TN 11th Business Maths and Statistics Operations Research Important Questions And Answers Study Material - QB365 Set A

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Questions:

Generic 5

10 x 5 = 50

The following data relate to advertisement expenditure(in lakh of rupees) and their corresponding sales( in crores of rupees)

Advertisement expenditure	40	50	38	60	65	50	35
Sales	38	60	55	70	60	48	30

Estimate the sales corresponding to advertising expenditure of Rs. 30 lakh.

The heights ( in cm.) of a group of fathers and sons are given below

Heights of fathers:	158	166	163	165	167	170	167	172	177	181
Heights of Sons:	163	158	167	170	160	180	170	175	172	175

Find the lines of regression and estimate the height of son when the height of the father is 164 cm.

A random sample of recent repair jobs was selected and estimated cost and actual cost were recorded.

Estimated cost	300	450	800	250	500	975	475	400
Actual cost	273	486	734	297	631	872	396	457

Calculate the value of spearman’s correlation coefficient.

Calculate correlation coefficient for the following data.

X	25	18	21	24	27	30	36	39	42	48
Y	26	35	48	28	20	36	25	40	43	39

Solve the following linear programming problem graphically.
Maximise Z = 4x₁+ x₂ subject to the constraints x₁+ x₂≤ 50; 3x₁+ x₂≤ 90 and x₁≥ 0, x₂≥ 0.

A Project has the following time schedule

Activity	1-2	1-6	2-3	2-4	3-5	4-5	6-7	5-8	7-8
Duration(in days)	7	6	14	5	11	7	11	4	18

Construct the network and calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.

A project schedule has the following characteristics

Activity	1-2	1-3	2-4	3-4	3-5	4-9	5-6	5-7	6-8	7-8	8-10	9-10
Time	4	1	1	1	6	5	4	8	1	2	5	7

By the principle of mathematical induction, prove the following.
3²ⁿ-1 is a divisible by 8, for all \(n\in N\) .

By the principle of mathematical induction, prove the following.
1³ + 2³+ 3³+ ....... + n³= \(\frac { { n }^{ 2 }(n+1)^{ 2 } }{ 4 } \) for all \(n\in N\).

10.

Find the equation of the circle passing through the points (0, 1) , (4 ,3) and (1, -1).

Multiple Choice Question

10 x 1 = 10

11.

The term regression was introduced by ________.

(a)

R.A Fisher

(b)

Sir Francis Galton

(c)

Karl Pearson

(d)

Croxton and Cowden

12.

Scatter diagram of the variate values (X,Y) give the idea about ________.

(a)

functional relationship

(b)

regression model

(c)

distribution of errors

(d)

no relation

13.

The regression coefficient of Y on X ________.

(a)

b_xy =\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dy^{ 2 }-(\Sigma dy)^{ 2 } } \)

(b)

b_yx=\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dy^{ 2 }-(\Sigma dy)^{ 2 } } \)

(c)

b_yx =\(\frac { N\Sigma dxdy-(\Sigma dx)(\Sigma dy) }{ N\Sigma dx^{ 2 }-(\Sigma dx)^{ 2 } } \)

(d)

b_xy =\(\frac { N\Sigma xy-(\Sigma x)(\Sigma y) }{ \sqrt { N\Sigma { x }^{ 2 }-(\Sigma x)^{ 2 }\times \sqrt { N\Sigma { y }^{ 2 }-(\Sigma y)^{ 2 } } } } \)

14.

From the following data, N = 11, ΣX = 117, ΣY = 260, ΣX² = 1313, ΣY² = 6580, ΣXY = 2827 the correlation coefficient is ________.

(a)

0.3566

(b)

-0.3566

(c)

(d)

0.4566

15.

Correlation co-efficient lies between ______.

(a)

0 to ∞

(b)

-1 to +1

(c)

-1 to 0

(d)

-1 to ∞

16.

If the values of two variables move in same direction then the correlation is said to be ______.

(a)

Negative

(b)

positive

(c)

Perfect positive

(d)

No correlation

17.

Given an L.P.P maximize Z = 2x₁ + 3x₂ subject to the constrains x₁ + x₂ ≤ 1, 5x₁ + 5x₂≥ 0 and x₁ ≥ 0, x₂ ≥ 0 using graphical method, we observe ______.

(a)

No feasible solution

(b)

unique optimum solution

(c)

multiple optimum solution

(d)

none of these

18.

Which of the following is not correct?

(a)

Objective that we aim to maximize or minimize

(b)

Constraints that we need to specify

(c)

Decision variables that we need to determine

(d)

Decision variables are to be unrestricted

19.

In the given graph the coordinates of M₁ are

(a)

x₁ = 5, x₂= 30

(b)

x₁ = 20, x₂= 16

(c)

x₁ = 10, x₂= 20

(d)

x₁ = 20, x₂ = 30

20.

Maximize: z = 3x₁+ 4x₂ subject to 2x₁+ x₂≤ 40, 2x₁+ 5x₂≤ 180, x₁, x₂≥ 0. In the LPP, which one of the following is feasible corner point?

(a)

x₁ = 18, x₂= 24

(b)

x₁= 15, x₂= 30

(c)

x₁= 2.5, x₂= 35

(d)

x₁= 20.5, x₂= 19

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Answers Key:

Generic 5

10 x 5 = 50

X	Y	dx = X-48	dy = Y-52	dx²	dy²	dx dy
40	38	-8	-14	64	196	112
50	60	2	8	4	64	16
38	55	-10	3	100	9	-30
60	70	12	18	144	324	216
65	60	17	8	289	64	136
50	48	2	-4	4	16	-8
35	30	-13	-22	169	484	286
\(\Sigma X\) = 338	\(\Sigma Y\) = 361	\(\Sigma dx\) = 2	\(\Sigma dy\) = -3	\(\Sigma dx^2\) = 774	\(\Sigma dy^2\) = 1157	\(\Sigma dxdy\) = 728

\(\bar{X} =\frac{\Sigma X}{N}=\frac{338}{7}=48.29 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{361}{7}=51.57 \)
\(b_{y x} =\frac{N \Sigma d x d y-\Sigma d x \Sigma d y}{N \Sigma d x^2-(\Sigma d x)^2} \)
\(=\frac{7(728)+6}{7(774)-4} \)
\(=\frac{5102}{5414}=0.942\)
Regression equation of Y on X is
\(Y-\overset{-}{Y}=b_{yx}(X-\overset{-}{X})\)
Y - 51.57 = 0.942(X - 48.29)
Y = 0.942X - 45.49 + 51.57
= 0.942X + 6.08
Y = 0.942(30) + 6.08
Y = 34.34 (In Crores of rupees)

X	Y	dx = X-168	dy = Y-169	dx²	dy²	dxdy
158	163	-10	-6	100	36	60
166	158	-2	-11	4	121	22
163	167	-5	-2	25	4	10
165	170	-3	1	9	1	-3
167	160	-1	9	1	81	-9
170	180	2	11	4	121	22
167	170	-1	1	1	1	-1
172	175	4	6	16	36	24
177	172	9	3	25	9	27
181	175	13	6	169	36	78
\(\Sigma X\) = 1686	\(\Sigma Y\) = 1690	\(\Sigma dx\) = 6	\(\Sigma dy\) = 0	\(\Sigma dx^2\) = 410	\(\Sigma dy^2\)= 446	\(\Sigma dxdy\) = 248

\(\bar{X} =\frac{\Sigma X}{N}=\frac{1686}{10}=168.6 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{1690}{10}=169 \)
\(b_{x y} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d x)}{N \Sigma d y^2-(\Sigma d y)^2} \)
\(=\frac{10(248)-0}{10(446)-0}=\frac{248}{446}=0.556 \)
\(b_{y x} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d y)}{N \Sigma d x^2-(\Sigma d x)^2} \)
\(=\frac{2480}{4100-36}=\frac{2480}{4064}=0.6102\)
Regression equation of X on Y
\(X-\bar{X}=b_{x y}(Y-\bar{Y}) \)
X - 168.6 = 0.556(Y - 169)
X = 0.556 Y + 168.6-93.964
X = 0.556 Y + 74.64
Regression equation of Y on X
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 169 = 0.6102(X - 168.6)
Y = 0.6102 X - 102.8 + 169
Y = 0.6102X + 66.12
If X = 164
Y= 100.07 + 66.12
= 166.19
Height of son is 166.19

X	Y	R_x	R_y	d=R_x-R_y	d²
300	273	2	1	1	1
450	486	4	5	-1	1
800	734	7	7	0	0
250	297	1	2	-1	1
500	631	6	6	0	0
975	872	8	8	0	0
475	396	5	3	2	4
400	457	3	4	-1	1
					\(\sum\)d²= 8

\(\rho =1-\frac{6 \Sigma d^2}{n\left(n^2-1\right)} \)
\(=1-\frac{6(8)}{8(63)} \)
= 1-0.0952
= 0.905

X	Y	x²	y²	xy
25	26	625	676	650
18	35	324	1225	630
21	48	441	2304	1008
24	28	576	784	672
27	20	729	400	540
30	36	900	1296	1080
36	25	1296	625	900
39	40	1521	1600	1560
42	43	1764	1849	1806
48	39	2304	1521	1872
\(\sum\)X = 310	\(\sum\)Y = 340	\(\sum\)X² = 10480	\(\sum\)Y² = 12280	\(\sum\)XY = 10718

\(r(x, y)=\frac{N \Sigma X Y-\left(\sum X\right)\left(\sum Y\right)}{\sqrt{N \Sigma X^2-(\Sigma Y)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(10718)-(310)(340)}{\sqrt{10(10480)-(310)^2} \sqrt{10(12280)-(340)^2}} \)
\(=\frac{107180-105400}{\sqrt{104800-96100} \sqrt{122800-115600}} \)
\(=\frac{1780}{\sqrt{8700 \times 7200}}=\frac{1780}{7914.54}=0.2249\)

First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant write all the inequalities of the constraints in the form of equations.
∴ We have the lines \(x_1+x_2 \leq 50; 3 x_1+x_2 \leq 90\)
x₁ + x₂ = 50 is a line passing through the points (0,50) and (50,0).
[(0,50) is obtained by taking x₁ = 0 in x₁ + x₂ = 50, (50,0) is obtained by taking x₂ = 0 in \(\left.x_1+x_2=50\right]\)
Any point lying on or below the line x₁ + x₂ = 50. Satisfies the constraint \(x_1+x_2 \leq 50\)
We follow the same steps for the following
x₁ + x₂ = 50

x₁	0	50
x₂	50	0

\({ 3x }_{ 1 }+{ x }_{ 2 }=90\)

x₁	0	30
x₂	90	0

Now we draw the graph

The feasible region satisfying all the conditions is OABC. The co-ordinates of the points are O(0,0), A(30, 0), B(20,30), C(0,50).

Corner points	\( Z=4{ x }_{ 1 }+{ x }_{ 2 }\)
O(0,0)	0
A(30,0)	120
B(20,30)	80 + 30 = 110
C(0,50)	50

Optimal solution is at A(30,0)
x₁= 30, x₂= 0 and Z_max = 120.
Verification
\(3 x_1+x_2 =90 \)
\(x_1+x_2 =50 \)
\(2 x_1 =40 \)
\(x_1 =20 x_2 =50-20=30 \)
B(20,30)

E₁= 0	L₈ = 35
E₂ = 0 + 7 + 7	L₇= 35 - 18 - 17
E₃ = 7 + 14 = 21	L₆ = 18 - 11 = 7
E₄ = 7 + 5 = 12	L₅ = 36 - 4 = 32
E₅ = max of {21 + 11, 12 + 7} = 32	L₄ = 32 - 7 = 25
E₆ = 0 + 6 = 6	L₃ = 32 - 11 = 21
E₇ = 6 + 11 = 17	L₂ = min of {21 - 14, 25 - 5} = 7
E₈ = max of {32 + 4, 17 + 18) = 36	L₁= 7 - 7 = 0

Activity	Duration t_ij	EST	EFT = EST + T_ij	LST = LFT - t_ij	LFT
1-2	7	0	7	0	7
1-6	6	0	6	6	7
2-3	14	7	21	7	21
2-4	5	7	12	20	25
3 -5	11	21	32	21	32
4 - 5	7	12	19	7	14
6 - 7	11	6	17	7	18
5- 8	4	32	36	32	36
7 - 8	19	17	36	17	36

Since EFT and LFT are same in 1 - 2, 2 - 3, 3 - 5 and 5 -8, The critical path is 1 - 2 - 3 - 5 - 8 and the duration of time is 36 days.

E₁ = 0	L₁₀= 22
E₂ = 0 + 4 = 4	L₉= 22 - 7 = 15
E₃ = 0 + 1 = 1	L₈ = 22 - 5 = 17
E₄ = Max of {4 + 1, 1 + 1} = 5	L₇ = 17 - 2 = 15
E₅ = 1 + 6 = 7	L₆ = 17 - 1 = 16
	L₅= Min of {16-4, 15-8} = 7
E₆= 7 + 4 = 11	L₄ = 15 - 5 = 10
E₈ = Max of {15 + 2, 11 + 1}= 17	L₃ = Min of {10-1, 7-6] = 1
E₉ = 5 + 5 = 10	L₂ = 10 - 1 = 9
E₁₀ = Max of {10 + 7, 17 + 5} = 22	L₁ = 0

Activity	Duration t_ij	EST	EFT = EST + T_ij	LST = LFT - t_ij	LFT
1-2	4	0	4	9-4 = 5	9
1-3	1	0	1	1-1 = 0	1
2-4	1	4	5	10-1 = 9	10
3-4	1	1	2	10-1 = 9	10
3-5	6	1	7	7-6 = 1	7
4-9	5	5	10	15-5 = 10	15
5-6	4	7	11	16-4 = 12	16
5-7	8	7	15	15-8 = 7	15
6-8	1	11	12	17-1 = 16	17
7-8	2	15	17	17-2 = 15	17
8-10	5	17	22	22-5 = 17	22
9-10	7	10	17	22-7 = 15	22

Since EFT and LFT is same on 1 - 3, 3 - 5, 5 -7 and 7 - 8 and 8 -10 the critical path is 1- 3 - 5 -7 - 8 - 10 and the duration is 22 time units.

Let P (n) denote the statement. 3²ⁿ - 1 is a divisible by 8
Step-1: Put n = 1
∴ P(1) : 3² - 1 = 9 - 1 = 8 is divisible by 8
∴ P(1) is true.
Step-2: Let us assume that P(k) is true.
∴ 3^2k - 1 is divisible by 8
⇒ 3^2k - 1 = 8C (where C is a constant)
⇒ 3^2k = 8C + 1 ....(1)
Step-3:
To prove that P(k + 1) is true
P(k + 1):3² (k + 1) - 1
= 3^{2k + 2}- 1 = 3^2k .3² - 1
= (8 m + 1) . 9 - 1
= 8 m . 9 + 9 - 1
= 8 m . 9 + 8
= 8 (9 m + 1)
which is divisible by 8.
∴ P (k + I) is true whenever P( k) is true.
∴ p(n) is true for all \(n\in N\).

Let P (n) denote the statement 1³+ 2³+ 3³+ ....... + n³= \(\frac { { n }^{ 2 }(n+1)^{ 2 } }{ 4 } \)
Put n = 1
LHS = 1³ = 1
\(=\frac { 1^2(2)^2 }{ 4 } \Rightarrow 1\)
LHS = RHS
\(\therefore\) P (1) is true
Let us assume that P(k) is true
p(k) : 1³+ 2³+ ..... + k³= \(\frac { { k }^{ 2 }(k+1)^{ 2 } }{ 4 } \)
To prove that P(k+1) IS true
p(k) : 1³ + 2³+ ..... + k³+ (k + 1)³
\(=P(k)+(k+1)^3\)
= \(\frac { { k }^{ 2 }(k+1)^{ 2 } }{ 4 } +(k+1)^{ 3 }\)
\(=\frac{k^2(k+1)^2+4(k+1)^3}{4}\)
\(=\frac{(k+1)^2\left(k^2+4(k+1)\right)}{4}\)
\(=\frac{(k+1)^2\left(k^2+4 k+4\right)}{4}=\frac{(k+1)^2(k+2)^2}{4}\)
∴ p(k + 1) is true if P(k) is true.
∴ p(n) is true for all \(n\in N\)

10.

Equation of the circle be x² + y²+ 2gx + 2fy + c = 0 ..(1)
It passes through (0, 1)
0 + 1 + 0 + 2f + c = 0   \(\Rightarrow \) 2f + c = -1 ..(2)
The circle passes through (4, 3)
16 + 9 + 8g + 6f + c = 0  \(\Rightarrow \) 8g + 6f + c = -25 ...(3)
The circle passes through (1,-1)
1 + 1 + 2g - 2f + c = 0  \(\Rightarrow \) 2g - 2f + c = -2 ....(4)
Solving (1), (2) and (3) we get
c = 1   \(\Rightarrow\)  g = -5/2   \(\Rightarrow\)  f = -1
Equation of circle is
x² + y²+ 2 \(\left( -\frac { 5 }{ 2 } \right) \)x + 2(-1)y + 1 = 0
x² + y²- 5x - 2y + 1 = 0