TU Solution

2002 -2 4 32 1024 - 64 16 256 - 32
- 22
2003 -1 1 40 1600 - 40 22 484
0
2004 0 0 24 576 0 36 1296 40
88
2005 1 1 40 1600 40 40 1600 144
170
2006 2 4 32 1024 64 44 1936

2007 3 9 88 7788 264 48 2304

28 272 13868 216 222 8132

Here, n = 7, Σx2 = 28, Σy1 = 272, Σy12 = 13868, Σxy1 = 216,
Σy2 = 222, Σy22 = 8132, Σxy2 = 170
Σxy1 216 Σxy2 170
Now, b1 = Σx2 = 28 = 7.71 and b2 = Σx2 = 28 = 6.07

Since, b1 = 7.71 > b2 = 6.07
Therefore, based on growth rate in profit we should invest in company A.

Again,

y̅1 = Σy1 = 272 = 38.86
n 7

= √ 12 − ( 1)2

= √138768 − (2772)2

= √1981.14 − 1509.88

= √471.26

= 21.70

( ) = × 100% = 21.70 × 100% = 55.84%
̅̅ 1̅ 38.86

̅̅ 2̅ = 2 = 222 = 31.71
7

= √ 22 − ( 2 2

)

200

= √81732 − (2272)2

= √1161.71 − 10005.80

= √155.91

= 12.49

( ) = × 100% = 12.49 × 100% = 39.38%
̅ ̅ 2̅ 31.71

Since, CV(A) = 55.84% > CV(B) = 39.38%

Therefore, based on consistency in the profit we should invest in company B.

201

Chapter - 8

Index Number

Brief Question Answer

QN. 1) (2078) Find the aggregative price index number for the year 2021.

Commodities A B C D E

Price in 2020 10 20 18 15 12

Price in 2021 15 22 20 22 18

Solution: Calculation of price index number

Commodity Price in 2020 (po) Price in 2021 (p1)

A 10 15

B 20 22

C 18 20

D 15 22

E 12 18

ΣPo = 75 ΣP1 = 97

Now, price index number = Σp1 × 100 = 75 × 100 = 77.32
ΣPo 97

QN 1 (2077, Back) Find the price index number for the year 2015 taking 2014 as the base year for
the following data by using simple aggregate method.

Commodities A B C D E

Price in 2014 16 17 19 18 12

Price in 2015 21 20 25 23 18

Solution: Calculation of price index number

Commodity Price in 2014 (po) Price in 2015 (p1)

A 16 21

B 17 20

C 19 25

D 18 23

E 12 18

ΣPo = 82 ΣP1 = 107

202

Now, price index number = Σp1 × 100 = 107 × 100 = 133.75
ΣPo 82

QN. 2) (2075) Reconstruct the following index number shifting base year as 2063.

Year 2059 2060 2061 2062 2063 2064 2065

Index No. 100 112 123 131 144 152 160

Solution:

Recast index number of any year = Index numbr of old base × 100
Index number of new base

Calculation of index numbers with new base as 2063
Year Old index number New index number

2059 100 69.4444

2060 112 77.7778

2061 123 85.4167

2062 131 90.9722

2063 144 100

2064 152 105.5556

2065 160 111.1111

QN. 3) (2074) If during a period, the price index number goes up from 100 to 200 and wages of a

worker are raised from Rs. 300 and Rs.500, does the worker really gain?

Solution: Changed in index number= 200 - 100 = 100

Percentage changed in price = 100 × 100% = 100%
100

and changed in wages = Rs. 500 - Rs. 300 = Rs200

Percentage changed in wages = 200 × 100% = 66.6667 %
300

Since the price is increased by 100% but wages is increased by only 66.67%, so the worker

is not really gained.

or, Real wage of current year
Real wage of base year
Real wages (or income)index numbers = × 100

= 500 × 100 = 166.67
300
Price index number goes to 100 to 200 but real wages is only 166.67, so the worker is not really

gained.

QN. 4) (2072) Given that ΣPW = 12610, ΣW = 100, where P and W have their usual meaning.

Calculate the cost of 'living index'. A man of middle class family of a remote area of Nepal was

203

earning Rs.50,000 in the base period. What should be the salary in the current period, if his living

standard is to remain same?

Solution:

Here given

= 12610 = 100

Earning in base year = Rs. 50,000

Now cost of living index = ΣPW = 12610 = 126.1
ΣW 100
In order to maintain his former standard, he should get
126.1
100 × . 50,000 = . 63050

QN. 5) (2072) From the information given below, calculate the real index number

Year 2010 2011 2012 2013

Wages 650 980 1200 1800

Index no. 100 120 130 150

Solution: Calculation of real wages index number

Year Wage (W) Index Number(I) Real wages Real wages index
2010 650 100
2011 980 120 650 × 100 = 650 650 × 100 = 100
2012 1200 130 100 650
2013 1800 150
980 × 100 = 816.667 816.667 × 100 = 125.641
120 650

1200 × 100 = 923.08 923 × 100 = 142.0123
130 650

1800 × 100 = 1200 1200 × 100 = 184.6154
150 650

Real wages = wages × 100
Index Number

Real wages (or income)index numbers = Real wage of current year × 100
Real wage of base year

QN. 6) (2071) Form the information of pieces and quantities in the base year 1983 and

current year 1984 and current year 1984, the following results are obtained:

Σpoq1 = 184, Σp1qo = 121, Σp1q1 = 192, Σpoq1 = 108, where po, p1, qo, and q1 have their usual

meaning. Calculate the price index by Fisher's formula.

Solution:

204

Here given 0 1 = 184, 1 0 = 121, 1 1 = 192, 0 0 = 108

Now, Fisher′s index number p01(F) = √pp10qq00 × p1q1 × 100
p0q1

= √112081 × 192 × 100 = 1.08.12 × 100 = 108.12
184

Descriptive Answer Question

QN. 7) (2078) The following table gives the changes in the price and consumption (quantity) of

certain major constituents of the consumption basket of the labor class.

Commodity Units 2020 2021

Price Quantity Price Quantity

Wheat Quintals 1000 10 1100 6

Rice Quintals 1500 15 1700 18

Cloth Meters 50 50 40 30

Calculate the cost of living index. If the average of a labor wage was Rs. 12000 per month in

2020, what should be the average wage per worker per month in 2021, so that the standard of

living of workers does not falls in the year 2020 level?

Solution: Calculation of Index Number

Commodities 2020 2021 poqo poq1 p1qo p1q1

po qo p1 q1

Wheat 1000 10 1100 6 10000 6000 11000 6600

Rice 1500 15 1700 18 22500 27000 25500 30600

Cloth 50 50 40 30 2500 1500 2000 1200

Total 35000 34500 38500 38400

Here, Σpoqo =35000, Σpoq1 =34500, Σp1qo =38500, and Σp1q1 = 38400
Σp1qo 38500
Now, cost of living index = Σpoqo × 100 = 35000 × 100 = 110

In order to maintain former standard, he should earn

110 × 12000 = . 13200
100

205

QN. 7) (2077, back) Write down characteristics of index number. From the following data,

calculate the price index number for 2015 using a) Laspeyre's mwthod b) Paasche's method and c)

Fisher's ideal index.

Commodities 2014 2015

Price (po) Quantities (qo) Price (p1) Quantities (q1)

A 10 100 20 120

B 11 150 25 180

C 12 180 30 210

D 14 200 40 250

Solution:

Calculation of index number

Commodities 2014 2015 poqo poq1 p1qo p1q1
po qo p1 q1
A 10 100 20 120 1000 1200 2000 2400
B 11 150 25 180 1650 1980 3750 4500
C 12 180 30 210 2160 2520 5400 6300
D 14 200 40 250 2800 3500 8000 1000
7610 9200 19150 14200
Total

Here, Σpoqo =7610, Σpoq1 =1395, Σp1qo =19150, and Σp1q1 = 14200

a) Laspeyre's method or base year method
1 0 19150
01( ) = 0 0 × 100 = 7610 × 100 = 251.64

b) Paasche's method or given year method
1 1 14200
01( ) = 0 1 × 100 = 9200 × 100 = 154.35

c) Fishers ideal index number

01( ) = √ 10 00 × 1 1 × 100
0 1

= √179611500 × 14200 × 100
9200

= √2.5164 × 1.5435 × 100
= 197.08

206

QN. 8) (2077) Calculate Fisher's index number for the following data and show that it satisfies i)

time reversal test ii) factor reversal test.

Commodities 2018 2019
Price (po) Quantities (qo) Price (p1) Quantities (q1)

A 6 50 10 56

B 2 100 2 120

C 4 60 6 60

D 10 30 12 24

Solution:

Calculation of time reversal test and factor reversal test

Commodities 2018 2019 poqo poq1 p1qo p1q1

po qo p1 q1 560
240
A 6 50 10 56 300 336 500 360
288
B 2 100 2 120 200 240 200 1448

C 4 60 6 60 240 240 360

D 10 30 12 24 300 240 360

Total 1040 1056 1420

Here, Σp0q0 = 1040, Σp0q1 = 1056, Σp1q0 = 1420, Σp1q1 = 1448
i) Time reversal test

01( ) × 10( ) = √ 10 00 × 1 1 × 0 1 × 0 0
0 1 1 1 1 0

= √11402400 × 1448 × 1056 × 1040 = 1
1056 1448 1420

Hence, Fisher ideal index satisfies the time reversal test.

ii) Factor reversal test

P01(F) × Qo1(F) = √ΣΣpp10qq00 × Σp1q1 × Σq1p0 × Σq1p1
Σp0q1 Σq0p0 Σq0p1

= √11042400 × 1448 × 1056 × 1448
1056 1040 1420

= 1448
1040

207

01 = 1 1 = 1448
0 0 1040

∴ P01(F) × Qo1(F) = V01

Hence, Fisher's ideal index satisfies the factor reversal test.

QN. 9) (2077, back) find the cost of living index from the following data. If Mr. Ram Prasad is

getting his salary Rs.35,000 per month in base year, how much his salary should be increased in

current year to maintain the same standard of living?

Groups Group Index Group Weight

Food 412 62

Clothing 545 5

Fuel & Light 388 7

House Rent 117 9

Miscellaneous 186 17

Solution: Calculation of living index number

Groups Group Index (I) Group Weight (W) IW

Food 412 62 25544

Clothing 545 5 2725

Fuel & Light 388 7 2716

House Rent 117 9 1593

Miscellaneous 186 17 3162

ΣW =100 ΣIW = 35740

Cost of living index number = ΣIW = 35740 = 357.40
ΣW 100

In order to maintain the same standard of living, Mr. Ram prasad should earn

457.40 × . 35,000 = . 1,25,090
100

in order to maintain the same standard.

Solution:

QN. 10) (2076) Show that Fisher's ideal index number satisfies both time reversal test factor

reversal test from the following information.

208

2010 2012

Commodities Price (po) Expenditure Price (p1) Expenditure
(p0qo) (p1q1)

W 4 32 5 50

X 5 50 6 72

Y 3 18 4 28

Z 8 40 10 40

Solution:

Calculation of time reversal test and factor reversal test

Commodities 2010 2012 poqo poq1 p1qo p1q1

po q0 p1 q1 40 50
60 72
W 4 8 5 10 32 20 24 28
50 40
X 5 10 6 12 50 30 174 190

Y 3647 18 12

Z 8 5 10 4 40 80

Total 140 142

Here, Σp0q0 = 140, Σp0q1 = 142, Σp1q0 = 174, Σp1q1 = 190
i) Time reversal test

P01(F) × P10(F) = √ΣΣpp10qq00 × Σp1q1 × Σp0q1 × Σp0q0
Σp0q1 Σp1q1 Σp1q0

= √117440 × 190 × 142 × 140 = 1
142 190 174

Hence, Fisher ideal index satisfies the time reversal test.

ii) Factor reversal test

P01(F) × Qo1(F) = √ΣΣpp10qq00 × Σp1q1 × Σq1p0 × Σq1p1
Σp0q1 Σq0p0 Σq0p1

= √114740 × 190 × 142 × 190
142 140 174

= 190
140

V01 = Σp1q1 = 190
Σp0q0 140

209

∴ 01( ) × 1( ) = 01

Hence, Fisher's ideal index satisfies the factor reversal test.
QN. 11) (2075) Compute Fisher's index number. Show that Fisher's ideal index number satisfies

time reversal test from the following information.

2012 2013

Commodities Quantity Value (p0qo) Quantity Value (p1q1)
(po) (p1)

A 75 150 80 240

B 125 625 140 560

C 50 350 30 180

Solution:

Calculation of Fisher's index number

Commodities 2012 2013 poqo poq1 p1qo p1q1

po q0 p1 q1

A 2 75 3 80 150 160 225 240

B 5 125 4 140 625 700 500 560

C 7 50 6 30 350 210 300 180

1125 1070 1025 980

Here, Σp0q0 = 1125, Σp0q1 = 1070, Σp1q0 = 1025, Σp1q1 = 980
i) Fisher's index numner

01( ) = √ 10 00 × 1 1 × 100
0 1

= √11012255 × 980 × 100
1070

= 91.35

ii) Time reversal test

01( ) × 10( ) = √ 10 00 × 1 1 × 0 1 × 0 0
0 1 1 1 1 0

= √11012255 × 980 × 1070 × 1125 = 1
1070 980 1025

210

Hence, Fisher ideal index satisfies the time reversal test.

QN. 12) (2074) Define index number and explain the various types of index number. The shajha

Bus provides transportation for Kathmandu and addition its sells bus to different cities of Nepal.

The company has collected the following data in order to analyze its sales for year 2011 and 2012.

Cities Average selling price per bus in Number of buses sold
lakhs

2011 (po) 2012 (p1) 2011 (q0) 2012 (q1)

Kathmandu 24 25 17 20

Pokhara 20 20 14 18

Biratnagar 25 30 21 25

Compute the suitable index number for the given data.

Solution:

Calculation of Fisher's index number

Cities 2011 2012 poqo poq1 p1qo p1q1

po q0 p1 q1

Kathmandu 24 17 25 20 408 480 425 500

Pokhara 20 14 20 18 280 360 280 360

Biratnagar 25 21 30 25 525 625 630 750

1213 1465 1335 1610

Here, Σp0q0 = 1213, Σp0q1 = 1465, Σp1q0 = 1335, Σp1q1 = 1610
i) Fisher's index numner

P01(F) = √ΣΣpp10qq00 × Σp1q1 × 100
Σp0q1

= √11233135 × 1610 × 100
1465

= 109.98

QN. 13) (2074, old) From the data given below, calculate the index number for 2017 taking 2016

as base year using (a) Laspeyre's method (b) Paasch's method (c) Fisher's ideal method

Commodities 2016 2017
Price (po) Quantity (qo) Price (p1) Quantity (q1)

211

A 10 100 20 120
B 11 150 25 180
C 12 180 30 210
D 14 200 40 250

Solution: Calculation of index number

Commodities 2016 2017 poqo poq1 p1qo p1q1
po qo p1 q1
20 120 1000 1200 2000 2400
A 10 100 25 180 1650 1980 3750 4500
30 210 2160 2520 5400 6300
B 11 150 40 250 2800 3500 8000 1000
7610 9200 19150 14200
C 12 180

D 14 200

Total

Here, Σp0q0 = 7610, Σp0q1 = 9180, Σp1q0 = 19150, Σp1q1 = 14200

a) Laspeyre's method or base year method
1 0 19150
01( ) = 0 0 × 100 = 7610 × 100 = 251.64

b) Paasche's method or given year method
1 1 14200
01( ) = 0 1 × 100 = 9200 × 100 = 154.35

c) Fishers ideal index number

01( ) = √ 10 00 × 1 1 × 100 = √179611500 × 14200 × 100
0 1 9200

√2.5164 × 1.5435 × 100 = 197.08
QN. 14) (2073) Define index number and discuss the various problems that are faced in the

construction of an index number. The data given below related to the workers in housing

Groups Cost of living index for 2013 upto 2010 % in expenditure

Food 260 50

Fuel 140 5

Housing 150 15

Clothing 190 18

212

Other items 110 12

If the monthly pay of the workers in 2010 was Rs. 20,000. What should be the average monthly

pay in 2013 in order that the workers may be able to maintain the same standard of the living as

that of 2010.

Solution: Calculation of living index number

Groups Group index(I) % Expenditure (W) IW

Food 260 50 13000

Fuel 140 5 700

Housing 150 15 2250

Clothing 190 18 3420

Other items 110 12 1320

ΣW =100 ΣIW 20690

Cost of living index number = ΣIW = 20690 = 206.90
ΣW 100

In order to maintain the same standard of living, he/she should earn

206.90 × . 20,000 = . 41,380
100

in order to maintain the same standard.

QN. 15) (2073, old) The following information is related to budget of middle class families of

Kathmandu

Items Percentage of expenditure Price in 2010 Price in 2011

Food 35 1500 1650

Fuel 10 1650 1900

Clothing 20 1180 1200

Rent 15 1200 1500

Miscellaneous 20 370 500

What is the change in cost of living index in 2011 as compared to the prices of 2010?

Solution: Calculation of cost of living index number

Items % Expenditure (W) p0 p1 = 1 × 100 PW
0 3850

Food 35 1500 1650 110

Fuel 10 1650 1900 115.15 1151.5

Clothing 20 1180 1200 101.69 2033.8

213

Rent 15 1200 1500 125 1875

Misc. 20 370 500 135.14 2702.8

ΣW = 100 ΣPW = 11613.1

Now, cost of living index number

= ΣPW = 11613.1 = 116.131
ΣP 100

The change in cost of living index in 2011 as compared to the prices of 2010 is

(116.131 – 100) = 16.13%

QN. 16) (2072) Calculate the index number by using suitable formula for 1983 on the basis of

1982 from the following information:

Article - I Article - II Article - III
Year
Price Exp. Price Exp.
Price Exp.

1982 5 50 8 48 6 18

1983 4 48 7 49 5 20

Solution: Calculation of Fisher's ideal index number

Articles 1982 1983 poqo poq1 p1qo p1q1

po q0 p1 q1 48
49
I 5 10 4 12 50 60 40 20
117
II 8677 48 56 42

III 6354 18 24 15

116 140 97

Here, Σp0q0 = 116, Σp0q1 = 140, Σp1q0 = 97, Σp1q1 = 117
Fisher's index numner

01( ) = √ 10 00 × 1 1 × 100
0 1

= √19176 × 117 × 100
140

= 83.60

QN. 17) (2072, old) Prepare the real wages index form the following data: 2006
Year 2000 2001 2002 2003 2004 2005

214

Wages in Rs. 180 230 340 360 365 370 375
Price indices 100 170 300 320 330 340 350

Solution: Computation of real wages index

Year Wage Index Real wages Real wages index
(W) Number(I)
2000 180 180 100
2001 230 100 135.29 75.16
2002 340 170 113.33 62.96
2003 360 300 112.5 62.5
2005 365 320 110.61 61.45
2006 370 330 108.82 60.454
2007 375 340 107.14 59.52
350

Working note:

Real wages = wages × 100
Index Number

Real wage of 2000 = 180 × 100 = 180
100

Real wage of 2001 = 230 × 100 = 135.29
170

Real wage of 2002 = 340 × 100 = 113.33 and so on
300

Real wages index = Real wages of current year × 100
Real wages of base year

Real wages index of 2000 = 180 × 100 = 100
180

Real wages index of 2001 = 135.29 × 100 = 75.16
180

Real wages index of 2002 = 113.33 × 100 = 62.96 and so on
180

QN. 18) (2072, old) Compute the suitable index number from the following data:

Product p Product q Product r
Year
Price Qty. Price Qty.
Price Qty.

215

2009 4 54 2 5 3 10
2010 10 40 4 588

Solution: Calculation of Fisher's ideal index number

Products 2009 2010 poqo poq1 p1qo p1q1

po q0 p1 q1 400
20
p 4 54 10 40 216 160 540 64
484
q 2545 10 10 20

r 3 10 8 8 30 24 80

Total 256 194 640

Here, Σp0q0 = 256, Σp0q1 = 194, Σp1q0 = 640, Σp1q1 = 484
Fisher's index numner

01( ) = √ 10 00 × 1 1 × 100
0 1

= √265460 × 484 × 100
194

= 249.74

QN. 19) (2071, old) Compute the suitable index number from the following data:

Product X Product Y Product Z
Year
Price Qty. Price Qty.
Price Qty.

2008 4 50 3 10 2 5

2010 10 40 8 844

If the income of the family in 2008 was Rs.10,000 what should be the income of the same family

in 2010 so that the living standard should be same?

Solution: Calculation of Fisher's ideal index number

Products 2008 2010 poqo poq1 p1qo p1q1

po q0 p1 q1

X 4 50 10 40 200 160 500 400

Y 3 10 8 8 30 24 80 64

Z 2544 10 8 20 16

Total 240 192 600 480

216

Here, Σp0q0 = 240, Σp0q1 = 192, Σp1q0 = 600, Σp1q1 = 480
Fisher's index number

01( ) = √ 10 00 × 1 1 × 100
0 1

= √264000 × 480 × 100
192

= 250

Required income of the family in 2010 in order that the living standard remain same is

250 × . 10,000 = . 25,000
100

QN.20) (2070) Construct the real wages index number from the following:

Year wages in Rs. Price index

2005 1400 100

2006 1500 140

2007 1800 160

2008 2000 190

2009 2200 210

2010 2500 220

Solution: Computation of real wages index

Year Wage Index Real wages (Rs.) Real wages index (Rs.)
(W) Number(I)
2005 1400 1400 100
2006 1500 100 1071.43 76.53
2007 1800 140 80.36
2008 2000 160 1125 75.19
2009 2200 190 1052.63 74.83
2010 2500 210 1047.62 81.17
220 1136.36

Working note:

Real wages = wages × 100
Index Number

Real wage of 2005 = 1400 × 100 = 1400
100

217

Real wage of 2006 = 1500 × 100 = 1071.43
140

Real wage of 2007 = 1800 × 100 = 1125 and so on
160

Real wages index = Real wages of current year × 100
Real wages of base year

Real wages index of 2005 = 1400 × 100 = 100
1400

Real wages index of 2006 = 1071.43 × 100 = 76.53
1400

Real wages index of 2007 = 1125 × 100 = 80.36 and so on
1400

QN.21) (2070) Construct the cost of living index of the following:

Items Percentage of expenditure Price relatives

A 25 125

B 35 140

C- 160

D 10 130

E 15 190

Solution: Calculation of cost of living index number

Items % Expenditure (W) Price relatives (P) PW

A 25 125 3125

B 35 140 4900

C 15 160 2400

D 10 130 1300

E 15 190 2850

ΣW = 100 ΣPW = 14575

Now, cost of living index number

= = 14575 = 145.75%
100

Analytic Answer Question
QN.22) (2072) What is index number? Why they are called economic barometer? You are
required to prove from the following data that Fisher's index number is an ideal index number.

218

Commodities 2012 2013

Price/Unit Expenditure Price/Unit Expenditure

A 50 600 60 840

B 40 840 80 2400

C 30 900 30 1500

D 20 600 20 1000

Solution: Computation of time reversal test and factor reversal test

Commodities 2012 2013 poqo poq1 p1qo p1q1

po q0 p1 q1 840
2400
A 50 12 60 14 600 700 720 1500
1000
B 40 21 80 30 840 1200 1680 5740

C 30 30 30 50 900 1500 900

D 20 30 20 50 600 1000 600

Total 2940 4400 3900

Here, Σp0q0 = 2940, Σp0q1 = 4400, Σp1q0 = 3900, Σp1q1 = 5740
i) Time reversal test

01( ) × 10( ) = √ 10 00 × 1 1 × 0 1 × 0 0
0 1 1 1 1 0

= √32994000 × 5740 × 4400 × 2940 = 1
4400 5740 3900

Hence, Fisher ideal index satisfies the time reversal test.

ii) Factor reversal test

P01(F) × Qo1(F) = √ΣΣpp10qq00 × Σp1q1 × Σq1p0 × Σq1p1
Σp0q1 Σq0p0 Σq0p1

= √32990400 × 5740 × 4400 × 5740
4400 2940 3900

= 5740
2940

V01 = Σp1q1 = 5740
Σp0q0 2940

∴ P01(F) × Qo1(F) = V01

219

Hence, Fisher's ideal index satisfies the factor reversal test.
As the Fisher's index number satisfies both time reversal test and factor reversal test, so it is

known as ideal index number.

QN.23) (2071) Using the following information, examine whether time reversal test and factor

reversal tests are satisfies by Paasch's and Fisher's index number formula or not.

Commodities Base year Current year

Price Quantity Price Quantity

A 10 6 12 10

B 12 12 15 15

C 15 25 18 30

D 20 40 25 40

E 13 17 20 15

Also mention the features of Fisher's and Paasch's method.
Solution: Computation of time reversal test and factor reversal test

Commodities Base year Current year poqo poq1 p1qo p1q1
po q0 p1 q1
100 72 120
A 10 6 12 10 60 180 180 225
450 450 540
B 12 12 15 15 144 800 1000 1000
195 340 300
C 15 25 18 30 375 1725 2042 2185

D 20 40 25 40 800

E 13 17 20 15 221

Total 1600

Here, Σp0q0 = 1600, Σp0q1 = 1725, Σp1q0 = 2042, Σp1q1 = 2185

a) For Paasch's formula
Time reversal test

P01(P) × P10(P) = Σp1q1 × Σp0q0 = 2185 × 1600 = 0.99 ≠ 1
Σp0q1 Σp1q0 1725 2042

Factor reversal test Σp1q1 Σq1p1 2185 2185
Σp0q1 Σq0p1 1725 2042
P01(P) × Qo1(P) = × = × = 1.36

V01(P) = Σp1q1 = 2185 = 1.27
Σp0q1 1725
P01(P) × Qo1(P) ≠ V01(P)

220

Therefore, Paasch's formula does not satisfies the time reversal test and facror reversal test.
b) For Fisher's formula
Time reversal test

P01(F) × P10(F) = √ΣΣpp10qq00 × Σp1q1 × Σp0q1 × Σp0q0
Σp0q1 Σp1q1 Σp1q0

= √12064002 × 2185 × 1725 × 1600 = 1
1725 2185 2042

Factor reversal test

P01(F) × Qo1(F) = √ΣΣpp10qq00 × Σp1q1 × Σq1p0 × Σq1p1
Σp0q1 Σq0p0 Σq0p1

= √21600402 × 2185 × 1725 × 2185
1725 1600 1725

= 2185
1600

01 = 1 1 = 2185
0 0 1600

∴ 01( ) × 1( ) = 01

Hence, Fisher's index number satisfies both the time reversal test and factor reversal
test.

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Chapter - 9

Probability

Brief Question Answer
QN. 1) (2078) If P(A) = 0.6, P(B) = 0.56, and P(A∪B) = 0.75, find P(A∩B), where A and

B are disjoint events.
Solution: Here, P(A) = 0.6, P(B) = 0.56, P(A∪B) = 0.75, and P(A∩B) =?
Now we have P(A∩B) = P(A) + P(B) – P(A∪B) =0.6 + 0.56 – 0.75 = 0.41

QN 2) (2077) In a single throw of two dice, find the probability that the sum of two numbers

on two faces is 7 or 11.
Solution: Here sample space S = {(1, 1), (1, 2), (1, 3), …………..., (6, 6)}

Let the event of sum 7 and sum 11 be denoted by A and B respectively.

A = {(3, 4), (4, 3), (2, 5), (5, 2), (1, 6), (6, 1)}

B = {(5, 6), (6, 5)}
A ∪ B = {(3, 4), (4, 3), (2, 5), (5, 2), (1, 6), (6, 1), (5, 6), (6, 5)}
n(A∪B) = 8

n(S) = 36

Now, Probability that the sum of two numbers on two faces is 7 or 11 is

( ∪ ) = ( ∪ ) = 8 = 2
( ) 36 9

QN 3) (2076) State the addition theorem of probability in case of two events A and B.

Solution: Addition theorem of probability for mutually exclusive events
( ∪ ) = ( ) + ( )

Addition theorem of probability for non-mutually exclusive events
( ∪ ) = ( ) + ( ) − ( ∩ )

QN 4) (2075) When two dice is rolled, what is the probability of getting sum of two faces is

10?
Solution: Here sample space S = {(1, 1), (1, 2), (1, 3), …………..., (6, 6)}

Let the event of sum 10 be denoted by A.

A = {(4, 6), (6, 4), (5, 5)}

Now, n(S) = 36 and n(A) = 3

Probability that the sum of two numbers on two faces is 10 is

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( ) = ( ) = 3 = 1
( ) 36 12

QN 5) (2074) Given P(A) = 3/14, P(B) = 1/6, P(C) = 1/3, P(A∩C) = 1/7, find the

probabilities of P(A/C) and P(C/A).

Solution: Now we have

P(A/C) = P(A ∩ C) = 1/7 = 3
P(C) 1/3 7

and P(C/A) = P(A ∩ C) = 1/7 = 2
P(A) 3/14 3

QN 6) (2073) Two events A and B are statically dependent. If P(A) = 0.35, P(B) = 0.20,

P(AorB) = 0.45, find the probability that a) neither A nor B will occur

b) B will occur, given that A has occurred.

Solution: Now,

a) neither A nor B will occur = 1 – P(AorB) = 1- 0.45 = 0.55
( ∪ ) = ( ) + ( ) − ( ∩ )

or 0.45 = 0.35 + 0.20 – P(A∩B)
∴ P(A∩B) = 0.10

b) B will occur, given that A has occurred is

( / ) = ( ∩ ) = 0.10 = 0.29
( ) 0.35

QN 7) (2072) The probability that a manufacturer will produce "brand X' product is 0.13, the

probability that he will produce 'brand Y' produce is 0.28 and the probability that he will

produce both brands is 0.60. What is the probability that manufacture who has produced

'brand Y' will also has produced 'brand X'?
Solution: Here given P(X) = 0.13 and P(Y) = 0.28 and P(X∩Y) = 0.60

Now, probability that manufacture who has produced 'brand Y' will also has produced

'brand X' is

P(Y/X) = P(X ∩ Y) = 0.060 = 0.2143
P(Y) 0.28

QN 8) (2072) A box contains 4 black balls and 4 red balls. If 2 balls are drawn at random,

find the probability that they are black.

Solution: Let the event of black ball and the red ball be denoted by B and R respectively. Here

given n(B) = 4, n(R) = 4, and total number of balls n(S) = 8

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Now, probability of drawing both black balls is

4 × 3 = 0.2143
8 7

QN 9) (2071) Out of 9 candidates 6 men and 3 women apply for two vacancies of a

manufacturing company, what is the probability that one man and women are selected in

that vacancies?

Solution: Let the event of man and woman be denoted by A and B respectively.

Here, n(A) = 6, n(B) = 3 and n(S) = 9

Now, probability of selecting one man and one woman is

6 1 × 3 1 = 6×3 = 1 = 0.5
9 2 36 2

Descriptive Question Answer

QN. 10) The probability that a boy will get a scholarship is 0.9 and that a girl will get is 0.8. What

is the probability that at least one of them will get the scholarship?

Solution: Let A and B denotes the event of boy and girl respectively. Here given

P(A) = 0.9 and P(B) = 0.8
Now, P(A∪B) = P(A) + P(B) – P(A∩B) =0.9 + 0.8 – 0.9×0.8 = 1.7 – 0.72 = 0.98

Therefore, the probability that at least one will get the scholarship is 0.98 i.e. 98%.

QN 11) (2077) The record of 500 workers of certain factory according to their sex and skills are

given below:

Male Female Total

Skilled 250 150 400

un-skilled 50 50 100

Total 300 200 500

If a worker is selected at random, what is the probability that the selected worker is
i) Female, if it is known that she is un-skilled?
ii) Skilled, if it is known that he is male?
Solution: Let the events
A = male worker
B = female worker
C = skilled worker
D = un-skilled worker

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i) The probability of female, if it is known that she is un-skilled

P(B/D) = P(B ∩ D) = 50/500 = 50 = 0.5
P(D) 100/500 100

ii) The probability of skilled, if it is known that he is male

P(C/A) = P(A ∩ C) = 250/300 = 250 = 0.83
P(A) 300/500 300

QN 12) (2076) Three candidates 'X', 'Y' and 'Z' are applying for the post of Vice-chancellor of

T.U. Their probabilities of selection is the post are 0.2, 0.5, and 0.3 respectively. The probability

of introducing of semester system in Bachelor level by 'X' is 0.7 and the corresponding

probabilities by 'Y' and 'Z' are 0.3 and 0.5. What is the probability that the semester system will

introduced by now appointed vice-chancellor?

Solution: Let the event of introducing semester system be denoted by S, then here given

P(X) = 0.2, P(Y) = 0.5, P(Z) = 0.3, P(S/A) = 0.7, P(S/B) = 0.3, P(S/C) = 0.5

Now, the probability that the semester system will introduced by now appointed vice-chancellor is

P(S) = P(X)P(S/A) + P(Y)P(S/Y) + P(Z)P(S/Z)

= 0.2×0.7 + 0.5×0.3 + 0.3×0.5

= 0.14 + 0.15 + 0.15

= 0.44

Therefore, probability of introducing semester system is 44%.

QN 13) (2075) A bag contains 30 balls numbered from 1 to 30. One ball is drawn at random, what

is the probability that the ball drawn will be multiple of 3 or 7?

Solution: Let A = event of multiple of 3 and B = event of multiple of 7

Here, S = {1, 2, 3, …………30}

A = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}

B = {7, 14, 21, 28}

A∩B = {21}
A∪B = {3, 6, 7, 9, 12, 14, 15, 18, 21, 24, 27, 28, 30}

n(S) = 30
n(A∪B) = 13

Now, probability of multiple of 3 or 7 is

P(A ∪ B) = n(A ∪ B) = 13
n(S) 30

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QN 14) (2072) Assuming that a factory has two machines M1 and M2. past records shows that

machine M1 produce 70% of the items. Further, 5% and 1% of the items produced by the machine

M1 and M2 respectively were defective. If a item is selected at random what is the probability that

it was produced by machine M2?

Solution: Let the defective item be denoted by D, then

P(M1) = 0.7, P(M2) = 0.3, P(D/M1) = 0.05, P(D/M2) = 0.01

Now, the probability of defected item produced by machine M2 is

P(M2/D) = P(M1) × P(M2) × P(D/M2) P(D/M2)
P(D/M1) + P(M2) ×

= 0.7 × 0.3 × 0.01 0.01
0.05 + 0.3 ×

= 0.003
0.035 + 0.003

= 0.003
0.038

= 0.08

QN 15) (2072) A factory has two machines M1 and M2. machine M1 produce 60% of the total

output and machine M2 produces 40% of the total output. If 8% of output M1 and 6% of output of

machine M2 respective were defective. If a defective item is selected at random from lot and

found to be defective. What is the probability that it was produced from machine M2?

Solution: Let the defective item be denoted by D, then

P(M1) = 0.6, P(M2) = 0.4, P(D/M1) = 0.08, P(D/M2) = 0.06

Now, the probability of defected item produced by machine M2 is

P(M2/D) = P(M1) × P(M2) × P(D/M2) P(D/M2)
P(D/M1) + P(M2) ×

= 0.6 × 0.4 × 0.24 0.06
0.08 + 0.4 ×

= 0.003
0.048 + 0.024

= 0.024
0.072

= 0.33

QN 16) (2071) A cloth store purchase 65% of the ready-made clothes from supplier S1 and 35%

are from supplier S2. The past record shows that the percentage of demand clothes purchased from

supplier S1 and S2 are 2% and 5% respectively. If a ready-made cloth is drawn at random from the

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total purchase and found to be damaged. What is the probability that the damage cloth is

purchased from supplier S1?

Solution: Let the defective item be denoted by D, then

P(S1) = 0.65, P(S2) = 0.35, P(D/S1) = 0.02, P(D/S2) = 0.05

Now, the probability of defected item produced by machine M2 is

P(S2/D) = P(S1) × P(S1) × P(D/S1) P(D/S2)
P(D/S1) + P(S2) ×

= 0.65 × 0.65 × 0.02 × 0.05
0.02 + 0.35

= 0.013
0.013 + 0.0175

= 0.013
0.0305

= 0.4262

QN 17) (2071, old) An urn contains 9 red, 7 white, and 4 black balls. A ball is drawn at random.

Find the probability that the ball drawn is a) red ball and b) not red ball.

Solution: Let number of red balls n(B) = 9, number of white balls n(W) = 7, number of black balls

n(B) = 4, and total numbers of balls n(S) = 20

Now, a) probability of red ball is

P(R) = n(R) = 9
n(S) 20

b) Probability of not red ball is

1 – P(R) = 1 − 9 = 11
20 20

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Chapter - 10

Sampling and Estimation

Brief Question Answer

QN 1) (2073) What do you mean by sampling? Discuss the various types of sampling techniques.

Answer: A small portion chosen from the population for studying its properties is called a

sampling and the number of units in the sample is known as the sample size. The method of

selecting for study a small potion of the population to draw conclusion about the characteristics of

the population (universe) is known as sampling. Sampling may be defined as the selection of part

of the population on the basis of which a judgement or inference about the universe is made. Is

say, rather, taking small portion of the population and considering it to be representative. Thus,

sampling procedure provide generalization on the basis of a relatively small proportion of the

population.

Sampling method can be classified as the following two parts:

a) Random sampling (Probability sampling)

b) Non-random sampling (non-probability sampling).

QN.2) (2072) Differentiate between point and interval estimate.

Answer:A single number that is used to estimate an unknown population parameter is called a

point estimate. A single value of the sample statistics that is used to estimate a population
parameter is known as estimator. For example, a sample mean (X̅ ) and a sample standard
deviation (S) are point estimators of the population mean (μ) and population standard deviation
(σ) respectively. If we wish to estimate the population mean μ, the average percentage mark of all
the students of 12 class students of 10 + 2 college of Sindhuli. We take a random sample of n
students and calculate the sample mean x̅ ; then x̅ is a point estimator.

A range of values used to estimate a population parameter is called an interval estimate.
There are two ways for indicating error by the extent of its range and by the probability of the true
population parameter lying within that range. In this case, the decision maker has a better idea of
the reliability of his estimate. For example, if the average salary per month of all professors of TU
lies between Rs.9000 to Rs.15000, then TU management can estimate the exactly salary that fall
within this interval, such estimate is an interval estimate.

Descriptive Question Answer
QN.3) (2012) Differentiate between “census” and “sampling” method of the data collection. Why
sampling method is suitable to collect from the large population.

Answer: If collect the information from each and every unit of the population associated
with the subject matter of the inquiry. A census is the procedure of systematically calculating,
acquiring and recording information about the members of a given population. This term is used

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mostly in connection with national population and housing censuses; other common censuses
include the census of agriculture, and other censuses such as the traditional culture, business,
supplies, and traffic censuses. A small portion chosen from the population for studying its
properties is called a sampling and the number of units in the sample is known as the sample size.
The method of selecting for study a small portion of the population to draw conclusion about the
characteristics of the population (universe) is known as sampling. Sampling may be defined as the
selection of part of the population on the basis of which a judgement or inference about the
universe is made.

Sampling method is suitable to collect from the large population because:
i) In sampling process only, a small part of the population is enumerated, it is much more

economical than the census method in terms of time, money, and manpower.
ii) Time, money and effort are saves in the collecting data as well as processing them. So,

this method is quite suitable for investigating the ever-changing social and economic
problems of the modern word.
iii) Since time and money are saved, we can employ exports or trained persons, provided
better supervision and use sophisticated techniques and equipment. Thus, the
possibility of error is minimized and the results obtained are generally more reliable
than what we get from the complete census.
iv) By using sampling technique, we can also check the results obtained from a census
inquiry.
v) Sampling is the only suitable method if the universe is too large or infinite or
hypothetical or if the units are totally destroyed in the course of testing.
vi) A sample could be more accurate than a (attempted) census if the fact of the exercise
being a census increases the bias from non-sampling error. This could come about, for
example, if the census generates an adverse political campaign advocating non-
response (something less likely to happen to a sample).
QN. 4) (2063) What is sampling? Discuss various sampling techniques with their merits and
demerits.
Answer: A small portion chosen from the population for studying its properties is called a
sampling and the number of units in the sample is known as the sample size. The method of
selecting for study a small portion of the population to draw conclusion about the characteristics
of the population (universe) is known as sampling. Sampling may be defined as the selection of
part of the population on the basis of which a judgement or inference about the universe is made.
Sampling methods can be classified into two board categories:
I. Random sampling (Probability Sampling)
II. Non-random sampling (Non-probability sampling)
I) Random Sampling: Random sampling is that method of drawing a portion

(sample) of a population so that all possible samples of fixed size (n) have the
same probability of being selected. In probability sampling each sample unit
has an equal chance of being selected. Some of the important methods of
selecting a random sampling are as follows

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a) Simple random sampling: In simple random sampling, the units are selected in
such a way that each and every unit of the population has an equal chance of being
selected. The selection of the item depends on chance and not on the discretion or
judgement of the investigator. If the unit selected in any draw is not replaced in the
population before making the next draw, then it is known as sampling random
sampling without replacement and if it is replace back before making the next
draw, then the sampling is known as simple random sampling with replacement.
Merits
i) As the selection of item depends only chance, it is free from personal bias
or prejudice of the investigator.
ii) It is relatively cheap and the sample can be easily drawn in a short period of
time.
iii) As the method is based on the theory of probability, sampling error can also
be calculated.

Demerits
i) The method can be used only when we have a complete list of the population
units. Such a list is not always available.
ii) The method becomes costly and time consuming when the field of inquiry is
very large and scattered.
iii) The results will be misleading if the size of the sample is not sufficiently large.

b) Stratified random sampling:
In this method the population is divided into various groups or strata on the basis of certain
characteristics so that each stratum is as homogeneous as possible. Then a sample is selected at
random from each stratum. For example, we may have to study the examination results of 3000
students of a campus. We fill first divide the students into different faculties are 1000, 1400, and
1600 respectively and we want to select 10% of each for the samples, we will select random
samples consisting of 100, 140, and 160 students from them.
Merits

i) The investigator first his judgement to divide the proportion into various strata. Then
he selects the samples by random method. So this method benefits of both the
judgement method and simple random method. It also balance the uncertainty of
random sampling against the bias of deliberate selection.

ii) As different strata are taken into account separately, each group of the population will
be adequately represented in the samples. The samples are therefore more
representative of the universe than simple random samples.

iii) Since the stratum consists of homogeneous units, the results obtained from this method
are more accurate than those given by simple random samples.

iv) We can select the units from different stratum in such a way that they are localized
efforts than the simple random method.

Demerits
i) The results will not be reliable if each stratum does not contain homogeneous units.
ii) The method will not be effective if different strata are overlapping.

230

c) Systematic Sampling
A systematic random sampling is such type of sampling in which only the first sample unit is
selected at random and the remaining units are automatically selected at fixed equal intervals from
one another. This method may be successfully used when the complete and up-to-date list of the
sampling units is available. The items are first arranged in numerical, alphabetical or some other
orders. Then they are numbers serially. If the total number of items is N and the sample size is n,
we divide N by n and find the sample interval.
Merits

i) This method is very simple and convenient.
ii) Compared to simple or stratified random sampling, this sample design saves quite a lot

of money, time and effort.
iii) Most of the results are also satisfactory. In fact, the method is more efficient then

simple random sampling if the list is complete and the units are arranged serially at
random.
Demerits
i) If the list of the units are not arranged in random order, the result may be misleading.
ii) In some cases the sample will not represent the universe.
d) Multistage Sampling:
This is also the case of random sampling. But sampling is done in various stages. At the first
stage, the universe is divided into large sample units and the sample is selected at random from
them. At the second stage, the sample selected at first stage are divided again into smaller units
from which a further random sample is taken. Sampling in any other stages may be done in the
same way till we get the required sample.
Merits
i) This method is quite convenient when the area of investigation is vast.
ii) As the sample size is reduced at each stage, the method saves time and cost.
iii) The method is also more flexible than other method.
Demerits
i) If sufficient care is taken in the different stages of sampling, the sample will not be true
representative of the universe.
ii) A multistage sample is generally less accurate than another sample of the same size
which has been selected by suitable single stage method.
e) Cluster Sampling
A cluster sampling is a method of random sampling in which the population is divided into
groups, called cluster, in such a way that the characteristics within the clusters are heterogeneous
and between the cluster are homogeneous so that the number of sampling units in each cluster
should be approximately same and then a simple random sampling of these clusters is selected.
Merits
i) Elements (units) selected be well-designed cluster sampling procedures in easier,
faster, cheaper and more convenient than of simple random sampling and stratified
random sampling.

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ii) It is useful when the population under the study is infinity, where a list of population
does not exists, when the geographical distribution of units is scattered or when
sampling of individual units is not convenient for several administrative reasons.

iii) This method of sampling is more economical when the cost of measuring an individual
is relatively small and cost of researching him relatively large.

Demerits
i) The sampling efficient of cluster sampling is likely to decrease with the increase in
cluster size.
ii) The cluster sampling is not be recommended for taking samples from private
residential houses, business and industrial complexes due to widely varying the number
of persons or households.
iii) For a given total number of sampling units, cluster sampling is less efficient than
sampling of individual units from the point of sampling variance.
iv) Estimates based on cluster samples are usually more reliable per unit cost.

QN 5) (2061) Suggest suitable sampling method, if you are interested in studying the ‘popularity
of the entertainment center in Bhrikuti Mandap’.
Answer: A small portion chosen from the population for studying its properties is called a
sampling and the number of units in the sample is known as the sample size. The method of
selecting for study a small portion of the population to draw conclusion about the characteristics
of the population (universe) is known as sampling. Sampling may be defined as the selection of
part of the population on the basis of which a judgement or inference about the universe is made.

The method of selecting samples from populations is of fundamental importance in the
theory of sampling and depends very much on the objectives and scope of inquiry, nature of
universe (population), the size of the sample, available resources etc. For the study of the
popularity of the entertainment center in Bhrikuti Mandap the simple random sampling method is
suitable because for this study we can the take information from any person who are in used to go
to the Bhirkuti Mandap. For this study we can choose any person randomly, it need not to choose
any particular person. In the simple random sampling, the units are selected in such a way that
each and every units of the population have equal chance of being selected. The selection of an
item depends on chance and not on the discretion or judgement of investigator.

The simple random is easier to form representative groups from an overall population. If
we select the sample randomly we can choose all people from different groups, like gender, age,
profession, occupation etc. To use the simple random sampling method, it need not to have any
specific skill, experience. This process can also remove the classification errors that can occur in
other forms of information collection. Simple random sampling offers researchers an opportunity
to perform data analysis and a way that creates a lower margin of error within the information
collected. The goal of collecting information in this way is to provide an unbiased representation
of the entire group. In the case of conduction of study by the less experience person like me this
method may yields better results so I use the sampling random sampling method.

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Chapter - 11

Quantitative Analysis

Brief Question Answer

QN 1) (2076) Prepare regret or loss table from the following pay-off table:

Demands (Units) Decision Alternatives

15 15 16 17 18

16 150 120 90 60

17 150 160 130 100

18 150 160 170 140

150 160 170 180

Solution: Regret or loss table 15 Decision Alternatives 18
Demands (Units) 0 16 17 90
15 10 30 60 60
16 20 0 30 40
17 30 10 0 0
18 20 10

QN 2) (2075) Prepare regret from the given conditional profit table:

Demands (Units) 15 Decision Alternatives 18
15 150 16 17 60
16 150 120 90 100
17 150 160 130 140
18 150 160 170 180
160 170

Solution: Regret or loss table Decision Alternatives
Demands (Units)
233

15 15 16 17 18

16 0 30 60 90

17 10 0 30 60

18 20 10 0 40

30 20 10 0

QN 3) (2074) Following are the pay-off table rooms available in the hotel with different

types of rooms

Strategies Demand

(Room) Low Medium High

R1 25 35 50

R2 10 40 70

R3 30 20 100

Find the best strategy if you were i) A complete optimistic ii) A complete pessimistic

Solution: Payoff Table

Strategies Demand Maximum Minimum

(Room) Low Medium High payoff payoff

R1 25 35 50 50 25

R2 10 40 70 70 10

R3 30 20 100 100 20

i) According to optimistic criterion one should choose strategy R3.

ii) According to optimistic criterion one should choose strategy R1.
QN. 4) (2071) From the following pay-off table, prepare regret table

N1 N2 N3

S1 200 50 40

S2 100 60 30

S3 40 30 10

Solution: Regret Table

N1 N2 N3

S1 0 10 0

S2 100 0 10

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S3 160 30 30

Descriptive Question Answer

QN. 5) The following table is the conditional payoff table:

Strategy State of nature

N1(10) N2(11) S3(13) S4(14) S5(15)

S1(10) 400 400 400 400 400

S2 (11) 380 440 440 440 440

S3(12) 360 420 480 480 480

S4(13) 340 400 460 520 520

S5(14) 320 380 440 500 560

What will be your decision if:

ii) Maximax criteria is used

iii) Maximin criteria is used

iv) Minimax regret criteria is used.

Solution: Payoff Table

Strategy State of nature Maximum Minimum

N1(10) N2(11) S3(13) S4(14) S5(15) Payoff Payoff

S1(10) 400 400 400 400 400 400 400

S2 (11) 380 440 440 440 440 440 380

S3(12) 360 420 480 480 480 480 360

S4(13) 340 400 460 520 520 520 340

S5(14) 320 380 440 500 560 560 320

According to Maximax criteria we should choose strategy S5(14) and according to Maximin

criteria strategy S1(10) should choose.

Regret Table

Strategy State of nature Maximum

N1(10) N2(11) S3(13) S4(14) S5(15) regret

S1(10) 0 40 80 120 160 160

S2 (11) 20 0 40 80 120 120

S3(12) 40 20 0 40 80 80

235

S4(13) 60 40 20 0 40 60

S5(14) 80 60 40 20 0 80

According to Minimax Regret approach one should choose strategy S4(13).

QN. 6) (2077) From the following pay-off table, give the decision according to

i) Maximax approach ii) Maximin approach iii) Minimax Regret approach

State of nature Strategies

S1 S2 S3

N1 200 100 40

N2 50 60 30

N3 40 30 10

Solution: Pay-off Table

Strategies State of nature Maximum Minimum

N1 N2 N3 payoff payoff

S1 200 50 40 200 40

S2 100 60 30 100 30

S3 40 30 10 40 10

i) According to both Maximax approach and Maximin approach one should strategy

S1.

ii) Regret Table

Strategies State of nature Maximum

N1 N2 N3 Regret

S1 0 10 0 10

S2 100 0 10 100

S3 160 30 30 160

According to Minimax Regret approach one should choose strategy S1.
QN. 7) (2073) Given the following information
Prepare payoff table and regret table
Units demands: 12, 13, 14, 15, 16, 17
Selling price per unit = Rs. 200
and cost price per unit = Rs. 100

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You are required to give decision on the basis of a) Maximax criteria b) Minimax regret criteria

and c) Maximin criteria

Solution: Since the seller would not buy less than 12 units and not more than 17 units of,
there are six strategies (alternatives) such as 1 = 12, 2 = 13, 3 = 14, 4 = 15, 5 =
16, 6 = 17 Similarly there are six possible state of nature i.e. the demand such as 1 =
12, 2 = 13, 3 = 14, 4 = 15, 5 = 15, 6 = 17.
Here, SP= Rs.200, CP= Rs. 100, and Marginal Profit = Rs.100
Now, Payoff
P11 = payoff when S1 = 12 & N1 = 12

Payoff = Rs. 100 × 12 = Rs. 1200
P12 = payoff when S1 = 12 & N2 = 13

Payoff = Rs. 100 × 12 = Rs. 1200
P13 = payoff when S1 = 12 & N3 = 14

Payoff = Rs. 100 × 12 = Rs. 1200 and so on.

Payoff Table

Strategies State of nature (demand of orange) MaximumMinimum

Payoff Payoff

N1=12 N2=13 N3=14 N4=15N5 N6=17
=16

S1=12 1200 1200 1200 1200 1200 1200 1200 1200

S2= 13 1100 1300 1300 1300 1300 1300 1300 1100

S3=14 1000 1200 1400 1400 1400 1400 1400 1000

S4 = 15 900 1100 1300 1500 1500 1500 1500 900

S5 = 16 800 1000 1200 1400 1600 1600 1600 800

S6 = 17 700 900 1100 1300 1500 1700 1700 700

Loss table or Regret table

Strategies State of nature (demand of orange) Maximum

Regret

N1=12 N2=13 N3=14 N4=15N5=16N6=17

S1=12 0 100 200 300 400 500 500

S2= 13 100 0 100 200 300 400 400

S3=14 200 100 0 100 200 300 200

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S4 = 15 300 200 100 0 100 200 300
S5 = 16 400 300 200 100 0 100 400
S6 = 17 500 400 300 200 100 0 500

Therefore, the decision based on

a) Maximax criteria is stock 17 units

b) Minimax regret criteria stock 14 units and

c) Maximin criteria stock 12 units.

QN. 8) (2072) A coka-cola company buys the buys the bottles for Rs6 and sells them for Rs.10
each. All the bottles leftover are worthless. His daily sales of cold drinks is never less
than 15 and not more than 20. Prepare pay-off table and loss table. What will be the
distributer's decision if the certain adopted be
i)maxi-max ii) maxi-min iii)mini-max regret

Solution: Since the company would not sells less than 15 bottles and not more than 20
bottles of Coca-Cola, it has six strategies (alternatives) such as 1 = 15, 2 = 16, 3 =
17, 4 = 18, 5 = 19, 6 = 20. Similarly there are three possible state of nature i.e.
the demand of coca-cola such as 1 = 15, 2 = 16, 3 = 17, 4 = 18, 5 = 19, 6 =
20.
Here, SP= Rs.10, CP= Rs. 6, MP = Rs.4

Now, payoff table
State of nature (demand of orange)

Strategies N1=15 N2=16 N3=17 N4=18 N5=19 N6=20 Maximum Minimum
payoff payoff

S1=15 60 60 60 60 60 60 60 60

S2=16 54 64 64 64 64 64 64 54

S3=17 48 58 68 68 68 68 68 48

S4=18 42 52 62 72 72 72 72 42

S5=19 36 46 56 66 76 76 76 36

S6=20 30 40 50 60 70 80 80 30

i. Decision according to maxi-max criteria best alternative is S6, i.e. stock = 20 units.
ii. Decision according to maxi-min criteria the best alternative is S1, i.e. stock = 15 units.

Strategies loss or regret table N4=18 N5=19 N6=20 Maximum
N1=15 N2=16 N3=17 value

238

S1=15 0 4 8 12 16 20 20

S2=16 6 0 4 8 12 16 16

S3=17 12 6 0 4 8 12 12

S4=18 18 12 6 0 4 8 18

S5=19 24 18 12 6 0 4 24

S6=20 30 24 18 12 6 0 30

iii) According to Mini-max criteria the best alternative is S3, i.e. stock = 17 units.

QN. 9) (2072, old) From the following payoff matrix, give decision based on a) maximax

approach b) maximin approach

Event A B C

Action

1 80 50 40

2 100 50 30

3 40 50 100

Solution:

Payoff Table

Event A B C Maximum Minimum

Action Payoff Payoff

1 80 50 40 80 40

2 100 50 30 100 30

3 40 50 110 110 40

Therefore, decision based on a) maximax approach we should choose action 3
and b) maximin approach we should choose action 1 or 3.
QN. 10) (2070) A merchant of Bishal Bazar Co. buys an article at the rate of Rs.300 and
sells them at the rate of Rs.500. The merchant knows that he cannot sell more than 50
articles in a day and minimum sales would no be no less than 45. Assuming that, unsold
articles do not have any salvage value, prepare payoff table and regret table and give
decision according to a) optimistic and pessimist approach b) Minimax regret approach.

239

Solution: Since the company would not sells less than 45 units and not more than 50

units, it has six strategies (alternatives) such as
1 = 45, 2 = 46, 3 = 47, 4 = 48, 5 = 49, 6 = 50.
Similarly, there are three possible state of nature i.e. the demand such as 1 = 45, 2 =
46, 3 = 47, 4 = 48, 5 = 49, 6 = 50.
Here, SP= Rs.500, CP= Rs. 300, Marginal Profit = Rs.200

Now, payoff table
State of nature (demand of orange)

Strategies N1=45 N2=46 N3=47 N4=48 N5=49 N6=50 Maximum Minimum
payoff payoff

S1=45 9000 9000 9000 9000 9000 9000 9000 9000

S2=46 8700 9200 9200 9200 9200 9200 9200 8700

S3=47 8400 8900 9400 9400 9400 9400 9400 8400

S4=48 8100 8600 9100 9600 9600 9600 9600 8100

S5=49 7800 8300 8800 9300 9800 9800 9800 7800

S6=50 7500 8000 8500 9000 9500 10000 10000 7500

i) Decision according to maxi-max criteria maximum payoff is is Rs.10,000, so the
merchant choose alternative S6, i.e. stock = 50 units.

ii) Decision according to maxi-min the maximum payoff is Rs. 9000, so the merchant
choose the alternative is S1, i.e. stock = 45 units.

Strategies Loss or Regret Table N4=48 N5=49 N6=50 Maximum
N1=45 N2=46 N3=47 value
S1=45
S2=46 0 200 400 600 800 1000 1000
S3=47
S4=48 300 0 200 400 600 800 800
S5=49
S6=50 600 300 0 200 400 600 600

900 600 300 0 200 400 900

1200 900 600 300 0 200 1200

1500 1200 900 600 300 0 1500

iii) According to Mini-max criteria the minimum loss is Rs. 600, so the merchant
choose alternative is S3, i.e. stock = 47 units.

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Analytical Question Answer

QN. 11) (2076) A beer distribution buys keys for Rs.30 each and sells then for Rs.50 each. All
the kegs left at the end of the day are worthless where salvage value is zero. The following
table gives the distribution of sales and corresponding probabilities.

Keys sold 10 11 12 13 14

Probability 0.10 0.15 0.20 0.25 0.30

a. Construct a pay-off table.
b. Calculate the expected monetary value(EMV) for each strategy.
c. Find the optimal quantity that can maximize the expected profit.
d. Calculate expected profit with perfect information (EPPI)
e. Calculate the expected value of the perfect information (EVPI)
Solution:
a. Payoff Table

Since the company would not sells less than 10 bottles and not more than 14 bottles of
keys, it has six strategies (alternatives) such as 1 = 10, 2 = 11, 3 = 12, 4 =
13, 5 = 14. Similarly there are three possible state of nature i.e. the demand of keys
such as 1 = 10, 2 = 11, 3 = 12, 4 = 13, 5 = 14.
Here, SP= Rs.50, CP= Rs. 30, MP = Rs.20

Strategies Payoff Table
State of Nature (demand of keys)
S1=10
S2=11 N1=10 N2=11 N3=12 N4=13 N5=14
S3=12 200 200 200 200 200
S4=13 170 220 220 220 220
S5=14 140 190 240 240 240
110 160 210 260 260
80 130 180 230 280

b. Expected monetary value (EMV) for each strategy.

State Prob. Payoff Values (Rs.) Expected Profit (Rs.)
of (1) Decision Alternatives
Nature Decision Alternatives

N1=10 S1=10 S2=11 S3=12 S4=13 S5=14 S1=10 S2=11 S3=12 S4=13 S5=14
N2=11 (2) (3) (4) (5) (6) (1× 2) (1× 3) (1× 4) (1× 5) (1× 6)
N3=12
N4=13 0.10 200 170 140 110 80 20 17 14 11 8
N5=14 130 30 33 28.5 14 19.5
0.15 200 220 190 160 180 40 44 48 42 36
230 50 55 60 65 57.5
0.20 200 220 240 210 280 60 66 72 78 84
0.25 200 220 240 260 220 205
250 215 222.5
0.30 200 220 240 260
Expected monetary value

241

a. Calculation of optimal quantity that can maximize the expected profit.
Therefore, the optimal quantity that can maximize the expected profit is 12 units.

b. Expected profit with prefect information;
EPPI = RS.(20 + 33 + 48 + 65 + 84) = Rs.250

c. Expected value of the perfect information (EVPI) = Rs.250 - Rs.222.50 = Rs.27.50

QN. 12) (2073) What do you mean by decision with uncertainty? What are the methods of

decision making under the situation of uncertainty? A Grocery shopkeeper order and

receives 1 liter Snow Fun Curd early in the morning each day. The cost of curd per liter is Rs.

100 and he sells it for Rs. 130, thus making profit of Rs. 30 per liter. Based on the previous

year's sales statistics the Grocery shopkeeper estimate his daily sales of curds as follows:

Demand 15 16 17 18 19

probability of demand 0.30 0.25 0.20 0.15 0.10

Any unsold curd have to be thrown away at the end of the day. How many liters of curd should be

shopkeeper stock each day so as to maximize his profit in the lung run? Also, compute the

expected value for the perfect information.

Solution:

Since the Grocery would not sells less than 15 liter and not more than 19 liter of curds,
it has five strategies (alternatives) such as 1 = 15, 2 = 16, 3 = 17, 4 = 18, 5 =
19. Similarly there are five possible state of nature i.e. the demand of curd such as
1 = 15, 2 = 16, 3 = 17, 4 = 18, 5 = 19.
Here, SP= Rs.130, CP= Rs. 100, MP = Rs.30

Strategies N1=15 State of Nature (demand of curd)
450 N2=16 N3=17 N4=18 N5=19
S1=15 350 450 450 450 450
S2=16 250 480 480 480 480
S3=17 150 380 510 510 510
S4=18 50 280 410 540 540
S5=19 180 310 440 730

Expected monetary value (EMV) for each strategy.

Prob. Payoff Values (Rs.) Expected Profit (Rs.)
(1) Decision Alternatives Decision Alternatives

242

State S1=15 S2=16 S3=17 S4=18 S5=19 S1=15 S2=16 S3=17 S4=18 S5=19
(1× 6)
of (2) (3) (4) (5) (6) (1× 2) (1× 3) (1× 4) (1× 5)
15
Nature 45
62
N1=15 0.30 450 350 250 150 50 135 105 75 45 66
57
N2=16 0.25 450 480 380 280 180 112.5 120 95 70 245

N3=17 0.20 450 480 510 410 310 90 96 102 82

N4=18 0.15 450 480 510 540 440 67.5 72 76.5 81

N5=19 0.10 450 480 510 540 570 45 48 51 54

Expected monetary value 450 441 399.5 332

As, the maximum expected profit is Rs. 450 at stock 15 liters. Thus, shopkeeper should

make stock of 15 liters each day to maximize his profit in the long run.

Also,

Expected profit with prefect information;

EPPI = RS.(135 + 120 + 102 + 81 + 57) = Rs.495

Expected value of the perfect information (EVPI) = Rs.495 - Rs.450 = Rs.45

QN. 13) (2072) Under an employment promotion programe, it is proposed to allow to sale

of newspaper on the buses during pick hours. A newspaper boy has the following

probability of selling magazine.

No. of copies sales 10 11 12 13 14

probability 0.10 0.15 0.20 0.25 0.30

Cost per copy of magazine is Rs.30 and sales price per copy is Rs. 50. He cannot return the

unsold copies whether salvage value is zero.

a) Construct pay-off table.

b) Calculate the expected monetary value (EMV) for each strategy.

c) How many should be ordered?

d) Compute the expected profit with perfect information (EPPI).

e) Also, calculate the expected value of perfect information (EVPI) and interpret.

Solution: Since the newspaper boy would not sells less than 10 copies and not more
than 14 copies of newspapers, it has five strategies (alternatives) such as 1 = 1, 2 =
11, 3 = 12, 4 = 13, 5 = 14. Similarly there are five possible state of nature i.e. the
demand of newspapers such as 1 = 10, 2 = 11, 3 = 12, 4 = 13, 5 = 14.
Here, SP= Rs.50, CP= Rs. 30, Marginal Profit = Rs.20

f) Construction of Payoff table

Strategies Payoff Table
State of Nature (demand of newspapers)

N1=10 N2=11 N3=12 N4=13 N5=14

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S1=10 200 200 200 200 200
S2=11 170 220 220 220 220
S3=12 140 190 240 240 240
S4=13 110 160 210 260 260
S5=14 80 130 180 230 280

g) Calculation of Expected monetary value (EMV) for each strategy.

State Prob. Payoff Values (Rs.) Expected Profit (Rs.)

of (1) Decision Alternatives Decision Alternatives
Nature
S1=10 S2=11 S3=12 S4=13 S5=14 S1=10 S2=11 S3=12 S4=13 S5=14
(1× 6)
(2) (3) (4) (5) (6) (1× 2) (1× 3) (1× 4) (1× 5)
8
N1=10 0.10 200 170 140 110 80 20 17 14 11 19.5
36
N2=11 0.15 200 220 190 160 130 30 33 28.5 24 57.5
84
N3=12 0.20 200 220 240 210 180 40 44 48 42 205

N4=13 0.25 200 220 240 260 230 50 55 60 65

N5=14 0.30 200 220 240 260 280 60 66 72 78

Expected monetary value 200 215 222.5 220

As, the maximum expected profit is Rs. 222.5 at stock 12 units.

h) The shopkeeper should make the order of 12 copies of newspaper each day to maximize his

profit.

i) Expected profit with prefect information;(Diagonal values)

EPPI = RS.(20 + 33 + 48 + 65 + 84) = Rs.250

j) Expected value of the perfect information (EVPI) = Rs.250 - Rs.222.50 = Rs.27.5

Since, the maximum expected profit is Rs. 222.5 and expected value for the perfect

information is Rs.250, so the newspaper boy should not spend more than

Rs.27.5(Rs.250 - Rs.222.50) to get the perfect information.

QN. 14) (2072) An Ice-cream retailor buys Ice-cream at Rs.5 per cup and sells it for
Rs.8 per cup. Any remaining unsold at the end of the day useless and thrown away.
Past sales have been ranged between 15 and 18 per day. Assuming the sales history
has the following probabilities.

Quantity sold 15 16 17 18
probability 0.10 0.2 0.40 0.30

a) What quantity should bought to maximize the expected profit?
b) What will be his maximum expected profit? If he has perfect information of the

market, what would be the expected profit?

244

c) Also, compute the expected value for perfect information.

Solution: Since the Ice-cream seller would not sells less than 15 cups and not more
than 18 cups of Ice-cream, it has four strategies (alternatives) such as 1 = 15, 2 =
16, 3 = 17, 4 = 18. Similarly there are four possible state of nature i.e. the demand
of newspapers such as 1 = 15, 2 = 16, 3 = 17, 4 = 18.
Here, SP= Rs.8, CP= Rs. 5, Marginal Profit = Rs.3

Construction of Payoff table

Strategies Payoff Table
State of Nature (demand of Ice-cream)
S1=15 N1=15 N2=16 N3=17 N4=18
S2=16 45 45 45 45
S3=17 40 48 48 48
S4=18 35 43 51 51
30 37 46 54

Calculation of Expected monetary value (EMV) for each strategy.

State Prob. Payoff Values (Rs.) Expected Profit (Rs.)

of (1) Decision Alternatives Decision Alternatives
Nature
S1=15 S2=16 S3=17 S4=18 S1=15 S2=16 S3=17 S4=18

(2) (3) (4) (5) (1× 2) (1× 3) (1× 4) (1× 5)

N1=15 0.10 45 40 35 30 4.5 4.0 3.5 3.0

N2=16 0.20 45 48 43 37 9.0 9.6 8.6 7.4

N3=17 0.40 45 48 51 46 18 19.2 20.4 18.4

N4=18 0.30 45 48 51 54 13.5 14.4 15.3 16.2

Expected monetary value 45 47.2 47.8 45

a) As, the maximum expected profit is Rs. 47.80 at stock 17 units.

To maximize the profit he should bought 17 cup of Ice-cream.

b) EPPI = RS.(4.5 + 9.6 + 20.4 + 16.2) = Rs.50.70

The maximum profit would be Rs.50.70 if he has perfect information.

c) Expected value of the perfect information (EVPI) = Rs.50.70 - Rs.47.80 = Rs.2.90

Since, the maximum expected profit is Rs. 47.80 and expected value for the perfect

information is Rs.50.70, so he should not spend more than Rs.2.90(Rs.50.70 - Rs.47.80)

to get the perfect information.

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Chapter – 12
Linear Programming Problem
Descriptive Question Answer

QN. 1)(2078) Solve the following Linear Programming problem Graphically:

Maximize: Z = 30x + 50y

Subject to constraints: x + y ≤ 30

x + 2y ≤ 40

x ≤ 0, y ≤ 0

Solution: Let associated equations of the given constraints are

x + y = 30 ………….. (i)

x + 2y = 40 ……….. (ii)

From (i), we get

If x = 0, y = 30

If y = 0, x = 30

Thus, the line (i) passes through the points (0, 30) and (30, 0).

From (ii)

If x = 0, y = 20

If y = 0, x = 40

Thus, line (ii) is passing through the points (0, 20)

and (40, 0).

To identity feasible region, we use test point (0, 0).

From inequality (i), we get 0 ≤ 30 which is true.

From inequality (ii), we get 0 ≤ 40 which is true.

As, we have given x ≥ 0, and y ≥ 0, the feasible

region lies on first quadrant only.

Plotting above points in the graph, we have

Calculation of Maximum Value

Vertices Z = 30x + 50y Remark

O(0, 0) Z = 30×0 + 50×0 = 0

A(30, 0) Z = 30×30 + 50×0 = 900

246

C(20, 10) Z = 30×20 + 50×10 = 1600 Maximum
E(0, 30) Z = 30×0 + 50×30 = 1500

Therefore, maximum value of Z is 1600 at C(20, 10).
QN. 2) (2077) Maximize: Z = 3x + 4y
subject to the constraints:
2x + y ≤ 40
2x + 5y ≤ 180
and x ≥ 0, y ≥ 0
Solution:
Let associated equations of the given constraints are
2x + y = 40 ………….. (i)
2x + 5y = 180 ……….. (ii)
From (i), we get
If x = 0, y = 40
If y = 0, x = 20
Thus, the line (i) passes through the points (0, 40) and (20, 0).
From (ii)
If x = 0, y = 36
If y = 0, x = 90
Thus, line (ii) is passing through the points (0, 36) and (90, 0).
To identity feasible region, we use test point (0, 0).
From inequality (i), we get 0 ≤ 40 which is true.
From inequality (ii), we get 0 ≤ 180 which is true.
As, we have given x ≥ 0, and y ≥ 0, the feasible region lies on first quadrant only.
Plotting above points in the graph, we have

247

Calculation of Maximum Value

Vertices Z = 3x + 4y Remark
Maximum
O(0, 0) Z = 3×0 + 4×0 = 0

A(20, 0) Z = 3×20 + 4×0 = 60

B(25, 35) Z = 3×25 + 4×35 = 215

C(0, 36) Z = 3×0 + 4×36 = 144

Therefore, minimum value cost is 215 at the point B(25, 35).
QN. 3) (2077, Back) Maximize: Z = 50x + 100y
subject to the constraints:
x + y ≤ 30
x + 2y ≤ 40
and x ≥ 0, y ≥ 0.
Solution:
Let associated equations of the given constraints are
x + y = 30 ………….. (i)
x + 2y = 40 ……….. (ii)
From (i), we get
If x = 0, y = 30
If y = 0, x = 30
Thus, the line (i) passes through the points (0, 30) and (30, 0).
From (ii)
If x = 0, y = 20
If y = 0, x = 40
Thus, line (ii) is passing through the points (0, 20) and (40, 0).
To identity feasible region, we use test point (0, 0).
From inequality (i), we get 0 ≤ 30 which is true.

248

From inequality (ii), we get 0 ≤ 40 which is true.
As, we have given x ≥ 0, and y ≥ 0, the feasible region lies on first quadrant only.
Plotting above points in the graph, we have

Calculation of Maximum Value

Vertices Z = 50x + 100y Remark

O(0, 0) Z = 50×0 + 100×0 = 0

A(30, 0) Z = 50×30 + 100×0 = 1500

B(20, 10) Z = 50×20 + 100×10 = 2000 Maximum

C(0, 20) Z = 50×0 + 100×20 = 2000 Maximum

Therefore, the maximum value is 2000 at B(20, 10) or C(0, 20).

QN. 4) (2077, Regular) Maximize: Z = 20x + 30y
subject to the constraints:
10x + 20y ≤ 100
20x + 10y ≤ 140
and x ≥ 0, y ≥ 0
Solution:
Let associated equations of the given constraints are
10x + 20y = 100 ………….. (i)
20x + 10y = 140 ……….. (ii)
From (i), we get
If x = 0, y = 5
If y = 0, x = 5
Thus, the line (i) passes through the points (0, 5) and (10, 0).
From (ii)
If x = 0, y = 14
If y = 0, x = 7

249