of a line that passes through (-4, 1), and a portion of it between the axes is divided by the point in the ratio 1:2.
Answer:
A straight line passes through the point (-4, 7) and the portion of the line intercepted between the axes is divided at this point in the ratio 3: 2. Find the equation of the line.

Let the required equation of the line be
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}\) = 1 …………(1)
Then it meets the axes at the points A (a, 0) and B (0, b). The coordinates of the point which divides the line AB in the ratio 3: 2 are
\(\left(\frac{2 \cdot a+3 \cdot 0}{2+3}, \frac{2 \cdot 0+3 \cdot b}{2+3}\right)=\left(\frac{2 a}{5}, \frac{3 b}{5}\right)\) …….(2)
This is the same as (-4, 7) ……………(3)
Comparing (2) and (3), \(\frac{2 a}{5}\) =-4, \(\frac{3b}{5}\) = 7 or a = -10, b = \(\frac{35}{3}\)
Substituting these values of a, b in (1) we get
\(\frac{x}{-10}+\frac{y}{\frac{35}{3}}\) = 1

Part – E

V. Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) Find the sum to n terms of the series 5 + 55 + 555 +………..n terms.
Answer:
Let S = 5 + 55 + 555 + ………. to n terms
\(\frac{S}{2}\) = 1 + 11 + 111 +……….to n terms
\(\frac{9S}{2}\) = 9 + 99 + 999 +…………….to n terms
(10 – 1) + (100 – 1) + (1000 – 1) +………… to n terms
= (10 + 100 + 1000 + to n terms) – (1 + 1 + to n terms)
= \(\frac{10\left(10^{n}-1\right)}{9}\) – 1 × n
S = \(\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(b) A manufacture produced and sells balloons at ₹ 8 per unit. His fixed cost is X 6500 and the variable cost per balloon is ₹ 3.50. Calculate
(i) Revenue function (ii) Cost function (iii) Profit function (iv) Break even point
Answer:
Manufacture produceçi and sells balloons at ₹ 8 per unit. His fixed cost is ₹ 6500 and the variable cost per Balloon is ₹ 3.50 calculate,

1. Revenue Function
2. Cost Function
3. Profit Function
4. Break-Even point.

i.e. (i) R(x) = 8x

(ii) C(x) = 3.5x + 6500

(iii) P(x) = 4.5x – 6500
i.e.,P(x) = 6500 = 0

(iv) At BEP P(A) = 0
4.5x – 6500 = 0
4.5x = 6500
x = 144.4 units

(c) The weight of 6 men are 90 kg, 70.5 kg, 56 kg, 45.5 kg, 85 kg and 78 kg. Find the average weight.
Answer:
Average weight of 6 men = \(\frac{90+70.5+56+45.5+85+78}{6}=\frac{425}{6}\) = 70.83 kg

Question 50.
(a) Find the equation of the straight line which passes through the point of intersection of 2x – 3y = 4 and 2x + 2y = 1 and perpendicular to the line x + 4y = 8.
Answer:

(b) If A = (1, 3, 5) B = {5} C = {7} verify A × (B – C) = (A × B) – (A × C)
Answer:
B – C = {5}
A × (B -C) = {1,3,5} × {5} ={(1,5), (3,5), (5,5)}…………………. (1)
A × B= {(1,5), (3,5), (5,5)}
A × C = {(1,7), (3,7), (5,7)}
(A × B) – (A × C) = {(1,5), (3,5), (5,5)} ………..(2)
From (1) and (2), A – (B – C) = (A × B) – (A × C)

(c) If the HCF of two numbers is 42 and their product is 52920, find their LCM.
Answer:
H.C.F × L.C.M = A × B
42 × LCM = 52920 ⇒ LCM = \(\frac{52920}{42}\) = 1260
∴ LCM = 1260

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2020 (North), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2020 (North)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the imaginary part of 3-i.
Answer:
Imaginary part = – 1.

Question 2.
If A has 4 elements, how many elements will P(A) have?
Answer:
P(A) = 24 = 16 element

Question 3.
Simplify: a^x+y a^2x-y
Answer:

Question 4.
Find x if logx 25 = 2
Answer:
x² = 25
∴ x = ±5

Question 5.
Find the sum to infinity of GP 3,\(\frac{1}{1}, \frac{1}{3}\)…………..
Answer:
a = 3, r = \(\frac{1}{3}\)
∴ S_∞ = \(\frac{a}{1-r}=\frac{3}{1-\frac{1}{3}}=\frac{9}{2}\)

Question 6.
Solve for x : 7(x – 2) + 8(x – 3) = x + 10.
Answer:
7x – 14 + 8x – 24 = n + 10
∴ 14x = 48

Question 7.
Calculate SI for 2 years for ₹ 6000 at 8% pa.
Answer:

Question 8.
Find the present value of a perpetuity of 3,000 to be received forever at 14% pa.
Answer:
Perpetuity P_∞ = \(\frac{A}{r}=\frac{3000}{0.14}\) = ₹21,428

Question 9.
Convert 0.32 into a percentage.
Answer:
0.32 = 0.32 × 100 = 32%

Question 10.
Convert 45° into radians.
Answer:

Question 11.
If A = 45° then show that sin 2 A = 2 sin A cos A.
Answer:
sin(2 × 45) = 2sin45°cos45°
sin90°= 2 × \(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\)
1 = 1

Question 12.
Find the slope of the line 2x + 5y – 11 = 0.
Answer:
Slope = m = \(\frac{-2}{5}\)

Part-B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.
Find the number of positive divisors of 3⁴ × 5³ × 7²
Answer:
3⁴ × 5³ × 7² = p₁^α₁× p₂^α₂ × p₃^α₃
p₁ = 3, p₂ = 5, p₃ = 7, α₁ = 4, α₂ = 3, α₃ = 2
T(n) = (1 + α₁)(1 + α₂)(1 + α₃) = (1 +4)(1 + 3)(1 + 2) = (5)(4)(3) = 60

Question 14.
Three bells call at intervals 30 sec, 40 sec, 50 sec respectively. They start together. After how many minutes will the next bell call together.
Answer:

Hence the bills will fall together after 600 second or \(\frac{600}{60}\) = 10 minutes

Question 15.
If (x + y, x – y) = (5, 1) find x and y.
Answer:

Question 16.
Prove that (xb-c)a.(xc-a)b.(xa-b)c = 1
Answer:

Question 17.
Which term ofan AP \(\frac{1}{2}\), 1, \(\frac{1}{2}\) is 5?
Answer:
a = \(\frac{1}{2}\), d = 1 – \(\frac{1}{2}\)= \(\frac{1}{2}\), an = 5, n =?
a_n = a + (n – 1)d
5 = \(\frac{1}{2}\) + (n – 1)\(\frac{1}{2}\)

Question 18.
Solve x + 2y = 1
3x – 2y = 5
Answer:

x = 3/2
x + 2y = 1
\(\frac{1}{2}\) + 2y = 1
2y = 1 – \(\frac{3}{2}\)
2y = –\(\frac{1}{2}\)
∴ y = – \(\frac{1}{4}\)

Question 19.
If a and b are the roots of the equation 3x² – 6x +4 = 0. Find α² + β².
Answer:
α + β = \(-\frac{b}{a}=\frac{-3}{-6}=\frac{1}{2}\)
∴ α² + β² = (α + β)² – αβ
αβ = \(\frac{c}{a}=\frac{4}{5}\)
= \(\left(\frac{1}{2}\right)^{2}-\frac{4}{3}=\frac{1}{4}-\frac{4}{3}-\frac{3-16}{12}=\frac{-13}{12}\)

Question 20.
Solve linear inequalities and represent on the number line
7x + 3 < 5x + 9, X∈ R
Answer:

7x + 3 < 5x + 9
7x – 5x < 9 – 3
2x < 6
∴ x < 3

Question 21.
Find the future value of an annuity of 600 at 4% pa payable for 2 years,
Answer:
F=? u= 600, r = 0.04, n = 2ys
F = \(\frac{a\left[(1+r)^{n}-1\right]}{r}=\frac{600\left[(1+0.04)^{2}-1\right]}{0.04}=\frac{600\left[(1.04)^{2}-1\right]}{0.04}\) = ₹ 1224

Question 22.
Prove that (i + tan² θ)(1 – sin² θ ) = 1
Answer:
LHS= (1 + tan²θ)(1 – sin²θ)

Question 23.
Find the value of cos 60° – sin 30° – cots 450
Answer:

Question 24.
One end of a diameter of a circle is (1, 3) and its center is (4, -2). Find the coordinates of the other end of this diameter.
Answer:
By midsection formula:
(4, -2) = \(\left[\frac{x_{1}+1}{2}, \frac{y_{1}+3}{2}\right]\)
∴ \(\frac{x+1}{2}\) = 4
∴ x = 7
\(\frac{y_{1}+3}{2}\) = -2
∴ y = -7
∴ the other end of diameter = (x, y) = (7, -7)

Question 25.
Find the equation of the locus of all points equidistant from the point (2,4) and y-axis.
Answer:
Locus = P(x,y).A(2,4) and y-sinB (0, y)
Given PA = PB
\(\sqrt{(x-2)^{2}+(y-4)^{2}}=\sqrt{(x-0)^{2}+(y-y)^{2}}\)

y2 – 4x – 8y + 20 = 0 equation of the locus

Part-C

III. Answer any TEN questions. (10 × 3 = 30)

Question 26.
Prove that 3 + \(\sqrt{5}\) is an irrational number.
Answer:
If possible let 3 + \(\sqrt{5}\)be a rational number
∴ 3 + \(\sqrt{5}\) = \(\frac{p}{q}\) where p, q ∈ Z, q ≠ 0
∴ \(\sqrt{5}\) = \(\frac{p}{q}\) – 3 = rational – rational = rational number
∴ \(\sqrt{5}\) is a rational number which is a contradiction.
∴ our assumption is wrong
∴ 3 + \(\sqrt{5}\) is irrational.

Question 27.
If f(x) = 2x + 1 g(x) = x² + 2x + 1.
Find (i) fog (2)
(ii) gof (3)
(iii) fog (1)
Answer:
(i) fog(1) = f[g(1)] = f[13+1] = f(2) = 2
(ii) fog(2) = f[g(2)] = f[23+1] = f(9) = 9
(iii) gof(1) = g[f(1)] = g[1] = 13 + 1 = 2
(iv) gof(-1) = g[f(-1)] = g[-1] = (-1)3 + 1 = -1 + 10

Question 28.
If a^x = b^y = c^x and b² = ac show that \(\frac{1}{x} \times \frac{1}{z}=\frac{2}{y}\).
Answer:
Let a = b = c = k (say)
∴ a^x = k ⇒ a = k^1/x, b^y = k ⇒ b = k^1/y, c^z = k⇒ c^z = k^1/z
Now, b² = ac
∴ (k^1/y)² = k^1/x.k^1/z
∴ k^2/x = k^1/x+1/z

Bases are same ∴ Equating powers on both the sides, we get.
\(\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\)

Question 29.
If x² + y² = 12xy show that 2log(x-y) = log2 + log5 + logx + logy.
Answer:
LHS = 2 log (x -y)
= log(x – y)² = log(x² + y² – 2xy) Put x² + y² = 12xy
= log(12xy – 2xy) = log(10xy) = log(2 × 5 × x × y)
= log2 + log5 + logx + logy = RHS

Question 30.
The 3 HM’s between \(\frac{1}{4}\) and \(\frac{1}{12}\).
Answer:
Insert 3 H.M.’s between and \(\frac{1}{4}\) and \(\frac{1}{12}\)
\(\frac{1}{4}\), h1, h2, \(\frac{1}{12}\)………. HP
4, \(\frac{1}{\mathrm{~h}_{1}}, \frac{1}{\mathrm{~h}_{2}}, \frac{1}{\mathrm{~h}_{3}}\), 12……… AP
a = 4 d =? T = 12 n = 5
Tn = a + (n – 1)d
12 = 4 + (5 – 1)d
12 – 4 = 4d
4d = 8d = 2
4,6,8,10,12,……… AP
\(\frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \frac{1}{12}\) …………..HP .
∴ Harmonic means are \(\frac{1}{6}, \frac{1}{8}, \frac{1}{10}\)

Question 31.
Five years ago, the father’s age was 5 times as old as his son and after 3 years he will be 3 times as old as his son. Find their present ages.
Answer:
Let the present age of the son = x ys
5 yrs ago son’s age = (x – 5) yr
Father’s age = y yrs
5 year ago father age = (y – 5)
Given (y – 5) = 5(x – 5)
y – 5 = 5x – 25
5x – y – 20 = 0 …(1)
After 3 year Son’s age = (x + 3)
Father’s age = (y + 3)
∴ (y + 3) = 3(x + 3)
y + 3 = 3x + 9
∴ 3x- y + 6 = 0 …(2)

∴ x = 13yrs
y = 45 yrs

Question 32.
Solve the following graphically x + y < 5.
Answer:
x + y < 5

Question 33.
If ₹ 500 amounts to ₹ 725 at 9% simple interest in sometimes what will ₹ 600 amount to at 11% in the same time.
Answer:
Given P = ₹ 500, A = ₹ 725
SI = 225, R = 9%, T = ?
SI = \(\frac{\mathrm{PTR}}{100}\) = 225 = \(\frac{500 \times 7 \times 9}{100}\)
∴ T = ₹ 5yrs
Again is P = 600, R = 11% T = 5yrs I = ?
I = \(\frac{600 \times 5 \times 4}{100}\) = ₹330
∴ A = P + I = (600 + 300) = 930

Question 34.
5 kg of Sugar at the rate of ₹ 15/kg
8 kg of Wheat at a rate of ₹ 22/kg
7 kg of Rice at a rate of ₹ 351kg and
4 kg of oil at a rate of ₹ 85/kg. What is the average price/kg of all the commodities?
Answer:
Average price /kg (x̄w) = \(\frac{n_{1} w_{1}+n_{2} w_{2}+n_{3} w_{3}+n_{4} w_{4}}{n_{1}+n_{2}+n_{3}+n_{4}}\)
n = Quality, w = weight
= \(\frac{(5 \times 15)+(8 \times 22)+(7 \times 35)+(4 \times 85)}{(5+8+7+4)}\)
x̄w = \(\frac{836}{24}\) = ₹ 34.83

Question 35.
By how many percent should the use of tea be increased if the price of tea is decreased by 10% so that expenditure remains unchanged.
Answer:
Let the price of tea = ₹ 100
Quantity = 100 unit
∴ Total expenditure = (100 × 100) = 10,000
New price = 100 – 10 = 90 Net quantity = y
∴ Total expenditure = 90 y
∴ 90y= 10,000
∴ y = \(\frac{10000}{90}\) = 111.11
∴ % Increase = \(\frac{\text { Increase }}{\mathrm{CP}}\) × 100 = \(\frac{11.11}{100}\) × 100 = 11.11%

Question 36.
Find x : x sin45° cos260° = \(\frac{\tan ^{2} 60^{\circ}{cosec} 30^{\circ}}{\sec 45^{\circ} \cdot \cot ^{2} 30^{\circ}}\)
Answer:

Question 37.
Three comers of a parallelogram ABCD taken in order are A(-1, -6) B(2, 5) C(7, 2). Find the fourth vertex.
Answer:
Mid point of the diagonal AC = midpt. of diagonal BD
Formal D (x, y) \(\left[\frac{-1+7}{2}, \frac{-6+2}{2}\right]=\left[\frac{x+2}{2}, \frac{y-5}{2}\right]\)

4th vertex D = (x, y) = (4,1)

Question 38.
Find the equation of a line that passes through the point (-4 5) and whose intercepts are equal in magnitude but opposite in sign.
Answer:
Let a = a, b = – a ∴ \(\frac{x}{a}+\frac{y}{b}\) = 1 (Intercept form)
at(-4, 5), \(\frac{-4}{a}+\frac{5}{a}\) = 1
∴ a = 1, b = -1

Equation of the line is \(\frac{x}{1}+\frac{y}{1}\) = 1
∴ x – y – 1 = 0

Part-D

IV. Answer any SIX questions. (6 × 5 = 30)

Question 39.
In a certain college with 500 students, 300 take the milk and 250 take tea. Find how many take (a) Milk only (b) tea only (c) both milk and tea. Also, represent this by Venn diagram.
Answer:
n(M∪T) = 500
n(M) = 300 n(T) = 250
n(M∪T)= n(M) + n(T) – n(M∩T)
n(M∪T) = n(M)+n(T) – n(M∪T) = 300 + 250 – 500
n (M ∩ T) = 50
n(M – T)=n(M) – n(M∩T) = 300 – 50= 250
n(T – M)= n(T) – n(M ∩ T) = 250 – 50 = 200

Question 40.
Solve using log table : \(\frac{5.6348 \times 25.645}{12.75 \times 19.78}\)
Answer:
Let x = \(\frac{5.6348 \times 25.645}{12.75 \times 19.78}\)
apply log on both sides
log x = log\(\left[\frac{5.6348 \times 25.845}{120+2 \times 1 \times 19.7}\right]\)
logx = log 5.6348 + log 25. 645 – log (12.72)- log (19.78)
= (0.7508) + (1.4090) -(1.1044) – (1.2962)
log x = 0.2408
x = 1.4237

Question 41.
The first and last term of the GP is 3 and 96 respectively sum to n terms is 189. Find the common ratio and the number of terms.
Answer:
Given a = 3, a_n = 96, S_n = 189, r = ?
a_n = a.r^n-1
96 = 3.(r)^n-1
∴ (r)^n-1 = 32 ………(1)
If S_n = \(\frac{a\left(l-r^{n}\right)}{(l-r)}\)
189 = \(\frac{3\left(1-r^{n}\right)}{1-r}=\frac{3\left(1-r \cdot r^{n-1}\right)}{1-r}\)
189 = \(\frac{3(1-r(32))}{1-r}\)

1 – 32r = 63 – 63r
63r – 32r = 63 – 1
31r = 62
∴ r = 62/31 = 2
∴ r^n-1 = 32 = 2^n-1 = 32
2^n-1 = 25
∴ n – 1 = 5
∴ n = 6

Question 42.
Find the quotient and remainder obtained x³ + 4x² – 7x – 10 by (x + 1) dividing and solve.
Answer:
Given x + 1 = 0

Remeinder = 0, Quotient ∴ x² – 3x -10

Question 43.
Find the difference between simple interest and compound interest on ₹ 18,000 invested for 4 years at 8% pa. where Cl is compounded annually.
Answer:
Given P = ₹ 18,000, T = 4yr, R = 8%
SI – \(\frac{\mathrm{PTR}}{100}=\frac{18,000 \times 4 \times 8}{100}\) = ₹5,760
CI = A = P[(1 + i)]ⁿ
= 18,000[1 + 0.08]⁴
A = 24,488.8
∴ CI = 24.488.8 – 18,000 = 6488.8
∴ Difference between Cl and SI = 6488.8 – 5760 = ₹728.8

Question 44.
Mrs. Geetha deposits 4,00,000 on retirement in a bank that pays 10% p.a. interest. How much can be drawn annually for a period of 10 years?
Answer:
P = 4,00,000, i = 0.10, n = 10 ys, a = ?
P = \(\frac{a\left[(1+i)^{n}-1\right]}{i(1+i)^{n}}\)
4,00,000 = \(\frac{a\left[(1+0.10)^{10}-1\right]}{0.1(1+0.1) 10}\)
4,00,000 = \(\frac{a[2.5937-1]}{0.1 \times 2.5937}\)
∴ a = \(\frac{4,00,000}{6.1445}\)
a = ₹65,098.8

Question 45.
A shopkeeper sold a watch at a 5% loss. Had he purchased it at 10% p.a. sold it for 140 more his gain would have been one-fourth of the original cost price. Find the cost price of the watch.
Answer:
Let x be the original C.P of the watch
then 5% of x = \(\frac{95 x}{100}\) = SP
given caritian
(x – 10 %g.x) + \(\frac{x}{4}=\frac{95 x}{100}\) + 140
\(\frac{90 x}{100}=\frac{x}{4}=\frac{95 x+14000}{100}\)

90x + 25x = 95x + 14000
20x = 14000
∴ x = \(\frac{14000}{20}\)
x = ₹700

Question 46.
If then prove that x = arsin AcosB, y = brsin AsinBr and z = crossA then prove that \(\) = r2
Answer:
\(\frac{\mathrm{x}}{\mathrm{a}}\) = rsinAcosB \(\frac{\mathrm{y}}{\mathrm{b}}\) = rsinAstnB
\(\frac{\mathrm{z}}{\mathrm{c}}\) = rcos A
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\) = r²2sin2Acos2 B + r² sin A sin2 B + r² cos2A
= r² sin2 A (cos² B + sin² B) + r² cos² A = r² sin² A + r² cos² A = r2 (sin² A + cos² A) = r² (1) = r2

Question 47.
Find the circumcenter of the triangle whose vertices are (-1,2) (3,0) (-2,4).
Answer:
Assume the circum centre in p(x, y)
A (-1,2) B(3, 0), C(-2, 4)
PA = PB = PC
If PA = PB
∴ \(\sqrt{(x+1)^{2}+(y-2)^{2}}=\sqrt{(x-3)^{2}+(y-0)^{2}}\)

∴ 2x – 4y + 5 = -6x + 3
∴ 8x – 4y – 4 = 0 + 4
∴ 2x – y – 1 = 0 ………….(1)
Let PB = PC
\(\sqrt{(x-3)^{2}+(y-0)^{2}}=\sqrt{(x+2)^{2}+(y-4)^{2}}\)
∴ (x + 1)² + (y-2)² = (x-3)² + y²
-6x – 4x + 8 v + 9 – 20 = 0
-10x + 8y- 11 = 0 …(2)
Solve (1) and (2)
2x – y – 1=0
xy by 8

x = 2
y = 3
the circum centre = (2,3)

Question 48.
Find the foot of the perpendicular drawn from the point (-2, -1) on the line 3x + 2y – 5 = 0.
Answer:

Part-E

V. Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) If A = {x : x² – 5x + 6 = 0; x ∈ N}
B = {x : x² – 7x + 12 = 0 x ∈ N}
Find (A – B) × B. 4
Answer:
A= {2, 3},B = {3,4}
∴ A – B ={2}
∴ (A – B) × B = {2} × {3 – 4} = {(2, 3), (2, 4)}

(b) Find the sum to V terms of a GP. 0.6 + 0.66 + 0.66 +
Answer:
Let S_n = 0.6 + 0.66 + 0.666 +………….n term
S_n = 6{0.1 + 0.11 + 0.111 +……………n term}
∴ \(\frac{S n}{6}\) = 0.1 +0.11 +0.11 +…………..+ n term
x^ly both side by 9
∴ \(\frac{9 S n}{6}\) = 0.9 + 0.99 + 0.999 +……………. + n term
∴ \(\frac{9 S n}{6}\) = (1-0.1) + (1 – 0.01) + (1 – 0.001) +………..+ n term
\(\frac{9 S n}{6}=\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)\) + ……+ n term

(c) The average marks of 15 students of a class is 45. A student also has secured 50 marks leaves the class room. Find the average marks of the remaining 14 students.
Answer:
A × B = LCM × HCF
HCF 16, hem = 160, A = 64

∴ another no. = 4

Question 50.
(a) If the cost function C(x) of producing ‘x’ unit of a product is given by C(x)= 500x² + 2500 x + 5000 and if each unit of the product is sold at 6000 then find BEP.
Answer:
Given C(x) = 500x² + 2500 x ± 5000 (x = output)
∴ Total Revenue = R(x) = price x quantity = 6000x
For B ∈ P: C(x) = R(x)
∴ 500x² + 2500x + 5000 = 6000x
∴ 500x² – 3500x + 50000
÷500 ∴ x² – 7x + 10 = 0
x = 5 or 2 units

(b) Find ’a’ so that the lines x – 6y + a = 0, 2x + 3y + 4 = 0 and x + 4y + 1 = 0 are concurrent.
Answer:
Solve 2x + 3y + 4 = 0 ….(1) and x + 4y + 1 = 0 xly by 2
2x + 8y + 2 = 0 …(2)
(1) and (2)

2x + 3(2/5) + 4 = 0
2x = \(\frac{-6}{5}\) – 4 = \(\frac{-26}{5}\)
∴ x = -13/5
Sub in x – 6y + a = 0
a = 6y – x
∴ a = 6(2/5) – (-13/5)
a = \(\frac{12+13}{5}\) = 5

(c) The average marks of 15 students of a class is 45. A student also has secured 50 marks leaves the class room. Find the average marks of the remaining 14 students.
Answer:
Total marks and A. 15 students = 45 × 15 = 675

∴ Average marks of 14 student = \(\frac{625}{14}\) = 44.64

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2020 (South), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2020 (South)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the imaginary part of the complex number \(\)
Answer:
Imaginary part = \(\).

Question 2.
If A = {a, b, c, d, e}, B = {a, b, e, f} find A ∩ B
Answer:
A∩B = { a, b, e]

Question 3.
Simplify \(\left(x^{\frac{1}{2}}+y^{\frac{1}{2}}\right)\left(x^{\frac{1}{2}}-y^{\frac{1}{2}}\right)\)
Answer:

Question 4.
Form the quadratic equation whose roots are (2, -3).
Answer:
x2 – (2-3)x + (2-3) = 0
x2 + x – 6 = 0

Question 5.
Find the 8th term of the sequence \(\frac{1}{2}, \frac{1}{4}, \frac{1}{6}\) ………….
Answer:
a₈ = \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^{8-1}=\left(\frac{1}{2}\right)^{8}=\frac{1}{256}\)

Question 6.
Write log₈ 16 = 4/3
Answer:

Question 7.
Define annuity.
Answer:
A certain amount that is paid periodically is known as an annuity.

Question 8.
Find the simple interest on Rs. 500 at 5% for 5 years.
Answer:

Question 9.
Express \(\frac{7 \pi}{6}\) radians into degrees.
Answer:

Question 10.
Convert 0.12 into a percentage.
Answer:
(0.12 × 100) = 12%

Question 11.
Find the slope of the line perpendicular to the line 3x – 4y + 5 = 0.
Answer:
Slope = m = \(\frac{-(3)}{4}=\frac{3}{4}\)

Question 12.
Find the value of sin60° + cos60°.
Answer:
= \(\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2}\)

Part-B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.
Find the number of positive divisors of 360.
Answer:

T(N) = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24

Question 14.
Find the HCF of 18 and 24.
Answer:
HCFof 18 and 24 = 6

Question 15.
If A = {a, b, c, d} and B = {d, e, f, g, h, i} find A – B and B – A.
Answer:
A – B = {a, b, c}
B – A = {e,f g, h, i}

Question 16.
Simplify \(\frac{2^{n+1}+2^{n-1}}{2^{n}+2^{n+2}}\)
Answer:

Question 17.
If K + 9, K – 6 and 4 are in GP. then find the value of k.
Answer:
Common ratio \(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{T}_{3}}{\mathrm{~T}_{2}}\)
\(\frac{K-6}{K+9}=\frac{4}{K-6}\)
∴ (K-6)² = 4(K + 9)

K² = 16K
∴ K = 16

Question 18.
If α & B are the roots of the equation x² + 5x + 6 = 0 then find the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\).
Answer:
α + β = -b/a = -5 αβ = c/a = 6
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}=\frac{(-5)^{2}-2(6)}{6}=\frac{13}{6}\)

Question 19.
The sum of two consecutive numbers is 23. Find the numbers.
Answer:
Assume the two consecutive no. are x and x + 1
∴ x + (x + 1) = 23
∴ 2x = 2
∴ x = 1
the numbers are 11 and 12

Question 20.
Solve 2x + 6 < 0, x ∈ z the inequality in one variable and represent the solution on the number line.
Answer:
2x + b < 0
∴ x < -3

Question 21.
Find the future value of an annuity of ₹ 5000 at 12% p.a. for 6 years.
Answer:
A = Rs.5,000
r = 0.12
n = 6
F = \(\frac{A\left[(1+r)^{n}-1\right]}{r}=\frac{5000\left[(1+0.12)^{6}-1\right]}{0.12}=\frac{4869.11}{0.12}\) = 40,575.9

Question 22.
The weight of 6 men are 90 kg, 70.5kg, 56 kg, 45.5kg, 85 kg and 78 kg. Find the average.
Answer:
Average weight = \(\frac{90+70.5+56+45.5+85+78}{6}\) = 70.83

Question 23.
Prove that \(\sqrt{\frac{1+\cos x}{1-\cos x}}\) = cosec x + cot x.
Answer:

Question 24.
Evaluate sin2 π/6 + cos2 π/3 = tan2 π/4 + cot2 π/4
Answer:

Question 25.
Find the distance between the parallel lines. 3x + 4y + 5 = 0 and 6x + 8y + 20 = 0.
Answer:
d = \(\left|\frac{C_{1}-C_{2}}{\sqrt{a^{2}+b^{2}}}\right|=\left|\frac{5-20}{\sqrt{(3)^{2}+(4)^{2}}}\right|=\left|\frac{-15}{5}\right|\) = 3

PART-C

III. Answer any ten questions. (10 × 3 = 30)

Question 26.
Prove that \(\sqrt{3}\) is an irrational number.
Answer:
If \(\sqrt{3}\) in an rational no.
∴ \(\sqrt{3}\) = \(\frac{p}{q}\),p.q ∈ z, q ≠ u
i.e., p and q do not have any common factor

Squaring on both side
∴ \(\sqrt{3}\) = \(\frac{p}{q}\)
∴ p² – 3q² ⇒ 3 divide p²
⇒ 3 divide p …(1)
Let p = 3k (k∈z, k ± 0)
(3k)² = 3q²
q² = 3k² ⇒ 3 divide q²
⇒ 3 divide q …(2)
From (1) and (2) p and q have a common factor
This is a contradiction to the assumption
∴ \(\sqrt{3}\) is irrational

Question 27.
If A = {1, 2} B = {2, 3} C = {3, 4} Find (A × B) ∩ (A × C).
Answer:
A × B = {(1,2). (1,3), (2,2), (2,3),}
A × C= {(1,3),(1,4),(2,3),(2,4),}
(A × B) ∩ (A × C) = {(1, 3), (2, 3)}

Question 28.
Solve 32x – 10 × 3x + 9 = 0
Answer:
Let 3x = a ⇒ (3x)² = a²
⇒ 32x = a²
∴ a² – 10a + 9 = 0
a² – 9a – a + 9 = 0
a(a – 9) – 1 (a – 9) = 0
(a – 9)(a – 1) = 0
a – 1 = 0
a – 9 = 0
a = 1
a = 9
3^x = 3⁰ 3^x = 3²
x = 0
x = 2

Question 29.
Prove that log₄ 8 × log₂ 32 × log₁₆ 4 = \(\frac{15}{4}\)
Answer:

Question 30.
Insert 4 arithmetic means between 5 and 10.
Answer:
Let 5, a₁, a₂, a₃, a₄, 10
a = 5
d = ?
h = 0
a₆ = 10
a₆ = a + 5d
10 = 5 + 5 d
5 = 5d
d = 1
∴ a₁, a₂, a₃ and a₄ are 6, 7, 8 & 9

Question 31.
Divide Rs. 1600 between s, y and z so that y may have Rs. 100 more than x and z may have Rs. 200 more than y.
Answer:
Assume x = ₹ a
∴ y = (100 + a) ₹
2 = (200+ 100 +a) ₹
Given x + y + z = 1600
a = (100 + a) + (300 + a) = 1600
3a + 400 = 1600
3a = 1200
∴ a = ₹ 400
∴ x = ₹ 400,
y = ₹ 500,
z = ₹ 700,

Question 32.
Solve the inequality 3x + 2y > 6 graphically.
Answer:

Question 33.
Preritha bought a car for Rs. 4,00,000 if it depreciates at the rate of 12% p.a. how much will it be worth after 10 years?
Answer:
Given p = ₹ 4,00,000, n= 10 year, i = \(\frac{12}{100}\) = 0.12 A=?
Formula A = p[1 – i]ⁿ
= 4,00,000 [1 – 0.12]¹⁰
= 4,00,000 (0.88)¹⁰
= 4,00,000 x
= ₹ 1,11,400.39.

Question 34.
The average weight of a group of boys and girls is 38 kg. The average weight of the boys is 42 kg and that of the girls is 33 kg. If the number of boys is 25. Find the number of girls.
Answer:
Let the no. of girls be = x
∴ combined average weight boys and girls = \(\frac{(42 \times 25)+(33 \times x)}{25+x}\)
38 = \(\frac{1050+33 x}{25+x}\)
38(25 + x) = 1050 + 33x
38x – 33x = 1050 – 950
5x = 100
x = 20 girls

Question 35.
A man buys an article at (3/4)th of its cost price and sells it for 20% more than its cost price. What is his profit percentage?
Answer:
Let the Actual price = ₹ 100
Cost price = \(\frac{3}{4}\) × 100 = ₹ 75
20% profit of SP= 120
∴ profit = (120 -75) = 45
∴ profit % = \(\frac{\text { Profit }}{\mathrm{CP}}\) × 100 = \(\frac{45}{75}\) × 100 = 60%

Question 36.
Find the value of 3 tan2 30° + \(\frac{4}{3}\)cos2 30°-\(\frac{1}{2}\)cot2 45°- \(\frac{2}{3}\)sin2 60° + \(\frac{1}{8}\)sec4 60°
Answer:
\(3\left(\frac{1}{\sqrt{3}}\right)^{2}+\frac{4}{3}\left(\frac{\sqrt{3}}{2}\right)^{2}-\frac{1}{2}(1)^{2}-\frac{2}{3}\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{8}(2)^{4}\)
= \(3 \cdot \frac{1}{3}+\frac{4}{3} \cdot \frac{3}{4}-\frac{1}{2}(1)-\frac{2}{3} \cdot \frac{3}{4}+\frac{1}{8}(16)\)
= 1 + 1 – \(\frac{1}{2}\) – \(\frac{1}{2}\) + 2 = 4 – 1 = 3

Question 37.
A point P which moves such that (PA)² = 3(PB)² If A = (5, 0). Find the equation of the locus of P.
Answer:
Let p = (x, y) Given PA² = 3PB²
(x – 5)² + y² = 3[(x + 5)² + y²]
x² + 25 – 10x + y² = 3 (x² + 25 + 10x = y²)
x² + 25 – 10x + y² = 3x² + 75 + 30x + 3y²
2x² + 2y² + 40x + 50 = 0
x² + y² + 20x + 25 = 0
this is the equation of the locus.

Question 38.
Find the equation of the line passing through (5,2) and cutting off intercepts that are equal in magnitude but opposite in sign.
Answer:
By intercept form of the equation of like
\(\frac{x}{a}+\frac{y}{b}\) = 1 ……(1)
Given b = -a and (x, y) = (5, 2)
∴ (1) becomes
\(\frac{5}{a}+\frac{2}{-a}\) = 1
∴ \(\frac{5-2}{a}\) = 1
∴ a = 3, b = -3
∴ (1) becomes
\(\frac{x}{3}+\frac{y}{-3}\) = 1
x – y – 3 = 0 equation of the required line

Part -D

IV. Answer any SIX questions. (6 × 5 = 30)

Question 39.
Out of 50 people, 20 people drink tea, 10 take both tea & coffee. How many take at least one of the two drinks?
Answer:
n(T) = 20, n(C) = ? n(T ∩ C) = 10 n(T ∪ C) = 50
50 = 20 + n(C) – 10
∴ n(C) = 40

∴ no. of people taking atleast one of the two drinks
n(only T) = 10 Total = 40
n(only C) = 30

Question 40.
Evaluate \(\frac{0.5679 \times 0.0789}{0.0073 \times 0.123}\)
Answer:
Let x = \(\frac{0.5679 \times 0.0789}{0.0073 \times 0.123}\)

Question 41.
Find the sum of n terms of a GL P. 5 + 55 + 555 + …………
Answer:
5 + 55 + 555+ …………….
Let S_n = 5+ 55 + 555 +……….. n terms
S_n = 5(1 + 11 + 111 +…………)
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}\) = 1 + 11 + 111 +…………..
Multiply both the sides by 9,
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}\) = 9 + 99 + 999 +……….
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}\) = (10 – 1) + (10² – 1) + (10³ – 1) + ………..
\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{5}\) = (10 + 10² + 10³ + …………..) – (1 + 1 + 1 + ………)
\(\frac{9 S_{n}}{5}=\frac{10\left(10^{n}-1\right)}{10-1}\) – n
\(\frac{9 S_{n}}{5}=\frac{10\left(10^{n}-1\right)}{9}\) – n
S_n = \(\frac{5}{9}\left[\frac{10\left(10^{\mathrm{n}}-1\right)}{9}-\mathrm{n}\right]=\frac{5}{9}\left[\frac{10}{9}\left(10^{\mathrm{n}}-1\right)-\mathrm{n}\right]\), n ∈N

Question 42.
Find the integral root between -3 and 3 by inspection and then using synthetic division. Solve the equation x³ – 2x² – 5x + 6 = 0.
Answer:
Given p(x) = x² – 2xz – 5x + 6
Put x = -2
∴ p(-2) – (-2)³ – 2(-2)² – 5(-2) + 6 = – 8 – 8 + 10 + 6 = 0
By synthetic division

∴ x² – 4x + 3 = 0 ⇒(x – 3)(x – 1) = 0
∴ x = 3 & 1
∴ x = 1, -2, 3

Question 43.
A sum of money amounts to Rs. 19,500 in 5 years and Rs. 22,200 after 8 years at the same rate of simple interest. Find the principal and the rate of interest.
Answer:
Principle = P = ? T₁ = 5 yrs, R₁ = R, A₁= ₹ 19.500
T₂ = 8 yrs, R₂ = R, A₂ = ₹ 22.200

74(1 + 5R) = 65 (1 + 8R)
74+ 370 R = 65 + 520R
74 – 65 = 520R – 370R
∴ 9 = 150R

R = 6%
Substitute
19,500 = P(1 + 5 × 0.06)
∴ P = ₹15,000

Question 44.
If Poornima deposits Rs. 600 at the beginning of every year for the next 15 years. Then how much will be accumulated at the end of 15 years if the interest rate is 7% p.a?
Answer:
a = ₹600, n = 15 yrs, i = 7% = 0.07, F = ?
F = \(\frac{a\left[(1+i)^{n}-1\right]}{i}\)(1 + i)
= \(\frac{600\left[(1.07)^{15}-1\right]}{0.07}\)(1 + 0.07)
= \(\frac{600[2.759-1]}{0.07}\)(1.07)
= \(\frac{600 \times 1.759 \times 1.07}{0.07}\) = ₹16,132.54

Question 45.
The price of a pair of trousers was decreased by 22% to Rs. 390 what was the original price of the trouser?
Answer:
Let the original price = ₹ x
then (x – \(\frac{22}{100}\)x) = 390
x(1 – \(\frac{22}{100}\)) = 390
x(\(\frac{78}{100}\)) = 390
x = \(\frac{390 \times 10}{78}\) = ₹500

Question 48.
If θ = \(\frac{5}{2}\) and θ is acute then prove that \(\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}=\frac{19}{7}\)
Answer:
\(\left[\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}\right]\) + by sinθ

∴ put cot θ = \(\frac{5}{2}=\left[\frac{3(5 / 2)+2}{3(5 / 2)-4}\right]=\left[\frac{15+4}{15-8}\right]\) = \(\frac{19}{7}\)

Question 47.
Show that the following points are the vertices of a square A (1, 1) B(4, 1), C(4, 4), D(l,4)
Answer:
AB = \(\sqrt{(1-4)^{2}+(1-1)^{2}}=\sqrt{(-3)^{2}}\) = 3
BC = \(\sqrt{(4-4)^{2}+(1-4)^{2}}=\sqrt{(-3)^{2}}\) = 3
CD = \(\sqrt{(4-1)^{2}+(4-4)^{2}}=\sqrt{3^{2}}\) = 3
AD = \(\sqrt{(1-1)^{2}+(1-4)^{2}}=\sqrt{(-3)^{2}}\) = 3
∴ AB = BC = CD = AD = 3A. ABCD forms and equal.

Question 48.
Find ‘a’ so that the lines x – 6y + a = 0,2x + 3y + 4 = 0 and x + 4y + 1 = 0 are concurrent.
Answer:
2x + 3y + 4 = 0 …(1)
x + 4y +1 = 0 xly by 2
2x + 8y + 2 = 0 …(1)
(1) – (3) -5y + 2 = 0
∴ y = 2/5 in (1)
2x + 3(215) + 4 = 0
∴ 2x = –\(\frac{6}{5}\) – 4 = –\(\frac{26}{5}\)

Substitute x – 6y + a = 0
∴ a = \(\frac{13}{5}+\frac{12}{5}\) = 5

–\(\frac{13}{5}\) – 6(\(\frac{2}{5}\)) – a = 0
∴ a = 5

Part-E

V. Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) If U = {1,2,3,4,5,6,7,8,9} A = {1,2,3,4,5} B = {3,4,5,6,7} show that (A∩B)’ = A’ ∪ B’.
Answer:
(a) A∩B = {3,4,5} A’ = {6,7,8,9}
B’ = {1,2,89}
LHS= (A∩B)’ = {3,4, 5}’ = {1,2,6, 7,8,9}
RHS = A’∪B’ = {6,7,8,9}∪{1, 2, 8, 9} = {1, 2, 6, 7, 8, 9}
∴ LHS = RHS

(b) Insert 3 Geometric means between and \(\frac{1}{4}\) and \(\frac{1}{64}\)
Answer:
\(\frac{1}{4}\), g₁, g₂, g₃, \(\frac{1}{64}\)
3 Gm’n = g₁, g₂, and g₃
a = \(\frac{1}{4}\) formula an = a.rn-1
a⁵ = \(\frac{1}{64}\)
n = 5
r = ?
\(\frac{1}{64}=\frac{1}{4}\) x r5-1
\(\frac{1}{16}\) = r4
∴ r = \(\frac{1}{2}\)
∴ g₁ = ar = \(\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}\)
g₂ = ar² = \(\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}\)
g₃ = ar³ = \(\frac{1}{4} \times \frac{1}{8}=\frac{1}{32}\)

(c) Find the L.C.M of \(\frac{1}{3}, \frac{5}{6}, \frac{5}{7}\)
Answer:
\(\frac{\mathrm{HCF} \text { of } \mathrm{Nr}}{\text { Lem of Dr }}=\frac{1}{42}\)

Question 50.
(a) A manufacture produces and sells balloons at Rs. 8 / unit His fixed cost is Rs. 6500 and the variable cost/balloon is Rs. 3.20 calculate.
(i) Revenue function
(ii) Cost function
(iii) profit function
(iv) Revenue at BEP
Answer:
(i) Revenue function = R(x) = Price × Quantity = 8x Quantity = x unit
(ii) Cost function : C(x) = TVC + TFC = 3.5x + 6500
(iii) Profit function : P(x) = R(x) – C(x)
= 8x – (3.5x + 6500)
= 4.5x – 6500
(iv) Revenue B ∈ P
R(x) = C(x)
4.5x – 6500 = 0
x = \(\frac{6500}{4.5}\) = 1445 unit

(b) Find the foot of the perpendcular drawn from the point (-2, -1) on the line 3x- + 2y-5 = 0.
Answer:
Let P(x₁, y₁) = (-2, -1) given pt.
Q (h, k) be the Foot of ⊥r to the line ax + by+ 1 = 0
Formula
\(\frac{h-x_{1}}{a}=\frac{x-y_{1}}{b}=\frac{-\left(a x_{1}+b y+1\right)}{a^{2}+b^{2}}\)
a = 3, b = 2, c = 5, (x₁, y₁) = (-2, -1)

\(\frac{h-(-2)}{3}=\frac{k-(-1)}{2}=\frac{-[3(-2)+2(-1)-5]}{3^{2}+2^{2}}\)
\(\frac{h+2}{3}=\frac{k+1}{2}=\frac{-(-6-2-5)}{13}\)

∴ \(\frac{h+2}{3}\) = + 1
\(\frac{k+1}{2}\) = + 1
∴ h = 1
k = 1
∴ (h, k) = (1, 1)

(c) Find the equation of the locus of the point which moves such that it is equidistant from (4, 2) and the x-axis.
Answer:
Let p(x, y) be a locus
pt-A(4, 2) x-axis B(x, 0)
Given PA = PB
\(\sqrt{(x-4)^{2}+(y-2)^{2}}=\sqrt{(x-x)^{2}+(y-0)^{2}}\)
(x – 4)2 + (y – 2)2 = y2

Students can Download 1st PUC Basic Maths Model Question Paper 6 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Model Question Paper 6 with Answers

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any ten questions (10 × 1 = 10)

Question 1.
Find the imaginary part of 1 + i
Answer:
1 + \(\frac{1}{i}\) = 1 + \(\frac{1}{i}\) × \(\frac{i}{i}\) = 1 + \(\frac{i}{i^{2}}\) = 1 + \(\frac{i}{-1}\) = 1 – i
∴ Imaginary part = -1

Question 2.
If A = {5, 6}, B = {7, 8} find B × A
Answer:
B × A = {(7, 5), (7, 6), (8, 5), (8, 6)}

Question 3.
If f: R → R is defined byf(x) = 5x + 3 then find f(1/5)
Answer:
f(1/5) = 5(1/5) + 3 = 1 + 3 = 4

Question 4.
If log2 \(\sqrt{32}\) = x then solve for x.
Answer:
log2 \(\sqrt{32}\) = x ⇒ \(\sqrt{32}\) = 2^x ⇒ (2⁵)^1/2 = 2^x ⇒ 2^5/2 = 2^x ⇒ x = \(\frac{5}{2}\)

Question 5.
Simplify \(\left(\frac{9}{4}\right)^{-3 / 2}\)
Answer:
\(\left(\frac{9}{4}\right)^{-3 / 2}=\left(\frac{4}{9}\right)^{3 / 2}=\left[\left(\frac{2}{3}\right)^{2}\right]^{3 / 2}=\left(\frac{2}{3}\right)^{3}=\frac{8}{27}\)

Question 6.
Find the 12th term of the A.P 1, 4, 7, ………..
Answer:
a = 1, d – 3, T₁₂ = ?, n = 12
T_n = a + (n – 1)d
T₁₂ = 1 + (12 – 1)(3) = 1 + 11(3) = 1 + 33 = 34

Question 7.
Solve for x. 3(x + 5) – 25 = 9 + 2(x – 7)
Answer:
3 (x + 5) – 25 = 9 + 2 (x – 7)
3x + 15 – 25 = 9 + 2x – 14
3x- 10 = 2x- 5
3x- 2x = 10 + 5
x = 5

Question 8.
Convert \(\frac{4}{5}\) into percentage
Answer:
\(\frac{4}{5}=\frac{4}{5}\) × 100% = 4 × 20 = 80%

Question 9.
Convert 300° into radians.
Answer:
300° = 300 × \(\frac{\pi}{180}=\frac{5 \pi^{C}}{3}\)

Question 10.
If the slope of the line AB is \(\frac{5}{2}\) and AB is perpendicular to- CD then find the slope of CD.
Answer:
Slope of = \(\frac{-2}{5}\)

Question 11.
Find the simple interest on Rs. 1500 at 4% p.a. for 145 days.
Answer:
P = 1500 t = \(\frac{145}{365}\) R = 4%
SI = \(\frac{\mathrm{PTR}}{100}=\frac{1500 \times \frac{145}{365} \times 4}{100}\) =23.83523.84

Question 12.
Write the formula for the present value of an annuity due.
Answer:
P = \(\frac{\mathrm{A}\left[(1+r)^{n}-1\right]}{r(1+r)^{n}}\)(1 + r)

PART-B

II. Answer any Ten questions. (10 × 2 = 20)

Question 13.
Find the H.C.F of 55 and 210.
Answer:

The last non-zero remainder is 5, (55, 210) = 5

Question 14.
U = {1,2,3,4, 5,6, 8,9}, A = {1,2,3,4, 5}, B = {3, 4,5, 6, 7} Prove that (A∪B)’ = A’∩B’.
Answer:
A∪B ={1,2,3,4,5,6,7}
(A∪B)’ = {8,9} …………….(1)
A’ = {6, 8, 9}
B’ ={1,2, 8, 9}
A’∩ B’ = {8, 9} ……………..(2)
From (1) and (2) (A∪B)’ = A’∩B’

Question 15.
Prove that \(\frac{1}{1+x^{p-q}}+\frac{1}{1+x^{q-p}}\)
Answer:
LHS = \(\frac{1}{1+\frac{x^{p}}{x^{q}}}+\frac{1}{1+\frac{x^{q}}{x^{p}}}=\frac{x^{q}}{x^{q}+x^{p}}+\frac{x^{p}}{x^{p}+x^{9}}=\frac{x^{p}+x^{q}}{x^{p}+x^{q}}\) = 1 = RHS

Question 16.
Prove that loga \(\) = log_a m – log_a n.
Answer:
Let log_a m = x ⇔ a^x = m
log_a n = y ⇔ a^y = n
log_a \(\left(\frac{\mathrm{m}}{\mathrm{n}}\right)\) = z ⇔ a^z = \(\frac{m}{n}\)
Consider a^z = \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{\mathrm{a}^{x}}{\mathrm{a}^{\mathrm{y}}}\) = a^x-y
= log_a\(\left(\frac{\mathrm{m}}{\mathrm{n}}\right)\) = log_am – log_an

Question 17.
Insert 3 means between \(\frac{1}{4}\) and \(\frac{1}{64}\)
Answer:
Let g₁, g₂, g₃ be the 3 GM’s between \(\frac{1}{4}\) and \(\frac{1}{64}\)
∴ \(\frac{1}{4}\) g₁, g₂, g₃ \(\frac{1}{64}\) are in GP.
Here T₅ = \(\frac{1}{64}\)
⇒ ar⁴ = \(\frac{1}{64}\)
\(\frac{1}{4}\) r⁴ = \(\frac{1}{64}\)
r⁴ = \(\frac{4}{64}=\frac{1}{16}=\left(\frac{1}{2}\right)^{4}\)
∴ r = \(\frac{1}{2}\)
∴ The 3 GM’s are \(\frac{1}{8}, \frac{1}{16}, \frac{1}{32}\)

Question 18.
Solve 3x – 2 < 2x + 1, x ∈ R, represent the solution on the number line.
Answer:
3x – 2 < 2x + 1
3x – 2x < 2 + 1
x < 3

Question 19.
Find the quotient and remainder when 4x³ + 3x² – 2x 1 is divided by x +1 using synthetic division.
Answer:
Here the multiplier is -1
x = -1


Quotient = 4x² – x – 1 = 0 and Remainder = 0.

Question 20.
By selling 8 erasers a trader gains the selling price of 1 eraser, percent.
Answer:
Let the selling Price of one eraser be x then SP of 8 erasers = 8x
Profit =x
C.P =S.P- Profit
= 8x – x = 7x
Profit% = \(\frac{\text { Profit }}{\text { C.P }}\) × 100
= \(\frac{x}{7 x}\) × 100 = 14\(\frac{2}{7}\)%

Question 21.
Find the effective rate of interest when a sum lent at 18% p.a. is compounded quarterly.
Answer:
i = 0.18 q = 4
r = \(\left(1+\frac{i}{q}\right)^{q}\) – 1
= \(\left(1+\frac{0.18}{4}\right)^{4}\) – 1 = 0.1925
∴ r = 19.25%

Question 22.
The average weight of 10 boys is 30 kg. If an 11th boy is added then the average weight is r increased by 2 kg. Find the weight of the 11th boy.
Answer:
Total weight of 10 boys = 30 × 10 = 300 Kg.
Total weight of (10 + 1) boys = (30 + 2) 11
= 32 × 11
= 352 Kgs.
The weight of 11th boy added = (352 – 300) kg.
= 52 Kgs.

Question 23.
P.T. (1 + cot A – cosec A) (1 + tanA + sec.A) = 2
Answer:
L.H.S. = (1 + cot A – cosec A) (1 + tanA + sec.A)
= \(\left(1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{1}{\cos A}\right)\)
= \(\left(\frac{\sin A+\cos A-1}{\sin A}\right)\left(\frac{\cos A+\sin A+1}{\cos A}\right)\)
= \(\frac{(\cos A+\sin A)^{2}-1}{\sin A \cos A}\)
= \(\frac{\cos ^{2} A+\sin ^{2} A+2 \cos A \sin A-1}{\sin A \cdot \cos A}\)
= \(\frac{1+2 \cos A \cdot \sin A-1}{\sin A \cdot \cos A}\)
= \(\frac{2 \cos A \cdot \sin A}{\sin A \cdot \cos A}\) = 2 = R.H.S

Question 24.
Show that points A(7, 9), B(3, -7), C(-3, 3) are the vertices of the right-angled isosceles triangle.
Answer:
AB² = (7 – 3)² + (9 + 7)² = 272
BC² = (-3 -3)² + (3 + 7)² =136
AC² = (-7 -3)² + (9-3 )² = 136
Clearly AB² = AC² + BC². Thus ABC is a right-angled triangle. Also, BC² =AC² which implies the BC = AC and so ABC is an isosceles triangle. Hence the given three points form a right-angled isosceles triangle.

Question 25.
Find the equation to the straight-line cutting of equal intercepts and passing through (-2,5)
Answer:
Let the straight line AB cut off equal Mercept with both the axes.
Let OA = OB = a
Hence equation of AB is given by \(\frac{x}{a}+\frac{y}{a}\) = 1
i.e. x + y = a
Since this line passes through (-2, 5) we get -2 + 5 = a i.e. a = 3
Hence the required equation of AB is x + y = 3

PART-C

III. Answer any ten questions. (10 × 3 = 30)

Question 26.
Prove that 2 + 3\(\sqrt{5}\) is an irrational number.
Answer:
If possible, let 2 + 3\(\sqrt{5}\) is rational
∴ 2 + 3\(\sqrt{5}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\), p, q ∈ z, q ≠ 0
3\(\sqrt{5}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\) – 2 = rational – rational = rational.

∴ \(\sqrt{5}\) is irrational
∴ 3\(\sqrt{5}\) is irrational
∴ irrational number = a rational number which is a contradiction.
∴ our assumption is wrong.
∴ 2+ 3\(\sqrt{5}\) is irrational.

Question 27.
Out of 50 people, 20 people drink tea, 10 take-ups both tea, and coffee. How many take one of the two drinks. Show the result by Venn diagram.
Answer:

n (T ∪C) = n(T) + n(C) – n(T∩C)
50 = 20 + n (C) – 10
40 = n (C)
Number of people taking atleast one of the two drinks
10+ 10 + 30 = 40 + 10 = 50

Question 28.
Prove that \(\left(\frac{x^{2}}{x^{b}}\right)^{a^{2}+a b+b^{2}}\left(\frac{x^{b}}{x^{c}}\right)^{b^{1}+b c+e^{2}}\left(\frac{x^{e}}{x^{a}}\right)^{c^{2}+\infty+a^{2}}\) = 1
Answer:

Question 29.
If log \(\left(\frac{\mathrm{a}-\mathbf{b}}{4}\right)\) = log\(\sqrt{a}\) + log\(\sqrt{b}\), Show that (a + b)² = 20ab
Answer:
log\(\left(\frac{\mathrm{a}-\mathbf{b}}{4}\right)\) = lo\(\sqrt{a}\) + log\(\sqrt{b}\)
log\(\left(\frac{\mathrm{a}-\mathbf{b}}{4}\right)\) = log\(\sqrt{a b}\)
\(\frac{a-b}{4}\) = \(\sqrt{a b}\)
a – b = 4\(\sqrt{a b}\)
Squaring both sides, we get
(a-b)² = 16 ab
a² + b² – 2ab = 16ab
a² + b² – 16ab = 2ab
a² + b² – 18ab
a² + b² + 2ab = 18 ab + 2ab
(a – b)² = 20 ab

Question 30.
Ankur Choudhary agrees to pay the rent ₹ 30,000 for the first year ₹ 32,000 for the second year and so on each year the rent is increased by ₹ 2,000/- Find the total amount he paid for 10 years.
Answer:
a = 30,000
d = 2,000
n= 10
Now S_n = \(\frac{n}{2}\)[2a + (n- 1)d]
S_n = \(\frac{10}{2}\)(2(30,000) + (l0-l)2,000)
= 5[60,000 + 9(2,000)]
= 5(60,000 + 18,000]
= 5 [78,000]
= 3,90,000
∴ He paid 3,90,000 for 10 years.

Question 31.
Solve x³ – 2x² – 5x + 6 = 0.
Answer:
Let f (x) = x² – 2x² – 5x + 6
f(1)= 1 – 2 – 5 + 6 = 7 – 7 = 0
∴ x = 1 is a root of the given equation. Let us remove this root by Synthetic division.

∴ The resulting equation is x² – x – 6 = 0 is the quotient and remaind = 0
x² – x – 6 = 0
(x – 3) (x + 2) = 0
x = 3 or -2
Thus x = 1, -2, 3 are the roots of the given equation.

Question 32.
₹ 9000 amounts to 10,418.652 in 3 years. Find the compound interest rate percent.
Answer:
P = 9000 A= 10,418.625 n = 3r = ?
i = Antilog\(\left[\frac{\log A-\log P}{n}\right]\) – 1
= Antilog\(\left[\frac{\log 10418.625-\log 9000}{3}\right]\) – 1
= Antilog\(\left[\frac{4.0174-3.9542}{3}\right]\) – 1
= Antilog[0.021067] – 1
= 1.050 – 1
= 0.05
R = i × 100
= 0.05 × 100 = 5%

Question 33.
A father wishes to divide 50,000 amongst his two daughters who are respectively 12 and 15 years old in such a way that the sum invested at 5% p.a compound interest will give the same amount to both of them when they attain the age of 18. How is the sum divided?
Answer:
Let the share of the older daughter = x then share of young one = 50,000 – x
For older daughter,
A = A₁ P = x, i = 0.05, n = 3
A₁ = x (1 + 0.05)³
For younger daughter
A₁ = A₂ P = 50,000 -x, i = 0.05 n = 6
Given A₁ = A₂
x(1.05)³ = (50,000 – x)(1.05)⁶
x = (50,000 – x)(1.05)³
x = 57881.25 = 1.157625 x
2.157625 x = 57881.25
x = 26826 (Approx.)
∴ Share of older daughter = ₹ 26826
Share of younger daughter = 50,000 – 26826 = ₹ 23174

Question 34.
A person repaid his loan in 10 equal annual installments starting from the beginning of the first year. If each installment was ₹ 6000 and compound interest charged was 12% p. a. What was the amount borrowed.
Answer:
a = 6000, n=10, i = 0.12. It is the present value of annuity due.
P = \(\frac{a\left[(1+i)^{n}-1\right]}{i(1+i)^{n}}\)(1 + i) = \(\frac{8000\left[(1+0.05)^{3}-1\right](1+0.05)}{(0.05)(1+0.05)^{3}}\)
= \(\frac{8000[(1 . .2763)-1](1.05)}{0.05 \times 1.2763}\) = ₹ 36369.50

Question 35.
The average age of 10 students in a class increases by 4.8 months. When a boy of age 6 years is replaced by a new boy. What is the age of the new boy?
Answer:
Let the average age of 10 boys = x years
Total age of 10 boys = 10x
After replacement of a boy of 6 years age and inclusion of new boys if his age is assumed to be y years.
Then the total age of 10 boys in a new group = (10x – 6) + y
Given the new average of 10 boys = (x + 4.8 months) or
= x + \(\frac{4.8}{12}\) years
= (x + 0.4) years
∴ Average of new group of 10 boys = \(\frac{\text { Total age of } 10 \text { boys }}{\text { No of boys }}\)
(x + 0.4) = \(\frac{(10 x-6)+y}{10}\)
10 (x + 0.4) = (10x – 6) + y
10x + 4 = 10x-6 + y
∴ y = 10
∴ Age of the new boy replaced = 10 years

Question 36.
P.T. (1 + sin θ + cosθ)² = 2(1 + sinθ) (1 + cosθ)
Answer:
LHS. = {(1 + sin θ) + (cos θ)}²
= 1 + sin2θ + 2sinθ + cos2θ + 2cosθ (1 + sin θ)
= 2 + 2 sin θ + 2 cos θ (1 + sin θ)
= 2(1 + sin θ) + 2 cos θ (1 + sin θ)
= (1 + sin θ) (2 + 2 cos θ)
= 2 (1 + sin θ) (1 + cos θ)

Question 37.
Show that the points (1 -1) (5, 2) and (9, 5) are coolinear.
Answer:
We know that the three points A, B and C taken in this order are collinear if and only if AB + BC = AC
Let A= (1, -1), B = (5, 2) C = (9, 5)
Consider
AB = \(\sqrt{(5-1)^{2}+(2+1)^{2}}=\sqrt{16+9}\) = 5
BC = \(\sqrt{(9-5)^{2}+(5-2)^{2}}=\sqrt{16+9}\) = 15
AC = \(\sqrt{(-9-1)^{2}+(5+1)^{2}}=\sqrt{64+36}\) = 10
Clearly AB + BC = AC
⇒ The points, A, B and C are collinear

Question 38.
If the lines 2x – y = 5, Kx – y = 6 and 4x – y = 7 are concurrent, find K.
Answer:
We have 2x – y = 5 ….(1)
Kx – y = 5 ………(2)
4x – y = 7 ……..(3)
Solving (1) and (3) we get x = 1 and y = – 3
Since the lines are concurrent put x = 1 and y = 3 in (2)
We get K (1) – (-3) = 6 ⇒ K + 3 = 6 ⇒ K = 3

PART – D

IV. Answer any six questions. (6 × 5 = 30)

Question 39.
f(x) = x + 1 and g(x) = x² + 1
Find (i) fog (1) (ii) fog (2) (iii) gof (1), (iv) gof (2) (v) fog (3)
Answer:
(i) fog(1) = f (g (1))
= f (2)
= 2 + 1 = 3
(ii) fog (2) = f(g(2))
= f(5)
= 5 + 1 = 6
(iii) gof (1) = gof(1)
= g(f(D)
= g (2) = 4 + 1 = 5
(iv) gof (2) = g(f(2))
= g (3)
= 9 + 1 = 10
(v) fog (3)
f(g(3))
f (10) = 10 + 1 = 11

Question 40.
Find the sum of 6 terms of the GP 1, 3, 9
Answer:
Here, a = 1, r = 3, n = 6
Sn = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
Sn = \(\frac{729-1}{2}=\frac{729-1}{2}\) = 364

Question 41.
If \(\frac{2}{3}\), x, \(\frac{1}{x}\) areinH.Pfindx
Answer:
Given
\(\frac{2}{3}\), x, \(\frac{1}{2}\) are in H.P.
⇒ \(\frac{3}{2}\), \(\frac{1}{x}\), 2 are in A.P.
⇒ \(\frac{1}{x}-\frac{3}{2}\) = 2 – \(\frac{1}{x}\)
\(\frac{1}{x}+\frac{1}{x}=\frac{3}{2}\) + 2
\(\frac{2}{x}=\frac{7}{2}\)
\(\frac{x}{2}=\frac{2}{7}\)
∴ x = \(\frac{4}{7}\)

Question 42.
A certain two digits number is 2 times the sum of the digits, if 63 is added to the number the digits get interchanged. Find the number.
Answer:
Let the digit in ten’s place be x and digit in units place be v.
∴ The number is 10x + y
Given 10x + y = 2 (x + y)
8x – y = 0 ……………(1)
Also given 10x + y + 63 = 10y + x
9x – 9y + 63 = 0 ⇒ x – y + 7 = 0 ………….(2)
Solving (1) and (2), we get: x = 1 and y = 8
∴ The required number is 10x + y = 10
1 + 8 = 10 + 8 = 18.

Students can Download 1st PUC Basic Maths Model Question Paper 5 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Model Question Paper 5 with Answers

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART – A

I. Answer any ten questions (10 × 1 = 10)

Question 1.
Write the canonical representation of 60.
Answer:

Question 2.
If A = {1, 2, 3}, B = {3, 4, 5, 6} find A – B.
Answer:
A – B = {1,2}

Question 3.
If f: R → R is defined by/(x) = 2x + 3 then find f(1/2).
Answer:
f(1/2) = 2 × \(\frac{1}{2}\) + 3 =1 + 3 = 4

Question 4.
Simplify : (3°)2 + (32)°
Answer:
= 12 + 9° = 1 + 1 = 2

Question 5.
Evaluate : log\(\sqrt{x}\) 4 = 2
Answer:
⇒ (\(\sqrt{2}\))2 = 4 ⇒ x = 4

Question 6.
Find the 11th term of the A.P. 3, 5, 7, 9…………
Answer:
Here a = 3, d = 5 – 3 = 2, n = 11
T_n = a + (n – 1 )d
T₁₁ = 3 + (11 – 1)2 = 3 + (10) = 3 + 20 = 23

Question 7.
The sum of 6 times a number and 5 times the number is 55. Which is that number?
Answer:
Let the no. be x.
Given 6x + 5x =55
11x =55
x = \(\frac{55}{11}\) = 5
Hence the number is 5.

Question 8.
Convert the ratio 2: 5 into percentages.
Answer:
2:5 = \(\frac{2}{5}\) × 100% = 40%

Question 9.
Write the formula for the future value of annuity immediate.
Answer:
F = \(\frac{\mathrm{A}\left[(1+r)^{n}-1\right]}{r}\)

Question 10.
Convert 225° into radians.
Answer:
225° = 225 × \(\frac{\pi}{180}=\frac{5 \pi}{4}\)

Question 11.
The average age of 10 girls ¡n a class in 17 years. What is the sum of their ages?
Answer:
Sum of their ages 10 × 17 = 170 years.

Question 12.
Find the slope of the line 3x + 4y – 29 = O
Answer:
Slope = \(\frac{-a}{b}=\frac{-3}{4}\)

PART-B

Answer any Ten questions. (10 × 2 = 20)

Question 13.
Find the real part of \(\frac{1+2 i}{3-4 i}\)
Answer:
\(\frac{1+2 i}{(3-4 i)} \times \frac{(3+4 i)}{(3+4 i)}=\frac{(1+2 i)(3+4 i)}{9+16}=\frac{(3-8)+i(10)}{25}=\frac{-5+10}{25}=-\frac{1}{5}+\frac{2 i}{5}\)
Real part = –\(\frac{1}{5}\) Imaginary part = \(\frac{2}{5}\)

Question 14.
Let f = {(1, 2), (2,3), (3, 4)} be a function from 2 to 2 where z is the set of integers defined by/W = ax + b v some integers a and b. Determine a and b.
Answer:
2 = a + b …(i) 3 = 2a + b …………. (ii)
equation (1) and (2)
-1 = -a
∴ a = 1 ∴ b = 1

Question 15.
Solve : 32x – 10. 31 + 9 = 0.
Answer:
32x – 10. 3x + 9 = 0
Let 3x = a ⇒ (3x)² = a²
⇒ 3^2x = a²
∴ a² – 10a + 9 = 0
a² – 9a – a + 9 = 0
a(a – 9) -1 (a – 9) = 0
(a – 9) (a – 1) = 0
a – 1 = 0
a – 9 = 0
a = 1 a = 9
3^x = 3°
3^x = 3²
x = 0 x = 2

Question 16.
Prove that log_a mn = log_a m + log_a n
Answer:
Let log_a m=x ⇔ a^x = m
log_a n = y ⇔ a^y = n
log_a mn = z ⇔ a^z = mn
Consider a^z = mn ⇔ a^z = a^x.a^y ⇒ a^z = a^x+y ⇒ z = x + y ⇒ log_a mn = log_a m + log_a n

Question 17.
If \(\frac{3}{5}\), K, \(\frac{13}{5}\) are inA.P., then find the value of K.
Answer:
k – \(\\frac{3}{5}=\frac{13}{5}\) – k
k + k = \(\frac{13}{5}+\frac{3}{5}=\frac{16}{5}\)
2k = \(\frac{16}{5}\) ⇒ k = \(\frac{8}{5}\)

Question 18.
Form cubic equation whose roots are 3, 5, 7
Answer:
3, 5, 7 are the roots
∴ x – 3, x – 5, x – 7 are the factors
Equation is (x – 3) (x – 5) (x – 7) = 0
⇒ x³ – 15x² + 71x – 105 = 0

Question 19.
Solve 3x – 4 > 7 – 2x (x∈R)
Answer:
3x – 4 > 7 – 2x
3x + 2x > 7 + 4
5x > 11
x > \(\frac{11}{5}\) or x ∈ (\(\frac{11}{5}\), ∞)

Question 20.
Find the simple interest on Rs. 800 at 8% p.a. for one year one month and 10 days.
Answer:
P = 800. R = 8 T = 1yr, I month and 10 days
T = (1 + \(\frac{1}{12}+\frac{10}{365}\))years (1 + 0.08 + 0.02)(1.1)year
SI = \(\frac{\mathrm{PTQ}}{100}=\frac{800 \times 8 \times 1.1}{100}\) = 71.11

Question 21.
Find the effective rate of interest when a sum lent at 12% is computed half yearly.
Answer:
R = Nominal rate of interest = 12% = \(\frac{12}{100}\) = 0.12
q = number of times interest is compounded in one year = 2
r = effective rate of interest
= \(\left(1+\frac{R}{q}\right)^{q}\) – 1 = \(\left(1+\frac{0.12}{2}\right)^{2}\) – 1 = (1 + 0.6)² – 1 = (1.06)² – 1 = 1.1236 – 1
= 0.1236 = 12.36%
∴ Effective rate of interest = 12.36%

Question 22.
Find the present value of a perpetuity of Rs.3000 to be received forever at 4% p.a.
Answer:
A = Rs.3000 r = 0.04
P∞ = \(\frac{\mathrm{A}}{\mathrm{r}}=\frac{3000}{0.04}\) = 75,000

Question 23.
The average mark of is students in a class is 45. A student who has scored 50 marks leaves the classroom. Find the average marks of the remaining 14 students.
Answer:
n= 15, x̄ =45, Σx = nx̄ = 15(45) = 675
Total marks of 14 students = 675 – 50 = 625
Average marks of 14 students = \(\frac{625}{14}\) =44.6

Question 24.
The angles of a triangle are in the ratio 3:4:5. Find them in radians and in degrees.
Answer:
3:4:5 or 3x, 4x, 5x
3x + 4x + 5x = 180°, 12x = 180
x = \(\frac{180}{12}\) = 15°
Angles are 3(15°), 4(15°), 5(15°)
45°, 60°, 75°,
45° = 45 × \(\frac{\pi}{180}=\frac{\pi}{4}\)
60° = 60 × \(\frac{\pi}{180}=\frac{\pi}{3}\)
75° = 75 × \(\frac{\pi}{180}=\frac{5 \pi}{12}\)

Question 25.
If the distance between (2, a) and (-1, 1) is \(\sqrt{13}\) find the value of a
Answer:
Let A = (2, a) B = (-1, 1)
∴ AB = \(\sqrt{(2+1)^{2}+(a-1)^{2}}\)
i.e \(\sqrt{13}=\sqrt{9+(a-1)^{2}}\)
squaring
∴ 13 = 9 + (a – 1)²
∴ (a – 1)² = 4
∴ (a – 1) = ±2
∴ a – 1 = 2 or a – 1 = -2
i.e., a = 3 or a = -1

PART-C

III. Answer any ten questions. (10 × 3 = 30)

Question 26.
Prove that 3 + \(\sqrt{5}\) is an irrational number.
Answer:
If possible let 3 + \(\sqrt{5}\) be a rational number
∴ 3 + \(\sqrt{5}\) = \(\frac{p}{q}\) where p, q ∈ z, q ≠ 0
∴ \(\sqrt{5}\) = \(\frac{p}{q}\) – 3 = rational – rational rational number
∴ \(\sqrt{5}\) is a rational number which is a contradiction.
∴ our assumption is wrong
∴ 3 + \(\sqrt{5}\) is irratjona1.

Question 27.
In a class of 150 students, It was found that 95 like burgers and 79 like pizzas. Assuming every student like at least one of the above, find the number of students who like both burgers and pizzas.
Answer:
n(B∪P) = n(B) + n(P) – n(B ∩ P)
150 = 95 + 79 – n(B ∩ P)
∴ n(B ∩ P) – 95 + 79 – 150 = 24
n(B ∩ P) = 24

Question 28.
If 45^1/x = 3^1/y = 5^1/z, prove that x = 2y + z.
Answer:
Let 45^1/x = 3^1/y = 5^1/z = k
(45)^1/x = k ⇒ 45 = k^x
3^1/y = k ⇒ 3 = k^y
5^1/z = k ⇒ 5 = k^z
Consider kx = 45 = 9 × 5 = 3² × 5 = (k^y)² k^z = k²y k^z = k^2y-z
∴ x = 2y + z

Question 29.
If x = log_2a a, y = log_3a 2a, z = log_4a 3a. Show that xyz + 1 = 2yz.
Answer:
xyz +1 = log_2a a. log_3a 2a. log_4a 3a + 1
= log_3a a. log_4a 3a. +1 (∵ log_b a. log_c b = log_c a)
= log_4a a +1
= log_4aa + log_4a 4a (∵ log_a a = 1)
= log_4a 4a²
= log_4a (23)²
= 2 log_4a 2a (∵ log an = m log a)
= 2log_3a 2a. log_4a 3a = 2 yz = RHS

Question 30.
The sum of three numbers in APis -18 and sum of their squares is 140. Find the numbers.
Answer:
(a – d) + a + (a + d) = -18
3a = -18 ⇒ a = -6
(a – d) + a² + (a + d)² = 140

3a² + 2d² = 140
3(-6)² + 2d² = 140
108 + 2d² = 140
2d² = 140- 108
2d² = 32
d² = 16
d = 4
∴ The numbers are
-6 – 4, -6, -6 + 4
-10, -6, -2

Question 31.
Find an integral root between -3 and 3 by inspersion and then using synthetic division solve the equation x³ + 2x² – 11x -12 = 0.
Answer:
Put x = – 1
(-1)³ + 2(- 1)² – 11 (- 1) – 12
-1 + 2 + 11 – 12 = 13 – 13 = 0
x = -1 is a root, x + 1 is a factor

Quotient = x² + x – 12
Remainder = 0
x³ + 2x² – 11x – 12 = 0
(x + 1)(x² + x – 12) = 0
x + 1 = 0, x² + x – 12 = 0
x = 3, -4 x = -1
The roots are -1. 3, -4

Question 32.
Pradeep invested equal amounts one at 6% SI and the other at 5% CI. If the former earns Rs.437.5 more as interest at the end of two years, find the total amount invested.
Answer:
Let the amount invested = x
P = x r = 6% T = 2years
SI = \(\frac{P \times T \times R}{100}=\frac{x \times 2 \times 6}{100}=\frac{12 x}{100}\) = 0.12x
P = x r = 5% = 0.05 n = 2
A =P(1 + r)ⁿ = x(1 + 0.05)² = x(1.05)² = 1.1025x
CI = A – P = 1.1025 x – x = 0.1025 x
Given 0.12x – 0.1025x = 437.5
0.0175x = 437.5
Amount invested at 6% SI = 25,000
Amount invested at 5% Cl = 25,000
∴ Total amount invested = 25,000 + 25,000
= Rs. 50,000

Question 33.
A school runs in the morning and afternoon shifts and employs 40 teachers. The average salary of 25 teachers working in the morning shift is Rs.2800/- and the average salary of teachers working the afternoon shift is Rs.3000/- find
(i) the average salary of the teachers in the school.
(ii) the average salary is 5 teachers shifted from morning to afternoon shift,
Answer:
(i) N₁ = 25 X₁ = 2800
N₂ = 15 X₂ = 3000
X̄ = \(\frac{\mathrm{X}_{1} \mathrm{~N}_{1}+\mathrm{X}_{2} \mathrm{~N}_{2}}{\mathrm{~N}_{1}+\mathrm{N}_{2}}=\frac{2800(25)+3000(15)}{25+15}=\frac{70000+45000}{40}\) = Rs. 2875

(ii) X₁ = 2800 N₁ = 20
X₂ = 3000 N₂ = 20
X̄ = \(\frac{X_{1} N_{1}+X_{2} N_{2}}{N_{1}+N_{2}}\)
= \(\frac{2800(20)+3000(20)}{20+20}=\frac{56000+60000}{40}\) = Rs. 2900

Question 34.
A wholesale dealer sold a machine to a shopkeeper at 20% profit. The shopkeeper sold it to a customer so as to get 25% profit for himself. The difference between the selling price of the shopkeeper and that of the wholesale dealer was found to be Rs.129. Find the initial price of the machine.
Answer:
Let x be C.P. of the machine
SP of wholesale dealer = x + 20% of x = \(\frac{120 x}{100}\)

SP of the shopkeeper = 25% of \(\frac{120 x}{100}+\frac{120 x}{100}\)
\(\frac{120 x}{100}+\frac{25}{100}\left(\frac{120 x}{100}\right)-\frac{120 x}{100}\) = 129
\(\frac{25}{100} \times \frac{120 x}{100}\) = 129
30x = 129 × 100
x = \(\frac{12900}{30}\) = Rs. 430

Question 35.
Find the orthocentre of the triangle formed by the vertices A (5, -2), B (-1, 2) C (1, 4).
Answer:
Let AD, BE be the altitudes.
Now slope of BC =\(\frac{4-2}{1+1}\) = 1
∴ Slope of AD = -1 since AD perpendicular to BC.
∴ Equation to the altitude AD is y + 2 = -1 (x – 5)
or x + y – 3 = 0
Again slope of AC = \(\frac{4+2}{1-5}=-\frac{3}{2}\)
∴ Slope of altitudes BE = \(-\frac{1}{3}=\frac{2}{3}\)
∴ Equation to the altitude BE is
y – 2 = \(\frac{2}{3}\)(x+ 1)
or 2x – 3y + 8 = 0 (2)
Solving (1)and (2), we get x = \(\frac{1}{5}\), y = \(\frac{14}{5}\)
∴ The orthocentre is \(\left(\frac{1}{5}, \frac{14}{5}\right)\).

Question 36.
Prove that tan 120° +3 sin2 3000_ 2 cosec2 240° – \(\frac{3}{4}\) cot2 60° = \(\frac{10}{3}\)
Answer:
tan 120 = tan(180 – 60) = -tan60° = \(\sqrt{3}\)
sin 300 = sin (360 – 60) = -sin 60 = \(-\frac{\sqrt{3}}{2}\)
cosec 240 = cosec (180 + 60) = —-cosec 60 = \(-\frac{2}{\sqrt{3}}\)
cot 60 = \(\frac{1}{\sqrt{3}}\)
LHS = \(\frac{4}{3}(-\sqrt{3})^{2}+3\left(-\frac{\sqrt{3}}{2}\right)^{2}-2\left(\frac{-2}{\sqrt{3}}\right)^{2}-\frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^{2}\)
= \(\frac{4}{3}(3)+3\left(\frac{3}{4}\right)-2\left(\frac{4}{3}\right)-\frac{3}{4}\left(\frac{1}{3}\right)\) = 4 + \(\frac{9}{4}-\frac{8}{3}-\frac{1}{4}=\frac{48+27-32-3}{12}=\frac{40}{12}=\frac{10}{3}\) = RHS

Question 37.
Find a point that is equidistant from the points (1, 2), (5, -6), and (3, -4).
Answer:
P = (x, y) A(1, 2) B(5, -6)
PA = PB = PC
PA = PB
\(\sqrt{(x-1)^{2}+(y-2)^{2}}=\sqrt{(x-5)^{2}+(y+6)^{2}}\)
(x – 1)² + (y – 2)² = (x – 5)² + (y + 6)²

-2x + 10x – 4y – 12y + 5 – 61 = 0, 8x – 16y -56 = 0
x – 2y -7 = 0 (1)
PB = PC
\(\sqrt{(x-5)^{2}+(y+6)^{2}}=\sqrt{(x-3)^{2}+(y+4)^{2}}\)
(x – 5)² + (y + 6)² = (x – 3)² + (y +4)²

-10x + 6x + 12y – 8y + 61 – 25 = 0
-4x + 4y + 36 = 0
x – y – 9 = 0 (2)
÷ -4

y = 2
x – y – 9 = 0
x – 2 – 9 = 0
x – 11 = 0
x = 11
Required Point = P = (11, 2)
Which is nothing but the circumcenter of the ∆le ABC.

Question 38.
Find the lengths of the altitudes of the triangle whose vertices are (5, 2) (3, -3) and (-4, 3).
Answer:
A = (5,2) B = (3, -3) C = (-4, 3)
Area of ∆ABC = \(\frac{1}{2}\) [x₁(y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\)[5 (-3 – 3) + 3 (3 – 2) -4 (2 + 3)] = \(\frac{1}{2}\)[5 (-6) + 3 (1) – 4 (5)]
= \(\frac{1}{2}\) [-30 + 3 – 20] =\(\frac{1}{2}\)[-47] = \(\left|\frac{-47}{2}\right|=\frac{47}{2}\) = Square units
Area of the ∆ABC = \(\frac{1}{2}\) × BC × AD
AD = \(\frac{2 \Delta \mathrm{ABC}}{\mathrm{BC}}\)
BC = \(\sqrt{(-4-3)^{2}+(3+3)^{2}}=\sqrt{49+36}=\sqrt{85}\)

AC = \(\sqrt{(-4-5)^{2}+(3-2)^{2}}=\sqrt{81+1}=\sqrt{82}\)
∆ABC = \(\frac{1}{2}\) × AC ×BE
BE = \(\frac{2 \times \Delta \mathrm{ABC}}{\mathrm{AC}}=\frac{2 \times 47}{2 \times \sqrt{82}}\)
BE = \(\frac{47}{\sqrt{82}}\)
AB = \(\sqrt{(3-5)^{2}+(-3-2)^{2}}=\sqrt{4+25}=\sqrt{29}\)
∆ABC = \(\frac{1}{2}\) × AB ×CF

PART – D

IV. Answer any six questions. (6 × 5 = 30)

Question 39.
If x^5/3 – y^1/3 – z = 0, then show that (x⁵ – y – z³)³ = 27 x⁵ yz³
Answer:
x^5/3 – y^1/3 – z = 0 ⇒ x^5/3 – y^1/3 = z
Cubing on both sides
(x^5/3 – y^1/3) = z³
(x^5/3)³ – (y^1/3)³ – 3x^5/3y^1/3(x^5/3 – y^1/3) = z³
x⁵ – y – 3x^5/3 y^1/3(z) = z³ ⇒ x⁵ – y – z³ = 3x^5/3 y^1/3 z
Cubing on both sides
(x⁵ – y – z3)³ = 27x⁵yz³

Question 40.
Using log tables find the value of \(\frac{\sqrt{14.5} \times \sqrt[3]{8.571}}{(16.751)^{2 / 3}}\)
Answer:
logx = log \(\frac{\sqrt{14.5} \times \sqrt[3]{8.571}}{(16.751)^{2 / 3}}\)
= log\(\sqrt{14.5}\) + log\(\sqrt[3]{8.571}\) – log(16.751)2/3
= \(\frac{1}{2}\)log14.54 + \(\frac{1}{3}\)log8.571 – \(\frac{2}{3}\)log(l6.751) .
= \(\frac{1}{2}\)(l.l614) + \(\frac{1}{3}\)(0.9331) – \(\frac{2}{3}\)(1.2240)
= 0.5807 + 0.3110 – 0.816 = 0.0757
x = antilog 0.0757= 1.191

Question 41.
Find the sum of all integers between 60 and 400, which are divisible by 13.
Answer:
S_n = 65 + 78 + 91 +…………..+ 390
a = 5 d = 13 n = ? Tn = 390
T_n = a + (n – 1)d
390 = 65 + (n – 1)13
390 = 65 + 13n- 13
13n = 338 n = 26
S_n = \(\frac{n}{2}\) [a + l] = \(\frac{26}{2}\)(65 +390) = 13 (455) = 5,915
∴ S = 5,915

Question 42.
Two brothers have their annual income in the ratio 8: 5, while their annual expenditures are in the ratio 5 : 3, if they save ₹ 1200/- and ₹ 1000/- per annum. Find their incomes.
Answer:
Let the income be x and expenditure be y.
So the income of two brothers would be 8x and 5x and expenditure would be 5y and 3y.
We knew that Income – Expenditure = Saving
∴ we get:
8x – 5y = 1200 …………..(1)
5x – 3y = 1000 …………..(2)
Solving (1) and (2), we get: x = 1400 and y – 2000
∴ Annual incomes of two brothers are 8x and 5x
= 8 × 1400 & 5 × 1400 = ₹ 11200 and ₹ 7000.

Question 43.
If the interest on 800 is more than the interest on ₹ 400 by Rs 40 in 2 years. Find the rate of interest.
Answer:
P = 800, T = 2, R = R
SI = \(\frac{800 \times 2 \times R}{100}\)
I₁ = 16R
P = 400, T = 2 R = R
SI = \(\frac{400 \times 2 \times \mathrm{R}}{100}\)
I₂ = 8R
I₁ = I₂ = 40
16R – 8R = 40
R = \(\frac{40}{8}\) = 5%

Question 44.
If the cost function C (x) of producing ‘x’ unit of a product is given by C(x) = 500x² + 2500x + 5000 and if each unit of the product is sold at ₹ 6000. then find BEP.
Answer:
Given C(x) = 500 x² + 2500x + 5000 (x = output)
Total Revenue = R(x) = Selling price X quantity = 6000 x
For BEP : C (x) = R (x)
500x² + 250Qx + 5000 = 6000x
500x² + 2500x – 6000x + 5000 = 0
500x² – 3500x – 5000x = 0
÷ 500
x² – 7x + 10 = 0
(x – 5) (x – 2) = 0
x = 5 or 2 units

Question 45.
\(\frac{\mathrm{x} \cdot{cosec}^{2} 30^{\circ} \cdot \sec ^{2} 45^{\circ}}{8 \cos 45^{\circ} \cdot \sin 60^{\circ}}\) = tan2 60° – tan2 30°
Answer:

Question 46.
Find the ratio in which the co-ordinate axes, divide the line joining the points (2, 5) and (1, 9). Find also the coordinates of the points of division.
Answer:
The co-ordinate of the point dividing the join of A (2, 5) and B (1, 9) in the ratio
K: 1 are \(\left(\frac{K+2}{K+1}, \frac{9 K+5}{K+1}\right)\) (1)
(i) If the point lies on x axis then its y co-ordinate is zero
i.e. \(\frac{9 \mathrm{~K}+5}{\mathrm{~K}=1}\) = 0
⇒ K = \(\frac{-5}{9}\)
∴ x axis divides AB externally in the ratio 5 : 9
Substituting K = \(\frac{-5}{9}\) in (1) the co-ordinates of the point of division is \(\left(\frac{K+2}{K+1}, \frac{9 K+5}{K+1}\right)=\left(\frac{\frac{-5}{9}+2}{\frac{-5}{9}+1}, 0\right)=\left(\frac{13}{4}, 0\right)\)

(ii) If the point lies on y axis then its abscissa (x coordinate) = 0
i.e \(\frac{\mathrm{K}+2}{\mathrm{~K}+1}\) = 0
⇒ K = -2
∴ y axis divides AB externally in the ratio 2 : 1
Coordinates of the point of division
= \(\left(\frac{K+2}{K+1}, \frac{9 K+5}{K+1}\right)=\left(0, \frac{9(-2)+5}{-2+1}\right)\) = (0, 13)

Question 47.
Find the ratio in which the line joining (1, 2) and (4, 3) is divided by the line joining the points (2, 3) and (4, 1)
Answer:
Let A = (2, 3) and B (4, 1). The equation of the line AB is given by
\(\frac{y-3}{x-2}=\frac{1-3}{4-2}\) ⇒ y – 3 = x + 2 ⇒ x + y = 5
Let the line joining the pins C (1, 2) and D (4, 3) cut the line AB at P (x, y) in the ratio r : 1. Then
P = \(\left(\frac{4 r+1}{r+1}, \frac{3 r+2}{r+1}\right)\)
This lies on the line x + y = 5
Thus the co-ordinates of P must satisfy this equation.
∴ \(\frac{4 r+1}{r+1}+\frac{3 r+2}{r+1}\) = 5
4r + 1 + 3r + 2 = 5 (r + 1)
7r + 3 = 5r + 5 ⇒ r = 1
Thus the required ratio is 1 : 1

Question 48.
Show that the following points are the vertices of a rectangle. (1, 6), (-1, -2), (4,1) (-4, 3)
Answer:
Let A = (1, 6), B = (-1, -2), C = (4, 1), D = (-4, 3)
AB = \(\sqrt{(1+1)^{2}+(6+2)^{2}}=\sqrt{4+64}=\sqrt{68}\)
BC = \(\sqrt{(-1-4)^{2}+(-2-1)^{2}}=\sqrt{25+9}=\sqrt{34}\)
CD = \(\sqrt{(4+4)^{2}+(1-3)^{2}}=\sqrt{64+4}=\sqrt{68}\)
DA = \(\sqrt{(1+4)^{2}+(6-3)^{2}}=\sqrt{25+9}=\sqrt{34}\)
∴ AB = CD and BC = AD
i.e., opposite sides are equal.
Further
AC = \(\sqrt{(1+4)^{2}+(6-1)^{2}}=\sqrt{9+25}=\sqrt{34}\)
BD = \(\sqrt{(-1+4)^{2}+(-2-3)^{2}}=\sqrt{9+25}=\sqrt{34}\)
∴ The diagonals AC and BD are equal.
∴ ABCD is a parallelogram with diagonals equal.
∴ ABCD is a rectangle.

PART-E

V. Answer any one equestion. (1 × 10 = 10)

Question 49.
(a) Derive the equation of the straight line is the form \(\frac{x}{a}+\frac{y}{b}\) = 1
Answer:
Let a line cut the x-axis at A and the y-axis at B. Let OA = a and OB = b. Then the length OA = a is called the x-intercept of the line and the length OB = b is called the y-intercept of the line. If ‘ the line passes the origin. Then both the x-intercept and y-intercepts are zero.

Consider the XOY plane. Let the line cut the x-axis at A and the y-axis at B. By data OA = a and OB = b. Thus A = (0, 0) and B = (0, b). Hence the required line is the line joining A and B, Its equation can be obtained by using a two-point form of the equation of a line.

Then the equation of AB is
\(\frac{y-0}{x-a}=\frac{0-b}{a-0}\)
⇒ \(\frac{y}{x-a}=\frac{-b}{a}\) ⇒ ay = -bx + ab ⇒ bx + ay = ab
Dividing throughout by ‘ab’ we have
\(\frac{b x}{a b}+\frac{a y}{a b}=\frac{a b}{a b}\)
This form of equation of the line is called Intercept form.
⇒ \(\frac{x}{a}+\frac{y}{b}\) = 1

(b) The angles of a triangle are in A.P. and the ratio of a number of degrees in the least to the number of radians in the greatest is 60: π. Find the angles of the triangle in radians.
Answer:
A – D, A, A + D
A – D + A + A + D= 180
3A = 180
A = 60°

A- D, A, A+ D Angles in degrees
(A – D)\(\frac{\pi}{180}\), A, \(\frac{\pi}{180}\)(A+D)\(\frac{\pi}{180}\) Angles in radians
Given
\(\frac{A-D}{(A+1) \frac{\pi}{180}}=\frac{60}{\pi}\)

(A – D) 180 = 60(A + D)
(A – D)3 = A + D
3A – 3D = A + D
3A – A = 3D + D
2A = 4D
2(60) = 4D
4D = 120
D = 30°
Angles in degrees are
60 – 30, 60, 60 + 30
A – D, A, A + D
300, 60°, 90°
or Angles in radians are
\(\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}\)

(c) Find the number of digits in 2S0. (log 2 = 0,3010)
Answer:
Let x = 250 log x = log 250 = 50 log 2 = 50 × 0.3010 = 15.05
As the characteristic of log x is 15, no. of digits are 15 + 1 = 16.

Question 50.
(a) Shanmukh buys every year Bank’s cash certificate of value exceeding the last year’s purchase by Rs. 500. After 15 years, he finds that the total value of the certificates purchased by him is Rs.82,500. Find the value of the certificates purchased by him
(a) in the first year and b in the 10lh year.
Answer:
d = Rs.500 n=15 S_n = 82,500 a = ? T₁₀ = ?
S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
82500 = \(\frac{15}{2}\) [2a+ 14(500)]
\(\frac{82500 \times 2}{15}\) = 2a + 7000
11000 = 2a + 7000
2a = 11000 – 7000 = 4,000
a = 2,000
a = Rs. 2,000
T₁₀ = a + (10 – 1) d = 2000 + 9 (500) = 2000 + 4500 = 6500
T₁₀ = 6,500

(b) x + 3y ≥ 3,2x + y ≥ 2, x ≥ 0, y ≥ 0

(c) Find the numbers which when divided by 36,40 and 48 leaves same remainder 5.
Answer:
Let us find out the LCM of 36, 40, 48
36 = 2² × 3²
40 = 2³ × 5
48 = 2⁴ × 3
LCM = 2⁴ × 3² × 5¹ .
= 16 × 45 = 720
Since 5 has to be the remainder, we have to add 5 to 720
i.e. 720 + 5 = 725

Students can Download 1st PUC Basic Maths Model Question Paper 4 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Model Question Paper 4 with Answers

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any ten questions (10 × 1 = 10)

Question 1.
Write the imaginary part of 2 + 3i.
Answer:
3

Question 2.
If A = {a, b, c}, B = {c, d, e, f}. Find A – B.
Answer:
A – B = {a, b}

Question 3.
f: R → R is defined by f(x) = 4x + 3 find f(3/4)
Answer:
f(3/4) = 4(3/4) + 3 = 3 + 3 = 6.

Question 4.
Simplify \(\left(\frac{16}{81}\right)^{-1 / 4}\)
Answer:

Question 5.
Evaluate x if logx 256 = 4
Answer:
x⁴ = 256
x⁴ = 44 ⇒ x = 4

Question 6.
Find the 10th term of the AP 1, 3, 5
Answer:
a = 1, d = 2, n = 10
Tn = o + (n – 1)d = 1 + (10 – 1)2 = 1 + 9(2) = 1 + 18 = 19

Question 7.
Solve for x.
Answer:
2(x- 3) = 9 + 3(x- 9).

Question 8.
Convert 0.12 into a percentage
Answer:
0.12 = 0.12 × 100%= \(\frac{12}{100}\) × 100% = 12%

Question 9.
Convert – 150° into radians.
Answer:
-150°= -150 × \(\frac{\pi}{180}=\frac{-5 \pi^{c}}{6}\)

Question 10.
Find the slope of a line parallel to the line 3.x + 5y -9 = 0 -3
Answer:
\(\frac{-3}{5}\)

Question 11.
If sin A = \(\frac{3}{5}\) and 90 < A < 180 find cos A
Answer:

Cos A = -4/5

Question 12.
Define perpetuity?
Answer:
If the annuity payments are made for an infinite period then it is called a perpetuity.

PART-B

II. Answer any Ten questions. (10 × 2 = 20)

Question 13.
Find the number which when divided by 16, 20 and 40 leaves the same remainder 4.
Answer:
16 = 2⁴, 20 = 2² × 5¹, 40 = 2³ × 5¹
LCM = 2⁴ × 5¹ = 16 × 5 = 80
∴ Required number 80 + 4 = 84.

Question 14.
IfA = {2, 3, 4) write all the proper subsets of A.
Answer:
Proper subsets ofA arc {2), {3}, (4), (2, 3), (2, 4), {3, 4}

Question 15.
Find the number of positive divisors and the sum of all positive divisors of 360.
Answer:
360 = 2³ × 3² × 5¹

n = P₁^α₁ × P₂^α₂ × P₃^α₃
P₁ = 2, α₁ = 3, P₂ = 3, α₂ = 2, P₃ = 5, α₃ = 1
T(n) = (1 + α₁)(1 + α₂)(1 + α₃)
T(360) = (3 + 1)(1 + 2)(1 + 1) = (4)(3)(2)=24
S(n) = \(\frac{P_{1}^{\alpha_{1}+1}-1}{P_{1}-1} \times \frac{P_{2}^{\alpha_{2}+1}-1}{P_{2}-1} \ldots \ldots \times \frac{P_{n}^{\alpha_{n}+1}-1}{P_{n}-1}\)
S(360) = \(\frac{2^{4}-1}{2-1} \times \frac{3^{3}-1}{3-1} \times \frac{5^{2}-1}{5-1}=\frac{15}{1} \times \frac{26}{2} \times \frac{24}{1}\) = 1170

Question 16.
Simplify 2(3-2) + \(\left(\frac{1}{3}\right)^{-3}\) + 32
Answer:
\(\frac{2}{3^{2}}\) + 32 + 32
= \(\frac{2+243+81}{9}=\frac{326}{9}\)

Question 17.
Solve for X: logx + log (x – 4) – log (x – 6) = 0
Answer:
logx + log(x – 4) – log(x – 6)= 0
logx(x – 4) – log(x – 6) = 0
log \(\frac{x^{2}-4 x}{x-6}\) = 0
(∵ log 1 = 0) ∴ \(\frac{x^{2}-4 x}{x-6}\) = 1
⇒ x² – 4x = x – 6
x² – 4x – x + 6 = 0
x² – 5x + 6 = 0
x² – 2x – 3x + 6 = 0
x (x – 2) -3 (x – 2) = 0
(x – 2) (x – 2) = 0
x = 2 or x = 3

Question 18.
If the second term of an A.P is 4 and 10th term is 20. Find the 15th term.
Answer:
T₂ – 4 T₁₀ = 20 T₁₅ = ?
T_n = a + (n – 1)d
T₂ = 4
a + (2 – 1) d = 4
a + d = 4 …………..(1)
T₁₀ = 20
a+ (10 – 1)d = 20
a + 9d = 20………..(2)
(2)-(1) gives
a + 9d = 20

∴ d = 2
a + d = 4
a + 2 = 4
a = 4 – 2 = 2
a = 2
T₁₅ = a + (15 – 1)d = a + 14d – 2 + 14(2) = 2 + 28 = 30
T₁₅ = 30

Question 19.
Divide ₹ 1600 between x, y, z so that y may have ₹ 100 morethan.x and z ₹ 200 more then y.
Answer:
Let x’s share be ₹ a.
∴ y’s share is ₹ 100 + a
& z’s share is ₹ 200 + 100 + a .
Given a + 100 + a + 200 + 100 + a = 1600
3a + 400= 1600 3a = 1200 ⇒ a = 40
∴ x, y & z’s share are ₹ 400, ₹500 and ₹700 respectively.

Question 20.
Calculate the simple on ₹ 4,000 at 4% from June 27 to Aug 29 in the same year.
Answer:
P = Rs. 4000R = 4%
June 27th to Aug 29
June : 3 days, July : 31 days, Aug : 29 days
Total = 63 days
T = \(\frac{63}{365}\) =0.1726
SI = \(\frac{4000 \times 0.1726 \times 4}{100}\) = Rs. 27.616

Question 21.
The average age of 10 students ¡n 14 years among them the average age of 4 students is 12 years. Find the average age of the remaining students.
Answer:
Given x̄₁ = i2yr. n₁ = 4
x̄₂ = 1 n=6
And Average of 10 students is 14 years.
∴ 14 = \(\frac{\bar{x}_{1} \mathrm{n}_{1}+\bar{x}_{2} \mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}\)
14 = \(\frac{(12 \times 4)+\left(\bar{x}_{2} \times 6\right)}{4+6}\)
14 = \(\frac{48+6 \bar{x}_{2}}{10}\)
∴ 48 + 6x̄₂ =140
6x̄₂ = 140 – 48
6x̄₂ = 92
∴ x̄₂ = \(\) =15.33yrs.

Question 22.
The sales of a company was 35,000 in June and 30,000 in July. Find the percentage decrease.
Answer:
% decrease = \(\frac{\text { Decrease }}{\text { Initial Value }}\) × 100
Decree = 35, 000 – 30,000 = 5000
% decrease = \(\frac{\text { Decrease }}{\text { Initial Value }}\) × 100
= \(\frac{5000}{35000}\) × 100 = 14.285 ≈ 14.29%

Question 23.
Prove that \(\frac{1-\cos \theta}{\sin \theta}=\frac{\sin \theta}{1+\cos \theta}\)
Answer:
L.H.S = \(\frac{1-\cos \theta}{\sin \theta}=\frac{1-\cos \theta}{\sin \theta} \times \frac{1+\cos \theta}{1+\cos \theta}\)
= \(\frac{1 \cos ^{2} \theta}{\sin \theta(1+\cos \theta)}=\frac{\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\frac{\sin \theta}{1+\cos \theta}\) = R.H.S

Question 24.
Find the value of \(\left(\frac{16 \pi}{3}\right)\)
Answer:
tan\(\left(\frac{16 \pi}{3}\right)\) = tan\(\left(4 \pi+\frac{4 \pi}{3}\right)\) = tan\(\frac{4 \pi}{3}\) = tan\(\left(\pi+\frac{\pi}{3}\right)\) = tan\(\frac{\pi}{3}\) = \(\sqrt{3}\)

Question 25.
Find the value of x so that the slope of the line joining the points A(2, 5) and B(x, 13) is 20.
Answer:
Let A =(2,5) B = (x, 13)
Slope m = 20
⇒ \(\frac{13-5}{x-2}\) = 20
⇒ 8 = 20 (x – 2)
⇒ 8 = 20x – 40
⇒ 48 = 20x
∴ x = \(\frac{48}{20}=\frac{12}{5}\)

PART – C

III. Answer any ten questions. (10 × 3 = 30)

Question 26.
In a college \(\left(\frac{2}{5}\right)^{\text {th }}\) of the students play Basket ball and \(\left(\frac{3}{4}\right)^{t h}\) play volley ball. If 50 students play none of these two games and 125 play both, to find the number of students in the college.
Answer:
Let the number of students in the college = x
No of students who play Basket ball = \(\frac{2}{5}\)x
No of students who play Valley ball = \(\frac{3}{4}\)x
No of students who play both = 125
50 students play none of the games.
x-50 = \(\frac{2}{5}\)x + \(\frac{3}{4}\) x – 125
x – \(\frac{2}{5}\) x + \(\frac{3}{4}\) x = -125 + 50
\(\frac{20 x-8 x-15 x}{20}\) = -75
\(\frac{-3 x}{20}\) = -75
x = \(\frac{20 \times 75}{3}\) = 20 × 25 = 500
∴ Total number of students in the college = 500

Question 27.
Prove that 2 + \(\sqrt{3}\) is an irrational number.
Answer:
Sol. If possible let 2 + \(\sqrt{3}\) is a rational number
∴ 2 + \(\sqrt{3}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\), p,q∈z, q ≠ 0
⇒ \(\sqrt{3}=\frac{\mathrm{p}}{\mathrm{q}}\) – 2 = rational – rational = rational number
But \(\sqrt{3}\) is an irrational number

A rational number cannot be equal to an irrational number.
∴ Our assumption is wrong
∴ 2+ \(\sqrt{3}\) is irrational.

Question 28.
Evaluate 3 +5 + 7 + ….+ 61.
Answer:
Here, a=3, d = 5 – 3 = 2, l = Tn = 61
Consider, Tn = a + (n – 1)d
61 = 3 + (n – 1)2
61 – 3 = 2n – 2
58 = 2n – 2
2n = 58 + 2
2n = 60
n = \(\frac{60}{2}\) = 30 Sn = \(\frac{\mathrm{n}}{2}\)(a +1)
Now, S = \(\frac{30}{2}\)(3+61)
= 15 (64)
= 960

Question 30.
A father is 28 years older than the son, after 5 years the father’s age will be 7 years more than twice that of the son. Find their present ages.
Answer:
Let father’s age be x years and son’s age be y years.
Given: x – y = 28 ……………..(1)
After five years father’s age is x + 5 and son’s age is y + 5.
Also given: x + 5 = 2(y + 5) + 7
x + 5 = 2y + 10 + 7
Solving(1) and (2), we get: x = 44 and y = 16
Thus father’s age is 44 years and the son’s age is 16 years.

Question 31.
Three equal principals amount to 3720 after 3, 4 and 5 years at simple interest 6% p.a Find the principal.
Answer:
Amount after 3 years,
P = x, T = 3, R = 6
I₁ = \(\frac{x \times 3 \times 6}{100}\)
A₁ = P+ I = x + 0.18x = 1.18x

Amount after 4 years,
P = x, T = 4, R = 6
I₂ = \(\frac{x \times 4 \times 6}{100}\) = 0.24x
A₂ = x + 0.3x = 1.3x

Amount after 5 years,
P = x, T = 4, R = 6
I₃ = \(\frac{x \times 5 \times 6}{100}\)
A₃ = x + 0.3x = 1.3x
Given A₁ + A₂ + A₃ = 3720
1.18x + 1.24x +1.3x = 3720
3.72x = 3720
x = \(\frac{3720}{3.72}\) = 1000

Each Principle = ₹ 1000
Total Principle = ₹ 3000

Question 32.
The difference between simple interest and Compound interest on a certain sum of money invested for 3 years at 6% p.a is 110.16 Find the sum.
Answer:
Let the sum = x
Simple Interest = \(\frac{x \times 3 \times 6}{100}\) = 0.18 x
Compound interest
A = P(1 + i)ⁿ
= x (1 + 0.06)³
= 1.191016x
CI = A – P
= 0.191016 x- x
= 1.191016 x
Given, Cl -SI =110.16
0.191016x – 0.18x = 110.16
x = \(\frac{110.16}{0.011016}\)
= 10,000 Rs

Question 33.
Find the present value of an annuity of ₹ 3000 for 12 years at 6% p.a. computed half yearly.
Answer:
effective rate of interest
r = \(\left(1+\frac{i}{q}\right)^{q}\) – 1
= \(\left(1+\frac{06}{2}\right)^{2}\) – 1 = 1.0609 – 1 = 0.0609
A = 3000 i = 0.0609 n = 12
P = \(\frac{a\left[(1+i)^{n}-1\right]}{i \times(1+i)^{n}}\)
= \(\frac{3000\left[(1+0.0609)^{12}-1\right]}{0.0609(1+0.0609)^{12}}\)
= \(\frac{3000(2.0328-1)}{0609 \times 2.0328}\) = 25027.96₹
(by direct calculation (without rounding in between values) with calculator, answer is 25027.89)

Question 34.
A Dental Clinic purchased a certain number of chairs at an average price of ₹ 190 each. The average price of 30 chairs was ₹ 175 and that of the remaining chairs was ₹ 200/-. Find the total number of chairs the clinic purchased.
Answer:
Assumed that the number of chair the clinic purchased = x.
Total average price = ₹ 190
Average Price No. of chair
x̄₁ = ₹ 175 N₁ = 10
x̄₂ = ₹ 200 N₂ = (x – 10)
Combined average price = \(\frac{\bar{x}_{1} N_{1}+\bar{x}_{2} \mathrm{~N}_{2}}{(10)+(x-10)}\)
190 = \(\frac{(175 \times 10)+(200 \times(x-10))}{x}\)
190 = \(\frac{1750+200 x-2000}{x}\)
∴ 190x = 200x – 250
250 = 200x- 190x
250= 10x
∴ x = \(\frac{250}{10}\) = ₹ 25
Total Number of chairs purchased by the clinic = 25

Question 35.
Savithri sold her bag at a loss of 7%. Had she been able to sell it at a gain of 9% it would have fetched 64 more than it did. What was the cost price of the bag?
Answer:
Let C.P. = 100
Loss at 7% ⇒ SP = 93
Gain of 9% ⇒ SP = 109
Diff = 109 – 93 = 16

Then cost price of the bag is = \(\frac{100 \times 64}{16}\) = ₹ 400

Question 36.
P.T. (1 + tan A – sec A) (1 + cot A + cosec A) = 2
Answer:
LHS = (1 + tan A – sec A) (1 + cot A + cosec A)
= 1 + cotA + cosecA + tanA + tanA × cotA + tanA × cosec A-sec A-sec A. cotA – secA . cosec A

= 2 + \(\left[\cot \mathrm{A}+{Tan} \mathrm{A}-\frac{1}{\cos \mathrm{A} \cdot \sin \mathrm{A}}\right]\) = 2 + \(\left[\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}-\frac{1}{\cos A \sin A}\right]\)
= 2 + \(\left[\frac{\cos ^{2} A+\sin ^{2} A-1}{\cos A \sin A}\right]\) = 2 + \(\left[\frac{1-1}{\cos A \sin A}\right]\) = 2 + \(\frac{0}{\cos A \sin A}\)
= 2 + 0 = 2

Question 37.
Find x if
\(\frac{x \sin ^{2} 300^{\circ} \cdot \sec ^{2} 240}{\cos ^{2} 225 \cdot {cosec}^{2} 240}\) = cot2 315° . tan2 300°
Answer:
sin 300 = sin (360 – 60) = -sin 60 = \(\frac{\sqrt{3}}{2}\)
sec 240 = sec (180 + 60) = -sec 60 = -2
cos225 = cos(180 + 45)= -cos45 = –\(\frac{1}{\sqrt{2}}\)
cosec 240 = cosec (180 + 60) = -cosec 60 = –\(\frac{2}{\sqrt{3}}\)
cot 315 = cot(360 – 45) – cot45 = – 1
Tan 300 = Tan(360 – 60) = -Tan60 = –\(\sqrt{3}\)

Substituting these values the given equation becomes
\(\frac{x\left(-\frac{\sqrt{3}}{2}\right)^{2}(-2)^{2}}{\left(\frac{-1}{\sqrt{2}}\right)^{2}\left(\frac{-2}{\sqrt{3}}\right)^{2}}\) = (-1)2 \((-\sqrt{3})^{2}\)
\(\frac{x \cdot \frac{3}{4} \cdot 4}{\frac{1}{2} \cdot \frac{4}{3}}\) = (1) (3)
\(\frac{12 \mathrm{x}}{\frac{4}{4}{6}}\) = 3 ⇒ \(\frac{12 x}{4} \times \frac{6}{4}\) = 3
\(\frac{72 \mathrm{x}}{16}\) = 3 ⇒ \(\frac{48}{12}=\frac{2}{3}\)
∴ x = \(\frac{2}{3}\)

Question 38.
Find the coordinates of the circumcentre of the triangle so formed by the points (1, 1) (2, -1) and (3, 2)
Answer:
Let A = (1, 1), B(2, -1) C(3, 2)
Let S(x, y) be the circumcentre of the triangle ABC.
Then SA = SB = SC.
⇒ SA² = SB² = SC²

Consider SA² = SB²
i.e.(x – 1)² + (y – 1)² = (x – 2)² + (y + 1)²
i.e. x² – 2x + 1 + y² – 2y + 1 = x² – 4x + 4 + y² + 2y + 1
⇒ 2x – 4y = 3 (1)

Again consider SA² = SC²
(x – 1)² + (y – 1)² = (x – 2)² + (y + 1)²
x² – 2x + 1 + y² – 2y + 1 = x² – 6x + 9 + y² – 4y + 4
⇒ 4x + 2y = 11 (2)
Solving (1) and (2) we get x = 5/2 and y = 1/2
Thus the circumcentre is given by \(\left(\frac{5}{2}, \frac{1}{2}\right)\)

PART-D

IV. Answer any six questions. 6 × 5 = 30

Question 39.
A relation R on a collection of a set of integers defined by R = {(x,y): x – y is a multiple of 3}. Show that R is an equivalence relation on Z.
Answer:
0 = x – x is a multiple of 3 ∴ x R x ∴ R is reflexive
Let x Ry ⇒ x-y is a multiple of 3 ⇒ x -y = 3k, k ∈ Z y – x = – (x – y)= -3k = 3(-k) is a multiple of 3 -k∈Z
∴ y R x ∴ R is symmetric Let x R y & y R z
∴ x-y is a multiple of 3 ∴ x-y = 3a, a ∈ Z
∴ y-z is a multiple of 3 ∴ y-z = 3b, b ∈ Z
(x-y) + (y-z) = 3a + 3b = 3(a + b)
∴ x-z = 3(a + b),a + b ∈ Z
∴ x – z is a multiple of 3 ∴ x R z ∴ R is transitive
∴ R is an equivalence relation on Z.

Question 40.
Find the sum of all numbers between 50 an 200 which are divisible by 11.
Answer:
AP is given by
55, 66, 77,… 198
Here a = 55, l = T_n =198, d = 11
Now, T_n = a + (n – 1) d
198 = 55 + (n- 1)11
198 – 55 = (n – 1) 11
\(\frac{143}{11}\) = n – 1
n – 1 = 13
n = 13 + 1
n = 14
Now consider:
S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₁₄ = \(\frac{14}{2}\)(55 + 198)
= 7 (253)
= 1,771

Question 41.
The income of three persons Anil, Akbar, and Antony is 6: 5: 4, and their expenditures are in the ratio of 3: 2: 1. If Anil saves ₹ 120 out of his income of ₹ 1500. Find the saving of Akbar and Antony.
Answer:
Let the income be x and the expenditure be y.
Incomes are 6x, 5x and 4x and their expenditures are 3y, 2y and y.
Given : Anil’s income = ₹ 1500/-
⇒ 6x = 1500 ⇒ x = 250
Anil’s expenditure = 3y = Income – Savings
3y = 1500 – 120
3y = 1380 y = ₹ 460
Thus Akbar’s incoe = 5x = 5 × 250 = ₹1250
Akbar’s expenditure = 2y = 2 × 460 = ₹920
Antony’s income = 4x = 4 × 250 = ₹1000
Antony’s expenditure = y = ₹1000

Question 42.
A sum of money lent at compound interest for 2 years at 20% p.a. would fetch Rs.482 more, if the interest was payable half yearly than if it was payable annually. Find the sum.
Answer:
A₁ = Amount wjjen the interest payable is annually
A₂ = Amount when the interest payable is half yearly
A₂ – A₁ = 482
Let the princible be P n = 2 years r = 20%
A₁ =P(1 +0.20)²
A₁ = P(1.2)² = 1.44 P
Half yearly R = 20% q = 2
r = \(\left(1+\frac{\mathrm{R}}{\mathrm{q}}\right)^{q}\) – 1
= \(\left(1+\frac{0.20}{2}\right)^{2}\) – 1 = 1.21 – 1 = 0.21 = 2.1%
A₂ = P(1 + 0.21)²
A₂ = P(1.21)² = 1.4641 P
A₂ – A₁ = Rs.482
1.4641 P- 1.44 P = 482
0.0241 P = 482
P = \(\frac{482}{0.0241}\) = Rs. 20,000

Question 43.
A person purchases a house for Rs.25 lakhs with Rs. 5 lakhs as a down payment. The rest of the amount he loans from a bank that offers 16% p.a compound interest and has to repay the loan in 20 equal annual installments. If the first installment is paid at the end of the third year, find how much he has to pay each year?
Answer:
P = 25,00,000 – 5,00,000 = 20,00,000
r = 16% = 0.16 d = 2 n = 20 A=?
A = \(\frac{\mathrm{P} \cdot \mathrm{r} \cdot(1+\mathrm{r})^{\mathrm{n}+\mathrm{d}}}{(1+\mathrm{r})^{\mathrm{n}}-1}=\frac{20,00,000(0.16)(1+0.16)^{22}}{(1+0.16)^{20}-1}=\frac{83,79,647}{18.46}\) = 4,53,935.37

Question 44.
Find the equation of the locus of a point that moves so that its distance from (3, 2) is equal to its distance from 2x + y = 3
Answer:
A = (3,2)
Let P (x, y) be any point on the locus
PA = distance of P from the line 2x + y – 3 = 0
\(\sqrt{(x-3)^{2}+(y-2)^{2}}=\left|\frac{2 x+y-3}{\sqrt{2^{2}+1^{2}}}\right|\)
(x – 3)² + (y – 2)² = \(\left(\frac{2 x+y-3}{\sqrt{5}}\right)^{2}\)
x² – 6x + 9 + y² – 4y + 4 = (4x² + y² + 9 + 4xy – 6y – 12x)/5
5x² – 30x + 45 + 5y² – 20y + 20 = 4x² + y² + 9 + 4xy – 6y – 12x
x² – 4xy + 4y² – 18x – 14y + 56 = 0

Question 45.
Evaluate using log tables \(\)
Answer:
Let x = \(\)
Log x = log\(\) = log \(\) + log 0.5789 – log(13.46)3/2
= log (6.43)1/2 + log0.5789-log (13.46)3/2 = \(\frac{1}{2}\)log(6.43) + log0.5789- \(\frac{3}{2}\) log(13.46)
= 0.4041 + 1̄.7626- 1.6935 = -1.5268 = -1 – 0.5268 = -1 -1 + 1 – 0.5268
= -2 + 0.4732 = 2̄ .4732
x̄ = Antilog (2̄ .4732) = 0.0297

Question 46.
Find the equation of a straight line passing through the point (3, 4) such that the sum of its intercepts on the axes is 14.
Answer:
Let the x and y intercepts of the line be a and b respectively so that a + b = 14 or b = 14 – a
Thus the equation of the line in die intercept form is
\(\frac{x}{a}+\frac{y}{14-a}\) = 1
But it passes through the point (3, 4).
therefore \(\frac{3}{a}+\frac{4}{14-a}\) = 1 a 14-a
3 (14 – a) + 4a = a(14 – a)
i.e., a² – 13a + 42 = 0
i.e., (a – 6) (a – 7) = 0 or a = 6 or a = 7
If a = 6 then b = 8; If a = 7 then b = 7
Therefore equations of the hoes are \(\frac{x}{6}+\frac{y}{8}\) = 1 or \(\frac{x}{7}+\frac{y}{7}\) = 1

Question 47.
Find the coordinates of the reflection of the point (1, 2) about the line 3x + 4y – 1 = 0. Sol. Let P (1, 2) be the given point and 3x + 4y – 1 = 0 be the given line. Draw PM perpendicular to the given line and produce it to P’ such that PM = MP’. Then M is the foot and P’ is the reflection of the point P. Also M is the mid-point of PP’ and PP’ is perpendicular to the given line.
Answer:
Let P’ = (a, b). We first find the coordinates of the foot M.
Now any line perpendicular to 3x + 4y – 1 = 0 is 4x – 3y + k = 0.
But this perpendicular line passes through P (1,2)
∴ 4(1) – 3(2) + k = 0
∴ Equation to the perpendicular line PP’ is 4x – 3y + 2 = 0
Now M is the point of intersection of the given line and the above.
Hence solving these two equations we get x = –\(\frac{1}{5}\), y = \(\frac{2}{5}\)

∴ The foot M is \(\left(-\frac{1}{5}, \frac{2}{5}\right)\).
But M = Mid-point of PF = \(\left(\frac{a+1}{2}, \frac{b+2}{2,}\right)\).
Therefore
\(\frac{a+1}{2}=-\frac{1}{5}, \frac{b+2}{2}=\frac{2}{5}\)
a = –\(\frac{7}{5}\); b = – \(\frac{6}{5}\)
Required reflection of P is P’ = \(\left(-\frac{7}{5},-\frac{6}{5}\right) \)

Question 48.
If sec a = \(\frac{13}{5}\) where 270° < a < 360° find the value of \(\frac{2 \sin \alpha-3 \cos \alpha}{4 \sin \alpha+9 \cos \alpha}\)
Answer:
α lies in 4th Quadrant .
cos α and sec α are positive and rest arernegative
sin α = \(\frac{-12}{13}\)
cos α = \(\frac{5}{13}\)
Given Expression

PART-E

V. Answer any one equestion. (1 × 10 = 10)

Question 49.
(a) Find the equation of the straigh&fine passing through the point of intersection x² + 2y + 3 = 0 and 3x + 4y + 7 = 0 and has a slope \(\)
Answer:
x + 2y+ 3 = 0 x 2
3x + 4y + 7 = 0 x 1

x = -1
x + 2y + 3 = 0
-1 + 2y + 3 = 0
2y + 2 = 0
y = -1
(x1, y1) = (-1,-1) = m= \(\frac{-3}{2}\)

(b) P.T \(\frac{\tan \mathbf{A}}{1-\cot \mathrm{A}}+\frac{\cot \mathbf{A}}{1-\tan \mathrm{A}}\) =1+SecA.CosecA
Answer:
LHS = \(\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}\)

= sec A. cosec A + 1 = RHS

(c) If log 5 = 0.6990, find the number of digits in the integral part of 5²³
Answer:
Consider log (5²³) =23 log 5
=23 (06990) = 16.077
Since the characteristic of log (5²³) is 16, there are 17 digits in an integral part of 5²³

Question 50.
(a) Find the sum to n terms of the GP 0.6 + 0.66 + 0.666 +……….
Answer:
Let S_n = 0.6 + 0.66 + 0.666 +……. to n terms
= 6[0.1 + 0.11 + 0.111 + ……… to n terms]
= \(\frac{6}{9}\)[0.9 + 0.99 + 0.999 +……… to n terms]
= \(=\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+\ldots \text { to } \mathrm{n} \text { terms }\right]\)

= \(\frac{6}{9}\)[(1 + 1 + 1 + ……… to n terms) – \(\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\ldots \text { to n terms }\right)\)]

= \(\frac{6}{9}\left[n-\frac{\frac{1}{10}\left(1-\frac{1}{10^{n}}\right)}{1-\frac{1}{10}}\right]\)
= \(\frac{6}{9}\left[\mathrm{n}-\frac{1}{10} \times \frac{10}{9}\left(1-\frac{1}{10^{n}}\right)\right]\)
S_n = \(\frac{6}{9}\left[\mathrm{n}-\frac{1}{9}\left(1-\frac{1}{10^{\mathrm{n}}}\right)\right]=\frac{2}{3}\left[\mathrm{n}-\frac{1}{9}\left(1-\frac{1}{10^{\mathrm{n}}}\right)\right]\)

(b) A batsman’s average score for a number of intlings was 21.75 runs per innings. In the next three inning he scored 28,34 and 37 runs. And his average for all the inning was revised by 1.125 runs. How many inning did he play?
Answer:
Let the number of innings he played = x
∴ Total runs in ’x’ innings = 21.75 x
Again total runs in next 3 innings = 28 + 34 + 37 = 99 runs.
Number of inning raised to = (x + 3)
New Average after 3 inning raised to = 21.75 + 1.125 = 22.875

∴ Average run after (x + 3) innings = \(\frac{\text { Total runs of all the innings }}{\text { Total no. of innings }}\)
22.875 = \(\frac{21.75 x+99}{x+3}\)
∴ (22875) (x+ 3) = 21.75x+ 99
22.87x + 68.625 = 21.75x + 99
(22.875x – 21.75x) = 99 – 68.625
1.125x = 30.375
∴ x = \(\frac{30.375}{1.125}\) = 27
∴ Total innings played = x + 3 = 27 + 3 = 30

(c) A confectioner makes and sells Chocolates. He sells one box of Chocolates at The cost of manufacturing is Rs.60/box as variable cost and Rs.2000 as Fixed C* (i) Revenue function (ii) the cost function.
Answer:
Let x be the number of boxes of chocolates
Total cost = variable cost + fixed cost = 60x + 2000
Revenue function = 180x

Students can Download 1st PUC Basic Maths Model Question Paper 3 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Model Question Paper 3 with Answers

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART – A

I. Answer any ten questions (10 × 1 = 10)

Question 1.
Give the canonical representation of 156.
Answer:

Canonical representation is 156 = 22 × 31 × 13′

Question 2.
If A = {1, 2, 3, 4, 5} B = {1, 2, 3, 4, 5, 6^ 7} find a relation from A to B defined by R = {(x, y)/x > y}
Answer:
R = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4)}

Question 3.
If f(x) = x + 1 and g(x) = x² + 1 find fog (1)
Answer:
(fog) (1) = f[g(l)] = f[l2 + 1] = f(2) = 2+1 = 3

Question 4.
Simplify \(\left(\frac{5 x^{3}}{2 y}\right)^{2}\)
Answer:
\(\frac{5^{2} x^{6}}{2^{2} y^{2}}=\frac{25 x^{6}}{4 . y^{2}}\)

Question 5.
Find x if log_x 625 = 4
Answer:
625 =x⁴
5⁴ = x⁴ ⇒ x = 5

Question 6.
Find the 6th term of the GP. 3, 6,12 …….
Answer:
a = 3, r = 6/3 = 2
T₆ = ar^6-1 = ar⁵ = 3.(2)⁵ = 3.32 = 96

Question 7.
Solve for x if (x + 2) (x + 3) = (x – 2) (x – 4) + 20.
Answer:

5x + 6 = -6x + 28
5x + 6x =28-6
11x = 22
x = 2

Question 8.
What percent is 64m of t 2km?
Answer:
12 km = 12 × 1000m = 12000m
Let the required number be x
x% of 12000 = 64
\(\frac{\mathrm{x}}{100}\) × 12000 = 64
120x = 64
x = \(\frac{64}{120}\) = 0.5333%

Question 9.
Define an annuity.
Answer:
An annuity is a fixed sum paid at regular intervals of time under certain conditions.
Example: LIC premia

Question 10.
Express 3TC/4 in degrees
Answer:
\(\frac{3 \pi^{c}}{4}=\frac{3 \pi}{4}\) × = 3 × 45 = 135°

Question 11.
The average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find the age of the 10th student.
Answer:
Average age of 10 students = 6 years
∴ Total age of 10 students = 10(6) = 60 years
Total age of 9 students = 52 years
∴ age of the 10th student = 60 – 52 = 8 years

Question 12.
Find the slope of the line joining the points (1, 2) and (-1, -2)
Answer:
m = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-2}{-1-1}=\frac{-4}{-2}\) = 2

PART – B

II. Answer any Ten questions. (10 × 2 = 20)

Question 13.
Find the greatest number which when divides 989 and i327 leaves the remainder of 5 and 7 respectively.
Answer:
989 – 5 = 984
1327 – 7= 1320
HCF of 984 and 1320
984 = 23 × 31 × 41


1320 = 2³ × 3¹ × 5¹ × 11¹
HCF = 2³ × 3¹ = 8 × 3 = 24

Question 14.
IfA={x: x ∈ N and x < 3} and B={x: x² – 16 = 0 and x < 0} find B × A
Answer:
A= {1, 2), B={-4}
B × A= {(-4, 1),(-4,2)}

Question 15.
Find the number of positive divisors and the sum of all positive divisors of 825.
Answer:
825 = 3¹ × 5² × 111
P₁ = 3, α₁ = 1, P₂ = 5, α₂ = 2, P₃ =11, α₃ = 11
T (n) = (1 + α₁) (1 + α₂) (1 + α₃)
=(1 + 1)(1 +2)(1 + 1)
= (2) (3) (2)
= 12

S(n) = \(\)
= \(\)
= 4 x 31 x 12 = 1488

Question 16.
Simplify \(\left(\frac{x^{b}}{x^{c}}\right)^{\frac{1}{b c}}\left(\frac{x^{e}}{x^{2}}\right)^{\frac{1}{c u}}\left(\frac{x^{a}}{x^{b}}\right)^{\frac{1}{2 b}}\)
Answer:

Question 17.
Find the value of k so that 2/3, k (5/3)k are three consecutive terms of A.P.
Answer:

Question 18.
Determine the nature of the roots of the equation 2x² – 9x + 7 = 0
Answer:
a = 2, b = -9, c = l
Δ = b² – 4ac = 81 – 4(2) (7) = 81 – 56 = 25 > 0 and is perfect square.
∴ The roots are rational and distinct.

Question 19.
Find the amount on ₹ 500 for 10 years at the rate of 15% compound interest?
Answer:
P = r, 500, n= 10, r = 15%
A = P(1 + r)ⁿ = 500(1 + 0.15)¹⁰ = 500(1.15)¹⁰ = ₹ 2022.77

Question 20.
Solve for x if 4x – 5 < 27 and represent on a number line.
Answer:
4x – 5 ≤ 27
4x ≤ 27 + 5
4x ≤ 32
x ≤ 8

Question 21.
The average score of 65 boys is 60 and the average score of 15 girls is 65. Find the combined average score.
Answer:
X₁ = 60, N₁ = 65
X₂ = 65, N₂ = 15
X̄ = \(\frac{\mathrm{X}_{1} \mathrm{~N}_{1}+\mathrm{X}_{2} \mathrm{~N}_{2}}{\mathrm{~N}_{1}+\mathrm{N}_{2}}=\frac{(60)(65)+(65)(15)}{65+15}=\frac{4875}{80}\) = 60.93

Question 22.
A shopkeeper buys 50 pencils for ₹ 80 and sells them at 40 pencils for ₹ 90. Find his gain or loss percent?
Answer:
CP of 1 pencil = \(\frac{80}{50}\) = ₹ 1.6
SPof 1 pencil = \(\frac{90}{40}\) = ₹ 2.25
Profit on 1 pencil = 2.25 – 1.6 = ₹ 0.65
Profit % = \(\frac{\text { Profit }}{\mathrm{CP}}\) × 100 = \(\frac{0.65}{1.6}\) × 100 = 40,625%

Question 23.
Prove that \(\frac{1}{1+\cos A}+\frac{1}{1-\cos A}\) = 2cosec2A
Answer:
\(\frac{1}{1+\cos \mathrm{A}}+\frac{1}{1-\cos \mathrm{A}}=\frac{1-\cos \mathrm{A}+1+\cos \mathrm{A}}{(1+\cos \mathrm{A})(1-\cos \mathrm{A})}=\frac{2}{1-\cos ^{2} \mathrm{~A}}=\frac{2}{\sin ^{2} \mathrm{~A}}\) =2cosec2A

Question 24.
If θ = 45 show that \(\frac{\tan \theta}{1+\tan \theta}-\frac{1+\tan \theta}{\tan \theta}=\frac{-3}{2}\)
Answer:
LHS = \(\frac{\tan 45^{\circ}}{1+\tan 45^{\circ}}-\frac{1+\tan 45}{\tan 45^{\circ}}=\frac{1}{1+1}-\frac{1+1}{1}=\frac{1}{2}-\frac{2}{1}=\frac{1-4}{2}=\frac{-3}{2}\) = RHS

Question 25.
Derive the equation of the straight line in the slope point form.
Answer:

Slope of AP = m = tan θ = \(\frac{P N}{A N}=\frac{P M-M N}{L M}=\frac{P M-M N}{O M-O L}\)
m = \(\frac{y-y_{1}}{x-x_{1}}\)
∴ y – y₁ = m(x – x₁)

PART – C

III. Answer any ten questions. (10 × 3 = 30)

Question 26.
In a group of 65 people, 40 were found to like hockey, 10 like both tennis and hockey. How many like only tennis but not hockey? How many like tennis?
Answer:
n(C∪H) = n(C) + n (H) -n(C ∩H)
65 = 40 + n(H) – 10
65 = 30 + n(H)
n(H) = 35
Since n (C ∩ H) = 10 number of people like hockey 35.

Question 27.
Given A = {2, 4, 6, 8} and R = {(2, 4), (4, 2), (4, 6), (6, 4)} Show that R is not reflexive, symmetric and not transitive.
Answer:
(2. 2) ∉ R ∴ R is not reflexive
(2, 4) ∈ R = (4, 2) ∈ R (4, 6) ∈ R = (6, 4) ∈ R
(a, b) ∈ R = (b, a) ∈ R ∴ R is symmetric
(2, 4) ∈ R, (4, 2) ∈ R But (2, 2) ∉ R ∴ (a, b) ∈ R, (b, c) ∈ R But (a, c) ∉ R
∴ R is not transitive

Question 28.
The sum of the two numbers is 528 and their HCF is 33. Find the number of pairs satisfying the given condition?
Answer:
Let the two numbers be 33a, 33b
33a + 33b = 529
-33 a + b= 16
The possible value of a and b are (1, 15), (3, 13), (5, 11), (7, 9)
Note that a and b must be relatively prime i.e., they do not have any common factor other than 1.
∴ The possible pairs of numbers are (33, 495), (99, 429), (165, 363), (231, 297)

Question 29.
Prove that zyz = 1 if \(\frac{\log x}{a-b}=\frac{\log y}{b-c}=\frac{\log z}{c-a}\)
Answer:
\(\frac{\log x}{a-b}=\frac{\log y}{b-c}=\frac{\log z}{c-a}\) = k
∴ log x = k(a – b)
logy = k(b – c)
logz = k(c – a)
∴ logx + logy + logz = k(a – b) + k(b – c) + k(c – a)
= k(a – b + b – c + c – a) = k( 0) = 0
∴ log (xyz) = 0 ∴ xyz = 1

Question 30.
If the first term of GP is 729 and the 7th term is 64 find the sum of first seven terms of the GP
Answer:
a = 729, T₇ = 64, S₇ = ?
T₇ = ar⁶
64 =729. r⁶ ⇒ r⁶ = \(\frac{64}{729}=\frac{2^{6}}{3^{6}}=\left(\frac{2}{3}\right)^{6}\) ⇒ r = \(\frac{2}{3}\)
S_n = \(\frac{a\left(1-r^{n}\right)}{1-r}\)
∴ S_n = \(=\frac{729\left(1-(2 / 3)^{7}\right)}{1-2 / 3}=\frac{729\left(1-\frac{128}{2187}\right)}{\frac{3-2}{3}}=\frac{729\left(\frac{2187-128}{2187}\right)}{\frac{1}{3}}=\frac{729\left(\frac{2059}{2187}\right)}{\frac{1}{3}}\)
= \(\frac{2059}{3}{\frac{1}{3}}\) = 2059

Question 31.
Find the present value of the annuity immediate for ₹ 3000 for 5 years at 10% p.a
Answer:
P = \(\frac{A\left[(1-r)^{n}-1\right]}{r(1+r)^{n}}\)
A = 3,000, n = 5, r = 0.10
∴ P = \(\frac{3000\left[(1+0.1)^{5}-1\right]}{0.1(1+0.1)^{5}}=\frac{3000\left[(1.1)^{5}-1\right]}{0.1(1.1)^{5}}=\frac{3000[1.61051-14]}{0.1[1.61051]}\)
= \(\frac{3000(0.61051)}{0.1(1.61051)}=\frac{1831.53}{0.161051}\)
P = 11,372.30

Question 32.
Solve the linear inequalities graphically
3x + 2y ≤ 6,
4x – y ≤ 6
Answer:
3x + 2y = 6
Put y = 0 3x = 6 ∴ x = 2 ∴ A = (2, 0)
x = 0 2y = 6 ∴ y = 3 B = (0, 3)

4x – y = 6
x = 0 -y = 6 ⇒ y = -6 C(0, -6)
y = 0 4x = 6 ⇒ x = 3/2 D(3/2, 0)

The shaded region is the feasible region which is the intersection of the solution set of each inequality 3x + 2y ≤ 6 and 4x – y ≤ 6.

Question 33.
The average temperature for the first four days of the week was 39°C. The average for the whole week was 40°C. What was the average temperature during the last three days of the week?
Answer:
The average temperature of first four days of the week = 39°C
∴ The total temperature of the first four days of the week = 39 × 4= 156
The average temperature of the whole week = 40°
∴ Total temperature of the whole week = 40 × 7 = 280°
∴ The total temperature of the last three days of the week = 280° – 156° = 124°
∴ Average temperature of the last three days of the week = \(\frac{124}{3}\) = 41°C.

Question 34.
Find the circumcentre of the triangle with vertices A(-3, 4), B (3, 4), and C(-4, 3). Also, find the circumradius and the area of the circle.
Answer:
Let S(x, y) be the circumcentre of ΔABC
SA = SB => SA² = SB²
(x – 3)² + (y – 4)² = (x + 4)² + (y + 3)²

-6x – 8x – 8y – 6y =0
-14x- 14y = 0 +-14
x + y =0
0 + y = 0
y = 0
∴ circumcentre = (0, 0)
Circumradius = \(\sqrt{(0+3)^{2}+(0-4)^{2}}=\sqrt{9+16}=\sqrt{25}\) = 5 r = 5
∴ Area of the circle = rcr² = TC(25) = 25K sq. units.

Question 35.
The rate of a movie ticket was Rs. 150. This was reduced by 20% due to the discount in price the revenue increased by 20%, what was the percentage increase in the number of viewers?
Answer:
Let the no. of initial viewers be 100
Cost of each ticket = Rs. 150
Revenue generated initially = 150 × 100 =Rs. 15000
After the reduction in price cost of each ticket
= 150 – 20% of 150 = 150 – \(\frac{20}{100}\)× 150 = 150 – 30 = Rs 120

New revenue generated = 15000 + 20% of 15000
= 15000+ \(\frac{20}{100}\) × 15000 = 15000 + 3000 = Rs. 18000
No. of viewers = \(\frac{\text { Total revenue }}{\text { Cost of each ticket }}\) = 150
Percentage increase in the no. of viewers = 150 – 100 = 50%

Question 36.
Find the value of x if x sin 45° tan 60° = \(\frac{\sin 30^{\circ} \cot 30^{\circ}}{\cos 60^{\circ}{cosec} 45}\)
Answer:

Question 37.
Find the third vertex of the triangle if two of its vertices are A(-2,4) and B(7, -3) and the centroid is (3, 2)
Answer:
A = (-2, 4), B = (7, -3), O = (a, b), G = (3, 2)
G = \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)
(3, 2) = \(\left(\frac{-2+7+a}{3}, \frac{4-3+b}{3}\right)\)
(3, 2) = \(\left(\frac{5+a}{3}, \frac{1+b}{3}\right)\)
\(\frac{5+a}{3}\) = 3 ⇒ 5 + a = 9 ⇒ a = 4
\(\frac{1+b}{3}\) = 2 ⇒ 1 + b = 6 ⇒ b = 5
∴ Third vertex = C = (4, 5)

Question 38.
A number which when decreased by 20 is equal to 69 times the reciprocal of the number. Find the number.
Answer:
Let the number be x
x – 20 = 69\(\frac{1}{x}\)
x² – 20x – 69 =0
x² – 23x + 3x -69 = 0
x(x – 23) + 3(x – 23) = 0
x = 23 or x = – 3

PART – D

IV. Answer any six questions. (6 × 5 = 30)

Question 39.
In a survey it was found that 31 people liked a product A, 36 liked a product B and 39 liked the product C. If 24<people liked products A and B, 22 people liked product C and A, 24 people liked products B and C, 18 liked all the three products, then find how many people liked product C only?
Answer:
n(A) = 31
n(B) = 36
n(C) = 39
n(A∩B) = 24 n(B∩C) = 24 n(C∩A) = 22 n(A∩B∩C) = 18
n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(A∩C) – n(C∩A) + n(A∩B∩C)
= 31 + 36 + 39 – 24 – 24 – 22 + 18 = 54

Number of people who like the product C only = 11

Question 40.
The sum of three numbers which are in GP is 43 and their product is 216. Find the numbers.
Answer:
\(\frac{a}{r}\), a, ar
\(\frac{a}{r}\) × a × ar = 216
a³ = 216 ⇒ a =6
\(\frac{a}{r}\)+ a + ar = 43
\(\frac{6}{r}\) + 6 + 6r = 43
\(\frac{6}{r}\) + 6r = 43 – 6
\(\frac{6}{r}\) + 6r = 37
6 + 6r² = 37r
6r – 37r + 6 =0
6r² – 36r – r + 6 = 0
6r(r – 6) – (r – 6) =0
(r- 6) (6r- 1) = 0
∴ r = 6 or r = \(\frac{1}{6}\)
∴ The numbers are \(\frac{6}{6}\), 6, 6×6
1, 6, 36

Question 41.
If α and β are the roots of the equation 3x² – 4x + 15 = 0 then find the value of \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}\)
Answer:
α + β = –\(\left(\frac{-4}{3}\right)=\frac{4}{3}\)
αβ = \(\frac{15}{3}\) = 5
\(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}=\frac{\alpha^{3}+\beta^{3}}{\alpha \beta}=\frac{(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)}{\alpha \beta}\)

= \(\frac{\left(\frac{4}{3}\right)^{3}-3(5)\left(\frac{4}{3}\right)}{5}=\frac{\frac{64}{27}-20}{5}=\frac{64-540}{27(5)}=\frac{-476}{135}\)

Question 42.
If the present price of a bike is ₹ 32,290. It is value decreases every year by 10% then find its value before 3 years?
Answer:
P = 32,290, r = 0.10, n = 3
A = P( 1 – r)ⁿ
A =32,290(1 -0.1)³
= 32290 (0.9)³
= 32290 (0.729)
= 23,539.41
∴ The value of the bike after 3 years = ₹ 23,539.41

Question 43.
In what time will a sum of ₹ 2000 becomes ₹ 3900 at 5% p.a compound interest payable half yearly?
Answer:
P = ₹ 2,000, A= 3900, n = ?, R= 5%, q = 2, r = ?
r = \(\left(1+\frac{\mathrm{R}}{q}\right)^{q}\) – 1 = \(\left(1+\frac{0.05}{2}\right)^{2}\) – 1
r = 0.050625
A = P(1 + r)ⁿ
3900 = 2000 (1 + 0.050625)ⁿ
3900 =2000(1.050625)
\(\frac{39}{20}\) = (1.050625)
log 39 – log 20 = n log 1.050625 ⇒ 1.5910 – 1.3010 = n(0.02 144)
⇒ 0.29 = n(0.02144)
n = \(\frac{0.29}{0.02144}\) =13.52years.

Question 44.
Find the lengths of the altitudes of the triangle whose vertices are (5,2) (3, -3) and (-4, 3).
Answer:
A = (5,2) B = (3, -3) C = (-4, 3)
Area of AABC = \(\frac{1}{2}\) [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= \(\frac{1}{2}\)[5 (-3-3) + 3(3-2)-4(2 + 3)]= \(\frac{1}{2}\) [5 (-6) + 3 (1)-4 (5)]
= \(\frac{1}{2}\)[-30 + 3 – 20] = \(\frac{1}{2}\)[-47] = \(\left|\frac{-47}{2}\right|=\frac{47}{2}\) Square units

Area of the ΔABC = \(\frac{1}{2}\) × BC × AD
AD = \(\frac{2 \Delta \mathrm{ABC}}{\mathrm{BC}}\)
BC = \(\sqrt{(-4-3)^{2}+(3+3)^{2}}=\sqrt{49+36}=\sqrt{85}\)

AC = \(\sqrt{(-4-5)^{2}+(3-2)^{2}}=\sqrt{81+1}=\sqrt{82}\)
ΔABC = \(\frac{1}{2}\) × AC × BE
BE = \(\frac{2 \times \Delta \mathrm{ABC}}{\mathrm{AC}}=\frac{2 \times 47}{2 \times \sqrt{82}}\)
BE = \(\)
AB = \(\sqrt{(3-5)^{2}+(-3-2)^{2}}=\sqrt{4+25}=\sqrt{29}\)
ΔABC = \(\frac{1}{2}\) × AB × CF

Question 45.
(a) Form a quadratic equation whose roots are -2 and 5
Answer:
(a) (x + 2) (x – 5) = 0
x² – 5x + 2x – 10 = 0
x² – 3x – 10 = 0

(b) Evaluate using log. tables \(\frac{(25.36)^{2} \times(0.4569)^{2}}{(847.5)}\)
Answer:
Let x = \(\frac{(25.36)^{2} \times(0.4569)^{2}}{(847.5)}\)
log x = 2 log 25.36 + 2 log 0.4569 – log 847.5
= 2(1.4041) + 2(1.6598) – (2.9281)
= 2.8082 + 2(-1 + 0.6598)-(2.9281)
=-0.8003 = -1 + 1 -0.8003 = -1 + 0.1997
log x = 1.1997
x = AL[1.1997] = 0.1584

Question 46.
Find the equation of the straight line passing through the point of intersection of 2x + 4y = 3 and x + 5y = 1 and making equal positive intercepts on the coordinate axes.
Answer:
2x + 4y = 3 × 1
x + 5y = 1 × 2

y = –\(\frac{1}{6}\)
x + 5y = 1
x + 5(-1/6) = 1
x = 1 + \(\frac{5}{6}=\frac{6+5}{6}=\frac{11}{6}\)

Point of intersection \(\left(\frac{11}{6}, \frac{-1}{6}\right)\)
\(\frac{x}{a}+\frac{y}{b}\) = 1
\(\frac{x}{a}+\frac{y}{a}\) = 1 ⇒ x + y = a (1)
It passes through\(\left(\frac{11}{6}, \frac{-1}{6}\right)\) put x = \(\frac{11}{6}\) , y = -1/6 in(1)
\(\frac{11}{6}-\frac{1}{6}\) = a ⇒ \(\frac{10}{6}\) = a
a = \(\frac{5}{3}\)
b = \(\frac{5}{3}\)
\(\frac{x}{a}+\frac{y}{b}\) = 1
\(\frac{x}{\frac{5}{3}}+\frac{y}{\frac{5}{3}}\) = 1
\(\frac{3 x}{5}+\frac{3 y}{5}\) = 1
3x + 3y = 5
3x + 3y – 5 = 0

Question 47.
Find the locus of a point equidistant from (1, 0) and (-1, 0).
Answer:
Let A = (I, O), B = (-1, 0)
Let P(x, y) be any point on the locus
∴ PA = PB
\(\)
(x – 1)² + y² = (x + 1)² + y²

-4x = 0
∴ x = 0

Question 48.
Find the sum of the following series: 1 + (1 + 2) + (1 + 2 + 3)………..n terms.
Answer:
t =1 + 2 + 3 + ………….. + n = \(\frac{n(n+1)}{2}\)
S_n = Σt_n = \(\frac{\Sigma n(n+1)}{2}=\frac{1}{2}\)[Σn² + Σn] = \(\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right]\)
= \(\frac{1}{2}\left[n(n+1)\left\{\frac{2 n+1}{6}+\frac{1}{2}\right\}\right]=\frac{n(n+1)}{2}\left\{\frac{2 n+1+3}{6}\right\}=\frac{n(n+1)\{2 n+4\}}{2.6}\)

PART-E

V. Answer any one equestion. (1 × 10 = 10)

Question 49.
(a) Prove that the lines x + y + 4 = 0, 2x = 3y + 7 and 3x + y + 6 = 0 are concurrent. Also find the point of concurrency.
Answer:
(a) x + y = -4 × 2
2x – 3y = 7 × 1

y = -3
x + y = -4
x – 3 = -4
x = -4 + 3
x = -1
∴ point of intersection = (-1, -3)
Put x = -1 y = -3 in 3x + y + 6 = 0
we get 3(-1) -3 + 6 = -3 – 3 + 6 = 0
∴ The three lines are concurrent
point of concurrency = (-1, -3)

(b) If 12 cot² A – 31 cosecA + 32 = 0, find the valu&of sinA
Answer:
12cot² A – 31 cosec A + 32 = 0
12(cose2c A – 1) – 31 cosec A + 32 = 0
12 cosec2A- 12 – 31 cosec A + 32 = 0
12 cosec2 A – 31 cosec A + 19 = 0
12 cosec2A- 12 cosecA- 19 cosec A + 19 = 0
12 cosecA (cosecA – 1) – 19 (cosecA – 1) = 0
(cosecA – 1) (12 cosecA – 19) = 0
cosec A – 1 =0
cosecA = 1
sinA = 1

12 cosecA – 19 = 0
cosec A = \(\frac{19}{12}\)
sin A = \(\frac{12}{19}\)

(c) Find the number of digits in 2M
Answer:
Let x = 2M
log x = log₂ 64 = 64 log2 = 64 (0.3010) = 19.264
As the characteristic of log x is 19 the number of digits = 19 + 1 = 20. 50

Question 50.
(a) Find the sum to n terms of the series:
.3 + .33 + .333 + .3333 +…….. n terms
Answer:
(a) Let S_n = 0.3 + 0.33 + 0.333 +………..n terms
= \(\frac{3}{10}+\frac{33}{100}+\frac{333}{1000}\) + ……………
\(\frac{S_{n}}{3}=\frac{1}{10}+\frac{11}{100}+\frac{111}{1000}\) + ……………
\(\frac{9 \mathrm{~S}_{n}}{3}=\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}\) + …………

\(\frac{9 \mathrm{~S}_{\mathrm{n}}}{3}\) = n – \(\frac{1}{9}\left(1-\frac{1}{10^{8}}\right)\)
S_n = \(\frac{3}{9}\left[n-\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)\right]=\frac{1}{3}\left[n-\frac{1}{9}\left(1-\frac{1}{10^{n}}\right)\right]\) n ∈ N

(b) A publishing house finds that the production of each book and the cost of the book are directly attributed, if the cost of each book is 30 and the fixed costs are 15000, selling price of each book is 45 then determine
(i) Revenue function
(ii) Cost function
(iii) Break even point function
Answer:
FC = ₹ 15,000, VC = 30xm
Total cost function = C(x) = VC + FC = 3Ox + 15,000
Revenue function = R(x) = 45x
For Break even point, C(x) = R(x)
30x+ 15000 = 45x
15000 = 45x – 30x 15000 = 15x
x = 1000 units.

(c) If n(∪) = 700, n(A) = 200, n(B) = 300 and n (A ∩ B) = 100 Find n (A’ ∩ B’)
Answer:
n(∪) = 700, n(A) – 200, n(B) = 300, n(A∩B) = 100
n(A’∩B’) = n(A∪B)’ – n(∪) – n(A∪B) = 700 – [n(A) + n(B) – n(A∩B)]
= 700 – [200 + 300 – 100] = 700 – [400] = 300