blog.surabooks.comblog.surabooks.com/.../2016/12/Business-Maths_October-2016_Final.pdf · SURA BOOKS PART - A. Note : (i) Answer . all. the questions. (ii) Choose and write the

SURA BOOKSPART - A

Note : (i) Answer all the questions. (ii) Choose and write the correct answer. [40 × 1 = 40]1. If A is a square matrix of order 3 then |Adj A| is :

a) |A|2 b) |A| c) |A|3 d) |A|4

2. The rank of an n × n matrix each of whose element is 2 is: a) 1 b) 2 c) n d) n2

3. A system of linear homogeneous equations has at least __________ solution(s).a) 1 b) 2 c) 3 d) 4

4. The number of Hawkins - Simon conditions for the Viability of an input - output model is :a) 1 b) 3 c) 4 d) 2

5. If

A BA 0.7 0.3T B 0.8

= x is a transition probability matrix, then the value of x is :

a) 0.3 b) 0.2 c) 0.7 d) 0.8

6. Length of the Latus rectum of an ellipse 2 2

2 2+x ya b

=1 (a>b) is :

a) 22a

b b)

2

2ab

c) 22b

a d)

2

2ba

7. The eccentricity of a conic is 12

. The conic is :

a) a parabola b) an ellipse c) a circle d) a hyperbola

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

9. The eccentricity of the rectangular hyperbola is:

a) 2 b) 12

c) 2 d) 12

10. The average fixed cost of the function C = 2x3−3x2 +4x +8 is :

a) 2x

b) 4x

c) 3−x

d) 8x

[1]

OCTOBER - 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 - October Question Paper with Answers

11. The demand for some commodity is given by q = −3p + 15 (0<p<5) where p is the unit price. The elasticity of demand is:

a) 29 15pp+ b) 9 45−p

p c) 15 9−p

p d)

5− +p

p

12. If y = 2x2+3x, the instantaneous rate of change of y at x =4 is :a) 16 b) 19 c) 30 d) 4

13. The slope of the tangent to the curve y = x2 − log x at x = 2 is:

a) 72

b) 27

c) 72

− d) 27

−

14. The slope of the curve x = y2 − 6y at the point where it crosses the y-axis is:

a) 5 b) −5 c) 16

d) 116

−

15. The stationary value of x for f(x) = 3(x−1) (x−2) is:

a) 3 b) 32

c) 23

d) 32−

16. y = x3 is always :a) an increasing function b) a decreasing functionc) a constant function d) None of these

17. If q1 = 2000 + 8p1− p2 then 1

1

qp

∂∂

is:

a) 8 b) –1 c) 2000 d) 0

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

19. 3

3

x dx−∫ is :

a) 0 b) 2 c) 1 d) –1

20. 2

2

cos x dxp

p−∫ is :

a) 2 b) –2 c) –1 d) 121. The Marginal revenue of the 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 e

dx−= is :

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

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

23. The solution of x dy + y dx = 0 is :

a) x + y = c b) x2 + y2 = c c) xy = c d) y = cx

24. The degree and order of the differential equation 2

2 6d y dydx dx

− = 0:

a) 2 and 1 b) 1 and 2 c) 2 and 2 d) 1 and 1

25. The solution of 2

2 0d y ydx

− = :

a) (A +B)ex b) (Ax+B)e–x c) BA xxe

e+ d) (A+Bx)e–x

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

27. The normal equations 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. The mean and variance of a binomial distribution are :a) np, npq b) pq, npq c) np, npq d) np, nq

29. If X is a Poisson variate with P(X = 1) = P(X=2), the mean of the Poisson variate is:a) 1 b) 2 c) –2 d) 3

30. The standard deviation of a Poisson variate is 2, the mean of the Poisson variate is :

a) 2 b) 4 c) 2 d) 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. A hypothesis complementary to null hypothesis is called:a) primary hypothesis b) statistical statementc) alternative hypothesis d) confidence hypothesis

33. 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

34. The 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

35. The Z-value that is used to establish a 95% confidence interval for the estimation of a population parameter is:a) 1.28 b) 1.65 c) 1.96 d) 2.58

36. A time series consists of __________ components.a) 2 b) 3 c) 4 d) 1

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37. Laspeyre’s index formula uses the weights of the :a) base year quantities b) current year pricesc) average of the weights of number of years d) none of these

38. Variation due to assignable causes in the product occur due to:a) faulty process b) carelessness of operatorsc) poor quality of raw material d) all the above

39. The range of correlation co-efficient is:a) 0 to ∞ b) –∞ to ∞ c) –1 to 1 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

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

41. Given A = 3 14 2

, B = 1 0

2 1−

verify that Adj(AB) = (Adj B) (Adj A).

42. Find the rank of the matrix A = 1 1 1 11 3 2 12 0 3 2

−

−

.

43. Find the eccentricity, foci and latus rectum of the ellipse 9x2 +16y2 = 144.

44. Find the elasticity of demand when the demand is 20

1q

p=

+ and p = 3. Interpret the result.

45. For what values of x, is the rate of increase of x3–5x2+5x+8 is twice the rate of increase of x?

46. If u = x3+3xy2+y3, prove that 2 2u u

x y y x∂ ∂=

∂ ∂ ∂ ∂.

47. The elasticity of demand with respect to price p for a commodity is 5xx− , x>5 when the

demand is ‘x’. Find the demand function if the price is 2 when demand is 7. Also find the revenue function.

48. Solve: 1cos sin 22

dy y x xdx

+ = .

49. Solve: x(y2+1)dx + y(x2+1)dy = 0.

50. Find the missing term from the following data:

x : 0 5 10 15 20 25y : 7 11 14 - 24 32

51. In a straight line of best fit find x-intercept when ∑x =10; ∑y = 16.9; ∑x2 = 30; ∑xy = 47.4 and n = 7.

52. Find E(2x–7) for the following :X : –3 –2 –1 0 1 2 3

p(x) : .05 .1 .3 0 .3 .15 .1

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53. A random sample of 500 apples was taken from a large consignment and 45 of them were found to be bad. Find the limits at which the bad apples lie at 99% confidence level.

54. Calculate the regression equation of X and Y from the following data:X : 10 12 13 12 16 15Y : 40 38 43 45 37 43

55. Calculate the cost of living index number using Family Budget method :

Commodity A B C D E F G HQuantity in Base year (unit) 20 50 50 20 40 50 60 40

Price in Base year (`) 10 30 40 200 25 100 20 150

Price in Current year (`) 12 35 50 300 50 150 25 180

PART - C

Note : Answer any ten questions. [10 × 10 = 100]56. Solve by matrix method for the equations.

x – 2y + 3z = 1; 3x –y + 4z = 3; 2x + y – 2z = –1. 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 14 6 8 28Q 7 18 11 36

Find the outputs when the final demand changes to 20 for P and 30 for Q.58. Find the equations of the asymptotes of the hyperbola

2x2 + 5xy + 2y2 – 11x – 7y – 4 = 0.59. If AR and MR denote the Average and Marginal Revenue at any output level, show that

elasticity of demand is equal to ARAR MR−

. Verify this for the linear demand law p = a + bx

where p is the price and x is the quantity.60. Show that the maximum value of the function f(x) = x3 – 27x +108 is 108 more than the

minimum value.61. A manufacturer has to supply his customer with 600 units of his products per year. Shortage

are not allowed and storage cost amounts to 60 paise per unit per year. When the set up cost is ` 80 find,

(i) the Economic Order Quantity (EOQ) (ii) the minimum average yearly cost (iii) the optimum number of orders per year. (iv) the optimum period of supply per optimum order

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62. Evaluate dx

x16

3

+∫ cotπ

π

.

63. In a perfect competition the demand and supply curves of a commodity are given by pd = 40 – x2

and ps = 3x2+ 8x+8. Find the Consumer’s Surplus and the Producer’s Surplus at the market

equilibrium price.64. Suppose that the quantity demanded

2

242 4 4ddp d pQ pdt dt

= − − + and quantity supplied

Qs = –6 + 8p where p is the price. Find the equilibrium price for market clearance.65. Apply Lagrange’s formula to find y when x = 5, given that :

x : 1 2 3 4 7y : 2 4 8 16 128

66. It is stated that 2% of razor blades supplied by a manufacturer are defective. A random sample of 200 blades is drawn from a lot. Find the probability that 3 or more blades are defective (e–4 = 0.01832).

67. In a sample of 1000 candidates the mean of certain test is 45 and SD is 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 ?

Z 0.33 1Area 0.1293 0.3413

68. The mean IQ of a sample of 1600 children was 99. Is it likely that this was a random sample from a population with mean IQ 100 and standard deviation 15 ?

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. Using the following data, construct Fisher’s Ideal index and show that it satisfies Factor Reversal

test and Time Reversal test.

CommodityPrice Quantity

Base year Current year Base year Current yearA 6 10 50 56B 2 2 100 120C 4 6 60 60D 10 12 30 24E 8 12 40 36

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Answers

Part -A

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

PART - B

41. AB = 3 14 2

−

1 02 1 =

− + +− + +

3 2 0 14 4 0 2 =

−

1 10 2

Adj (AB) = 2 10 1

−−

.........(1)

Adj A = 2 14 3

−−

; Adj B =

1 02 1− −

(Adj B) (Adj A) = 1 02 1− −

2 14 3

−−

=

2 0 1 04 4 2 3+ − +

− + −

= 2 10 1

−−

.........(2)

From (1) and (2), we getAdj (AB) = (Adj B) (Adj A). Hence verified.

42. Order of A is 3 × 4 ∴ ρ(A) ≤ 3Let us reduce the matrix A to a triangular form

A = 1 1 1 11 3 2 12 0 3 2

−−

Applying R2 → R2 – R1,R3 → R3 – 2R1

A~

1 1 1 10 2 3 00 2 5 0

−− −

Applying R3 → R3 + R2

A~1 1 1 10 2 3 00 0 8 0

−−

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This is now in a triangular form

We find, 1 1 10 2 30 0 8

−−

= 1 (−16+0)−1(0−0)+1(0−0)

= −16 ≠ 0There is a minor of order 3 which is non-zero∴ ρ(A) =3.

43. The equation of the ellipse is9x2 + 16y2 = 144 ÷ by 144

⇒ x y2 2

16 9+ = 1

The given equation is of the form xa

2

2 + y

b

2

2 = 1.

Here a2 = 16, b2 = 9

∴ e = 12

2−ba

= 1 916

− = 7

4The foci are S(ae, 0) and S’ (–ae, 0)

or, S( 7 , 0) and S’(– 7 , 0)

The latusrectum = 2 2ba

= 2 3

4

2( ) =

92

44. q = 20

1p + Differentiating with respect to p, we get,

dqdp

= 20 2

1( 2)p

− +

= 2

20( 1)p

−+

[∵ ddx

1x

= −1

2x]

Elasticity of demand = dy

xh −= dq

dp =

−

+

p

p20

1

−+

201 2( )p

= ( 20p ) ( +1p )

20 2 ( +1)p

= 1

pp +

When p = 3, ηd = +3

(3 1) =

34

= 0.75

(ie) when the price is increased by 1%, demand is decreased by 0.75% and when the price is decreased by 1% the demand is increased by 0.75% at p = 3.

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45. Let y = x3– 5x² + 5x +8Differentiating both sides with respect to t,

dydt

= 3x² dxdt

– 10x dxdt

+ 5 dxdt

+ 0

dydt

= (3x² – 10x + 5) dxdt

Given dydt

= 2 dxdt

∴ 2 dxdt

= (3x² – 10x + 5) dxdt

⇒ 3x² – 10x + 3 = 0⇒ (3x – 1) (x–3) = 0

x = 1 3

or x = 3.

46. 3 2 33= + +u x xy y Differentiate partially with to x, Differentiate partially with respect y,

2 23 3∂ = +

∂u x yx

26 3∂ = +∂u xy yy

Diff.p.w.r to y, Diff.p.w.r to x,

2

6 .....(1)∂ =∂ ∂

u yy x

2

6 .....(1)∂ =∂ ∂

u yy x

2

6 .....(2)∂ =∂ ∂

u yx y

2

6 .....(2)∂ =∂ ∂

u yx y

2 2

(1), (2) =>. .

∂ ∂=∂ ∂ ∂ ∂

u ux y y x

Hence Verified.47. Given that,

Elasticity of demand, ηd = 5xx−

i.e. Px

− 5dx x

dp x−=

5dx

x − = dp

p−

Integrating both sides,

5

dx dpx p

= −−∫ ∫ + log k

log (x –5) = –log p + log k log (x –5) + log p = log k

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log p (x –5) = log k p(x –5) = k when p = 2, x = 7 k = 4∴ The demand function is

p = 45x −

Revenue, R = px or R = 45

xx −

, x>5.

48. Here P = cosx; Q = 12

sin 2x

∫ pdx = ∫ cosx dx = sin x

I.F = pdx

e∫ = esinx

The general solution is

y(I.F) = ∫ Q(I.F) dx+c Let sin x=t then cos x dx = dt

= 12∫ sin 2x esinx dx+c [sin 2x = 2 sin x cos x]

= ∫ sinx cosx esinxdx+c

= ∫ t et dt+c = et (t–1)+c (integration by parts)

= esinx (sinx–1)+c49. x(y2+1)dx + y(x2+1)dy = 0

x(y2+1)dx = –y(x2+1)dy

2 1xdx

x +∫ = – 2 1ydy

y +∫ = 2

21

xx +∫ = – 2

2c

ydyy +∫

=> log(x2+1) = – log(y2+1)+logclog(x2+1) = – log(y2+1)+logclog(x2+1) + log(y2+1) = logc

=> log(x2+1).(y2+1) = logc∴ (x2+1) (y2+1) = c

50. Since 5 values are given we assume that the polynomial is of degree four.Hence fifth order difference are zero.

∆5 (y0) = 0(E–1)5 (y0) = 0

=> (E5 – 5E4 + 10E3 – 10E2 + 5E – 1) y0 = 0

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y5 – 5y4 + 10y3 – 10y2 + 5y1 – y0 = 0 Here y5 = 32

y4 = 24 y3 = to be found y2 = 14 y1 = 11 y0 = 732 – 5(24) + 10y3 – 10(14) + 5(11) – 7 = 0

10y3 = 120 + 140 – 55 – 32 + 7 = 267 – 87 10y3 = 180 y3 = 18∴ Missing number is 18.

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

y a x nb

xy a x b x

= +

= +

∑ ∑

∑ ∑∑ 2

⇒ + =+ = ⋅

10 7 16 9 130 10 47 4 2

a ba b

. ( )( )

..................

Solving (1) and (2) we get,

a = 1.48 and b = 0.3The line of best fit is y = 1.48x + 0.3

∴The x-intercept of the line of best fit is −0 31 48

..

52. X : –3 –2 –1 0 1 2 3

P(x) : .05 0.1 0.3 0 0.3 0.15 0.1

E(2X–7) = 2E(X)–7 → 1 E(X) = –3(0.05)– 2 (0.1)–1 (0.3) + 0 (0) + 1(0.3) + 2(0.15) + 3(0.1) = –0.15 –0.2 –0.3 + 0 + 0.3 + 0.3 + 0.3 = 0.25

∴ E(2X–7) = 2 (0.25) – 7 = 0.5 – 7 = –6.553. We shall find confidence limits for the proportion of bad apples.

Sample size, n = 500

Proportion of bad apples in the sample = 45

500 = 0.09

p = 0.09∴ Proportion of good apples in the sample q = 1 – p = 0.91The confidence limits for the population proportion P of bad apples are given by

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p ± (Zc) pqn

=> 0.09 ± (2.58) ( )0.09 (0.91)500

=> 0.09 ± 0.033The required interval is (0.057 , 0.123). Thus, the bad apples in the consignment lie between 5.7% and 12.3%.

54. X Y x = X X− y = Y Y− x2 y2 xy

10 40 –3 –1 9 1 3

12 38 –1 –3 1 9 313 43 0 2 0 4 012 45 –1 4 1 16 –416 37 3 –4 9 16 –1215 43 2 2 4 4 4

78 246 0 0 24 50 6

XX

n= ∑ =

786

= 13 ; Y

Yn

= ∑ = 246

6 = 41

26 0.12

50xyxy

by

−= = = −∑∑

∴ Regression equation of X on Y is ( X X− ) = bxy ( Y Y− ) X–13 = –0.12 (Y–41) X = 17.92–0.12Y

55. Calculation of Cost of Living Index by family budget method

Commodity p0 p1 Quantity V

P = 1

0

pp

× 100 PV

A 10 12 20 120.00 2400.00B 30 35 50 116.67 5833.50C 40 50 50 125.00 6250.00D 200 300 20 150.00 3000.00E 25 50 40 200.00 8000.00F 100 150 50 150.00 7500.00G 20 25 60 125.00 7500.00H 150 180 40 120.00 4800.00

330 45283.50

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Cost of living index = PVV

∑∑

= 45283.50330

= 137.22

Hence, there is 37.22% increase in cost of living in current year compared to base year.PART - C

56. x−2y+3z=13x−y+4z=32x+y−2z=−1

=> 1 2 33 1 42 1 2

131

−−

−

=

−

xyz

∴

xyz

=

−−

−

−

−1 2 33 1 42 1 2

131

1

To find: 1 2 33 1 42 1 2

1−−

−

−

Let A = 1 2 33 1 42 1 2

−−

−

|A| =

1 2 33 1 42 1 2

−−

− = 1(2−4) +2(−6−8) +3(3+2)

= −2−28+15 =−15 ≠ 0

Matrices of co-factors =

−−

−−

−

−−

− −−

−

−−

−−−

1 41 2

3 42 2

3 12 1

2 31 2

1 32 2

1 22 1

2 31 4

1 33 4

1 23 1

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= 2 141 8 55 5 5− − −−

− 5

∴ Adj A = − − −

−−

2 1 514 8 55 5 5

Inverse of A = −

− − −−−

115

2 1 514 8 55 5 5

Hence

xyz

= −

− − −−−

−

115

2 1 514 8 55 5 5

131

= −

−−

=

115

01515

011

=> x = 0, y = 1, z = 157. With the usual notation we have,

a11 = 14, a12 = 6, x1 = 28 a21 = 7, a22 = 18, x2 = 36

Now, b11 = 11

1

14 128 2

ax

= =

b12 = 12

2

6 136 6

ax

= =

b21 = 21

1

7 128 4

ax

= =

b22 = 22

2

18 3 136 6 2

ax

= = =

∴ The technology matrix is

B =

1 12 61 14 2

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I − B =

1 1 1 11 01 0 2 6 2 60 1 1 1 1 10 1

4 2 4 22 1 1 1 1

2 6 2 61 2 1 1 1

4 2 4 2

− − − =

− − − − −

= =

− − −

The main diagonal of I − B are positive.

Also, |I − B| =

1 11 1 6 12 6

1 1 4 24 244 2

524

− −= − =

−

=

which is positive

∴ The two Hawkins - simon conditions are satisfied. ∴ The system is viable.

Now, (I − B)−1 =

1 1 1 11242 6 2 65

1 1 1 124 54 2 4 2

12 416 125

=

=

Now, X = (I − B)−1 D

12 4 2016 12 305240 1201120 3605360 721480 965

=

+

= +

= =

The output of P should be 72 crores and that of Q should be 96 crores.58. The combined equation of the asymptotes differs from the equation of the hyperbola only by

a constant.So, the combined equation of the asymptotes is2x2 + 5xy + 2y2 –11x –7y + K = 0 ----------which is a pair of straight lines satisfying the conditionabc + 2fgh –af2 –bg2 –ch2 = 0

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In the given equation,

a = 2, h = 52

, b = 2, f = −72

, g = −112

, C = K

substituting in (2) we get K = 5So the combined equation of the asymptotes is

2x2 + 5xy + 2y2 –11x –7y + 5 = 0 ----------(From ) To find seperate equation of asymptotes consider 2x2 + 5xy + 2y2 (of case )= 2x2 + 4xy + xy + 2y2

= (2x + y) (x + 2y)let for some l, m seperate equation of asymptotes are an 2x + y + l = 0 ----------

x + 2y + m = 0 ----------Now combined equation is (2x + y + l) (x + 2y + m) = 2x2 + 5xy + 2y2 − 11 x - 7y + 5 = 0 (from )

Comparing the co-efficients of xl + 2m = −11 ---------- on both side

Comparing the co-efficients of y2l + m = –7 ---------- on both side

=> l = –1, m = –5The equations of the asymptotes are

2x + y –1 = 0 and x + 2y –5 = 059. Total Revenue R = px

Average Revenue AR = P

Marginal Revenue MR = ddx

(R) = ddx

(px)

= p + x dpdx

[product rule for differentiation

ddx

(uv) = uv’+ vu’]

Now, −

AR(AR MR)

= P

p x dpdx

p − +

= − −

Pdpp p x dx

= − px

. dxdp

= Elasticity demand ηd

∴ −

AR(AR MR)

= ηd

Given p = a + bx

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Differentiating with respect to x,dpdx

= b

R= px = ax + bx²AR = a + bx (AR = price).

MR = ddx

(ax + bx²)

= a + 2bx

∴ ARAR MR−( )

= a+bxa bx a bx+ − −2

= −+( )a bxbx

− (1)

h d = − px

. dxdp

= −+( )a bxx

. 1b

= −+( )a bxbx

− (2)

From (1) and (2) we get that ARAR MR−( )

= h d

60. 3( ) 27 108= − +f x x x

2( ) 3 27= −f x x'

( ) 6=f x x"

Condition for maximum and minimum

1( ) 0=f x

23 27 0− =x (÷) by 3

2 9 0− =x

2 9=x

3= ±x

11When x 3 ( ) 6( 3) 18= − ⇒ = − = − = −f x ve" ( )

∴ f ”(x) < 0 ∴f(x) is maximum 11when x 3 ( ) 6(3) 18= ⇒ = = = +f x ve" ( ) f ”(x)>0 ∴ f(x) is minimum The maximum value when x=–3 is

f(–3) = (–3)3–27(–3)+108 = –27+81+108

f(–3) = 162

SURA BOOKS


The minimum value when x=3 is f(3) = (3)3–27(3)+108

= 27–81+108=54 ∴ f(3) = 54Maximum value – Minimum Value = 162–54 = 108Hence the maximum value of f(x) is 108 more than the minimum value.

61. Given R = 600, C1 = Rs 0.60, C3 = Rs 80

i) EOQ, 3

1

2 2 80 6000.6

× ×= =C Rq0 C= 400

ii) Minimum Average yearly cost,

0 1 32 2 0.6 80 600= = × × ×C C C R = Rs.240iii) Number of orders per year,

1

3

600 0.62 2 80

×= =×

RCC

n036 0=16 0

64

=

32

=n0 orders per year.

iv) Time between orders (ie) optimum period of supply per optimum order.

00

400 4600 6

qt

R= = =

23

= of a year.

62. π

π

6

3

1∫ + √dx

cotx =

π

π

6

3

1∫

+

dxcos xsin x

Let I = π

π

6

3

1∫ + √dx

cot x =

π

π

6

3

∫ +sin x

sin x cos xdx .....(1)

Since ( )b

a

f x∫ dx = ( )b

a

f a b x+ −∫ dx

I = π

π π π

π π π π6

3 6 3

6 3 6 3

∫+ −

+ −

+ + −

sin

sin cos

x

x xdx

SURA BOOKS


= π

π π

π π6

3 2

2 2

∫−

−

+ −

sin

sin cos

x

x xdx

I = π

π

6

3

∫ +cosx

sinx cosxdx .....(2)

Add (1) and (2)

2I = π

π

π

π

π

π

6

3

6

3

6

3∫ ∫++

= = ( )sin x cos xsin x cos x

dx dx x

2I = π3

– π6

= π6

I = π

1263. At market equilibrium Pd = Ps

40 –x2 = 3x2 + 8x + 84x2 + 8x –32 = 0

x2 + 2x –8 = 0(x + 4) (x –2) = 0 (Since x cannot be –ve)

=>x = –4, x = 2At x = 2, P = 40 –4 = 36

∴ x0 = 2, p0 = 36 => x0p0 = 720

0 00

CS ( ) x

f x dx p x= −∫

CS = 2

0

(40 −∫ x2) dx –p0x0 = 403

3

0

2

x −

x –72

= 8803

− – 72 = − 88

3

= 163

units

0

0 00

PS = P X ( )x

g x dx− ∫

SURA BOOKS


PS = p0x0 –0x

2

0

(3x∫ + 8x + 8) dx

= 72 –2

2

0

(3x∫ + 8x + 8) dx = 72 –23 2

0

3x 8x 8x3 2

+ +

= 72 4 83 2

0

2− + + x x x

= 72 –(8 + 16 + 16) = 32 units.64. For market clearance, the required condition is

Qd = Qs

⇒ 42 – 4p–4dpdt

+2d p

dt = –6 + 8p

⇒ 48 – 12p – 4dpdt

+ 2d p

dt = 0

2

2

d pdt

⇒ – 4dpdt

– 12p = –48

The auxiliary equation is m2–4m–12=0m=6, –2

C.F = Ae6t + Be–2t

P.I. = 2

1D 4D 12 − −

(–48)e0t

= 112−

(–48) = 4

∴ The general solution isP = C.F+P.IP = Ae6t + Be–2t + 4

65. By data we havex0 = 1, x1 = 2, x2 = 3, x3 = 4, x4 = 7y0 = 2, y1 = 4, y2 = 8, y3 = 16, y4 = 128x = 5 to find y

Lagrange’s formula

y =−( ) −( ) −( ) −( )−( ) −( ) −( ) −( ) ( ) +

−( ) −( ) −5 2 5 3 5 4 5 71 2 1 3 1 4 1 7

25 1 5 3 5 44 5 72 1 2 3 2 4 2 7

4

5 1 5 2 5 4 5 73 1

( ) −( )−( ) −( ) −( ) −( ) ( )

+−( ) −( ) −( ) −( )−( ) 33 2 3 4 3 7

85 1 5 2 5 3 5 74 1 4 2 4 3 4 7−( ) −( ) −( ) ( ) +

−( ) −( ) −( ) −( )−( ) −( ) −( ) −(( ) ( )

+−( ) −( ) −( ) −( )−( ) −( ) −( ) −( ) ( )

16

5 1 5 2 5 3 5 47 1 7 2 7 3 7 4

128

SURA BOOKS


=( )( )( ) −( )−( ) −( ) −( ) −( ) ( ) + ( )( )( ) −( )

( ) −( ) −( )3 2 1 2

1 2 3 62

4 2 1 21 1 2 −−( ) ( )

+( )( )( ) −( )( )( ) −( ) −( ) ( ) + ( )( )( ) −( )

( )(

54

4 3 1 22 1 1 4

84 3 2 23 2))( ) −( ) ( )

1 316

+( )( )( )( )( )( )( )( ) ( )4 3 2 16 5 4 3

128

= –0.666 + 6.4 – 24 + 42.666 + 8.5333y = 32.9333.

66. Let x = Number of defective blades.

p = probability that a fuse is defective.

= 2

100

n = 200, λ=np = 200. 2

100 =4

x follows poisson distribution.

∴ P (3 or more defective blades) = P (x≥3) = 1–p(X<3) and .P(X )!

xexx

l l−= =

=1–[P(x=0)+P(x=1)+P(x=2)]

4 1 24 44 41

1! 2!ee e

−− −

⇒ − + +

=1–e–4.[1+4+8]=1–(0.01832).(13)

=1–0.238=0.762.

67. 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)

SURA BOOKS


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

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

> Z

ZZ 0= 13

= 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.

68. Sample size, n = 1600Sample Mean X = 99µ = 100, σ = 15, H0 : µ = 100 H1 : µ ≠ 100Test statistic, Z is the standard normal variate under H0

Z = X−µ

σn

= 99 – 1 00

151600

= – 2.67

|Z| = 2.67Since |Z| = 2.67 > 1.96, acceptance region is |Z| < 1.96, H0 is rejected at 5% level of significance.∴ The sample is not from this population.

69. 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)

SURA BOOKS


∴ 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 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. Commodity p0 q0 p1 q1 P1q0 P0q0 P1q1 P0q1

A 6 50 10 56 500 300 560 336B 2 100 2 120 200 200 240 240C 4 60 6 60 360 240 360 240D 10 30 12 24 360 300 288 240E 8 40 12 36 480 320 432 288

1900 1360 1880 1344

Fisher’s Ideal Index = 01FP = ∑

∑∑∑

p qp q

p qp q

1 0

0 0

1 1

0 1

× × 100

SURA BOOKS


= 19001360

18801344

× × 100

= 1.3979 × 100 = 139.79Time Reversal Test: Test is satisfied when P01 × P10 = 1

P01 = ∑∑

∑∑

p qp q

p qp q

1 0

0 0

1 1

0 1

× = 19001360

18801344

×

P10 = ∑∑

∑∑

p qp q

p qp q

0 1

1 1

0 0

1 1

× = 13441880

13601900

×

P01 × P10 = 19001360

18801344

13441880

13601900

× × × = 1 = 1

Hence Fisher’s Ideal Index satisfies Time reversal test.Factor reversal test:Test is satisfied when P01 × Q01 = 1 1

0 0

p qp q

∑∑

Q01 = ∑∑

∑∑

q pq p

q pq p

1 0

0 0

1 1

0 1

× = 13441360

18801900

×

∴ P01 × Q01 = 19001360

18801344

13441360

18801900

× × ×

= 18801360

= 1 1

0 0

p qp q

∑∑

Hence Fisher’s Ideal Index satisfies Factor reversal test.

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