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

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 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) = x^{2} + 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 =x^{4}

5^{4} = x^{4} ⇒ x = 5

Question 6.

Find the 6th term of the GP. 3, 6,12 …….

Answer:

a = 3, r = 6/3 = 2

T_{6} = ar^{6-1} = ar^{5} = 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 = 2^{3} × 3^{1} × 5^{1} × 11^{1}

HCF = 2^{3} × 3^{1} = 8 × 3 = 24

Question 14.

IfA={x: x ∈ N and x < 3} and B={x: x^{2} – 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 = 3^{1} × 5^{2} × 111

P_{1} = 3, α_{1} = 1, P_{2} = 5, α_{2} = 2, P_{3} =11, α_{3} = 11

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

=(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 2x^{2} – 9x + 7 = 0

Answer:

a = 2, b = -9, c = l

Δ = b^{2} – 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:

X_{1} = 60, N_{1} = 65

X_{2} = 65, N_{2} = 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 – y_{1} = m(x – x_{1})

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, T_{7} = 64, S_{7} = ?

T_{7} = ar^{6}

64 =729. r^{6} ⇒ r^{6} = \(\frac{64}{729}=\frac{2^{6}}{3^{6}}=\left(\frac{2}{3}\right)^{6}\) ⇒ 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 => SA^{2} = SB^{2}

(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 = rcr^{2} = 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\(\frac{1}{x}\)

x^{2} – 20x – 69 =0

x^{2} – 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

a^{3} = 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 + 6r^{2} = 37r

6r – 37r + 6 =0

6r^{2} – 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 3x^{2} – 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

x^{2} – 5x + 2x – 10 = 0

x^{2} – 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} + y^{2} = (x + 1)^{2} + y^{2}

-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} = \(\frac{\Sigma n(n+1)}{2}=\frac{1}{2}\)[Σn^{2} + 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 cot^{2} A – 31 cosecA + 32 = 0, find the valu&of sinA

Answer:

12cot^{2} 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 = log_{2} 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