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SURA BOOKSPART - A

Note : (i) Answer all the questions. (ii) Choose and write the correct answer from the four choices given. [40 × 1 = 40]1. If AB = BA = |A|I then the matrix B is :

a) The inverse of A b) The transpose of A c) The adjoint of A d) 2A

2. If A = 0 8 0 60 6 0 8. .. .−

then A−1 is :

a) −−

0 8 0 60 6 0 8

. .

. . b) 0 8 0 6

0 6 0 8. .. .

−

c) 0 8 0 6

0 6 0 8. .. .

d) 0 2 0 4

0 4 0 2. .. .−

3. The rank of a non-singular matrix of order n × n is :a) n b) n2 c) 0 d) 1

4. The equations AX = B can be solved by Cramer’s rule only when :a) |A| = 0 b) |A| ≠ 0 c) A = B d) A ≠ B

5. The number of Hawkins-Simon conditions for the viability of an input-output model isa) 1 b) 3 c) 4 d) 2

6. Equation of the directrix of x2 = 4ay is :a) x + a = 0 b) x –a = 0 c) y + a = 0 d) y –a = 0

7. The length of latus rectum of 4x2 + 9y2 = 36 is :a) 4

3 b) 8

3 c) 4

9 d) 8

98. The sum of Focal distances of any point on the ellipse is equal to length of its :

a) Minor axis b) Semi minor axis c) Major axis d) Semi major axis9. If a is the length of the semi transverse axis of rectangular hyperbola xy = c2 then the value of c2 is :

a) a2 b) 2a2 c) a2

2 d) a

4

2

10. For the cost function C = 110

2e x , the marginal cost is :

a) 110

b) 15

e2x c) 110

e2x d) 1

10 ex

11. If the rate of change of y with respect to x is 6 and x is changing at 4 units/sec, then the rate of change of y per sec is :a) 24 units/sec b) 10 units/sec c) 2 units/sec d) 22 units/sec

[1]

JULY - 2016 BUSINESS MATHEMATICS [Time : 3 Hours] (With Answers) [Max. Marks : 200]

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2 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers

12. For the curve y=1+ax–x2 the tangent at (1, –2) is parallel to x-axis. The value of ‘a’ is :a) –2 b) 2 c) 1 d) –1

13. The slope of the tangent to the curve y = cos t, x = sint at t = p4

is :

a) 1 b) 0 c) 12

d) –1

14. The tangent to the curve y = 2x2–x+1 at (1, 2) is parallel to the line :a) y = 3x b) y = 2x+4 c) 2x+y+7 = 0 d) y = 5x–7

15. The curve y = 4 – 2x – x2 is :a) Concave upward b) Concave downwardc) Straight line d) None of these

16. If f(x, y) = 2x+ye–x, then fy(1,0) is equal to :

a) e b) 1e c) e2 d) 2

1e

17. If z = x3+3xy2+y3 then the marginal productivity of x is :a) x2+y2 b) 6xy+3y2 c) 3(x2+y2) d) (x2+y2)2

18. The cost function y=40–4x+x2 is minimum when :a) x = 2 b) x=–2 c) x=4 d) x=–4

19. x dx4

2

2

−∫ is :

a) 325

b) 645

c) 165

d) 85

20. The area bounded by the curve y = ex, the x-axis and the lines x = 0 and x = 2 is :a) e2 – 1 b) e2 + 1 c) e2 d) e2 – 2

21. The Marginal revenue of a firm is MR = 15 –8x. Then the revenue function is :

a) 15x –4x2 + k b) 15 8x

− c) –8 d) 15x–8

22. The solution of x ydy edx

−= is :

a) ey ex = c b) y = log cex c) ( )log xy e c= + d) ex +y = ce

23. The order and degree of 2

2 3 2

21 dy d ydx dx

+ = are :

a) 3 and 2 b) 2 and 3 c) 3 and 3 d) 2 and 2

24. The integrating factor of (1+x2) dydx

+xy=(1+x2)3 is :

a) 21 x+ b) log(1+x2) c) etan-1x d) log(tan-1x)

SURA BOOKS

Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 3

25. The complementary function of the differential equation(D2 – D)y=ex is :a) A +Bex b) (Ax+B)ex c) A+Be–x d) (A+Bx)e–x

26. when h = 1, ∆(x2) =a) 2x b) 2x – 1 c) 2x + 1 d) 1

27. The normal equation of fitting a straight line y = ax+ b are 10a + 5b = 15 and 30a + 10b = 43. The slope of the line of best fit is :a) 1.2 b) 1.3 c) 13 d) 12

28. If the probability density function of a variable X is defined as f(x) = Cx (2 – x), 0 < x < 2 then the value of C is :

a) 4 3

b) 64

c) 34

d) 35

29. The standard deviation of a poisson variate is 2, the mean of the poisson variate is :

a) 2 b) 4 c) 2 d) 1 2

30. The mean and variance of a Binomial distribution are 8 and 4 respectively. Then P (X = 1) is equal to :

a) 12

12

b) 4

12

c) 6

12

d) 10

12

31. If X ~ N (8, 64), the standard normal variate Z will be :

a) X − 648

b) X − 864

c) X − 88

d) X − 88

32. The theory of sampling is based on :a) Sample size b) Sample unit c) Principle of statistical regularity d) Population size

33. Probability of rejecting, the null-hypothesis when it is true is :a) Type I error b) Type II error c) Sampling error d) Standard error

34. The number of ways in which one can select 2 customers out of 10 customers is :a) 90 b) 60 c) 45 d) 50

35. The critical region for z at 1% level is :a) |z| ≤ 1.96 b) |z| ≥ 2.58 c) |z| < 1.96 d) |z| > 2.58

36. A time series consists of :a) Two components b) Three components c) Four components d) None of these

37. Most frequently used index number formulae are :a) Weighted formulae b) Unweighted formulaec) Fixed weighted formulae d) None of these

38. Variation in the items produced in a factory may be due to :a) Chance causes b) Assignable causes c) Both (a) and (b) d) Neither (a) nor (b)

39. If X and Y are two variates, there can be at the most :a) One regression line b) Two regression linesc) Three regression lines d) None of these

40. The term regression was introduced by :a) R.A. Fisher b) Sir Francis Galton c) Karl Pearson d) None of these

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PART - BNote : (i) Answer any ten questions. [10 × 6 = 60]

41. Given A = 1 1 12 1 13 1 1

−

−

, verify that |Adj A| = |A|2.

42. Find k if the equations 2x+3y−z=5, 3x−y+4z=2, x+7y−6z=k are consistent.43. Find the equation of ellipse whose focus is (2, –1) directrix is x – 5 = 0 and eccentricity is 1

244. Find the Equilibrium Price and Equilibrium Quantity for the following demand and supply

functions qd = 4 – 0.05P and qs = 0.8 + 0.11P

45. For the cost function y= 2x xx

++

43

+3, prove that the marginal cost falls continuously as the output x increases.

46. Find the points of inflection of the curve y = x4 – 4x3 + 2x + 3.47. The marginal cost function of manufacturing x units of a commodity is MC = 6 + 10x – 6x2. Find

the total cost and average cost, given that the total cost of producing 1 unit is 15.48. Solve (1–ex) sec2 y dy + 3ex tan y dx = 0.

49. Solve logx dydx

yx

+ = sin2x.

50. If f (0)=5, f(1)=6, f(3)=50,f (4)=105, find f (2) by using Legrange’s formula.

51. Fit a straight line for the following data.

x : 0 1 2 3 4y : 1 1 3 4 6

52. Ten coins are thrown simultaneously. Find the probability of getting atleast 7 heads.53. Out of 1000 TV viewers, 320 watched a particular programme. Find 95% confidence limits

for TV viewers who watched this programme.54. Find trend values to the following data by the method of semi-averages :

Year 1980 1981 1982 1983 1984 1985 1986Sales 102 105 114 110 108 116 112

55. Construct cost of living index for 2000 taking 1999 as the base year from the following data using Aggregate Expenditure method :

Commodity Quantity (kg) 1999

Price1999 2000

A 6 5.75 6.00B 1 5.00 8.00C 6 6.00 9.00D 4 8.00 10.00E 2 2.00 1.80F 1 20.00 15.00

PART - C

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Note : (i) Answer any ten questions. [10 × 10 = 100]56. A salesman has the following record of sales during three months for three items A, B and C

which have different rates of commission.

MonthsSales of units Total commission

drawn (in `)A B CJanuary 90 100 20 800

February 130 50 40 900March 60 100 30 850

Find out the rates of commission on the items A, B and C. Solve by Cramer’s rule.57. The data below are about an economy of two industries P and Q. The values are in crores of

rupees.

ProducerUser

Final Demand Total OutputP Q

P 50 75 75 200Q 100 50 50 200

Find the outputs when the final demand changes to 300 for P and 600 for Q.58. Find the equations of the asymptotes of the hyperbola

3x2 – 5xy – 2y2 +17x + y + 14 = 059. At what points on the circle x2 + y2 – 2x – 4y + 1 = 0, the tangent is parallel to (i) x axis

(ii)y axis?60. The total cost and total revenue of a firm are given by C = x3 – 12x2 + 48x + 11 and

R = 83x – 4x2 – 21. Find the output. (i) When the revenue is maximum (ii) When profit is maximum

61. The demand function for a commodity y is q1 = 12 – p12 +p1 + p2. Find the partial elasticities

when p1 = 10 and p2 = 4.

62. Evaluate a x b xx x

dxsin cossin cos

++∫

0

2π

.

63. The demand and supply curves are given by Pd = 164x +

and Ps = x2

. Find the Consumers’ Surplus

and Producers’ Surplus at the Market Equilibrium Price.

64. Solve (15D2 – 2D – 1) y ex

= +3 5

65. Find y when x = 0.2 given that :

x : 0 1 2 3 4y : 176 185 194 202 212

(Using Gregory - Newton’s formula)66. Fine the Mean and Variance for the following Probability distribution.

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f x e x

x

x( )

,,

=≥<

−2 00 0

2

67. In a sample of 1000 candidates the Mean of Certain text is 45 and S.D 15. Assuming the normality of the distribution find the following:

(i) How many candidates score between 40 and 60 ? (ii) How many candidates score above 50 ? (iii) How many candidates score below 30 ?

Z 0.33 1Area 0.1293 0.3413

68. A Sample of 400 students is found to have a Mean height of 171.38 cms. Can it reasonably be regarded as a sample from a large population with Mean height of 171.17 cms and standard deviation of 3.3 cms. (Test at 5% level).

69. Solve the following, using graphical method Maximize z = 45x1 + 80x2 Subject to the constraints

5x1 + 20x2 ≤ 400 10x1 + 15x2 ≤ 450 x1 , x2 ≥ 0 70. Marks obtained by 10 students in Economics and Statistics are given below:

Marks in Economics X : 25 28 35 32 31 36 29 38 34 32

Marks in Statistics Y : 43 46 49 41 36 32 31 30 33 39 Find (i) The Regression Equation of Y on X. (ii) Estimate the marks in Statistics when the marks in Economics is 30.

Answers

Part -I

1. (c); 2. (b); 3. (a); 4. (b); 5. (d); 6. (c); 7. (b); 8. (c); 9. (c); 10. (b); 11. (a); 12. (b); 13. (d); 14. (a); 15. (b); 16. (b); 17. (c); 18. (a); 19. (b); 20. (a); 21. (a); 22. (c); 23. (b); 24. (a); 25. (a); 26. (c); 27. (b); 28. (c); 29. (b); 30. (a); 31. (c); 32. (c); 33. (a); 34. (c); 35. (b); 36. (c); 37. (a); 38. (c); 39. (b); 40. (b);

PART - B

41. Adj A = Act

A = 1 1 12 1 13 1 1

−

−

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C11 = + 1 11 1-

= −1 − 1 = −2 C21 = − -

-1 1

1 1 = –(1 − 1) = 0

C12 = − 2 13 1-

= −(−2 −3) = 5 C22 = + 1 13 1-

= (−1 − 3) = −4

C13 = + 2 13 1

= 2 − 3 = −1 C23 = − 1 13 1

- = −(1 + 3) = −4

C31 = +-1 11 1

= (−1 −1) = −2

C32 = −1 12 1

= −(1−2) = 1

C33 = +1 12 1

- = (1+2) = 3

Ac = − −

− −−

2 5 10 4 42 1 3

Adj A = − −

−− −

2 0 25 4 11 4 3

|Adj A| = − −

−− −

2 0 25 4 11 4 3

= −2 --

4 14 3

−0 5 11 3-

+ (−2) 5 41 4

-- -

= −2 (−12+4) −0 −2 (−20−4)= −2 (−8) −2 (−24)= 16 + 48= 64 ............... (1)

|A| = 1 1 12 1 13 1 1

−

−

= 1 1 11 1−

+1 2 13 1-

+ 1 2 13 1

= 1 (−1 −1) +1 (−2 −3) +1 (2 −3)= −2 −5 −1 = −8

|A|2 = (−8)2 = 64 ............... (2)(1) and (2) => |Adj A| = |A|2

Hence it is proved.

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42. (A,B) = 2 3 1 53 1 4 21 7 6

−−

−

k

, A = 2 3 13 1 41 7 6

−−

−

|A| = 2 3 13 1 41 7 6

−−

−

=> 2 -

-1 4

7 6 −3

3 41 6-

−1 3 11 7

-

=> 2 (6−28) −3 (−18−4) −1 (21+1)=> 2(−22) −3 (−22) −1 (22)=> −44 + 66 – 22=> 0

2 32 1

= (−2 −9) = −11 ≠ 0

Obviously, ρ(A) = 2For the equations to be consistent, ρ(A,B) should also be 2.Hence every minor of (A,B) of order 3 should be zero.

3 1 51 4 27 6

−−

−

k

= 0

=> 3 4 26- k

+1 -1 27 k

+5 --

1 47 6

= 0

=> 3 (4k + 12) +1 (−k – 14) +5 (6 – 28) = 0=> 3 (4k + 12) +1 (−k −14) +5 (−22) = 0=> 12k + 36 – k – 14 – 110 = 0=> 11k − 88 = 0=> 11k = 88=> k = 8

43. Focus is (2, –1), eccentricity is 12

and directrix is x – 5 = 0

For an ellipse with focus S, SPPM

2

2 = e2

Let p(x,y) be any point on the ellipse and PM be the perpendicular distance of P from the directrix.

∴ Equation of ellipse is

∴ x y x−( ) + +( ) =

−

2 1 12

51

2 22 2

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⇒ x x y y x x2 2 24 4 2 1 14

10 25− + + + + = − +[ ]

⇒ 4 4 2 5 10 252 2 2x y x y x x+ − + + = − +

⇒ 4 4 16 8 20 10 25 02 2 2x y x y x x+ − + + − + − = ⇒ 3x2 + 4y2 – 6x + 8y – 5 = 0.

44. At equilibrium dq = sq (ie) 4 – 0.05p = 0.8 + 0.11p 4 – 0.8 = 0.11p + 0.05p (ie) 3.2 = 0.16p

∴ p = 3.20.16

= 32010

= 20

At p = 20, q = 4 – 0.05 (20) = 4 –1 = 3∴ At equilibrium q = 3 and p = 20

45. We have y = 2x xx

++

43

+3

y = 2 83

2x xx

++

+3 ...... (1)

Marginal cost is dydx

Differentiating (1) with respect to x, we get

1 1

2

Quotient ruledy u vu uvdx v v

− =

dydx

= x x x x

x

+( ) +( ) − +( )( )+( )

3 4 8 2 8 1

3

2

2 +0

= 2 6 12

3

2

2

x x

x

+ +( )+( )

= 2 6 9 3

3

2

2

x x

x

+ + +( )+( )

= 2 x

x

+( ) +

+( )

3 3

3

2

2=2 1 3

3 2++( )

x

This shows that as x increases, the marginal cost dydx

decreases.

46. 4 34 2 3= − + +y x x x

3 24 12 2= − +dy x x

dx

22

2 12 24= −d y x xdx

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3

3 24 24= −d y xdx

Condition for points of inflection is

2

2 0 and=d ydx

3

3 0 ≠d ydx

212 24 0∴ − =x x

12 ( 2) 0− =x x

0=x (or) 2 0− =x ⇒ 2=x

3

3when x 0 24(0) 24 24 0= ⇒ = − = − ≠d ydx

3

3when x 2 24(2) 24 48 24 0= ⇒ = − = − ≠d ydx

∴ Points of inflection exist.

4 3when x 0 (0) 4(0) 2(0) 3 3= ⇒ = − + + =y

4 3when x 2 (2) 4(2) 2(2) 3= ⇒ = − + +y

y = 16 – 32 + 4 + 3 = –9 ∴ The points of inflection are (0,3) and (2,–9)

47. Solution: Given that

MC = 6 + 10x –6x2

C = (MC)∫ dx + k

= (6∫ + 10x –6x2) dx + k

= 6x + 210x

2 –

36x3

+ k

= 6x + 5x2 –2x3 + k .....(1)Given that x = 1, C = 15∴ (1) =>15 = 6 + 5 –2 + k =>k = 6∴ Total cost function, C = 6x + 5x2 –2x3 + 6

Average cost function, AC = cx

, x≠0

= 6 + 5x –2x2 + 6x

48. Solution: (1–ex) sec2y dy + 3ex tany dx = 0 (1–ex) sec2y dy = –3ex tany dx

sec2

tanyy∫ dy = x

31

xee −∫ dx

log tan y = 3 log(ex–1)+logclog tan y = log (ex–1)3+logc

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=> log tan y = log [c(ex–1)3]=> tan y = c(ex–1)3

49. Solution: log x dydx

+ yx

= sin 2x

÷ by logx dydx

+ y

x logx = 2sin x

logx

Here P = xlogx

; Q = 2sin x

logx

1 log

pdx dxx x

=∫ ∫ t = log x ⇒ dt = 1x

dx

pdx∫ = 1xlogx∫ dx = pdx∫ = 1

t∫ dt = log t = log(log x)

log(log ) logpdx xe e x∫ = =

Hence solution is

y(log x) = sin2xlogx∫ logx dx = 2 sin x dx∫

y(log x) = – cos22

x + c

50. Solution: By data we have x0 = 0, x1 = 1, x2 = 3, x3 = 4 y0 = 5, y1 = 6, y2 = 50, y3 = 105 x = 2 to find y. By Lagrange’s formula

yx x x x x x

x x x x x xy

x x x x x

=−( ) −( ) −( )−( ) −( ) −( )

+−( ) −( ) −

1 2 3

0 1 0 2 0 30

0 2 xxx x x x x x

y

x x x x x xx x x x

3

1 0 1 2 1 31

0 1 3

2 0 2

( )−( ) −( ) −( )

+−( ) −( ) −( )−( ) − 11 2 3

2

0 1 2

3 0 3 1 3 23

( ) −( )

+−( ) −( ) −( )−( ) −( ) −( )

x xy

x x x x x xx x x x x x

y

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=−( ) −( ) −( )−( ) −( ) −( ) +

−( ) −( ) −( )−( ) −( )

2 1 2 3 2 40 1 0 3 0 4

52 0 2 3 2 41 0 1 3 11 4

6

2 0 2 1 2 43 0 3 1 3 4

502 0 2 1 2 34

−( )

+−( ) −( ) −( )−( ) −( ) −( ) +

−( ) −( ) −( )− 00 4 1 4 3

105( ) −( ) −( )

=( ) −( ) −( )−( ) −( ) −( )

+( ) −( ) −( )( ) −( ) −( )

1 1 21 3 4

52 1 21 2 3

6

+( )( ) −2 1 2(( )( )( ) −( )

+( )( ) −( )( )( )( )3 2 1

502 1 14 3 1

105

= –0.833 + 4 + 33.333 – 17.5y = 19.

51. Solution:The line of best fit is y = ax + bThe normal equations are

a x nb y

a x b x xy

+ =

+ =

∑ ∑∑∑ ∑

………

………

( )

( )

1

22

Now from the data

x y x2 xy0 1 0 01 1 1 12 3 4 63 4 9 124 6 16 24

Total 10 15 30 43

Here x y x y= = = =∑ ∑∑∑10 15 30 432, , , xy

By substituting these values in (1) and (2) we get,10a + 5b = 15 …..… (3)30a + 10b = 43 …..… (4)

Solving (3) and (4) we get,a = 1.3 and b = 0.4The line of best fit is y = 1.3x + 0.4

52. Let x = number of heads. When a coin is thrown, p = prob (head) = 1

2 q=1–p=prob (Tail) = 1

2 ∴P(atleast 7) = P(x≥7)=P(x=7)+P(x=8)+P(x=9)+P(x=10)

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P[X=x]= Cn x n xx p q −

As x follows Binomial distribution,

p(x≥7) = 7 3 8 2 9 1 10 0

10 10 10 107 8 9 10

1 1 1 1 1 1 1 1. . . . . .2 2 2 2 2 2 2 2

C C C C + +

10 10 10 10

7 8 9 10102

C C C C+ + += = 120 45 10 1 176 11

1024 1024 64+ + + = =

53. Solution: Sample size n = 1000

Sample proportion of TV viewers p = xn

= 3201000

= .32

∴ q = 1 – p = 0.68

S.E (p) = pqn

= 0.0147The 95% confidence limits for population proportion P are given by

p ± (1.96) S.E (p) = 0.32 ± 0.028=> 0.292 and 0.348

∴ TV viewers of this programme lie between 29.2% and 34.8%.54. Solution: No of years = 7 (odd no). By omitting the middle year (1983) we have

Year Sales Semi–total Semi–average1980 102

321321 107

3=1981 105

1982 1141983 1101984 108

336336 112

3=1985 116

1986 112 Difference between middle periods = 1985–1981 = 4 Difference between semi averages = 112–107= 5

Annual increase in trend = 54

= 1.25

Year 1980 1981 1982 1983 1984 1985 1986Trend 105.75 107 108.25 109.50 110.75 112 113.25

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55. Solution:

Commodity Quantity 1999

Price

1999 2000q0 p0 p1 p1q0 p0q0

A 6 5.75 6.00 36.00 34.50B 1 5.00 8.00 8.00 5.00C 6 6.00 9.00 54.00 36.00D 4 8.00 10.00 40.00 32.00E 2 2.00 1.80 3.60 4.00F 1 20.00 15.00 15.00 20.00

156.60 131.50

C.L.I = 1 0

0 0

p qp q

∑∑

× 100

= 156.60131.50

× 100 = 1.19087 × 100

= 119.087 = 119.09PART - C

56. Solution:Let x, y and z be the rate of commission in Rs. per unit for A,B and C items respectively.According to the problem

90x + 100y + 20z = 800130x + 50y + 40z = 90060x +100y + 30z =850

Dividing each of the equations by 10 throughout,9x+10y+2z=8013x+5y+4z=906x+10y+3z=85

Now, ∆ = 9 10 2

13 5 46 10 3

= − − − + −⇒ − − +⇒ − ≠

9 15 40 10 39 24 2 130 30225 150 200175 0

( ) ( ) ( )

∆x = 80 10 290 5 485 10 3

= − − − + −⇒ − + +⇒ −

80 15 40 10 270 340 2 900 4252000 700 950350

( ) ( ) ( )

∆y = 9 80 2

13 90 46 85 3

= − − − + −⇒ − − +⇒ −

9 270 340 80 39 24 2 1105 540630 1200 1130700

( ) ( ) ( )

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∆z= 9 10 80

13 5 906 10 85

= − − − + −⇒ − − +⇒

9 425 900 10 1105 540 80 130 304275 5650 8000

1 925

( ) ( ) ( )

,

By Cramer’s rule

∴ x = ∆∆

x = --

350175

= 2

y = ∆

∆y =

--

700175

= 4

z = ∆∆

z = --1925175

=11

Hence the rates of commission for A,B and C are Rs.2, Rs.4 and Rs. 11 respectively.57. Solution: With the usual notation,

a11=50, a12=75, x1=200a21=100, a22=50, x2=200.

Now, b11 = ax

11

1

= 50200

14

=

b12 = ax

12

2

= 75200

38

=

b21 = ax

21

1

= 100200

12

=

b22 = ax

22

2

= 50200

14

=

∴ The technology matrix B =

14

38

12

14

I−B = 1 00 1

−

14

38

12

14

= 34

38

12

34

−

−

|I−B| = 916

316

616

− = (+ve) = 38

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Now, (I−B)−1 =

138

34

38

12

34

∴ X = (I−B)−1 D

= 83

34

38

12

34

300600

= 83

450600

=

12001600

The output is Rs. 1200 crores and Rs. 1600 crores.58. Solution:

3x2 –5xy –2y2 + 17x + y + 14 = 0The combined equation of the asymptotes differs from the standard equation of a hyperbola by a constant.∴ The combined equation of the asymptotes is

3x2 –5xy –2y2 + 17x + y + k = 0Taking the quadratic factor

3x2 –5xy –2y2 = 3x2 –6xy + xy –2y2

= 3x(x –2y) + y(x –2y) = (3x +y) (x –2y)

∴ The combined equation is for some l, m seperate equation of asymptotes are 3x + y + l = 0, x – 2y + m = 0.3x2 –5xy –2y2 + 17x + y + k = (3x + y + l) (x –2y + m)Equating the co-efficients,

l + 3m = 17 → (1)–2l + m = 1 → (2)

lm = k → (3)Solving (1) & (2),We get l = 2, m = 5∴ K = l × m = 2 × 5 = 10Seperate equation of asymptotes are

3x + y + 2 = 0 &x –2y + 5 = 0

The combined equation of the asymptotes is3x2 –5xy –2y2 + 17x + y + 10 = 0

59. Solution: Given circle is x2+y2 – 2x–4y+1 = 0 ........ (1)

Differentiating with respect to x,

2x+2y dydx

–2(1) –4 dydx

+0 = 0

SURA BOOKS


(2y–4) dydx

+ (2x–2) = 0

dydx

= – ( )2 22 4

xy

−−

= − −−

2 12 2

( )( )

xy

dydx

= − −−

( )xy

12

⇒ dxdy

= yx−

− −21( )

i) Tangent is parallel to x–axis

when dydx

= 0

− −−

( )xy

12

= 0

x–1 = 0x = 1

Substituting x =1 in equation (1)12+y2–2(1) – 4y+1 = 0y2–4y = 0y(y–4) =0y = 0 (or) y–4 = 0 ⇒ y = 4

At (1, 0) and (1, 4) tangent of the circle are parallel to x-axis.ii) Tangent is parallel to y- axis

dydx

= ∝ (or) dxdy

= 0

yx−

− −21( )

= 0

y–2 = 0 ⇒ y = 2Substituting y=2 in equation (1)

x2+22 – 2x–4(2)+1 = 0x2–2x–3 = 0 ⇒ (x–3) (x+1) = 0x–3 = 0 (or) x+1 = 0x = 3 (or) x = –1

∴ Required points are (3, 2) and (–1, 2)60. Solution:

i) Revenue 283 4 21= − −R x x Differentiating with respect to x,

ddx

xR = −83 8

2

2 8= −d Rdx

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Revenue is maximum when 0=dRdx

and 2

2

830 0 83 3 08

d R dR x xdx dx

< = ⇒ − = ∴ =

Also 2

8 0= − < ∴d Rdx2

R is maximum

∴ When the output 838

=x units, revenue is maximum

ii) Profit P = R–C

2 3 2(83 4 21) ( 12 48 11)= − − − − + +x x x x x

3 28 35 32= − + + −x x x

Differentiating with respect to x

23 16 35= − + +dp x xdx

2

2 6 16= − +d P xdx

Profit is maximum when 0=dPdx

and 2

2 0<d Pdx

20 3 16 35 0∴ = ⇒ − + + =dP x xdx

23 16 35 0⇒ − − =x x

(3 5)( 7) 0⇒ + − =x x

53

−=x (or) x = 7

when 2

2

5 5, 6 163 3

− − = = − + d Pxdx

= 26 > 0 ∴ P is minimum

when x = 7, 2

6(7) 16 26 0= − + = − <d pdx

∴ P is maximum

∴ When x = 7 units, profit is maximum61. Solution :

Required partial elasticities are

1 1 1

1 1 1

− ∂=∂

Eq p qEp q p

→ (1)

1 2 1

2 1 2

− ∂=∂

Eq p qEp q p

→ (2)

2

1 1 1 212= − −q p p p

Diff.p.w.r to p1

1

1 22

2∂ = − +∂

q p pp

Diff.p.w.r to p2

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∂∂

=qp

p1

21

( )1 11 22

1 1 1 2

212

−= − +− +

Eq p p pEp p p p

from (1)

21 1 2

21 1 2

212

−=− +p p p

p p p When p1=10 and p2=4

21

21

2(10) 10(4) 200 4012 10 10(4) 12 100 40

− −= =− + − +

EqEp

160 10

48 3−= =

−

( )1 2 1 21 22

2 1 1 2 1 1 212 12− −= =

− + − +Eq p p ppEp p p p p p p from (2)

When p1=10 and p2=4

EqEp

1

22

10 412 10 10 4

4012 100 40

4048

56

=−

− +

=−

− +=−−

=

( )( ) ( )( )

62. Solution:

Let I = 0

2π

∫++

a x b xx x

dxsin cossin cos

.....(1)

I = 0

2 2 2

2 2

π π π

π π∫−

+ −

−

+ −

a x b cos x

x cos x

sin

sin

dx ( ) ( )a a

0 0

f x f a x dx= −∫ ∫

I = 0

2π

∫++

acosx b xx x

dxsincos sin

.....(2)

2I = 0

2π

∫+( ) + +( )

+a x x b x x

x cos xdx

sin cos sin cossin

= 0

2π

∫ +( )a b dx = ( )a b x+[ ]02π

2I = (a + b) π2

I = (a + b) π4

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63. Solution: At market equilibrium Pd = Ps

16 x 4

=+

= x2

=> 32 = x (x + 4)

=> x2 + 4x –32 = 0(x + 8) (x –4) = 0x = –8 or x = 4

At x = 4, p = 4 22

=

∴ x0 = 4 p0 = 2 => x0P0 = 80

0 00

CS ( ) x

f x dx p x= −∫

CS = 164

8 160

4

xdx

+− =∫ 4

0log( + 4) 8 − x

= 16 [log 8 –log 4] –8 = 16 [log2] –8 units

0

0 00

PS P ( ) x

x g x dx= − ∫

PS = 8 – x dx x2

84

2

0

4

0

4

= −

∫

= 8 –4 = 4 units.

64. Solution: (15D2 – 2D – 1) y ex

= +3 5

Auxiliary equation is 15m2 – 2m – 1 = 0 ⇒ (3m–1) (5m+1) = 0

⇒ m = 13

, – 15

∴ C.F. = 1 13 5

x xAe Be

−

+

P.I1 = e ex x3 3

3 1 5 1 15 13

15D D D D−( ) +( )

=−( ) +( )

−2

−15

−515

31515

−13

= xe xe xex x x3 3 3

15 1315 15 815

8+( ) = ( ) =

PI2 = 53 1 5 1

515 1

315

515 1

315

5. .e eox ox

D D D D−( ) +( )=

−( ) +( ) =−( ) ( ) = −

y = CF + PI1 + PI2

y = A Be e xex xx

13

15

3

85+ + −

− .

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65. Solution:0.2 lies in the first interval (x0, x1) ie in (0, 1) so, we can use Gregory – Newton’s forward interpolation formula. Since five values are given the interpolation formula is

y y u yu u

yu u u

y

u u u

= + ∆ +−( )

∆ + +−( ) −( )

∆

+−( ) −( )

0 02

03

011

21 23

1 2! ! !

………

uuy where u x x

h−( )

∆ =−3

44

00

!Here h = 1, x0 = 0 and x = 0.2

∴ =−

=u 0 2 01

0 2. .

The forward difference table:

x y ∆y ∆2y ∆3y ∆4y0

1

2

3

4

176

185

194

202

212

9

(185–176)9

(194–185)8

(202–194)10

(212–202)

0

(9–9)

–1 (8–9)

2 (10–8)

–1

(–1–0)

3 (2+1)

4

(3+1)

∴ = + ( ) +−( ) ( ) +

−( ) −( )−( )

+

y 176 0 21

90 2 0 2 1

20

0 2 0 2 1 0 2 23

1

0

.!

. .!

. . .!

.22 0 2 1 0 2 2 0 2 34

4. . .

!−( ) −( ) −( ) ( )

= 176 + 1.8 – 0.048 – 0.1344 = 177.6176ie, when x = 0.2 y = 177.6176.

66. Solution: E(x) = 2

0

( ). .2 .xx f x dx x e dx∞

−=∫ ∫∞

−∞

Using Bernoulli’s formula [Bernoulli’s formula

2 2

0

2 . 1.2 4

x xe ex∞− −

= − −

∫udv = uv – u′v1 + u″ v2 – u′″ v3...]

1 1 12 (0) 0 2.4 4 2

= − − = =

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22 2 2

0

( ) . ( ). 2 xE x x f x dx x e dx∞

− = =∫ ∫X

Using Bernoulli’s formula,

2 2 22

0

2 . 2.2 .2 4 8

x x xe x e ex∞− − − −= − + − −

2 2 12 (0) 0 0 28 8 2

= − − + = = − ∴ variance = E(x2)–[E(x)]2 =

21 1 1 1 12 2 2 4 4

− = − = 67. Solution: Mean = μ = 45, and S.D = σ =15

Then Z = X µσ− = 45

15X−

i) P (40 < X < 60) = P 40 45 60 4515 15

z− − < ≤ Z

Z1Z 0=13

−

1 13

p z− = < < P

= P 1 0 p(0 1)3

z z− ≤ ≤ + ≤ ≤ P

= P (0 ≤ z ≤ 0.33) + P(0 ≤ z ≤ 1) = 0.1293+0.3413 (from tables) P(40 < X < 60) = 0.4706 Hence number of candidates scoring between 40 and 60 out of 1000 candidates =1000 × 0.4706 = 470.6 471

ZZ 0= 13

ii) P(X > 50)=P 1z3

> Z

= 0.5–P 10 z3

< < Z

= 0.5 – P (0 < z < 0.33) = 0.5–0.1293=0.3707 (from tables)Hence number of candidates scoring above 50 = 1000 × 0.3707 = 371.iii) P(X < 30)=P(Z < –1)

ZZ 0=1−

= 0.5–p(–1 ≤ z ≤ 0) = 0.5–p(0 ≤ z ≤ 1) symmetry = 0.5–0.3413=0.1587 (from tables)∴ Number of candidates scoring less than 30. = 1000 × 0.1587 =159

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68. Solution: Sample size, n = 400Sample Mean, X = 171.38Sample Standard deviation, σ = 3.3Set H0 = µ = 171.38

The test statistic Z = X−µσn

~ N (0, 1)

= X−µσn

since the sample is large, S = σ

= 171.38 – 1 71.17

33400

= 1.273

Since |Z| = 1.273 < 1.96, we accept the null hypothesis at 5% level of significance.Thus the sample of 400 has come from the population with mean height of 171.17 cms.

69. Solution: 5x1 + 20x2 = 400

D (45, 0)

X2 ≥ 0

(0, 20)

X1 ≥ 0

(8B 0, 0)

10x1+15x2 ≤ 45 5x1+20x2 ≤ 400

If x1 = 0, 20, x2 = 400 => x2 = 40020

= 20, A(0, 20)

If x2 = 0, 5x1 = 400 => x1 = 4005

= 80, B(80, 0)

∴ A(0, 20), B(80, 0) are points of the line 5x1 + 20x2 = 400 To draw 10x1 + 15x2 = 450

If x1 = 0, 15, x2 = 450 => x2 = 45015

= 30, C(0, 30)

If x2 = 0, 10x1 = 450 => x1 = 45010

= 45, D(45, 0)

∴ C(0, 30), D(45, 0) are points of the line 10x1 + 15x2 = 450

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To find E:5x1 + 20x2 = 400 …… (1)

10x1 + 15x2 = 450 …… (2)(1) x 2 10x1 + 40x2 = 800 …… (3)(2) – (3) –25x2 = –350

x2 = 35025

−−

= 14

Substituting x2 = 14 in equation (1) 5x1 + 20(14) = 400 5x1 = 400–280 = 120; ∴ x1 = 120

5 = 24

AODE form feasible region. ∴ The coordinate of extreme points are O = (0, 0), D = (45, 0), A = (0, 20), E = (24, 14) At O, Z = 45x1 + 80x2 = 45(0) + 80(0) = 0 At C, Z = 45(45) + 80(0) = 2025 + 0 = 2025 At A, Z = 45(0) + 80(20) = 0 + 1600 = 1600 At E, Z = 45(24) + 80(14) = 1080 + 1120 = 2200 ∴ Maximum Z = 2200 at x1 = 24, x2 = 14

70. Solution : Let the marks in Economics be denoted by X and statistics by Y.

X Y x = X X− y = Y Y− x2 y2 xy

25 43 –7 5 49 25 –3528 46 –4 8 16 64 3235 49 3 11 9 121 3332 41 0 3 0 9 031 36 –1 –2 1 4 236 32 4 –6 16 36 –2429 31 –3 –7 9 49 2138 30 6 –8 36 64 –4834 33 2 –5 4 25 –1032 39 0 1 0 1 0

320 380 0 0 140 398 –93

XX

n= ∑ = 320

10 = 32 Y

Yn

= ∑ = 38010

= 38 byx = 2

xy

x∑∑

= 93140− = – 0.664

(i) Regression equation of Y on X is Y Y− = byx ( X X− ) Y – 38 = – 0.664 (X – 32) => Y = 59.25 – 0.664X(ii) To estimate the marks in statistics (Y) for a given marks in the Economics (X), put

X = 30, in the above equation we get, Y = 59.25 – 0.664 (30) = 59.25–19.92 = 39.33 or 39

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