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)
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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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4 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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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Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 5
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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6 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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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Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 7
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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8 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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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Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 9
⇒ 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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10 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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
SURA BOOKS
Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 11
=> 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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12 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
=−( ) −( ) −( )−( ) −( ) −( ) +
−( ) −( ) −( )−( ) −( )
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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Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 13
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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14 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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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Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 15
∆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
SURA BOOKS
16 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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
Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 17
(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
SURA BOOKS
18 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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
SURA BOOKS
Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 19
∂∂
=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
SURA BOOKS
20 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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+ + −
− .
SURA BOOKS
Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 21
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
= − − = =
SURA BOOKS
22 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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
SURA BOOKS
Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers 23
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
SURA BOOKS
24 Sura’s n XII Std n Business Mathematics n 2016 - July Question Paper with Answers
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