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

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.

Write the imaginary part of 3 – Imaginary part of 3 – i is -1.

Answer:

The imaginary part of 3 – i is -1

Question 2.

If A has 4 elements, how many elements, how many elements will P(A) have.

Answer:

P(A) will have 16 Elements.

Question 3.

If R-1 = {(2, 4) (1, 2) (3, 1) (3, 2)} find R.

Answer:

R= {(4, 2) (2,1) (1,3) (2, 3)}.

Question 4.

Simplify: (2^{2})^{2} – 2^{(3°)}

Answer:

= 1 – 2′ = 1 – 2 = -1

Question 5.

Solve for x if log_{7} x = 2.

Answer:

x = 7^{2} = 49.

Question 6.

Find the sum to infinity of the GP 3,1, \(\frac{1}{3}\), ……………

Answer:

S_{n} = \(\frac{a}{1-r}=\frac{3}{1-\frac{1}{3}}=\frac{9}{3-1}=\frac{9}{2}\)

Question 7.

Form the quadratic equation whose roots are 2, 3.

Answer:

(x – 2)(x – 3) = 0 ⇒ x^{2} – 5x + 6 = 0.

Question 8.

What is the simple interest on 650 for 14 weeks at 6% p.a?

Answer:

S.I = 650 × \(\frac{14}{52} \times \frac{6}{100}\) = 10.5

Question 9.

The average score of 35 girls is 80 and the average score of 25 boys ¡s 68. Find the average score of both boys and girls together.

Answer:

Question 10.

By selling a book at’ 250 the profit made is ₹ 50. What is the cost price of the book?

Answer:

C.P = S.P – Profit = 250 – 50 = 200₹.

Question 11.

Convent 315° to radians.

Answer:

Question 12.

Find the slope of the line 2x + 5y – 11 = 0.

Answer:

Slope = –\(\frac{2}{5}\)

Part – B

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

Question 13.

Find the number of positive divisors of 360.

Answer:

360 = 2^{3} × 3^{2} × 5^{1}

n = P_{1}^{α1}. P_{2}^{α2}. P_{3}^{α3}

T(n) = T(360) = (1 + α_{1}) (1 + α_{2}) (1 + α_{3}) = (3 + 1) (2 + 1) (1 + 1) = 24.

∴ If has 24 positive divisors.

Question 14.

Find the HCF of two numbers if their LCM is 1260 and product is 52920.

Answer:

H.C.F = \(\frac{a \times b}{\mathrm{LCM}}=\frac{52920}{1260}=\frac{5292}{126}\) = 42

Question 15.

If f(x) = x, g(x) = x^{3} + 1 find (a) fog (2) (b) gof (1)

Answer:

fog (2) = f(g (2)) = f( 2^{3} + 1) = f( 9) = 9

gof(1) – g (f(1)) = g (1) = 1^{3} + 1 = 1 + 1 = 2.

Question 16.

Find the domain and range of the relation R = {(x, y): y = x^{2}, x is a positive prime number less than 10}

Answer:

R= {(2,4)'(3, 9) (5,25) (7, 49)}

Domain – {2, 31 5, 7}, Range = {4,9,25, 40}.

Question 17.

If a = 3^{x} b = 3^{y} c = 3z and ab – c^{2}. P.T. x + y = 2z.

Answer:

Given ab = c^{2}

3^{x}.3^{y} = (3^{2z}) ⇒ 3^{x+y} = 3^{2z}

x + y = 2z.

Question 18.

Insert 3 Arithmetic means between -2 and -10.

Answer:

Let A_{1}, A_{2}, A_{3} are 3 A.m between -2 and -10

∴ -2, A_{1}, A_{2}, A_{3}………. -10 are in A.P .

a a + d a + 2d a + 3d a + Ad are in A.P

∴ a = – 2 and a + 4d = – 10 ⇒ 4d = -10 + 2 = -8

d = – 2 .

Hence A_{1} = -2 – 2 = -4,

A_{2} = a + 2d = – 2 – 4 = – 6,

A_{3} = a + 3d = -2 – 6 = -8

Hence 3 Ams are -4, -6, and -8.

Question 19.

If a and b are the roots of the equation 2x^{2} + 5x + 5 = 0 the find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)

Answer:

Given 2x^{2} + 5x + 5 = 0

α + β = \(\frac{-b}{a}=\frac{-5}{2}\)

αβ = \(\frac{c}{a}=\frac{5}{2}\)

\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-5 / 2}{5 / 2}\) = -1

Question 20.

Solve 7x + 3 < 5x + 9, x∈R and represent on number line.

Answer:

7x + 3 < 5x + 9

7x – 5x <9 – 3

2x < 6

x < 3

∴ Solution set is (- ∞, 3)

Question 21.

Find the effective rate of interest when a sum lent at 12% is computed half yearly.

Answer:

r = \(\left(1+\frac{i}{q}\right)^{q}\) – 1 = \(\left(1+\frac{0.12}{2}\right)^{2}\) – 1

= (1 + 0.06)2 – 1 =(1.06)2 – 1 = 12.36.

Question 22.

The average age of 12 boys is 8 years. Another boy of 21 years joins the group. Find the average age of the new group.

Answer:

Ave age of 12 boys = 8 yrs.

∴ Total age of boys = 8 x 12 = 96 yrs.

When a boy of 21 yrs joins the group

∴ Total age of the 13 boys = 96 + 21 = 117 yrs.

∴ Average age of the new group = \(\frac{117}{13}\) = 9 yrs.

Question 23.

Find the value of sin (480°) + tan 135°.

Answer:

sin (480°) + tan (135°) = sin (360 + 120) + tan (180 – 45)

= sin 120-tan 45

= sin (180 – 60) – 1 = sin 60 – 1 = \(\frac{\sqrt{3}}{2}\) – 1

Question 24.

The CEO the triangle ABC is the point (2, 3). The coordinates of A are (5, 6) and B(-l, 4). Find the coordinates of C.

Answer:

G(x, y) = \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\), C(x_{3}, y_{3}) = ?

(2, 3) = \(\left(\frac{5-1+x_{3}}{3}, \frac{6+4+y_{3}}{3}\right)\)

(2, 3) = \(\left(\frac{4+x_{3}}{3}, \frac{10+y_{3}}{3}\right)\) ⇒ 6 = 4 + x_{3},

10 = y_{3} = 9

⇒ x_{3} – 6 – 4, y_{3} = 9 – 10

⇒ x_{3} = 2, y_{3} = -1

Hence co-ordinates of C(x_{3}, y_{3}) = C(2, – 1)

Question 25.

Find equation of line passing through (0, -4) and making an angle of 30° with the x-axis.

Answer:

Required equation is y – y_{1} = m(x – x_{1})

i.e., (y – (-4) = tan 30° (x = 0)

y + 4 = \(\)(x)

x = \(\sqrt{3}\)y + 4\(\sqrt{3}\) = 0

Part-C

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

Question 26.

Find the number which when divided by 36, 40 and 48 leaves the same remainder 5.

Answer:

Let us find out the L.C.M of 36,40 and 48

36 = 2^{2} × 3^{2}, 40 = 2^{3} × 5^{1}, 48 = 2^{4} × 3

∴ LCM = 2^{4} × 3^{2} × 5^{1}

= 16 × 9 × 5 = 720

Since 5 has to be the remainder. We have to add 5 to 720 i.e., 720 + 5 = 725

Question 27.

If A = {1, 2, 3} and R = {(1, 1) (1, 2) (2,1) (2, 2) (3, 3)} Prove that R is an equivalence relation on A.

Answer:

R = { (1, 1) (1, 2) (2, 1) (2, 2) (3, 3)}

R is reflexive ∵ (1, 1) (2, 2) (3, 3) ∈ R

R is Symmetric ∵ (1, 2) ∈ R => (2, 1) ∈ R

R is transitive ∵ (1, 2) (2, 1) ∈ R => (1, 1) ∈ R

Question 28.

If a^{1/3} + b^{1/3} = c^{1/3} Prove that (a + b + c)^{3} = 27 abc.

Answer:

Given a^{1/3} + b^{1/3} = -c^{1/3}

Cubing both sides we get

(a^{1/3} + b^{1/3})^{3} = -c

a + b + 3a^{1/3}b^{1/3}(a^{1/3} + b^{1/3})^{3} = -c

a + b + c = -3a^{1/3}, b^{1/3}, c^{1/3}.

Cubing both sides we get

{a + b + c)^{3} = 27 abc.

Question 29.

Solve: log_{2} x + log_{4} x = 3.

Answer:

\(\frac{\log x}{\log 2}+\frac{\log x}{2 \log 2}\) = 3

\(\frac{2 \log x+\log x}{2 \log 2}\) = 3

⇒ 3 log x = 6 log 2 ⇒ log x^{3} = log 2^{6} ⇒ x^{3} =(2^{2})^{3} ⇒ x = 4

Question 30.

The sum of 3 numbers in A.P. is 15 and their product is 105. Find the three numbers.

Answer:

Let the 3 numbers be a – d, a, a + d

Given

3a = 15 ⇒ a = 5

Also (a – d) (a) (a + d) = 105

5(25 – d^{2}) = 105

25 – d^{2} = 21 ⇒ 25 – 21 = d^{2} ⇒ d^{2} = 4 ⇒ d = 2

The 3 numbers are 5 – 2, 5, 5 + 2 = 3, 5, 7.

Question 31.

The age of a father is 5 times that of a son. 3 years ago the age of the father was 8 times that of his son. Find their present ages.

Answer:

Let the present age of the son be x yrs.

Let the present age of father be 5x yrs.

3yr ago, the age of son is x – 3 and that of father is 5x – 3

Given 5x – 3 = 8 (x – 3)

5x – 3 = 8x – 24

24 – 3 = 8x – 5x = 3x

x = 21 ⇒ x = 7

∴ The present age of son = 7 yrs and father is 35 yrs.

Question 32.

Solve graphically:

x + 3y ≥ 3; 2x + y ≤ 2; x ≥ 0, y ≥ 0.

Answer:

Let x + 3y = 3,

Put x = 0, y= 1, (0,1)

Put y = 0 x = 0, x = 3, (3,0)

2x + y = 2,

Put x = 0, y = 2, (0,2)

Put y = 0, x = 1, (1, 0)

Plot the two lines in the graph we get

Question 33.

Find the future value of an annuity of 5000 at 12% p.a. for 6 years.

Answer:

F = \(\frac{a\left[(1+i)^{n}-1\right]}{i}\) = 500\(\frac{\left[(1+0.12)^{6}-1\right]}{0.12}\) = 500\(\frac{\left[(1.12)^{6}-1\right]}{0.12}\) = 40575.9 Rs

Future value = 4.575.9

Question 34.

Ritu’s salary was increased by 10% and then again by 5%. If the present salary is ₹ 9,240. What was Ritu’s previous salary?

Answer:

Let his salary be x

\(\frac{220 x+11 x}{200}\) = 9240 ⇒ 231x = 9240 × 200

x = \(\frac{9240 \times 200}{231}\) = 8000

Question 35.

Prove that \(\frac{1}{1+\cos A}+\frac{1}{1-\cos A}\) = 2cosec^{2}A

Answer:

\(\frac{1}{1+\cos A}+\frac{1}{1-\cos A}\)

Question 36.

Find area of the quadrilateral whose vertices are A(1, 2) B(6, 2) C(5, 3) and D(3, 4) in order.

Answer:

Area of the Quadrilateral ABCD

= Area of triangle ∆ ABC + Area of ∆ACD

= \(\frac{5}{2}\) + 3 = \(\frac{11}{2}\) sq units

But Area of ∆ABC = \(\frac{1}{2}\)Σx_{1} (y_{2} – y_{3})

= | Area of ∆ABC| + |Area of ∆ACD| = \(\left|\frac{5}{2}\right|\) + |3| = \(\frac{5}{2}\) + 3 = \(\frac{5+6}{2}=\frac{11}{2}\) sq.units

Question 37.

Find the equation of the locus of the point which moves such that it ¡s equidistant from (4, 2) and x-axis.

Answer:

Let P(x, y) be the point on the locus and A = (4,2)

∴ PA = Y = (PA)^{2} = y^{2}

⇒ (x – y)^{2} + (y – 2)^{2} = y^{2}

⇒ x^{2} – 8x – 4y + 20 = 0 is the required equation.

Question 38.

Find k so that the distance from (2, 3) to the line 8x + 15y + k = 0. may be equal to 4 units.

Answer:

\(\left|\frac{8(2)+15(3)+k}{\sqrt{8^{2}+15^{2}}}\right|\) = 4

\(\left|\frac{16+45+k}{\sqrt{64+225}}\right|\) = 4

16 + k = 4\(\sqrt{289}\) = 4 × 17 = 68

k = 4(17) – 61 = 68 – 61 = 7. ∴ k = 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 need magazines A and B, 10 read magazines A and C, 5 read magazines B and C while 3 read all the three magazines.

Find: (a) How many read none of the three magazines

(b) How many read-only magazines

(c) How many read exactly one magazine only.

Answer:

Given U = 100, n(A) = 28, n(B) = 30, n(C) = 42

n(A∩B) = 8, n(A∩C) = 10, n(B∩C) = 5, n(A∩B∩C) = 3

(i) Number of people who read none of the magazines

= 100 – (13 + 5 + 20 + 7 + 3 + 2 + 30)

= 100 – 80 = 20

(ii) No. of people who read only magazines C = 30

(iii) No. of people who read exactly one magazines only = 13 + 20 + 30 = 63

Question 40.

Using log table find the value of \(\frac{21.54 \times 72.6}{6.5}\).

Answer:

Let log = log\(\left(\frac{21.54 \times 72.6}{6.5}\right)\)

logx = log 21.54 + log 72.6 – log 6.5

= 1.3332+ 1.8609-0.8129

x = A.L (2.3812) = 240.58.

Question 41.

Find the sum of all even integers from 40 to 160.

Answer:

S_{n} = 40, 42, 44……………160

a = 40, d = 2, T_{n} = 160

T_{n} = a + (n – 1)d

160 = 40 + (n – 1)2

120 = 2n – 2

122 = 2n ⇒ n = 61

Consider Sn = \(\frac{n}{2}\)(a + l) = \(\frac{61}{2}\)(40+160)= \(\frac{61 \times 200}{2}\) =6100

∴ Sum of all even integers = 6100.

Question 42.

Solve x^{3} – 10x^{2} + 29x – 20 = 0. Using synthetic division given that it has an integral root between -3 and 3.

Answer:

f(x) = x^{3} – 10x^{2} + 29x – 20

f(1)= 1 – 10 + 29 – 20 = 0

x = 1 is a root of the given equation. Let us remove this root by synthetic division.

∴ The resulting equation is x^{2} – 9x + 20 = 0 is the quotient and remainder is 0.

x^{2} – 9x + 20 = 0 (x – 5) (x – 4) = 0

x = 4 or 5

Thus x = 1, 4, 5 are the roots of the given equation.

Question 43.

The difference between simple interest and compound interest on a certain sum of money invested for 3 years at 6% p.a. is f 110.16. Find the sum.

Answer:

Let the sum = x

S.I = \(\frac{x \times 3 \times 6}{100}\) = 0.18x

C.I. = A – P where A = P(1 + i)^{n}

= x(1 +0.06)^{3} = 1.191016x

C.I.= A – P = 1.191016x – 1x = 0.191016x

Given C.I. – S.I = 110.16

0.191016x – 0.18x = 110.16

0.011016x = 110.16

x = \(\frac{110.16}{0.011016}\) = 10,000

∴ The sum= 10,000.

Question 44.

Find x if x^{3} sin45°cos 60° = \(\frac{\tan ^{2} 60^{\circ} \cdot \cos e c 30^{\circ}}{\sec 45^{\circ} \cdot \cot ^{2} 30^{\circ}}\)

Answer:

x^{3}. \(\frac{1}{\sqrt{2}} \cdot\left(\frac{1}{2}\right)^{2}=\frac{(\sqrt{3})^{2} \cdot 2}{\sqrt{2} \cdot\left(\frac{1}{\sqrt{3}}\right)^{2}}\)

x^{3}.\(\frac{1}{\sqrt{2}} \cdot \frac{1}{4}=\frac{3.2}{\sqrt{2} \cdot \frac{1}{3}}\)

x^{3}. \(\frac{1}{4 \sqrt{2}}=\frac{18}{\sqrt{2}}\) ⇒ x3 = \(\frac{72 \sqrt{2}}{2}\) ⇒ x = 3\(\sqrt{72}\)

Question 45.

Raj wants to invest a lump-sum amount in the bank so that he can get an annual income of Rs.15,000 every year for the next 10 years. If the bank offers 16% p.a. compound interest, what is the amount he should invest today?

Answer:

Given, n = 10yrs, a = 15,000, i = 0.16, P = ?

P = \(\frac{a\left[(1+i)^{n}-1\right]}{i(1+i)^{n}}\)

= \(\frac{15,000\left[(1+0.16)^{10}-1\right]}{0.16(1+0.16)^{10}}=\frac{15,000\left[(1.16)^{10}-1\right]}{0.16 \times(1.16)^{10}}\) = 72498.2

∴ He should invert today 7249.8

Question 46.

A person spent 30% of his wealth and thereafter Rs. 20,000 and further 10% of the remainder. If Rs. 29,250 is still remaining, what was his total wealth?

Answer:

Let his wealth be x Rs

Again he spent 20,000 \(\frac{7 x}{10}\) – 20,000 \(\frac{7 x-2,00,000}{10}\)

Again he spent 10% = \(\left(\frac{7 x-2,00,000}{10}\right)-\frac{7 x-2,00,000}{10} \times \frac{1}{10}\)x

\(\left(\frac{7 x-2 ; 00,000}{10}\right)\left(1-\frac{x}{10}\right)=\frac{(7 x-2,00,000)(10-x)}{100}\)

\(\frac{(7 x-2,00,000)(10-x)}{100}\) = 29,250 =x 75,000

∴ Total wealth = 75,000.

Question 47.

Prove that A(-3, 6) B(-2,11) C(3,12) and D(2, 7) are the vertices of a rhombus. Also And its area.

Answer:

A(-3, 6) B(-2, 11) C(3, 12) and D(2, 7)

AB = \(\sqrt{(-2+3)^{2}+(11-6)^{2}}=\sqrt{1^{2}+5^{2}}=\sqrt{1+25}=\sqrt{26}\)

BC = \(\sqrt{(3+2)^{2}+(12-11)^{2}}=\sqrt{5^{2}+1^{2}}=\sqrt{25+1}=\sqrt{26}\)

CD = \(\sqrt{(2-3)^{2}+(7-12)^{2}}=\sqrt{(-1)^{2}+(-5)^{2}}=\sqrt{1+25}=\sqrt{26}\)

DA = \(\sqrt{(2+3)^{2}+(7-6)^{2}}=\sqrt{5^{2}+1^{2}}=\sqrt{25+1}=\sqrt{26}\)

d_{1} = AC\(\sqrt{(3+3)^{2}+(12-6)^{2}}=\sqrt{(6)^{2}+(6)^{2}}=\sqrt{36+36}=\sqrt{72}\) = 6\(\sqrt{2}\)

d_{2} = BD\(\sqrt{(3+3)^{2}+(12-6)^{2}}=\sqrt{(6)^{2}+(6)^{2}}=\sqrt{36+36}=\sqrt{72}\) = 4\(\sqrt{2}\)

All sides are equal = \(\sqrt{26}\) ∴ ABCD forms Rhombus.

Area = \(\frac{1}{2}\) d_{1}d_{2} = \(\frac{1}{2}\).6\(\sqrt{2}\).4\(\sqrt{2}\) = 24 sq. units

Question 48.

Find the equation of a straight line passing through the point (2, 2) such that the sum of its intercepts on the axes is 9.

Answer:

Intercept equation is \(\frac{x}{a}+\frac{y}{b}\) = 1

Also given a + b = 9

b = 9 – a

∴ Equation becomes \(\frac{x}{a}+\frac{y}{9-a}\) = 1. But this equation passes through (2,2)

\(\frac{2}{a}+\frac{2}{9-a}\) = 1

18 = 9a – a^{2}

a^{2} – 9a + 18 = 0

(a-6) (a-3) = 0

a = 6 or 3,

b = 3 or 6

The required equations are \(\frac{x}{6}+\frac{y}{3}\) = 1 or \(\frac{x}{3}+\frac{y}{6}\) = 1

Part-E

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

(a) If A = {3, 5, 7, 8} B = {4, 5, 7, 9} C = {1, 2, 3, 4, 5}

Find (i) (A – B) × (B – C)

(ii) (A ∩ B) × (B ∩ C)

Answer:

(A – B)= {3, 8}, B – C = {7, 9}

(i) (A – B) × (B – C) = {(3, 7) (3, 9) (8, 7) (8, 9)}

(ii) (A ∩ B) × (B ∩ C) = {5, 7} × {4, 5}

= {5, 4) (5, 5) (7, 4) (7, 5)}.

(b) A manufactures sells his product at Rs. 8.35 per unit, he is able to sell his entire production. His fixed cost is Rs. 2116 and his variable cost per unit is ₹ 7.20 Find

(i) Level of production at which he can make a profit of ₹ 4, 600

(ii) Level of output at which he will incur a loss of ₹1150

(iii) Break-even level of production.

Answer:

C(x)= 7.20x + 2116,

R(x)= 8.35x

P(x) = R(x) – C(x)

(i) P(x) = 4,600,

4600 = 8.35x – [7.20x + 2116] = 1.15x – 2116

1.15x = 4600 + 2116 = 6716 = 5840 .

(ii) C(x) – R(x) = 1150,

7.20x + 2116 – 8.35x = 1150

115x = 966,

x = 840 units

(iii) C(x) = R(x)

720x + 2116 = 8.35x,

2116 = 8.35x – 7.20x

2116 = 1.1 5x

x = 1840 units BEP

(c) Using log find the number of digits in 350.

Answer:

Consider log3^{20} =20 log 3 = 20 × 0.477 1 = 9.542

Since the characteristic is 9, there are 10 digits in 3^{20}.

Question 50.

(a) If the lines 2x – y = 5,

Kx-y = 6 and 4x – y = 1 are concurrent find k.

Answer:

Solving equations 2x – y = 5, & 4x – y = 7 we get

∴ y = 2x – 5 = 2 – 5 = -3

Since lines are concurrent put x = 1 and y = -3 in kx – y = 6

Kx – y = 6 ⇒ K(1) – (-3) = 6

⇒ K + 3 = 6

⇒ K = 6 – 3

⇒ K = 3

(b) Find sum of 7 + 77 + 777 + ……… n terms.

Answer:

Let S_{n} = 7 + 77 + 777 +…………….n terms

= 7 [1 + 11 + 111 + –—–— to n terms] = \(\frac{7}{9}\)[9 + 99 + 999 + …………. to n terms]

= \(\frac{7}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1)+……………to n terms]

= \(\frac{7}{9}\)[(10 + 10^{2} + 10^{3} +…………to n terms) – (1 + 1 + 1 …………to n terms]

= \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]\)[… 10 + 10^{2} + 10^{3}……….. to n terms is in GP Where a = 10, r = 10]

S_{n} = \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(c) A scholarship of Rs. 2000 every year has to be instituted. How much should be invested today ¡f the rate of interest, is 8% p.a. .

Answer:

P = \(\frac{a}{i}=\frac{2000}{0.08}\) = 25,000