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## Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2017 (North)

Time: 3.15 Hours

Max. Marks: 100

Instructions:

- The questions paper consists of five parts A, B, C, D, and E.
- 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.
- 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.

Answer:

825 = 3^{1} × 5^{2} × 11.

Question 2.

Convert the set A = {a, e, i, o, u} from roster form into rule form.

Answer:

A = {x/x is a vowel in English alphabet}

Question 3.

If A = {a, b, c}; B = {c, d}. Find B × A.

Answer:

B × A = {c, d) x {a, b, c}

= c, a) (c, b) (c, c) (d, a) (d, b) (d, c)}.

Question 4.

Simplify: \(\frac{a^{m+n} \cdot a^{2 m-n}}{a^{m-n}}\)

Answer:

Question 5.

Express Y = 27 in the .Lograrithmic Form.

Answer:

log_{3} 27 = 3

Question 6.

Find the 8th term of an A.P -2, -4, -6 ………..

Answer:

T = a + (n – 1)d

T_{8} = -2 + (8 – 1)(-2)

T_{8} = -2 – 14 = -16

Question 7.

Solve: \(\frac{x+2}{x-2}=\frac{5}{2}\)

Answer:

2(x + 2) = 5(x – 1)

2x + 4 = 5x – 5

4 + 5 = 5x – 2x ⇒ 3x = 9 ⇒x = 3.

Question 8.

Define perpetuity.

Answer:

An annuity that is payable forever is called a perpetuity.

Question 9.

Convert 42% into a decimal.

Answer:

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.

Answer:

X̄_{12} = \(\frac{35 \times 80+25 \times 68}{35+25}=\frac{2800+1700}{60}\) = 75

Question 11.

Convent 75° into radians.

Answer:

Question 12.

If the distance between the points (3, -2) and (-1, a) is 5 units. Find the values of

Answer:

\(\sqrt{(3+1)^{2}+(-2-a)^{2}}\) = 5

42 + 4 + a^{2} + 4a = 5

a^{2} + 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.

Answer:

∴ L.C.M = 2^{3} × 3 × 7 = 24 × 7 = 168

Question 14.

Find the H.C.F of 18 and 24.

Answer:

18 = 2 × 3^{2}, 24 = 2^{3} × 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.

Answer:

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.

Answer:

Given a = 6 and T_{12} = a + 11d = 72

6 + 11d = 72,

11d = 72 – 6 = 66 ⇒ d = 6.

Question 18.

The sum of two consecutive numbers is 23. Find them.

Answer:

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.

Answer:

Answer:

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.

Answer:

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

Answer:

⇒ sec^{2} θ – tan^{2} 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.

Answer:

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.

Answer:

Diff in the amount = 8000-5000 = 3000₹

Question 24.

Prove that tan^{2} A(1 – sin^{2}A) = sin^{2} A

Answer:

Question 25.

Find the area of the triangle whose vertices are A(3, 4); B(2, -1), and C(4, -6).

Answer:

Area of ∆ABC = \(\frac{1}{2}\)[x_{1}(y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})]

= \(\frac{1}{2}\)[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.

Answer:

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).

Answer:

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’.

Answer:

Proper subsets of A

X = P(A) = {{Q} (2) (3) (4) (2, 3) (3, 4) (4, 2)}.

Question 29.

Prove that (x^{b-c})^{a} . (x^{c-a})^{b} . (x^{a-b})^{c} = 1.

Answer:

Question 30.

If the second term of the G.P is 6 and 5th term is 162. Then find the GP.

Answer:

T_{2} = ar = 6 ….(1), T_{5} = ar^{4} = 162 ….(2), G.P. = ?

ar = 6 ⇒ 3a = 6 ⇒ a = 2

Terms of G.P are a, ar ar^{2}, ar^{3}

2, 2 × 3, 2 × 3^{2}, 2 × 3^{3}………….

2, 6, 18…………..

Question 31.

The sum of 6 times a number and 5 times the number is 55. Which is that number?

Answer:

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

Answer:

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.

Answer:

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.

Answer:

X̄_{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.

Answer:

∴ Total number of votes = 6000

Question 36.

If tan θ = \(\frac{a}{b}\) show that \(\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\)

Answer:

Question 37.

A point ‘P’ moves such that PA^{2} = 3PB^{2}. If A = (5, 0) and B = (-5, 0). Find the equation of the Locus of ‘P’

Answer:

Let P(x, y) be the point on the locus

Given PA^{2} = 3PB^{2} and A = (5, 0), B = (-5, 0)

(x – 5)^{2} + (y – 0)^{2} = 3[(x + 5)^{2} + (y – 0)^{2}]

x^{2} + 25- 10x + y^{2} = 3[x^{2} + 25 + 10x + y^{2}]

3x^{2 }+ 75 + 30x + 3y^{2} = x^{2} + y^{2} – 10x + 25

2x^{2} + 2y^{2} + 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.

Answer:

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.

Answer:

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.

Answer:

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

a^{2} + b^{2} – 2ab = 16ab (Add 4ab both sides)

a^{2} + b^{2} + 2ab = 20ab,

(a + b)^{2} = 20ab.

Question 41.

Find the sum of n terms of GP. 5 + 55 + 555 +……………

Answer:

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 x^{3} – 2x^{2} – 5x + 6.

Answer:

Let f(x) = x^{3} – 2x^{2} – 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 x^{2} – x – 6 = 0 is the quotient and remainder is 0.

x^{2} – 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?

Answer:

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?

Answer:

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 \(\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}=\frac{19}{7}\)

Answer:

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?

Answer:

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).

Answer:

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) = \(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)

⇒ x_{2} + x_{3} = 8, y_{2} + y_{3} = 2 …………….(1)

E = Mid point of CA

(7, 9) = \(\left(\frac{x_{3}+x_{1}}{2}, \frac{y_{3}+y_{1}}{2}\right)\)

⇒ x_{3} + x_{1} = 14, y_{3} + y_{1} = 18 ……………..(2)

F = Mid point of AB

(4,11) = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

⇒ x_{1} + x_{2} = 8, y_{1} + y_{2} = 22 ………………..(3)

Solving equations 1,2 and 3 we get

x_{1} = 7, x_{2} = 1, x_{3} = 7

y_{1} =21, y_{2} = 1, y_{3} = -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.

Answer:

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’

Answer:

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 \(\frac{1}{64}\).

Answer:

G.M = \(\sqrt{\frac{1}{4} \times \frac{1}{64}}=\sqrt{\frac{1}{4^{4}}}=\frac{1}{4^{2}}=\frac{1}{16}\)

(c) If a^{x} = b; & b^{y} = c; c^{z} = a. Show that xyz = 1.

Answer:

Consider a^{x} = b’

(c^{z})^{x} = b

c^{zx} = b

(b^{y})^{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).

Answer:

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(x^{2} – 2x + 1 + y^{2} + 2y + 1) = 9x^{2} + 16y^{2} + 1 – 24xy – 8y + 6x

16x^{2} + 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.

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

(c) What is the present value of an income of 3000 to be received forever if the interest rate if 14% p.a.

Answer:

P_{∞} = \(\) = ₹21428.5