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## Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2017 (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.
Give the canonical representation of 825.
825 = 31 × 52 × 11.

Question 2.
Convert the set A = {a, e, i, o, u} from roster form into rule form.
A = {x/x is a vowel in English alphabet}

Question 3.
If A = {a, b, c}; B = {c, d}. Find B × A.
B × A = {c, d) x {a, b, c}
= c, a) (c, b) (c, c) (d, a) (d, b) (d, c)}.

Question 4.
Simplify: $$\frac{a^{m+n} \cdot a^{2 m-n}}{a^{m-n}}$$ Question 5.
Express Y = 27 in the .Lograrithmic Form.
log3 27 = 3 Question 6.
Find the 8th term of an A.P -2, -4, -6 ………..
T = a + (n – 1)d
T8 = -2 + (8 – 1)(-2)
T8 = -2 – 14 = -16

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

Question 8.
Define perpetuity.
An annuity that is payable forever is called a perpetuity.

Question 9.
Convert 42% into a decimal.
42% = $$\frac{42}{100}$$ = 0.42.

Question 10.
The average score of 35 girls is 80 and the average score of 25 boys is 68. Find the average score of both boys and girls together.
12 = $$\frac{35 \times 80+25 \times 68}{35+25}=\frac{2800+1700}{60}$$ = 75

Question 11. Question 12.
If the distance between the points (3, -2) and (-1, a) is 5 units. Find the values of
$$\sqrt{(3+1)^{2}+(-2-a)^{2}}$$ = 5
42 + 4 + a2 + 4a = 5
a2 + 4a + 15 = 0
Using formula method we get the values of a.

Part – B

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

Question 13.
Find the L.C.M of 12, 21 and 24. ∴ L.C.M = 23 × 3 × 7 = 24 × 7 = 168

Question 14.
Find the H.C.F of 18 and 24.
18 = 2 × 32, 24 = 23 × 3
H.C.F = 2 × 3 = 6 Question 15.
If A = {a, b,c, d}\ B = {d, e, f, g, h, i}. Find A – B & B – A.
Ans.wer:
A – B = {a, b, c}, B – A = {e, f g, h, i}

Question 16.
Find the value of x so that the slope of the line joining the points (2, 5) and (x 13) is 20.
Slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$
$$\frac{13-5}{x-2}$$ = 20 ⇒ $$\frac{8}{x-2}=\frac{20}{1}$$
⇒ 20x – 40 = 8 ⇒ 20x = 48
⇒ x = $$\frac{48}{20}=\frac{12}{5}$$ = 2.4

Question 17.
Find the common difference of an A.P. whose first term is 6 and 12th term is 72.
Given a = 6 and T12 = a + 11d = 72
6 + 11d = 72,
11d = 72 – 6 = 66 ⇒ d = 6.

Question 18.
The sum of two consecutive numbers is 23. Find them.
Let the 2 consecutive numbers be x and x + 1
Given sum = 23
x + x + 1 = 23
2x = 22 ⇒ x = 11
∴ The 2 consecutive numbers are 11 & 12.

Question 19.
Solve 2x + 6 < 0; x ∈ Z inequalities in one variable and represent the solution on the number line.
Given 2x + 6 < 0
∴ 2x ≤ -6 ⇒ x ≤ -3 Solution set is [-∞, -3]. Question 20.
Find the interest on Rs. 1,500 at 4% pa. for 145 days.
S.I.= 1500 × $$\frac{145}{365} \times \frac{4}{100}=\frac{870000}{36500}$$ =23.836.

Question 21.
If x = a secθ; y = btanθ prove that $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}$$ = 1 ⇒ sec2 θ – tan2 θ = 1 = R.H.S

Question 22.
The weight of 6 men is 90 Kg, 70.5, 56 Kg, 45.5Kg, 85 Kg, and 78 Kg. Find average weight.
X̄ = $$\frac{90+70.5+56+45.5+85+78}{6}=\frac{425}{6}$$ = 70.83

Question 23.
Karthik received a scholarship of ₹ 5,000 in 2011 and ₹ 8,000 in 2012. Find the percentage increase.
Diff in the amount = 8000-5000 = 3000₹  Question 24.
Prove that tan2 A(1 – sin2A) = sin2 A Question 25.
Find the area of the triangle whose vertices are A(3, 4); B(2, -1), and C(4, -6).
Area of ∆ABC = $$\frac{1}{2}$$[x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= $$\frac{1}{2}$$[3 (-1 + 6) + 2 (-6 – 4)+4 (4 – (-1)].
A = $$\frac{1}{2}$$[3(5) + 2(-10) + 4(5)] Part – C

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

Question 26.
If the product of two numbers is 216 and their L.C.M is 36. Find their H.C.F.
Given L.C.M of 36 and Product = 216, H.C.F = ?
H.C.F = $$\frac{a \times b}{L . C . M}=\frac{216}{36}$$ = 6

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

Question 28.
If A = {2, 3, 4} write ail the proper subsets of ‘A’.
Proper subsets of A
X = P(A) = {{Q} (2) (3) (4) (2, 3) (3, 4) (4, 2)}. Question 29.
Prove that (xb-c)a . (xc-a)b . (xa-b)c = 1. Question 30.
If the second term of the G.P is 6 and 5th term is 162. Then find the GP.
T2 = ar = 6 ….(1), T5 = ar4 = 162 ….(2), G.P. = ? ar = 6 ⇒ 3a = 6 ⇒ a = 2
Terms of G.P are a, ar ar2, ar3
2, 2 × 3, 2 × 32, 2 × 33………….
2, 6, 18…………..

Question 31.
The sum of 6 times a number and 5 times the number is 55. Which is that number?
Let the number be x
Given 6x + 5x = 55
11x = 55 ⇒ x = 5
∴ The number is 5.

Question 32.
Solve graphically:
3x – 7 > 2(x – 6) and 6 – x > 11 – 2x; x∈R
3x – 7 > 2x – 12
3x – 2x > -12 + 7
x > -5

6 – x > 11 – 2x
-x + 2x >11 – 6
x > 5

Question 33.
Find the present value of an annuity due of 8000 for 5 years at 5% p.a.
P = $$\frac{a\left[(1+i)^{n}-1\right]}{i(1+i)^{n}}$$(1 + i)
P = $$\frac{8000\left[(1+0.05)^{5}-1\right]}{0.05(1+0.05)^{5}}$$ × (1 + 0.05)
= $$\frac{8000\left[(1.05)^{5}-1\right](1.05)}{0.05 \times(1.05)^{5}}=\frac{8000[(1.2763-1)](1.05)}{0.05 \times 1.2763}$$ = ₹ 36369.50

Question 34.
Calculate the arithmetic average mark from the following data. w = $$\frac{45 \times 11+75 \times 10+60 \times 15+55 \times 12+93 \times 2}{11+10+15+12+2}$$
= $$\frac{495+750+900+660+186}{50}=\frac{2991}{50}$$ = 59.82 ≈ 60

Question 35.
In an election, the winning candidate got 4,800 votes which are 80% of the total votes. Calculate the total number of votes. ∴ Total number of votes = 6000

Question 36.
If tan θ = $$\frac{a}{b}$$ show that $$\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$$ Question 37.
A point ‘P’ moves such that PA2 = 3PB2. If A = (5, 0) and B = (-5, 0). Find the equation of the Locus of ‘P’
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 38.
Find the point of intersection of the lines 3x = 2y – 5 = 0 and 4x – y – 3 = 0.
3x + 2y – 5 = 0
4x – y – 3 = 0 → Multiplying by 2 11x – 11 = 0 ⇒ x = $$\frac{11}{11}$$ = 1
y = 4x – 3 = 4 – 3 = 1.
∴ Point of intersection is (1,1).

Part – D

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

Question 39.
Out of 50 people, 20 people drink tea, 10 take both tea and coffee. How many take at least one of the two drinks.
Given n(T) = 20, n(T∪C) = 50, n(T∩C) = 10, n(C) = ? n(T∪C) = n(T) + n(C) – n(T∩C)
50 = 20 + n(C)-10
n(C) = 40
∴ No. of people taking atleast one of the two drinks = 10 + 10 + 30 = 50.

Question 40.
If log$$\left(\frac{a-b}{4}\right)$$ = log$$\sqrt{a}$$ + log$$\sqrt{b}$$ show that (a + b)2 = 20ab.
Given log$$\left(\frac{a-b}{4}\right)$$ = log$$\sqrt{a}$$ + log$$\sqrt{b}$$
log$$\left(\frac{a-b}{4}\right)$$ = log$$\sqrt{ab}$$
$$\frac{a-b}{4}=\sqrt{a b}$$ S.B.S
(a – b)2 = 16ab
a2 + b2 – 2ab = 16ab (Add 4ab both sides)
a2 + b2 + 2ab = 20ab,
(a + b)2 = 20ab.

Question 41.
Find the sum of n terms of GP. 5 + 55 + 555 +……………
Let S = 5 + 55 + 555 +…………… to n terms
$$\frac{\mathrm{S}}{5}$$ = 1+ 11 + 111+………..to n terms
$$\frac{9 \mathrm{~S}}{5}$$ = 9 + 99 + 999 +…………… to n terms
=(10 – 1) + (100 – 1) + (1000 – 1) +……….. to n terms
= (10 + 100 + 1000 +………… to n terms) – (1 + 1 + 1………. to n ternis)
= $$\frac{10\left(10^{n}-1\right)}{9}$$ – 1 × n
S = $$\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]$$

Question 42.
Find the integral root between -3 and 3 by inspection and then using synthetic division, solve the equation x3 – 2x2 – 5x + 6.
Let f(x) = x3 – 2x2 – 5x + 6,
f(1)= 1 – 2 – 5 + 1 = 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 remainder is 0.
x2 – x – 6 = 0
(x – 3) (x + 2) = 0 ⇒ x = -2 or 3
Thus x = 1, -2, 3 are the roots of the given equation.

Question 43.
At what time a sum of ₹1,200 will earn ₹573 as compound interest at the rate of 5% p.a. if the interest is added annually?
Given P=1200, C.I = 573, A = 1200 + 573 = 1773
i = $$\frac{5}{100}$$ = 0.05, n = ?
A = P(1 + i)n
1773 = 1200(1 +0.05)n
$$\frac{1773}{1200}$$1.4775 = (1.05)n
(1.05)n = (1.05)8 ⇒ n = 8. Question 44.
Preritha wants to buy a house after 5years when it is expected to cost 50 lakhs. How much should she save annually if her savings earn a compound interest of 12 percent?
Given F = 50,00,000, n = 5, i = 0.12, a = ?
F = $$\frac{a\left[(1+i)^{n}-1\right.}{i}$$ ⇒ 50,00,000 = $$\frac{a\left[(1+0.12)^{5}-1\right]}{0.12}$$
50,00,000 = 6.3528a ⇒ a = $$\frac{50,00,000}{6.3528}$$ = ₹ 787054.5

Question 45.
If θ = $$\frac{5}{2}$$ and θ is acute then prove that $$\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}=\frac{19}{7}$$
Given cotθ = $$\frac{5}{2}=\frac{a d j}{o p p}$$
∴ Hyp = $$\sqrt{5^{2}+2^{2}}=\sqrt{29}$$
cosθ = $$\frac{5}{\sqrt{29}}$$, sinθ = $$\frac{2}{\sqrt{29}}$$
L.H.S = $$\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}$$ Question 46.
The price of a pair of trousers was decreased by 22% to 390. Whast was the original price of the trousers?
Let the price of trouser = x
It was decreased by 22% to ₹ 390
∴ x – $$\frac{22 x}{100}$$ = 390
$$\frac{100 x-22 x}{100}$$ = 390 ⇒ 78x = 390 × 100
x = $$\frac{39,000}{78}$$ =500
∴ Original price of the trouser = ₹500.

Question 47.
Find the coordinates of the vertices of the triangle. Given the midpoints of the sides as (4, -1); (7, 9) & (4, 11).
Let A, B, C be the vertices of the A,e, and D, E, and F be the mid-points of the sides BC, CA, and AB respectively.
D = Midpoint of BC
(4, -1) = $$\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)$$

⇒ x2 + x3 = 8, y2 + y3 = 2 …………….(1)
E = Mid point of CA
(7, 9) = $$\left(\frac{x_{3}+x_{1}}{2}, \frac{y_{3}+y_{1}}{2}\right)$$ ⇒ x3 + x1 = 14, y3 + y1 = 18 ……………..(2)
F = Mid point of AB
(4,11) = $$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$$
⇒ x1 + x2 = 8, y1 + y2 = 22 ………………..(3)

Solving equations 1,2 and 3 we get
x1 = 7, x2 = 1, x3 = 7
y1 =21, y2 = 1, y3 = -3
Thus A = (7, 21), B = (1, 1) and C = (7, -3). Question 48.
Find ‘a’ so that the lines x – 6y + a = 0; 2x + 3y + 4 = 0 and x + 4y + 1 = 0 concurrent.
Given 2x + 3y + 4 = 0 ………….. (1)
x + 4y + 1 = 0 …………. (2) × 2
x – 6y + a = 0 …………. (3)
Solving 1 and 2 we get = $$\frac{-8}{5}$$ – 1 = $$\frac{-8-5}{5}=\frac{-13}{5}$$

Put x = $$\frac{-13}{5}$$ and y = $$\frac{2}{5}$$ in equation 3
We get x – 6y + a = 0
a = 6y – x
a = 6.$$\frac{2}{5}-\left(\frac{-13}{5}\right)$$
= $$\frac{12}{5}+\frac{13}{5}=\frac{25}{5}$$ = 5
∴ 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’
A ∩ B = {3,4, 5}
(A ∩ B)’ = {1,2, 6, 7, 8,9} ………….(1)
A’∪B’ = {6,7, 8,9} ∪ {1,2, 8, 9}
= {1,2, 6,7, 8, 9} ………………(2)
From 1 and 2 we get A’∪B’ = (A ∩ B)’.

(b) Insert G. Means [Geometric Mean] between $$\frac{1}{4}$$ and $$\frac{1}{64}$$.
G.M = $$\sqrt{\frac{1}{4} \times \frac{1}{64}}=\sqrt{\frac{1}{4^{4}}}=\frac{1}{4^{2}}=\frac{1}{16}$$

(c) If ax = b; & by = c; cz = a. Show that xyz = 1.
Consider ax = b’
(cz)x = b
czx = b
(by)zx = b’ ⇒ & bxyz = b’⇒ xyz = 1. Question 50.
(a) Find the equation of the locus of the point which moves such that its distance from 3x – 4y + 1 = 0 is equal to its distance from (1, -1).
Let A = (1, -1) and P(x, y) be any point on the locus, then PA = distance from the line 3x – 4y +1 = 0
$$\sqrt{(x-1)^{2}+(y+1)^{2}}=\left|\frac{3 x-4 y+1}{\sqrt{9^{2}+16^{2}}}\right|$$ S.B.S

(x – 1)2 + (y + 1)2 = $$\frac{(3 x-4 y+1)^{2}}{25}$$
25(x2 – 2x + 1 + y2 + 2y + 1) = 9x2 + 16y2 + 1 – 24xy – 8y + 6x
16x2 + 24xy + 9y – 56x + 58y + 49 =0 is the required equation of the locus.

(b) A manufacturer produced and sells balloons at 8 per unit. His fixed cost is ₹6,500 and the variable cost per balloon is ₹3.50. Calculate
(i) Revenue Function (ii) Cost Function
(iii) Profi Function (iv) 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) What is the present value of an income of 3000 to be received forever if the interest rate if 14% p.a.
P =  = ₹21428.5