Arihant CBSE Class 11 Book

CBSE Term II Applied Mathematics XII 133

Solved Examples

Example 1. Sankar invests ` 9900 on ` 100 shares at Example 4. Deepa invested ` 30000 in a mutual fund in

` 90. If the company pays him 20% dividend, find 2015. The value of mutual fund increased to
his rate of return. ` 48000 in year 2020. Calculate the compound
annual growth rate of his investment.
Sol. Given, cost of investment = ` 9900 (given (1.6)1/5 = 1.099)
Market value of one share = ` 90
\ Number of shares bought = 9900 = 110 Sol. Given, beginning value of the investment (BV) = ` 30000
90
Number of years = 5
Total dividend earned by Sankar
= Number of shares ´ rate of dividend Ending value of the investment (EV) = ` 48000
´ face value of one share \Compound annual growth rate (CAGR) = æç EV ö÷1/ n - 1
= 110 ´ 20 ´ 100 = ` 2200
100 è BV ø

So, rate of return = 2200 ´ 100 = 22.22% 1
9900
= æç 48000 ö÷ 5 -1
è 30000 ø

Example 2. Which gives better yields 7.5% 1

compounded semi-annually or 7% compounded = (1.6)5 - 1
monthly? (given (1.0058)12 = 1.07186) and
(1.0375)2 = 1.07640) = 1.099 - 1
= 0.099

Hence, CAGR% is 0.099 ´ 100% = 9.9%

Sol. The effective rate corresponding to 7.5% compounded Example 5. A Lenevo computer, whose cost is ` 600000
semi-annually is
will depreciate to a scrap value of ` 60000 in 5 yr.
re = æèç1 + 0.075 ö÷ 2 - 1 é r = 7.5 = 0.075 and m = 2ûúù Using linear method of depreciation, find the book
2ø ëêQ 100 value of the computer at the end of third year.

Þ re = (1.0375)2 - 1 = 1.07640 - 1 = 0.07640 Sol. We know, C = Original cost of computer = ` 600000

So, the effective rate is 7.6% per annum. S = Scrap value of computer = ` 60000
n = useful life of computer = 5 yr
The effective rate corresponding to 7% compounded \ Annual depreciation
monthly is
= Original cost of computer - Scrap value
re¢ = æèç1 + 0.07 ÷ö12 -1 é r = 7 = 0.07 and m = 12ûúù Useful life of computer in years
12 ø ëêQ 100
= 600000 - 60000 = 540000
Þ re¢ = (1.0058)12 - 1 = 1.07186 - 1 = 0.07186 55

So, the effective rate is 7.18% per annum. = ` 108000

Hence, the investment with second option gives better The book value at the end of third year
= 600000 - 3 ´ 108000 = ` 276000
yields.

Example 3. To what sum will ` 5000 accumulate in 6 yr, Example 6. The value of a machine purchased 2 yr ago,

if invested at an effective rate of 6%? depreciate at the annual rate of 10%. If its present
(given (1.06)6 = 1.41851) value is ` 97200, then find the value of machine
after 3 yr.
Sol. We know that, an effective rate is the actual rate compounded
annually, So, the sum S accumulated is the compound amount Sol. Given, P = ` 97200, r = 10% per annum
of the sum ` 5000 invested at 6% compounded annually.
We have, P = ` 5000, i = 6 = 0.06% and n = 6 yr and i = 10 = 0.1
100 100
\ S = P(1 + i)n
Value of machine after 3 yr = P (1 - i)n
Þ S = 5000(1 + 0.06)6 = 5000 (1.06)6
= 97200 (1 - 0.1)3
= 5000 ´1.4185 = 7092.55
Hence, required sum = ` 7092.55 = 97200 ( 0.9)3

= 97200 ´ 0.729
= ` 70859

134 CBSE Term II Applied Mathematics XII

Example 7. The value of a car depreciate 10% every Example 9. The cost of a mobile phone is ` 9000. Its

year. By what percent will the value of the car value depreciates at the rate of 5% per annum. Find

decrease after 3 yr. the total depreciation in its value at the end of 2 yr.

Sol. Let the present value of the car be ` P, then value of the car Sol. The present cost of mobile phone ( P) = ` 9000
after n yr = P(1 - i)n, where i = r
100 Rate of depreciation, i = 5 = 0.05 é i = rù
Given, n = 3 yr and r = 10% 100 ëêQ 100 úû
= P(1 - i)3 = Pæç1 - 10 ÷ö 3
è 100ø n = Number of years = 2

S = Scrap value of mobile

= Pæç1 - 1 ÷ö 3 = Pæç 9 ÷ö 3 = ` 729 P \ S = P(1 - i)n
è 10ø è 10ø 1000
S = 9000 (1 - 0.05)2 = 9000 ( 0.95)2

Decrease in the value of car =` çæè P - 729P öø÷ = 9000 ´ 0.9025 = ` 8122.50
1000 Total depreciation = Original value - Scrap value

= ` 271P = 9000 - 8122.50
1000 = 877.50

\Percentage decrease in the value of the car after 3 yr Hence, total depreciation is ` 877.50.

= èçæ Decrease in value ´ 100÷öø % Example 10. Amit purchased an old motorcycle for
Present value
` 16000. If the cost of the scooter after 2 yr
= èæçç 271 P ´ 100ö÷ø÷ % = 27.1% depreciates to ` 14440. Find the rate of
1000 P
depreciation.

Sol. Present value of motorcycle ( P) = ` 16000

Example 8. A machine costing ` 30000 has a scrap Scrap value of motorcycle (S) = ` 14440
n = Number of years = 2
value of ` 4000, if annual depreciation charge is
` 2000. Find its useful life in years. Let rate of depreciation is r %.
\ S = P(1 - i)n
Sol. Given, original cost of machine = ` 30000
Annual depreciation = ` 2000 Þ 14440 = 16000(1 - r )2
Scrap value of machine = ` 4000
Þ (1 - i)2 = 14440
Let useful life of machine is n yr. 16000
\Annual depreciation = Original Cost - Scrap value
Þ (1 - i)2 = 0.9025
Useful life of machine in years
Þ 1 - i = 0.9025
Þ 2000 = 30000 - 4000
n Þ 1 - i = 0.95
Þ i = 1 - 0.95 = 0.05
Þ n = 26000 = 13 \ Rate of depreciation = ( 0.05 ´ 100)% = 5%
2000

Hence, useful life is 13 yr.

CBSE Term II Applied Mathematics XII 135

Chapter
Practice

PART 1 7. Varun made an investment of ` 225000 in a no-fee
Objective Questions
fund for one year. At the end of the year the value of

the investment increases to ` 250000. The nominal

rate of return percent on his investment is

l Multiple Choice Questions (a) 11% (b) 11.11%

(c) 11.33% (d) None of these

1. The effective rate of interest which is equivalent to 8. A person invested ` 20000 in a mutual fund in year
2016.The value of mutual fund in increased to
a nominal rate of 8% compounded semi-annually is ` 32000 in year 2015. The compound annual growth
rate of his investment is (given, (1.6)1/5 = 1.098)
(a) 8.24% (b) 8.33%

(c) 8.16% (d) 8.06%

2. The declared rate of return compounded (a) 9.5% (b) 9.8%
(c) 19.8% (d) None of these
semi-annually which is equivalent to 10.25%

effective rate of return, is 9. A machine costing ` 50000 has a useful life of 4 yr.

(a) 10.13% (b) 10.05% The estimated scrap value is ` 10000. The rate of

(c) 10% (d) 9.89% depreciation per annum is

3. A person invested ` 200000 in a fund for one year. (a) 20% (b) 30%

At the end of the year the investment was worth (c) 25% (d) 15%

` 216000. The nominal rate of return on his 10. Mr. Malik invested ` 16500 on ` 100 shares at

investment is premium of ` 10 paying 15% dividend. At the end

(a) 10% (b) 8% of the year, he sells the shares at a premium of ` 20.

(c) 6% (d) 5% The rate of return is

4. A man invests a sum of money in ` 100 shares (a) 22.66% (b) 20%

paying 15% dividend quoted at 20% premium. If (c) 22.33% (d) 22.72%

his annual dividend is ` 540. The rate of return on 11. The annual depreciation of a car is ` 30000, if the

his investment is scrap value of the car after 15 yr is ` 50000, then

(a) 12% (b) 15% original cost of the car, when depreciation is linear.

(c) 10% (d) 12.5% (a) ` 500000 (b) ` 450000

5. A money-lender charges interest at the rate of ` 10 (c) ` 600000 (d) None of these

per ` 100 per half-year, payable in advance. The 12. A machine costing ` 150000 has scrap value of

effective rate of interest per annum is ` 22500. If annual depreciation charge is ` 8500,

(a) 22% (b) 23% then useful life of the machine is

(c) 23.30% (d) 23.45% (a) 10 yr (b) 12 yr

6. Radha purchased 4000 shares of ` 100 each at a (c) 15 yr (d) 16 yr

premium of ` 25. If the company declares a 13. An asset costing ` 80000 has a useful life of 8 yr, if

dividend of 10%, then her rate of return is annual depreciation is ` 9000, then scrap value of

(a) 6% (b) 7% asset is

(c) 8% (d) 10% (a) ` 9000 (b) ` 8000

(c) ` 10000 (d) ` 7000

136 CBSE Term II Applied Mathematics XII

l Case Based MCQs 5. Find the declared rate of return compounded
semi-annually which is equivalent to 6% effective
14. The cost of AC depreciates by ` 722 during the rate of return. (given, (1.06)1/2 = 1.0296)
third year and by ` 685.90 during the fourth year.
Based on above information, answer the following 6. Mr. Rathi holds ` 1500, if ` 100 shares of a
questions. company paying 15% dividend annually quoted at
30% premium. Calculate rate of return of his
(i) The rate of depreciation per annum is investment.

(a) 5% (b) 6% 7. Which gives better yields:

(c) 4% (d) 8% (i) 8.4% per annum compounded monthly.

(ii) The original cost of AC is (ii) 8.5% per annum compounded semi-annually.
(a) ` 15000 [given, (1.007)12 = 1.0873 and (1.0425)2 = 1.0868]
(b) ` 20000
(c) ` 16000 8. How many years will it take for money to double at
(d) ` 18000 the effective rate of 6%?

(iii) The value of the AC at the end of fourth year is 9. Vishal invested ` 32000 in a mutual fund in year
(given, ( 0.95 )4 = 0.8145) 2015. The value of mutual fund increased to
` 57600 in year 2021. Calculate the compound
(a) ` 13000 growth rate of his investment.
(b) ` 13032 (given, (1.8)1/6 = 1.1029).
(c) ` 12000
(d) ` 14032

(iv) The total depreciation in its value at the end of 4 yr 10. Mr. Ahmed invested ` 10000 in a company’s fund.

is His yearly investment values are shown in the table
given below
(a) ` 3500 (b) ` 3000

(c) ` 2868 (d) ` 2968

(v) The scrap value of AC of the estimated useful life Year 0 1 2 3
is 15 yr is (given, ( 0.95 )15 = 0.4633) 14000 14700 22056
Amount (in `) 10000

(a) ` 7400 (b) ` 7413 Calculate CAGR of his investment.
(c) ` 7500 (d) ` 7450 (given (2.205)1/3 = 1.3016)

PART 2 11. An asset costing ` 150000 is expected to have a
Subjective Questions useful life of 5 yr and scrap value of ` 30000. Find
the annual depreciation charge and the
depreciation rate by using the linear depreciation
method.

l Short Answer Type Questions l Long Answer Type Questions

1. Find the effective rate that is equivalent to a 12. A money lender charges interest at the rate of 10
nominal rate of 8% compounded quarterly. paise per rupee per month, payable in advance.
What effective rate of interest does he charge per
2. To what amount will ` 12000 accumulate in 12 yr, if annum?
invested at an effective rate of 5%?
(given, (1.05)12 = 1.7958) 13. A piece of machinery costing ` 100000 is expected to
have a useful life of 5 yr and scrap value ` 20000.
3. ` 100 shares of a company are sold at a discount of Using the straight line method, find the annual
` 20. If the return on the investment is 15%. Find depreciation and construct a schedule for
the rate of dividend declared. depreciation. Also, find the depreciation rate
percent.
4. Mr. Taneja purchase 100 shares of a company that
cost ` 250 each. After one year the price of each 14. The cost of a machine purchased 2 yr ago
share rose to ` 300. Assuming that there no trading depreciate at the rate of 20% every year. If its
costs and no dividends. Find the nominal rate of present worth is ` 315600, find
return on the investment.
(i) its purchase price
(ii) its value after 3 yr

CBSE Term II Applied Mathematics XII 137

SOLUTIONS

Objective Questions 5. (d) It is given that the money-lender charges interest at the
rate of ` 10 per ` 100 half-year, payable in advance.
1. (c) We have, r = 8 = 0.08 and m = 2
100 This means that ` 10 is the interest of ` 90 for half year.

çæ1 r ÷ö m So, the interest rate per half-year is i = 10 = 1 and the
è m ø 90 9
\ re = + -1
annual interest rate is 2 compounded half-yearly.
çèæ1 0.08 øö÷ 2 9
2
Þ re = + -1 Thus, we have

= (1.04)2 - 1 = 1.0816 - 1 = 0.0816 r = 2 and m = 2
9
\ Effective rate of interest = ( 0.0816 ´ 100)%

= 8.16% \ re = æèç1 + r ÷ö m -1
mø
2. (c) We have, re = 10.25 = 0.1025 and m = 2
100
çæ1 1 ÷ö 2
Þ = çæè1 + r ö÷ 2 -1 é = èçæ1 + r ö÷ m ù Þ re = è + 9ø - 1
2ø êQ mø -1ú
re êë re
ûú
Þ re = 100 -1 = 19
Þ 0.1025 = çæ1 + r ö÷ 2 - 1 81 81
è 2ø
So, the effective rate is çèæ 19 ´ 100øö÷ % = 23.45%
Þ çæ1 + r ÷ö 2 = 1 + 0.1025 = 1.1025 81
è 2ø
6. (c) Here, Market value of each shares = ` 125

Þ 1 + r = 1.1025 = 1.05 \Market value of 4000 shares = ` ( 4000 ´ 125) = ` 500000
2
Face value of each shares = ` 100

Þ r = 1.05 - 1 Þ r = 0.05 \ Face value of 4000 share = ` ( 4000 ´ 100) = ` 400000
22
Rate of dividend = 10%
Þ r = 0.10
Income from 4000 shares =` æçè 10 ´ 400000øö÷ =` 40000
So, the declared rate of interest = ( 0.10 ´ 100)% = 10% 100

3. (b) Given, current value of investment = ` 216000 Hence, rate of return = æç 40000 ´100÷ö % = 8%
è 500000 ø
Initial investment = ` 200000

\ Nominal rate of return percent 7. (b) It is given that,

é çæ Current market value of investment ÷ö ù Original investment = ` 225000
ê ç - Original investment value ÷ ú
= ê ç ´100úú % Market value = ` 250000
ê èç Original investment value ÷ úû
ëê ÷ø \Nominal rate of return percent

èæç 216000 - 200000 ´100öø÷ çæè 16000 ´100ø÷ö % = é æç 250000 - 225000÷ö ´ 100úùû %
200000 200000 ëê è 225000 ø
= % = = 8%
çèæ ´ 100÷öø %
\Nominal rate of return is 8%. = 25000 = 11.11%
4. (d) Here, rate of dividend = 15% 225000

Face value of shares = ` 100 8. (b) Given, beginning value of investment (BV) = ` 20000
Ending value of an investment (EV) = ` 32000
and annual dividend = ` 540 Number of years, n = 5

\Number of shares = Rate of Annual dividend of share 1
dividend ´ Face value
\ CAGR = æç EV ö÷ n - 1
= 540 = 36 è BV ø
15 ´ 100
100 = æèç 32000 ÷øö1/5 -1
20000
Market value of share = ` 120
= (1.6)1/5 - 1
Investment on 36 shares = ` (120 ´ 36) = ` 4320
= 1.098 - 1 = 0.098
So, rate of return = çèæ 540 ´ 100ö÷ø % = 12.5% Hence, CAGR is ( 0.098 ´ 100)% = 9.8%
4325

138 CBSE Term II Applied Mathematics XII

9. (c) Annual depreciation amount = ` 10000 14. Let the original cost of AC = ` P
and rate of depreciation = r %
Original cost of machine - Salvage value of machine The value of machine after n yr = P(1 - i)n,

= ` (50000 - 10000) = ` 40000

æç ö÷ where i = r .
ç Annual Depreciation ÷ 100
\Depreciation rate percent = ç ÷
çèç Original Cost of Machine ÷÷ø (i) (a) The cost of AC depreciate by ` 722 during third
- Scrap Value of Machine
years

çæè 10000 ´ 100øö÷ % \ P(1 - i)2 - P(1 - i)3 = 722 …(i)
40000
= = 25% The cost of AC depreciate by ` 685.90 during
fourth years.
10. (d) Here, market value of each share = ` 110 …(ii)
\ P(1 - i)3 - P(1 - i)4 = 685.90
Cost of investment ( P0 ) = ` 16500
\Number of shares bought = 16500 = 150 or P(1 - i)2 i = 722 …(iii)
P(1 - i)3i = 685.90 …(iv)
110

Annual dividend ( D1) =` çæè150 ´ 15 ´ 100÷øö = ` 2250 From Eqs. (iii) and (iv), we get
100 1 - i = 685.90 = 0.95
722
As Mr. Malik sell his shares at premium of ` 20.
Þ i = 1 - 0.95 = 0.05
Market value of each share = ` 120 r = i ´ 100
= ( 0.05 ´ 100)%
The selling value of 150 shares ( P1) = ` (120 ´ 150) = 5%

= ` 18000 (ii) (c)\ P(1 - i)2 × i = 722

\Rate of return = D1 + ( P1 - P0 ) ´ 100 [from Eq. (iii)]
P0

= æçè 2250 + (18000 - 16500) ´ 100÷øö %
16500
Þ P = 722
( 0.95 )2 ´ 0.05
= æèç 225016+51500øö÷ % = çèæ 3750 ø÷ö =
165 % 22.72% 722
0.9025 ´
= 0.05

11. (a) Here, Annual depreciation = ` 30000

Scrap value of car = ` 50000 = ` 16000
\ Original cost of AC is ` 16000.
Useful life of car = 15 yr

Let the original cost of car is ` x. (iii) (b) The value of AC of the end of fourth years
\Annual depreciation = Original cost - Scrap value = P(1 - i)4

Useful life of car in year = 16000(1 - 0.05)4

30000 = x - 50000 = 16000 ( 0.95)4
15

Þ x = 450000 + 50000Þ x = 500000 = 16000 ´ 0.8145

Hence, the original cost of car was ` 500000. = ` 13032
(iv) (d) Original cost of AC = ` 16000
12. (c) Here, original cost of machine = ` 150000
Scrap value of AC at the end of fourth year = ` 13032
Scrap value of machine = ` 22500 \Total depreciation value in 4 yr = ` (16000 - 13032)

Annual depreciation = ` 8500 = ` 2968
\ Annual depreciation = Original cost - Scrap value (v) (b) Original cost of AC, P = ` 16000

Useful life of machine in year Rate of depreciation, r = 5%

Useful life of machine = 150000 - 22500 = 127500 = 15 yr
8500 8500
Number of years, n = 15
13. (b) Here, original cost = ` 80000 \ Scrap value in 15 yr = P(1 - r )15

Annual depreciation = ` 9000 = 16000(1 - 0.05)15

Useful life = 8 yr = 16000( 0.95)15
\ Annual depreciation = Original cost - Scrap value
= 16000( 0.4633)
Useful life of asset in year = 7412.80
= ` 7413
Þ 9000 = 80000 - Scrap value
8

Þ Scrap value = 80000 - 72000 Hence, scrap value of AC estmiated useful life in 15 yr
is ` 7413.
= ` 8000

CBSE Term II Applied Mathematics XII 139

Subjective Questions Þ æç1 + r ÷ö 2 = 1 + 0.06
è 2ø
1. We have, r = 8% = 8 = 0.08 and m = 4
100 Þ 1 + r = (1.06)1/ 2
2
æèç1 r ö÷ m - 1
\ re = + mø Þ 1 + r = 1.0296
2
4
çèæ1 0.08 øö÷ Þ r = 1.0296 - 1 = 0.0296
= + 4 -1 2

= (1.02)4 - 1 Þ r = 2 ´ 0.0296 = 0.0592

= 1.0824 - 1 = 0.0824 \Declared rate of interest = ( 0.0592 ´ 100)%
\Effective rate of interest = ( 0.0824 ´ 100)%
= 5.92%
= 8.24%
2. We have, P = ` 12000, i = 5 = 0.05 and n = 12 6. Given, market value of share = ` 130

100 Total value of 1500 shares = ` (1500 ´ 130)
\ S = P(1 + i)n
= ` 195000

Face value of a share = ` 100

= 12000 (1 + 0.05)12 Rate of dividend = 15%

= 12000(1.05)12 Total dividend earn by Mr. Rathi = ` çèæ 15 ´ 1500 ´ 100÷øö
100
= 12000 ´ 1.7958 = 21549.60
= ` 22500
Hence, required sum is ` 21549.60.
So, rate of return = æçè 22500 ´ 100øö÷ %
3. Given, face value of a share = ` 100 195000
Market value of a share = ` 80
Let, Rate of dividend = x% = 11.53%
Income of a share from dividend = 8x
10 7. (i) The effective rate corresponding to 8.4% compounded
and rate of return = 15
\ 8x = 15 monthly is
10
Þ x = 150 = 18.75 re = èæç1 + r ö÷ m -1
8 mø

re = èçæ1 + 8.4 ö÷ø12 -1 é r = 8.4 and m = 12ûùú
1200 êëQ 100

Hence, the declared rate of dividend is 18.75%. re = (1.007)12 - 1

4. We have, = 1.0873 - 1 = 0.0873

Cost of each share = ` 250 So, the effective rate = ( 0.0873 ´ 100)% = 8.73%

Cost of 100 share = ` ( 250 ´ 100) = ` 25000 (ii) The effective rate corresponding to 8.5% compounded

Current market value of each share = ` 300 semi-annually is

çæè1 8.5 ÷øö 2 é 8.5 2ûúù
200 ëêQ 100
Market value of 100 shares = ` ( 300 ´ 100) = ` 30000 r ¢e = + -1 r = and m =

\ Nominal rate of interest re ¢ = (1.0425)2 -1

= é ççæè Current market value - Original value ö÷÷ø ´ ù % =1.0868 - 1 = 0.0868
ê Original value 100ú
ë û
So, the effective rate = ( 0.0868 ´100)% = 8.68%
é æèç 3000205-00205000 ø÷ö ´ 100ûùú
= êë % Hence, the option (i) gives better yields.

æçè 5000 ´ 100÷öø % 8. Let the sum of money be ` P.
25000
= = 20% Since, the effective rate is actual rate compounded annually.

Hence, the nominal rate of return on investment is 20%. Therefore, we have
5. We have, effective rate of interest = 6% Principal = ` P,

Here, re = 6 = 0.06 and m = 2 Compound amount = ` A = 2P
\ 100 and i = 6 = 0.06

çèæ1 r ÷øö m 100
m \ A = P(1 + i)n
re = + -1

æçè1 r ÷öø 2 Þ 2P = P(1 + 0.06)n
2
Þ 0.06 = + -1 Þ 2 = (1 + 0.06)n

140 CBSE Term II Applied Mathematics XII

Taking log on both sides, we get 12. The money-lender charges interest at the rate of 10 paise
Þ log2 = n log1.06
Þ n = log2 per rupee per month payable in advance. This means that 10

log1.06 paise is the interest on 90 paise for one month. So, the
interest rate per month is i = 10 = 1.
= 0.3010 = 11.89
0.0253 90 9

Hence, it will take 11.89 yr. Thus, the annual interest rate, r = 12 = 4 compounded
9. Given, beginning value of investment (BV) = ` 32000 93

Ending value of the investment (EV) = ` 57600 monthly.
Number of years (n) = 6
Here, r = 4 and m = 12
3

1 Let re be the effective rate of interest.

So, CAGR = æç EV ö÷ n - 1 Then, re = æèç1 + r ÷ö m -1
è BV ø mø

1 Þ re = èçæ1 + 4 ø÷ö12 -1
36
= æç 57600 ÷ö 6 - 1
è 32000ø Þ re = æèç 10 ÷øö12 -1
9
= (1.8)1/ 6 - 1

= 1.1029 - 1 = 0.1029 Þ re + 1 = çæ 10÷ö12
Hence, CAGR is ( 0.1029 ´ 100)% = 10.29% è 9ø
10. Given, beginning value of an investment (BV) = ` 10000
Ending value of an investment (EV) = ` 22050 Taking log on both sides, we get
Number of years (n) = 3
Þ log( re + 1) = 12(log10 - log9)
So, 1 é = èçæ EV 1 ù Þ log( re + 1) = 12(1 - 2log3)
êêQCAGR BV - 1úú Þ log( re + 1) = 12(1 - 2( 0.4771)) [Q log3 = 0.4771]
CAGR = æç EV ö÷ 3 - 1 ëê ö÷ n Þ log( re + 1) = 12(1 - 0.9542)
è BV ø ø úû Þ log( re + 1) = 12( 0.0458)
Þ log( re + 1) = 0.5496
çèæ 22050 öø÷1/ 3 Þ re + 1 = Antilog (0.5496)
10000 Þ re + 1 = 3.545
= - 1 Þ re = 3.545 - 1
Þ re = 2.545
= ( 2.205)1/ 3 - 1 \ Effective rate of interest = [( 2.545) ´ 100]%

= 1.3016 - 1 = 0.3016 = 254.5%
Þ CAGR = ( 0.3016 ´ 100)% = 30.16%
13. We have,
Hence, CAGR is 30.16%. C = Original value = ` 100000
11. Given, original cost of an assets = ` 150000 S = Scrap value = ` 20000

Scrap value of an assets = ` 30000 n = Useful life in years = 5 yr
Useful life = 5 yr
\ Annual depreciation The annual depreciation D is given by

= Original cost of an asset - Scrap value \ D=C -S
Useful life of asset in years n

= 150000 - 30000 Þ D = ` æçè 100000 - 20000 øö÷
5 5

= 120000 = ` 16000
5
At the begining of the first year, the book value of the
= 24000 machine is ` 100000. At the end of the first year, the
accumulated depreciation is ` 16000. Hence, the
Hence, annual depreciation is ` 24000. depreciation charge for the first year is ` 16000.
\ Depreciation rate percent

= é èæçç Annual depreciation ÷öø÷ ´ ù %
ê Original value - Scrap value 100ú
ë û
The book value at the end of the first year or in the begining
= é èæç 12240000000 øö÷ ´ 100úûù % of the second year is ` (100000 - 16000) = ` 84000
êë
At the end of second year, we have
= 20% Accumulated depreciation = ` 32000

CBSE Term II Applied Mathematics XII 141

\Depreciation charge = ` ( 32000 - 16000) = ` 16000 Original cost of machine - Scrap value of machine
Þ Book value at the end of second year = `( 84000 - 16000)
= `(100000 - 20000)
= ` 68000
= ` 80000
At the end of third year, we have
Accumulated depreciation = ` 48000 \Depreciation rate percent = æç 16000 ´ 100÷ö % = 20%
\Depreciation charge = ` 16000 è 80000 ø
Þ Book value at the end of third year = ` ( 68000 - 16000)
14. Let the cost value of machine 2 yr ago = `P
= ` 52000 Rate of depreciation, r = 20% per annum
\ i = 20 = 0.2
Similarly, at the end of fourth and fifth year. 100
Present value of machine = ` 315600
Book values are ` 36000 and ` 20000 respectively. (i) Value of machine after 2 yr = P(1 - i)2

Clearly, ` 20000 is the scrap value of the machine. Þ 315600 = P(1 - 0.2)2

These values can be represented in the following tabular Þ P = 315600
form which is known as the depreciation schedule of the ( 0.8)2
machine.
= 315600
Depreciation Schedule 0.64

Year Book value Depreciation End of Year = ` 493125
(Beginning of Year) (ii) Now, present worth of machine = ` 315600

1 ` 100000 ` 16000 ` 84000 \ i = 0.2
\ Value of after 3 yr = P (1 - i)3
2 ` 84000 ` 16000 ` 68000

3 ` 68000 ` 16000 ` 52000 = 315600 (1 - 0.2)3

4 ` 52000 ` 16000 ` 36000 = 315600 ( 0.8)3

5 ` 36000 ` 16000 ` 20000 = 315600 ´ 0.512
= 161587.20
We find that,
Annual depreciation amount = ` 16000 Hence, value of the machine after 3 yr is ` 161587.20.

Chapter Test

Multiple Choice Questions Short Answer Type Questions

1. Depreciation arise due to 9. A man purchased 300 shares of the face value of
` 100 each from the market at ` 800 per share.
(a) wear and tear of assets If the company paid a dividend of 40%. Find his rate
(b) reduction in the value of assets of return.
(c) increase in the value of liabilities
(d) reduction in capital 10. How many years will it take for a sum of money
to treble at the effective rate of 4%?
2. The total amount of depreciation of an asset cannot (given, log3 = 0.4771 and log(1.04) = 0.0170)
exceed its
11. Mr. Lal has two investment options either at 7.5%.
(a) scrap value compounded quarterly or 8% compounded
(b) market value semi-annually. Which option is better for Mr. Lal?
(c) depreciation value (given, (1.04)2 = 1.0816 and (1.01875)4 = 1.0771)
(d) None of the above
12. The value of a car depreciates by 12.5% every year.
3. The effective rate of return, which is equivalent rate of By what percent will the value of the car decrease
10% compounded semi-annually is after 3 yr.

(a) 10.11% (b) 10.25% 13. A mainframe computer whose cost is ` 500000 will
(c) 10.38% (d) 9.75% depreciate to a scrap value of ` 50000 in 5 yr. Using
linear method of depreciation, find the book value
4. The declared rate of return compounded of the computer at the end of third year.
semi-annually which is equivalent to 12.36%, then
effective rate of return is Long Answer Type Questions

(a) 12.12% (b) 12% 14. Mr. Gupta took a loan of ` 3000. Find the effective
(c) 12.20% (d) 11.90% rate of interest charged by the lender, if while lending
the lender deducts in advance ` 100 as interest of
5. A man invested ` 18000 on ` 100 shares at a discount 3 months.
of ` 25 paying 12% dividend. At the end of the year,
he sells the shares at a discount of ` 10.Then, his rate 15. The cost of a machine depreciates ` 1050 during the
of return is second year and ` 918.75 during the third years.
Then, calculate
(a) 36% (b) 40%
(c) 30% (d) 35% (i) the rate of depreciation per annum

6. Mrs. Jyothsha invested ` 35000 in a mutual fund. In (ii) the original cost of machine.
2015, the value of mutual fund increased to ` 56000.
In 2020, the CAGR is

(a) 12% (b) 10%
(c) 9% (d) 9.8%

7. A machine costing ` 150000 has a useful life of 15 yr. Answers 8. (a)
If the scrap value of the asset is ` 22500. Then, annual
depreciation charge is 1. (a) 2. (c) 3. (b) 4. (b) 5. (a) 6. (d) 7. (b)
9. 5% 10. 28.06 yr 11. 8.16% (second option)
(a) ` 8000 (b) ` 8500 12. 33% 13. ` 230000 14. 14.52%
(c) ` 8800 (d) ` 9000 15. (i) 12.5% (ii) ` 9600

8. A machine costing ` C would reduce to ` 10000 in 7 yr, For Detailed Solutions
if annual depreciation charge is ` 10000, then the value Scan the code
of C is

(a) ` 80000 (b) ` 70000
(c) ` 60000 (d) ` 75000

CBSE Term II Applied Mathematics XII 143

CHAPTER 10

Linear
Programming

In this Chapter...

l Linear Programming Problem (LPP)
l Graphical Method for Solving (LPP)
l Mathematical Formation of LPP

An LPP is one that is concerned with finding the optimal value (iv) The maximum or minimum value of an objective
(maximum or minimum value) of a linear function of several function is known as the optimal value of LPP.
variables, subject to the conditions that the variables are
non-negative and satisfy a set of linear inequalities. (v) The common region determined by all the
constraints including non-negative constraints
Mathematical Form of LPP x, y ³ 0 of a linear programming is called the
feasible region or solution region. The region
The general mathematical form of a linear programming problem other than feasible region is called an infeasible
region.
may be written as =aarc1e1xxx++³bc102y,yy£
(vi) A feasible region of a system of linear inequalities
Maximise or Minimise Z ³d10,. a2x + b2y £ d 2 , etc. and is said to be bounded, if it can be enclosed within
subject to constraints are a circle. Otherwise, it is said to be unbounded
non-negative restrictions region, i.e. the feasible region does extend
indefinitely in any direction.
Important Terms Related to LPP
(vii) Points within and on the boundary of the feasible
(i) The linear inequations or inequalities or restrictions on the region represent feasible solution of the
variables of a linear programming problem are called constraints.
constraints. The conditions x ³ 0, y ³ 0 are called non-negative
(viii) A feasible solution at which the objective function
restrictions. has optimal value (maximum or minimum) is
called the optimal solution or optimal feasible
(ii) A problem which seeks to maximise or minimise a linear solution of the linear programming problem.
function subject to certain constraints as determined by a set of
linear inequalities is called an optimisation problem. (ix) The process of obtaining the optimal solution of
the linear programming problem is called
(iii) A linear function of two or more variables which has to be optimisation technique.
maximised or minimised under the given restrictions is called
an objective function. The variables used in the objective
function are called decision variables.

144 CBSE Term II Applied Mathematics XII

Important Theorems form. There is no particular method to formulate linear
programming problem but following steps will be helpful in
(i) Theorem 1 Let R be the feasible region (convex polygon) the formulation of linear programming problems.
for a linear programming problem and Z = ax + by be the
objective function. I. Identify the decision variables of the problem, i.e.
identify the unknown quantities, whose values are to be
When Z has an optimal value (maximum or minimum), determined and denote them by x, y, etc.
where the variables x and y are subject to constraints
described by linear inequalities, this optimal value must II. Identify the objective function to be optimised
occur at a corner point (vertex) of the feasible region. (maximised or minimised) and express it as a linear
function of the decision variables obtained in step I.
(A corner point of a feasible region is a point of
intersection of two boundary lines in the region). III. Identify the set of constraints in terms of decision
variables and express them as linear inequations or
(ii) Theorem 2 Let R be the feasible region for a linear equations.
programming problem and Z = ax + by be the objective
function. If R is bounded, then the objective function Z IV. State the non-negative restrictions, so that negative
has both a maximum and a minimum value on R and each values of the decision variables do not have any valid
of these occurs at a corner point (vertex) of R. physical interpretation. Thus, objective function, the set
of constraints and the non-negative restrictions or
Graphical Method (or Corner Point Method) constraints together form the linear programming
for Solving LPP problem.

(i) Find the feasible region of the LPP and determine its Various Problems Based on LPP
corner points (vertices) either by inspection or by solving
the two equations of the lines intersecting at that point. (i) Diet Problems

(ii) Find the value of objective function Z = ax + by at each In diet problems, we have to determine the amount of
corner point. Let M and m respectively denote the largest different kinds of constituents/nutrients which should be
and the smallest values at these points. included in a diet, so as to minimise the cost of the desired
diet such that it contains a certain minimum amount of each
(a) When the feasible region is bounded, M and m are constituent/nutrient.
the maximum and minimum values of Z.
(ii) Manufacturing Problems
(b) When the feasible region is unbounded, then
In manufacturing problems, we have to determine the
l M is the maximum value of Z, if the open half plane number of units of different products which should be
determined by ax + by > M has no point in common produced and sold by a firm, when each product requires a
with the feasible region. Otherwise, Z has no fixed manpower, machine hours, labour hour per unit of
maximum value. product, warehouse space per unit of the output, etc, in order
to make maximum profit.
l m is the minimum value of Z, if the open half plane
determined by ax + by < m has no point in common (iii) Transportation Problems
with the feasible region. Otherwise, Z has no
minimum value. In transportation problems, we have to determine a
transportation schedule in order to find the minimum cost of
Mathematical Formulation of LPP transporting a product from plants/factories situated at
different locations to different markets.
Formulation of the problem is the process of transforming the
verbal description of a decision problem into a mathematical

CBSE Term II Applied Mathematics XII 145

Solved Examples

Example 1. The corner points of the feasible region Sol. The feasible region is bounded. Therefore, maximum of Z
must occurs at the corner points of the feasible region
for an LPP are (1, 0), (4, 0), (3, 4), (0, 6) and (0, 3).
Let Z = 7x - 3y be the objective function. Find the Corner Point Value of Z
minimum value of Z. O(0, 0) 4 ( 0) + 3 ( 0) = 0
A(25, 0) 4 ( 25) + 3 ( 0) = 100
Sol. Given, Z = 7x - 3y B(16, 16) 4 (16) + 3 (16) = 112 ¬ Maximum
C(0, 24) 4 ( 0) + 3 ( 24) = 72
Value of Z at corner points is given by the following table.

Corner points Corresponding value of Hence, the maximum value of Z is 112.
Z = 7x - 3y Y
(1, 0) 7 ´1 - 3 ´0 = 7
(4, 0)
(3, 4) 7 ´ 4 - 3 ´ 0 = 28 (Maximum)
(0, 6)
(0, 3) 7´3-3´4 = 9

7 ´ 0 - 3 ´ 6 = -18

7 ´ 0 - 3 ´ 3 = -9

Hence, the maximum value of Z is 28. (0,40)
(0,24)C
Example 2. Corner points of the feasible region
B (16,16)
determined by the system of linear constraints are
(0, 3), (1, 1) and (3, 0). Let Z = px + qy, where
p, q > 0. Find the condition on p and q, so that the
minimum of Z occurs at (3, 0) and (1, 1).

Sol. O X
A (48,0)
Corner points Corresponding value of (25,0)
Z = px + qy; p, q > 0

(0, 3) p ´ 0 + q ´ 3 = 3q Example 4. Find the maximum value of Z for the

(1, 1) p ´1 + q ´1 = p + q problem maximise Z = 11x + 7y,
subject to constraints x £ 3, y £ 2, x, y ³ 0.
(3, 0) p ´ 3 + q ´ 0 = 3p
Sol. We have, maximise Z = 11 x + 7y,
So, condition of p and q, so that the minimum of Z occurs at
Subject to the constraints are
(3, 0) and (1, 1) is x £ 3, y £ 2, x ³ 0, y ³ 0
p + q = 3p Þ 2p = q
Y
or p = q
2 C B
(0, 2) (3, 2) y = 2
Example 3. Find the maximum value of Z = 4x + 3y, if
X¢ A X
the feasible region for an LPP is shown in following O (0, 0) (3, 0)
figure.

Y

(0, 40)

C (0, 24) B (16, 16) x=3
Y¢
(48, 0) X
The shaded region as shown in the figure as OABC is
OA bounded and the coordinates of corner points are (0, 0),
(25, 0) (3, 0), (3, 2) and (0, 2), respectively.

146 CBSE Term II Applied Mathematics XII

Corner points Corresponding value of Z Corner points Corresponding value of Z = 400x + 200y
3000
(0, 0) 0 (0, 15)
3000
(3, 0) 33 (5, 5) 400 ´ 30 + 200 ´ 30 = 18000

(3, 2) 47 (Maximum) çèæ 30 , 370 ÷öø 7 77
7 = 2571 .43 (Minimum)

(0, 2) 14

Hence, Z is maximum at (3, 2) and its maximum value is 47. Hence, the minimum cost is ` 2571.43.

Example 5. Find the maximum value of Z = 4x + 2y, Example 7. A furniture dealer deals in only two

subject to the constraints items–tables and chairs. He has ` 50000 to invest
2x + 3y £ 18, x + y ³ 10, x, y ³ 0. and has storage space of at most 60 pieces. A table
costs ` 2500 and a chair ` 500. Then, find the
Sol. The given LPP is, Maximise Z = 4x + 2y constraints of the above problem (where, x is
number of tables and y is number of chairs).
Subject to the constraints are
2x + 3y £ 18 Sol. The constraints in the given problem are
x + y ³ 10
x, y³0 x ³ 0, y ³ 0,
2500x + 500y £ 50000
The graph of the above LPP is or 5x + y £ 100

Y x + y £ 60
(0,10)

x + y =10 Example 8. A manufacturer produces two models of

(0,6) (9,0) (10,0) X bikes-model X and model Y. Model X takes a 6
X¢ O 2x + 3y = 18 man-hours to make per unit, while model Y takes
10 man hours per unit. There is a total of 450
Y¢ man-hour available per week. Handling and
marketing costs are ` 2000 and ` 1000 per unit for
From the above graph, we see that for the feasible solution, models X and Y, respectively. The total funds
there is no common area in the first quadrant. available for these purposes are ` 80000 per week.
Hence, the objective function Z cannot be maximised. Profits per unit for models X and Y are ` 1000 and
` 500, respectively. How many bikes of each model
Example 6. Find the minimum value of Z for the should the manufacturer produce, so as to yield a
maximum profit? Find the maximum profit.
problem minimise Z = 400x + 200y, subject to
constraints 5x + 2y ³ 30, 2x + y £ 15, x £ y, x, y ³ 0. Sol. Let the manufacturer produces x number of models X and y
number of model Y bikes.
Sol. We have, minimise Z = 400x + 200y,
Model X takes a 6 man-hours to make per unit and model Y
Subject to the constraints are takes a 10 man-hours to make per unit.
5x + 2y ³ 30, 2x + y £ 15, x £ y, x ³ 0, y ³ 0.
There is total of 450 man-hour available per week.
On solving x - y = 0 and 5x + 2y = 30, we get
y = 30, x = 30 \ 6x + 10y £ 450
77
Þ 3x + 5y £ 225
…(i)

(0, 15) (5, 5) For models X and Y, handling and marketing costs are
(y =
x) ` 2000 and `1000, respectively, total funds available for

these purposes are ` 80000 per week.

30, 30 5x + 2x + y = 15 \ 2000x + 1000y £ 80000
77 2y =
Þ 2x + y £ 80 …(ii)

30 Also, x ³ 0, y ³ 0

On solving x - y = 0 and 2x + y = 15, we get Since, the profits per unit for models X and Y are ` 1000 and
` 500, respectively.
x = 5, y = 5 \Required LPP is
Maximise Z =1000x + 500y
So, from the shaded feasible region it is clear that çèæ øö÷.
Subject to constraints,
coordinates of corner points are ( 0, 15), (5, 5) and 30 , 30 3x + 5y £ 225, 2x + y £ 80, x ³ 0, y ³ 0
7 7

CBSE Term II Applied Mathematics XII 147

From the shaded feasible region, it is clear that coordinates The given problem can be represented diagrammatically as
of corner points are (0, 0), (40, 0), (25, 30) and (0, 45). follows
On solving 3x + 5y = 225 and 2x + y = 80, we get
Godown A
x = 25, y = 30
Y 100 quintals ` 120.500–(x+y)

(0, 80) x `3 y
`6

Shop D Shop DE Shop F
60 quintals 650 quintals 40 quintals

(0, 45) 60 – ` 4 ` 2 50 – y 4`03–{o1r0x0+– y(x–+60y)}
x
(25, 30) Godown B
50 quintals

(40, 0) (75, 0) Let Z be the total cost of transportation, then
X

2x+y=80 3x+5y=225 Z = 6x + 3y + 2.50 (100 - x - y) + 4 ( 60 - x)

+ 2 (50 - y) + 3 [( x + y) - 60]

Corner points Value of Z =1000x + 500y = 6x + 3y + 250 - 2.50x - 2.50y + 240 - 4x
(0, 0) 0
(40, 0) + 100 - 2y + 3x + 3y - 180
(25, 30) 40000 (Maximum)
(0, 45) 25000 + 15000 = 40000 (Maximum) = 2.50 x + 1.50y + 410 ...(i)

22500 Subject to constraints are ... (ii)
60 - x ³ 0
So, the manufacturer should produce 25 bikes of model X ... (iii)
and 30 bikes of model Y to get a maximum profit of ` 40000. or x £ 60
50 - y ³ 0 ... (iv)
Since, in question it is asked that each model bikes should
be produced and the maximum profit is ` 40000. or y £ 50 ... (v)
100 - ( x + y) ³ 0 ... (vi)
Example 9. Two godowns A and B have a grain storage
or x + y £ 100
capacity of 100 quintals and 50 quintals, x + y - 60 ³ 0
respectively. They supply it to 3 ration shops D, E
and F, whose requirements are 60, 50 and 40 or x + y ³ 60
quintals, respectively. The cost of transportation per and x ³ 0, y ³ 0
quintal from godowns to the shops are given in the
following table Example 10. One kind of cake requires 200 g of flour

From Transportation cost per quintal (in `) and 25 g of fat and another kind of cake requires
To 100 g of flour and 50 g of fat. Then, find the
AB maximum number of cakes which can be made from
D 6.00 4.00 5 kg of flour and 1kg of fat assuming that there is no
E storage of the other ingredients used in making the
F 3.00 2.00 cakes.
2.50 3.00
Sol. Let x be the number of cakes of one kind and y be the
number of cakes of other kind. Construct the following table

Kind Number Flour required Fat required

of cakes (in g) (in g)

Formulate the LPP to find how should the supplies I x 200x 25x
be transported in order that the transport cost is
minimum. II y 100y 50y

Sol. Let godown A supplies x and y quintals of grain to the shops Total x + y 200x + 100y 25x + 50y
D and E, respectively.
Requirement 5000 1000
Then, (100 - x - y) will be supplied to shop F.
Our problem is to maximise Z = x + y …(i)
The requirement at shop D is 60 quintals, since x quintals are Subject to constraints are
transported from godown A. …(ii)
200x + 100y £ 5000 …(iii)
Therefore, the remaining ( 60 - x) quintals are transported Û 2x + y £ 50 …(iv)
from godown B. Similarly, (50 - y) quintals and 25x + 50y £ 1000 Û x + 2y £ 40
40 - (100 - x - y) = ( x + y - 60) quintals will be transported and x, y ³ 0
from godown B to shops E and F, respectively.

148 CBSE Term II Applied Mathematics XII

Firstly, draw the graph of the line 2x + y = 50 Putting (0, 0) in the inequality x + 2y £ 40, we have
0 + 2 ´ 0 £ 40
x 0 25
y 50 0 Þ 0 £ 40 (which is true)

Putting (0, 0) in the inequality 2x + y £ 50, we have So, the half plane is towards the origin.

2 ´ 0 + 0 £ 50 Þ 0 £ 50 (which is true) Since, x, y ³ 0

So, the half plane is towards the origin. So, the feasible region lies in the first quadrant.
Secondly, draw the graph of the line x + 2y = 40 On solving equations 2x + y = 50 and x + 2y = 40, we get

x 0 40 B ( 20, 10)
y 20 0 \ Feasible region is OABCO.

Y The corner points of the feasible region are
O ( 0, 0), A( 25, 0), B( 20, 10) and C( 0, 20).

The values of Z at these points are as follows

60 Corner point Z=x+ y

50 O(0, 0) 0
A( 25, 0) 25
40 B( 20, 10) 30 (Maximum)
C( 0, 20) 20
30
20 C(0, 20) X
10 B(20, 10)
Thus, the maximum number of cakes that can be made is 30
(25, 0)A i.e. 20 of one kind and 10 of the other kind.
X¢ O 10 20 30 40 50 60 70 80

(0, 0)
Y¢ 2x + y = 50 x + 2y = 40

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CBSE Term II Applied Mathematics XII 149

Chapter
Practice

PART 1 4. The corner points of the feasible region determined
Objective Questions by the system of linear constraints are (0, 0), (0, 40),
(20, 40), (60, 20), (60, 0). The objective function is
l Multiple Choice Questions Z = 4x + 3y. Compare the quantity in column A and
1. The variables x and y in a linear programming column B.

problem are called Column A Column B
Maximum of Z 325
(a) decision variables
(b) linear variables (a) The quantity in column A is greater
(c) optimal variables (b) The quantity in column B is greater
(d) None of the above (c) The two quantities are equal
(d) The relationship cannot be determined on the basis of the
2. The linear inequality in the graph is shown below
information supplied.
Y
5. The corner points of the feasible region determined
B (0, 6) by the system of linear constraints are (0, 10), (5, 5)
(15, 15), (0, 20). Let Z = px + qy , where p, q > 0.
Then, the condition on p and q so that the maximum

of Z occurs at both the points (15, 15) and (0, 20), is

(a) p = q (b) p = 2q

(0, 3) C 9, 3 (c) q = 2p (d) q = 3p
X¢ A 22
6. Z = 3x + 4y, subject to the constraints
O E (9, 0) x + y £ 1, x ³ 0, y ³ 0.
DX

(6, 0) x + 3y ³ 9 The shaded region shown in the figure as OAB is
bounded and the coordinates of corner points
Y¢ x + y £ 6 O, A and B are (0, 0), (1, 0) and (0, 1), respectively.

The feasible region is ……… . Y

(a) OCAD (b) ADE
(c) ABC (d) None of these

3. Corner points of the feasible region for an LPP are (0, 1) B
(0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y
be the objective function. The minimum value of F X¢ OA X
occurs at (0, 0) (1, 0)

(a) Only (0, 2) y£1
x+
(b) Only (3, 0)
Y¢
(c) the mid-point of the line segment joining the points (0, 2)
and (3, 0) The maximum value of Z is

(d) any point on the line segment joining the points (0, 2) and (a) 2 (b) 4
(3, 0) (c) 1 (d) 0

150 CBSE Term II Applied Mathematics XII

7. The feasible region for an LPP is shown in the 10. The feasible region (shaded) for a LPP is shown in
given figure. following figure.

Y Y

Q 3, 15
24

P 3 ,24 7 , 3
13 13 2
4
R

X¢ O 5 x + 3y = 9 X O X
Y¢ y=
x+ 18, 2
S 7 7

Then, the minimum value of Z = 11 x + 7y is Then, the maximum value of Z = x + 2y is,

(a) 21 (b) 47 (a) 1 (b) 5 (c) 9 (d) 8

(c) 20 (d) 31

8. The feasible solution for a LPP is shown in 11. The maximum value of Z for the problem maximise
following figure. Let Z = 3x - 4y be the objective Z = 2x + y subject to constraints
function. Minimum of Z occurs at x + y £ 2, x ³ 0, y ³ 0 is

Y (a) 4 (b) 3 (c) 1 (d) 0

(4, 10) 12. The linear programming problem minimize
Z = 3x + 2y subject to constraints x + y ³ 8,
(0, 8) (6, 8) 3x + 5y £ 15, x ³ 0 and y ³ 0, has
(6, 5)
(a) one solution (b) no feasible solution

(c) two solutions (d) infinitely many solutions

(0, 0) 13. The minimum value of the objective function
Z = 13x - 15y, subject to the constraints
x + y £ 7, 2x - 3y + 6 ³ 0, x ³ 0, y ³ 0 is

(5, 0) X (a) –32 (b) –30 (c) 20 (d) 30

' 14. The maximum value of the objective function
Z = 3x - 4y, subject to constraints
(a) (0, 0) x + y £ 8, x - y £ 6, x, y ³ 0 is
(b) (0, 8)
(c) (5, 0) (a) 0 (b) 18 (c) 17 (d) 9
(d) (4, 10)
15. The feasible region (shaded) for a LPP is shown in
9. The feasible region for an LPP is shown in the following figure.
following figure.
Y

(12, 6)

B

(0, 4) D x + 2y=76

E

(12, 0) O AC X
2x + y=104
Let F =3x - 4 y be the objective function. Maximum

value of F is Then, the maximum value of Z = 3x + 4y is

(a) 0 (b) 8 (a) 156 (b) 196 (c) 152 (d) 180

(c) 12 (d) -18

CBSE Term II Applied Mathematics XII 151

16. The maximum value of Z = 11x + 7y, She produces only two items M and N each
Subject of constraints 2x + y £ 6, x £ 2, x, y ³ 0 is requiring the use of all the three machines.

(a) 36 (b) 42 The number of hours required for producing 1 unit
(c) 22 (d) 10 of each of M and N on the three machines are given
in the following table.
17. If a furniture dealer estimates that from the sale of
one table he can make a profit of ` 250 and from Items Number of hours required on Machine
the sale of one chair a profit of ` 75 and if x is the
number of chairs and y is the number of tables, then I II III
its linear objective function is
M1 2 1
(a) Z = 75x + 250y
(b) Z = 75x + 25y N2 1 1.25
(c) Z = 250x + 75y
(d) Z = 25x + 75y She makes a profit of ` 600 and ` 400 on items M
and N, respectively.

l Case Based MCQs Let x and y be the number of items M and N,

18. A dietician has to develop a special diet using two respectively and assume that she can sell all the
foods P and Q. Each packet (containing 30 g) of items that she produced.
food P contains 12 units of calcium, 4 units of iron,
6 units of cholesterol and 6 units of vitamin A. Each On the basis of above information, answer the
packet of the same quantity of food Q contains following questions.
3 units of calcium, 20 units of iron, 4 units of
cholesterol and 3 units of vitamin A. The diet (i) Constraints on machine I is given by
required atleast 240 units of calcium, atleast 460
units of iron and at most 300 units of cholesterol. (a) x + 2y £ 12 (b) 2x + y £ 12

(c) x + 5y ³ 1 (d) None of these

Let x and y be the number of packets of food P and (ii) Objective function is given by
Q, respectively. (a) Maximise Z = 600x + 400y
(b) Maximise Z = 400x + 600y
(c) Minimise Z = 600x + 400y

(d) None of the above

On the basis of above information, answer the (iii) How many units of item M should she produce so
following questions. as to maximise her profit?

(i) Constraints on calcium is given by (a) 1 (b) 4
(c) 3 (d) 9
(a) 4x + 3y ³ 240 (b) 12x + 3y ³ 240

(c) 6x + 4y £ 300 (d) None of these (iv) How many units of item N should she produce so
as to maximise her profit?
(ii) Constraints on iron is given by

(a) 4x + 20y ³ 460 (b) 12x + 2y ³ 240 (a) 1 (b) 2
(c) 3 (d) 4
(c) 6x + 4y £ 300 (d) None of these

(iii) Constraints on cholestrol is given by (v) Number of corner points of feasible region is

(a) 6x + 4y £ 300 (b) 4x + 20y £ 460 (a) 5 (b) 4
(c) 3 (d) 2
(c) 3x + y ³ 500 (d) None of these

(iv) How many packets of food P should be used to

minimise the amount of vitamin A in the diet? PART 2
Subjective Questions
(a) 10 (b) 20

(c) 15 (d) 18

(v) How many packets of food Q should be used to

minimise the amount of vitamin A in the diet? l Short Answer Type Questions

(a) 10 (b) 20 1. Let x and y are the number of tables and chairs
respectively, on which a furniture dealer wants to
(c) 25 (d) 5 make profit for the constraints
Maximise Z = 250x + 75y, subject to constraints
19. A manufacturer has three machine I, II and III 5x + y £ 100, x + y £ 60, x ³ 0, y ³ 0.
installed in his factory. Machines I and II are
capable of being operated for at most 12 h, whereas
machine III must be operated for atleast 5 h a day.

152 CBSE Term II Applied Mathematics XII

Consider the following graph 11. Find the maximum value of Z for the problem
maximise Z = 5x - 7y, subject to the constraints
Y x + y £ 7, 2x - 3y + 6 ³ 0, x ³ 0 and y ³ 0.

100 12. Find the maximum value of Z for the problem
maximise Z = 100x + 300y, subject to constraints
90 x + y £ 24, x + 2y £ 32, x, y ³ 0.

80 13. Find the maximum value of Z for the problem
70 C (0, 60) maximise Z = 0.7x + y, subject to constraints
2x + 3y £ 120, 2x + y £ 80, x, y ³ 0.
60 B (10, 50)
50 14. Find the minimum value of Z for the problem,
minimise Z = x + 2y, subject to constraints
40 2x + y ³ 3, x + 2y ³ 6, x, y ³ 0.

30 15. Solve the following linear programming problem
graphically. Maximise Z = 34x + 45y,
20 A(20, 0) Subject to constraints x + y £ 300, 2x + 3y £ 70;
X¢ 10 x, y ³ 0
O X
16. Solve the following linear programming problems
10 20 30 40 50 60 70 graphically.
Minimise Z = x + y, subject to constraints
Y¢ 5x + y = 100 x + y = 60 3x + 2y ³ 12, x + 3y ³ 11, and x ³ 0, y ³ 0.

Then, find the maximum profit to the dealer results 17. If Z = 3x + 5y, subject to the constraints
from buying.
2. Find the maximum value of the objective function x + 2y ³ 10, x + y ³ 6, 3x + y ³ 8, x, y ³ 0
Z = 3x + 4y, subject to the constraints
x + y £ 4, x ³ 0, y ³ 0. Then, find minimum value of Z.
3. If the given constraints are
18. Find the minimum value of z for the problem,
5x + 4y ³ 2, x £ 6, y £ 7, x, y ³ 0 minimise Z = - 3x + 4y, subject to constraints
x + 2y £ 8, 3x + 2y £ 12, x ³ 0, y ³ 0.
Then, find the maximum value of the function
Z = x + 2y. 19. A manufacturer of electronic circuits has a stock
4. Find the maximum value of Z for the problem of 200 resistors, 120 transistors and 150 capacitors
maximise Z = x + y, subject to constraints and is required to produce two types of circuits A
x - y £ - 1, - x + y £ 0, x, y ³ 0. and B. Type A requires 20 resistors, 10 transistors
5. Find the maximum value of Z for the problem and 10 capacitors. Type B requires 10 resistors, 20
maximise Z = 100x + 170y, subject to constraints transistors and 30 capacitors. If the profit on type
3x + 2y £ 3600, x + 4y £ 1800, x, y ³ 0. A circuit is ` 50 and that on type B circuit is ` 60,
6. Find the maximum value of Z for the problem formulate this problem as an LPP, so that the
maximise Z = x + y, subject to constraints manufacturer can maximise his profit.
x + 4y £ 8, 2x + 3y £ 12, 3x + y £ 9, x ³ 0 and y ³ 0.
7. Find the minimum value of Z for the problem 20. A firm has to transport 1200 packages using large
minimise Z = 2x + y, subject to constraints vans which can carry 200 packages each and small
3x + y ³ 9, x + y ³ 7, x + 2y ³ 8, x, y ³ 0. vans which can take 80 packages each. The cost
8. Find the maximum value of Z for the problem for engaging each large van is ` 400 and each
maximise Z = 200x + 120y, subject to constraints small van is ` 200. Not more than ` 3000 is to be
x + y £ 300, 3x + y £ 600, x - y ³ - 100, x, y ³ 0. spent on the job and the number of large vans
9. Determine the maximum value of Z = 11x + 7y cannot exceed the number of small vans.
Subject to the constraints Formulate this problem as a LPP given that the
2x + y £ 6, x £ 2 , x ³ 0, y ³ 0. objective is to minimise cost.
10. Find the maximum value of Z for the problem
maximise Z = x + y, subject to constraints
2x + 3y £ 120, 8x + 5y £ 400, x, y ³ 0.

CBSE Term II Applied Mathematics XII 153

21. A company manufactures two types of screws A and Assuming that he can sell all the screws he
B. All the screws have to pass through a threading manufactures, formulate LPP to find how many
machine and a slotting machine. A box of type A packets of each type should the factory owner
screws requires 2 min on the threading machine produce in a day in order to maximise his profit.
and 3 min on the slotting machine. A box of type B
screws requires 8 min on the threading machine l Long Answer Type Questions
and 2 min on the slotting machine. In a week, each
machine is available for 60 h. On selling these 26. Solve the following problem graphically. [NCERT]
screws, the company gets a profit of ` 100 per box Minimise and maximise Z = 3x + 9y,
on type A screws and ` 170 per box on type B screws. Subject to constraints
x + 3y £ 60, x + y ³ 10, x £ y; x ³ 0, y ³ 0
Formulate this problem as a LPP given that the
objective is to maximise profit. [NCERT Exemplar] 27. Solve the following LPP graphically.
Minimise Z = 5x + 10y,
22. A company manufactures two types of sweaters Subject to the constraints
x + 2y £ 120, x + y ³ 60, x - 2y ³ 0 and x, y ³ 0
type A and type B. It costs ` 360 to make a type A
sweater and ` 120 to make a type B sweater. The 28. Find graphically, the maximum value of
company can make atmost 300 sweaters and spend Z = 2x + 5y,
Subject to constraints given below.
atmost ` 72000 a day. The number of sweaters of 2x + 4y £ 8, 3x + y £ 6, x + y £ 4, x ³ 0, y ³ 0 .

type B cannot exceed the number of sweaters of 29. Maximise and minimise Z = x + 2y,

type A by more than 100. The company makes a Subject to the constraints
profit of ` 200 for each sweater of type A and ` 120 x + 2y ³ 100, 2x - y £ 0, 2x + y £ 200 and x, y ³ 0

for every sweater of type B. Solve the above LPP graphically.

Formulate this problem as a LPP to maximise the

profit to the company. [NCERT Exemplar]

23. A cooperative society of farmers has 50 hectare of 30. Determine graphically the minimum value of the
land to grow two crops X and Y. The profit from objective function Z = - 50x + 20y,
crops X and Y per hectare are estimated as `10500 Subject to constraints
and ` 9000, respectively. To control weeds, a liquid 2x - y ³ - 5, 3x + y ³ 3, 2x - 3y £ 12
herbicide has to be used for crops X and Y at rates and x ³ 0, y ³ 0.
of 20 L and 10 L per hectare. Further no more than
800 L of herbicide should be used in order to 31. Solve the linear programming problems graphically.
protect fish and wildlife using a pond which collects Maximise Z = - x + 2y,
drainage from his land. Subject to constraints are x ³ 3, x + y ³ 5, x + 2y ³ 6
and x, y ³ 0.
To maximise the total profit of the society, find how
much hectare of land should be allocated to each 32. Solve the following LPP graphically.
crop? Maximise Z = 3x + 2y,
Subject to constraints are x + 2y £ 10, 3x + y £ 15
24. Anil wants to invest atmost ` 12000 in bonds A and and x ³ 0, y ³ 0.
B. According to the rules, he has to invest atleast
` 2000 in bond A and atleast ` 4000 in bond B. If Also, determine the area of the feasible region.
the rate of interest in bond A is 8% per annum and
on bond B is 10% per annum, then to maximise the 33. Find the maximum value of Z for the problem
interest, then find the investment in bond A and B. maximise Z = 2x + 3y,
Subject to constraints x + 2y £ 10, 2x + y £ 14, x, y ³ 0.
25. A factory manufactures two types of screws A and
B, each type requiring the use of two machines, an 34. A dietician wishes to mix two types of foods f1 and
automatic and a hand-operated. It takes 4 min on f 2 in such a way that the vitamin contents of the
the automatic and 6 min on the hand-operated mixture contain atleast 6 units of vitamin A and 8
machines to manufacture a packet of screw ‘A’ while units of vitamin B. Food f1 contains 2 units/kg of
it takes 6 min on the automatic and 3 min on the vitamin A and 3 units/kg of vitamin B while food f 2
hand-operated machine to manufacture a packet of contains 3 units/kg of vitamin A and 2 units/kg of
screw ‘B’. Each machine is available for atmost 4 h vitamin B. Food f1 costs ` 50 per kg and food f 2
on any day. The manufacturer can sell a packet of costs ` 75 per kg. Formulate the problem as an LPP
screw ‘A’ at a profit of 70 paise and screw ‘B’ at a to minimise the cost of mixture.
profit of `1.

154 CBSE Term II Applied Mathematics XII

35. A manufacturer produces two products A and B. makes a profit of ` 80 on each piece of type A and
Both the products are processed on two different ` 120 on each piece of type B. How many pieces of
machines. The available capacity of first machine is type A and type B should be manufactured per week
12 h and that of second machine is 9 h per day. to get a maximum profit? Make it as an LPP and
solve it graphically. What is the maximum profit per
Each unit of product A requires 3 h on both week?
machines and each unit of product B requires 2 h
on first machine and 1 h on second machine. Each 38. There are two types of fertilisers F1 and F2, F1
unit of product A is sold at a profit of ` 7 and B at a consists of 10% nitrogen and 6% phosphoric acid
profit of ` 4. Find the production level per day for and F2 consists of 5% nitrogen and 10% of
maximum profit graphically. phosphoric acid. After testing the soil conditions, a
farmer finds that she needs atleast 14 kg of nitrogen
36. There is a factory located at each of the two places P and 14 kg of phosphoric acid for her crop. If F1
and Q. From these locations, certain commodity is costs ` 6 per kg and F2 costs ` 5 per kg, determine
delivered to each of the three depots situated at how much of each type of fertiliser should be used
A, B and C. The weekly requirements of the depots so that nutrient requirements are met at a minimum
are respectively 5, 5 and 4 units of the commodity cost? What is the minimum cost?
while the production capacity of the factory at P and
Q are respectively 8 and 6 units. The cost of l Case Based Questions
transportation per unit is given below
39. A company manufactures three kinds of calculators:
To A Cost (in `) C A, B and C in its two factories I and II. The
From B company has got an order for manufacturing atleast
6400 calculators of kind A, 4000 of kind B and 4800
P 16 10 15 of kind C. The daily output of factory I is of 50
Q 10 12 10 calculators of kind A, 50 calculators of kind B and
30 calculators of kind C. The daily output of factory
How many units should be transported from each II is 40 calculators of kind A, 20 of kind B and 40 of
factory to each depot in order that the kind C. The cost per day to run factory I is ` 12000
transportation cost is minimum? What will be the and of factory II is ` 15000.
minimum transportation cost?
Let factory I run x days and factory II run y days.
37. A manufacturing company makes two types of On the basis of above information, answer the
teaching aids A and B of Mathematics for Class XII. following questions.
Each type of A requires 9 labour hours of
fabricating and 1 labour hour for finishing. Each (i) Formulate this problem as an LPP.
type of B requires 12 labour hours for fabricating
and 3 labour hours for finishing. For fabricating and (ii) Check whether feasible region is bounded or not
finishing, the maximum labour hours available per
week are 180 and 30, respectively. The company (iii) How many days do the factory I has to be in
operation to produce the order with the minimum
cost.

CBSE Term II Applied Mathematics XII 155

SOLUTIONS

Objective Questions

1. (a) The variables x and y in a linear programming problem Y
are called decision variables.

2. (c) The common region determined by all the constraints C
including non-negative constraints x, y ³ 0 of a linear
programming problem is called the feasible region (or BA
solution region) for the problem. In the given figure, the
region ABC (shaded) is the feasible region for the problem.

3. (d)

Corner points Corresponding value of X¢ x+y=5 x + 3y = 9 X
F = 4x + 6y O
Y¢
(0, 2) 4 ´ 0 + 6 ´ 2 = 12 (Minimum)

(3, 0) 4 ´ 3 + 6 ´ 0 = 12 (Minimum) Let A, B and C be the corner points of the feasible region
which is bounded.
(6, 0) 4 ´ 6 + 6 ´ 0 = 24 \ The coordinates of A are (3, 2).

(6, 8) 4 ´ 6 + 6 ´ 8 = 72 (Maximum) The coordinates of B are (0, 3).
and the coordinates of C are (0, 5).
(0, 5) 4 ´ 0 + 6 ´5 = 30 The given objective function is Z = 11x + 7y.
The values of Z at the corner points are given by
Hence, minimum value of F occurs at any points on the line
segment joining the points (0, 2) and (3, 0).

4. (b)

Corner points Corresponding value of Corner point Corresponding value of Z
Z = 4x + 3y (3, 2) 11 ´ 3 + 7 ´ 2 = 47
(0, 3)
(0, 0) 4´0+ 3´0 =0 (0, 5) 11 ´ 0 + 7 ´ 3 = 21 (Minimum)
11 ´ 0 + 7 ´5 = 35
(0, 40) 4 ´ 0 + 3 ´ 40 = 120

(20, 40) 4 ´ 20 + 3 ´ 40 = 200 From the above table, we see that the minimum value of Z
is 21.
(60, 20) 4 ´ 60 + 3 ´ 20 = 300 (Maximum)
8. (b)

(60, 0) 4 ´ 60 + 3 ´ 0 = 240 Corner points Corresponding value of
Z = 3x - 4y
Hence, maximum value of Z = 300 < 325.
So, the quantity in column B is greater. (0, 0) 3´0-4´0 =0

5. (d) The corner points of the feasible region are (5, 0) 3 ´5 - 4 ´ 0 = 15 (Maximum)

(0, 10), (5, 5), (15, 15), (0, 20) (6, 5) 3 ´ 6 - 4 ´5 = -2
The given objective function is Z = px + qy, where p, q > 0
(6, 8) 3 ´ 6 - 4 ´ 8 = -14

It is given that the maximum value of Z occurs at the points (4, 10) 3 ´ 4 - 4 ´10 = -28

(15, 15) and (0, 20). (0, 8) 3 ´ 0 - 4 ´ 8 = -32 (Minimum)
\ We have, p×15 + q ×15 = p× 0 + q ×20
Þ ( p + q) 15 = q ×20 Þ 3( p + q) = 4 q Hence, the minimum of Z occurs at (0, 8) and its minimum
\ 3p = q value is (-32).
9. (c) The feasible region as shown in the figure, has objective
6. (b) function F = 3x - 4y.

Corner points Corresponding value of Corner points Corresponding value of
Z = 3x + 4y F = 3x - 4y
(0, 0)
(0, 0) 3´0+ 4´0 =0 (12, 6) 3´0-4´0 =0
(0, 4)
(1, 0) 3 ´1 + 4 ´0 = 3 3 ´12 - 4 ´ 6 = 12 (Maximum)

(0, 1) 3 ´ 0 + 4 ´1 = 4 (Maximum) 3 ´ 0 - 4 ´ 4 = -16 (Minimum)

Hence, the maximum value of Z is 4 at (0, 1). Hence, the maximum value of F is 12.
7. (a) The given feasible region is

156 CBSE Term II Applied Mathematics XII

10. (c) From the shaded bounded region, it is clear that the Y

coordinates of corner points are èæç 3, 24 ÷øö, æçè 18 , 2 ø÷ö, çæè 7, 43 ÷öø
13 13 7 7 2

and çèæ 3, 15 öø÷. 8
2 4 7
6
Also, we have to determine maximum value of Z = x + 2y. 5
4
Corner points Corresponding value of Z 3
3 + 48 = 51 = 312 2
çæè 3 , 24 ÷öø 13 13 13 13 1
13 13
18 + 4 = 22 = 31 (Minimum)
æèç 18 , 72 ö÷ø 777 7 X¢ O 1 2 3 4 5 6 7 8 X
7
7 + 6 = 20 = 5 Y¢
çæè 7, 43 ÷öø 24 4
2 3 + 30 = 36 = 9 (Maximum) It can be concluded from the graph, that there is no point,
24 4 which can satisfy all the constraints simultaneously.
çæè 3, 15 ö÷ø
2 4 Therefore, the problem has no feasible solution.

Hence, the maximum value of Z is 9. 13. (b) The given LPP is
11. (a) We have, maximise Z = 2x + y, Minimise Z = 13x - 15y

Subject to the constraints are Subject to the constraints are
x + y £ 2, x ³ 0, y ³ 0 x+ y£7

The shaded region shown in the figure as OAB is bounded 2x - 3y + 6 ³ 0
and the coordinates of corner points O, A and B are (0, 0), x, y³0
(2, 0) and (0, 2), respectively.
The graph of the given LPP is
Y
Y

(0, 2) B 2x – 3y =
–6

X¢ OA X B(3,4)
(0, 0) (2, 0)

2) (0,2) C
y=
(x +

Y¢

Corner points Corresponding value of Z X¢ O A (7,0)
Y¢ X

x+y=7

(0, 0) 0

(2, 0) 4 (Maximum) From the above graph, we see that the shaded region is the
feasible OABC which is bounded.
(0, 2) 2
Now, the values of Z at the corner points are given by

Hence, the maximum value of Z is 4 at (2, 0). Corner point Corresponding value of Z
12. (b) Table for equation x + y = 8 is (0, 0) 0
(7, 0) 91
x 0 8 (3, 4)
y=8-x 8 0 (0, 2) -21
-30
Table for equation 3x + 5y = 15 is

x 0 5 From the above table, we see that the minimum value of Z
y = 15 - 3x 3 0 is -30.

5

CBSE Term II Applied Mathematics XII 157

14. (b) The given LPP is Maximise Z = 3x - 4y From the above graph, we see that the shaded region is the
Subject to constraints are feasible region OABC which is bounded.
x + y £ 8, x - y £ 6, x, y ³ 0 \ The maximum value of the objective function Z occurs at
Y the corner points.

(0,8) The corner points are O (0, 0), A (0, 6), B (2, 2), C (2, 0).

The values of Z at these corner points are given by

(7,1) Corner point Corresponding value of
Z =11x + 7 y
X¢ (6,0) X (0, 0)
(0,0) x+y=8 (0, 6) 0
(2, 2) 42 (Maximum)
Y¢ x – y = 6 (2, 0)
36
\We get the bounded feasible region 22
Corner Points
Z = 3x - 4y Thus, the maximum value of Z is 42 which occurs at the
point (0, 6).
(0, 0) 0
(6, 0) 18 17. (a) Z = 75x + 250y is the linear objective function for the
(7, 1) 17
(0, 8) - 32 given problem.

18. (i) (b) Constraint on calcium is given by
12x + 3y ³ 240

Hence, the maximum value of Z is 18. (ii) (a) Constraints on iron is given by

15. (b) As clear from the graph, corner points are O, A, E and D 4x + 20y ³ 460

with coordinates (0, 0), (52, 0), (44, 16) and (0, 38), (iii) (a) Constraints on Cholesterol is given by
6x + 4y £ 300
respectively. Also, given region is bounded.

Here, Z = 3x + 4y (iv) (c) Mathematical formulation of the given problem is as
follows :
Q 2x + y = 104 and 2x + 4y = 152
Þ -3y = - 48 Þ y = 16 and x = 44 Minimise Z = 6x + 3y (Vitamin A)

Corner points Corresponding value of Z Subject to constraints are …(i)
(0, 0) 0 12x + 3y ³ 240 or 4x + y ³ 80
(52, 0) 156
(44, 16) 196 (Maximum) 4x + 20y ³ 460 or x + 5y ³ 115 …(ii)
(0, 38)
152 6x + 4y £ 300 or 3x + 2y £ 150 …(iii)
x, y ³ 0 …(iv)

Let us graph the inequalities (i) to (iv).

Hence, Z at (44, 16) is maximum and its maximum value The feasible region (shaded determined by the
is 196. constraints (i) to (iv) is shown below and note that it is
16. (b) The given LPP is Maximise Z = 11x + 7y bounded.

Subject to the constraints are Y
2x + y £ 6, x £ 2, x ³ 0 , y ³ 0
80
The corresponding graph of the above LPP is L(2, 72)

Y 60

A (0,6) (0, 23) 40 (15, 20)
X¢
20 N(40, 15) (115, 0)
M 40 60 80 X

O 20 100 120 x + 5y = 115

B (2,2) Y¢ 4x + y = 80
3x + 2y = 150

X¢ O (2,0) C (3,0) X The coordinates of the corner points L, M and N are
Y¢ (2, 72), (15, 20) and (40, 15), respectively.

158 CBSE Term II Applied Mathematics XII

Let us evaluate Z at these points We see that the point (4, 4) is giving the maximum
value of Z.
Corner point Z = 6x + 3y
(2, 72) 6 ´ 2 + 3 ´ 72 = 228 Hence, 4 units of item M should be produced to
(15, 20) 6 ´ 15 + 3 ´ 20 =150 (Minimum) maximise the profit.
(40, 15) 6 ´ 40 + 3 ´ 15 = 285
(iv) (d)Q The point (4, 4) is giving the maximum value of Z.
From the table, we find that Z is minimum at the point \ 4 units of item N should be produced to maximise the
(15, 20). profit.

Hence, 15 packets of food P should be used to minimise (v) (a)Q Corner points of the feasible region are (5, 0),
the amount of vitamin A in the diet. (6, 0), (4, 4), (0, 6) and (0, 4).
\ Required answer is 5.
(v) (b) Since Z is minimum at the point (15, 20).
\ 20 packets of food Q should be used to minimise the Subjective Questions
the amount of vitamin A in the diet.
1. The graph of the given constraints is
19. (i) (a) Constraints on machine I is given by x + 2y £ 12.
Y

(ii) (a) Objective function is given by 100
maximise Z = 600x + 400y
90

(iii) (b) Mathematical formulation of the given problem is as 80
follows 70 C (0, 60)
Maximise Z = 600x + 400y
60 B (10, 50)
50

Subject to the constraints are 40
x + 2y £ 12
2x + y £ 12 …(i) 30
x + 5 y³5 …(ii)
4 x ³ 0, y ³ 0 20 A(20, 0)
…(iii) X¢ 10
…(iv) O X

10 20 30 40 50 60 70

Let us draw the graph of constraints (i) to (iv). ABCDE Y¢ 5x + y = 100 x + y = 60
is the feasible region (shaded) as shown below. Observe
that the feasible region is bounded, coordinates of The corner points (vertices) of the bounded (feasible) region
corner points A, B,C, D and E are (5, 0), (6, 0), (4, 4), are O, A, B and C and it is easy to find their coordinates as
(0, 6) and (0, 4), respectively. (0, 0), (20, 0), (10, 50) and (0, 60), respectively.

2x + y = 12 Let us now compute the values of Z at these points.
Y We have, Z = 250x + 75y

12 (0, 12) Vertex of the Corresponding value

10 feasible region of Z (in `)

8 O (0, 0) 0

x + 5 y = 5 6 D(0, 6) C (0, 60) 4500
4
B (10, 50) 6250 (Maximum)
C(4, 4)
4
A (20, 0) 5000
E(0, 4)
2

X¢ B(6, 0) We observe that the maximum profit to the dealer results

O X from the investment strategy (10, 50), i.e. buying 10 tables
2 4 6 8 10 12
and 50 chairs and the maximum profit is ` 6250.

Y¢ A(5, 0) 2. Our problem is to maximise …(i)
Let us evaluate Z = 600x + 400y at these corner points. Z = 3x + 4y

Corner point Z = 600x + 400y Subject to the constraints are …(ii)
(5, 0) 600 ´ 5 + 400 ´ 0 = 3000 x+ y£4 …(iii)
(6, 0) 600 ´ 6 + 400 ´ 0 = 3600
(4, 4) 600 ´ 4 + 400 ´ 4 = 4000 (Maximum) x ³ 0, y ³ 0
(0, 6) 600 ´ 0 + 400 ´ 6 = 2400 Draw the graph of the line x + y = 4.
(0, 4) 600 ´ 0 + 400 ´ 4 = 1600
x04

y40

CBSE Term II Applied Mathematics XII 159

Putting (0, 0) in the inequality x + y £ 4, we have Also, the maximum value of Z will occur at corner points.
The values of Z at corner points are given by
0+ 0£4 Þ 0£4 (which is true)

Y Corner point Corresponding value of Z
2
6 çæè 2 , 0ö÷ø 5
5
5

4 B(0,4) (6, 0) 6
(6, 7) 20 ¬ Maximum
3

2 (0,7) 14
æèç 0, 12÷øö 1
1 A(4,0)

X¢ O X
(0,0)
123456

Y¢ From the above table, we see that the maximum value of Z is
20 which occurs at (6, 7).
So, the half plane is towards the origin. Since, x, y ³ 0.
4. Our problem is to maximise ...(i)
So, the feasible region lies in the first quadrant. Z=x+ y
\ Feasible region is OABO.
Subject to constraints are x - y £ - 1 ...(ii)
The corner points of the feasible region are O(0, 0), A(4,0) -x+ y£0 ...(iii)
and B(0, 4). The values of Z at these points are as follows ...(iv)
and x ³ 0, y ³ 0
Corner point Z = 3x + 4y
O( 0, 0) Table for line x - y = - 1 is
A( 4, 0) 3´0+ 4´0 = 0
B( 0, 4) 3 ´ 4 + 4 ´ 0 = 12 x 0 –1
3 ´ 0 + 4 ´ 4 = 16 (Maximum)
y10
Therefore, the maximum value of Z is 16 at the point B( 0, 4).
So, the line x - y = -1 is passing through the points ( 0, 1)
3. The given LPP is and ( - 1, 0).
Maximise Z = x + 2y On putting (0, 0) in the inequality x - y £ - 1, we get

Subject to the constraints are 0 - 0 £ - 1 Þ 0 £ - 1 , which is not true.
5x + 4y ³ 2 So, the half plane is away from the origin.
x£6 Table for line - x + y = 0 is
y£7
x, y³0 x01
y01
The graph of the above LPP is
So, the line -x + y = 0 is passing through ( 0, 0) and (1, 1).
Y On putting (2, 0) in the inequality - x + y £ 0, we get

D(0,7) C (6,7) -2+ 0£0
y=7 Þ - 2 £ 0 , which is true.

0 , 1 E So, the half plane is towards the X-axis.
2 Also, x, y ³ 0.
X¢ B (6,0)
O A x=6 X So, the region lies in the I quadrant.

5x + 4y = 22 , 0 Y x – y = –1
5
6 –x + y = 0
Y¢ 5
4
From the above graph we see that the shaded region is the 3
feasible region ABCDE which is bounded.
2
1
X¢ –1 O 1 2 3 4 5 6 7 8 X

Y¢

From the above graph it is clearly shown that there is no
common region.

Hence, there is no feasible region and thus Z has no
maximum value.

160 CBSE Term II Applied Mathematics XII

5. We have, Corner points Value of Z = x + y
Maximise Z =100 x + 170y (0, 0) 0
Subject to the constraints are
3x + 2y £ 3600, (3, 0) 3
x + 4y £ 1800, 43 = 3 10 (Maximum)
x ³ 0, y ³ 0 èçæ 28 , 15 ø÷ö 11 11
From the shaded feasible region it is clear that the 11 11
coordinates of corner points are (0, 0), (1200, 0), (1080, 180)
and (0, 450). (0, 2) 2
On solving x + 4y = 1800 and 3x + 2y = 3600, we get
x = 1080 and y = 180 Hence, the maximum value is 3 10.
Y 11

(0, 1800) 7. We have, minimise Z = 2x + y,

(0, 450) (1080, 180) Subject to the constraints are
O (0, 0) 3x + y ³ 9, x + y ³ 7, x + 2y ³ 8, x ³ 0, y ³ 0
(1200, 0) x+4y=1800 X
(1800, 0) From the shaded graph, we see that for the shown
3x+2y=3600 unbounded region, we have coordinates of corner points A,
B, C and D as (8, 0), (6, 1), (1, 6), and (0, 9), respectively.

[on solving x + 2y = 8 and x + y = 7, we get x = 6, y= 1 and
on solving 3x + y = 9 and x + y = 7, we get x = 1, y = 6 ]

(0, 9)

Corner points Corresponding value of Z =100x + 170y (0, 7)
(0, 0) 0 (1, 6)
C
(1200, 0) 1200 ´ 100 = 120000
(1080, 180) 100 ´ 1080 + 170 ´ 180 = 138600 (Maximum) (0, 4)

(0, 450) 0 + 170 ´ 450 = 76500 (6, 1)B x+2y=8

(3, 0) (4, 0)(7, 0) A (8, 0)

3x+y=9 2x+y=8 x+y=7

Hence, the maximum value of Z is 138600. Corner points Value of Z =2x + y
(8, 0) 16
6. Here, the given LPP is, (6, 1) 13
Y (1, 6) 8 (Minimum)
(0, 9) 9
(0, 9)

(0, 4) 28, 15 Thus, we see that 8 is the minimum value of Z at the corner
(0, 2) 11 11 point (1, 6). Here, we see that the feasible region is
unbounded.
(0, 0) (3, 0) (6, 0) (8, 0) X
3x+y=9 x+4y=8 Therefore, 8 may or may not be the minimum value of Z. To
decide this issue, we graph the inequality
2x+3y=12
2x + y < 8
Maximise Z = x + y,
and check whether the resulting open half has points in
Subject to the constraints are common with feasible region or not. If it has common point,
then 8 will not be the minimum value of Z, otherwise 8 will
x + 4y £ 8, 2x + 3y £ 12, 3x + y £ 9, x ³ 0, y ³ 0 be the minimum value of Z.

On solving x + 4y = 8 and 3x + y = 9, we get Thus, from the graph it is clear that, it has no common point.
Therefore, Z = 2x + y has 8 as minimum value subject to the
x = 28, y = 15 given constraints.
11 11 8. We have, maximise Z = 200x + 120 y

From the feasible region, it is clear that coordinates of Subject to the constaints are
x + y £ 300, 3x + y £ 600, x - y ³ - 100, x ³ 0, y ³ 0
corner points are (0, 0), (3, 0), çèæ 28 , 15 ö÷ø and (0, 2).
11 11

CBSE Term II Applied Mathematics XII 161

On solving x + y = 300 and 3x + y = 600, we get Corner points Corresponding value of Z =11x + 7 y
x = 150, y = 150 (0, 0) 0
(2, 0) 22
On solving x - y = - 100 and x + y = 300, we get (2, 2) 36
x = 100, y = 200 (0, 6) 42 (Maximum)

Y

(0, 600)

Hence, the maximum value of Z is 42 at (0, 6).

10. We have,

Y

(0, 300) (150, 150) (0, 80)
(100, 200)

(0, 100)

(–100, 0) (0, 0) (200, 0) (300, 0) X (0, 40) 300, 80
x – y=–100 x+y=300 (0, 0) 77

3x+y=600 (60, 0) X

(50, 0) 2x+3y=120

From the shaded feasible region it is clear that coordinates 8x+5y=400
of corner points are (0, 0), (200, 0), (150, 150), (100, 200) and
(0, 100). Maximise Z = x + y,

Corner points Corresponding value of Z = 200x + 120y Subject to the constraints are
2x + 3y £ 120, 8x + 5y £ 400, x ³ 0,y ³ 0
(0, 0) 0
On solving, we get
(200, 0) 40000 8x + 5y = 400 and 2x + 3y = 120
(150, 150) 150 ´ 200 + 120 ´ 150 = 48000 (Maximum) x = 300, y = 80
(100, 200) 77
(0, 100) 100 ´ 200 + 120 ´ 200 = 44000
120 ´ 100 = 12000 From the shaded feasible region, it is clear that coordinates

Hence, the maximum value of Z is 48000. of corner points are (0, 0), (50, 0), çæè 300 , 870 öø÷ and (0, 40).
9. We have, maximise Z = 11x + 7y 7

...(i) Corner points Corresponding value of Z = x + y
(0, 0) 0
Subject to the constraints are (50, 0) 50

2x + y £ 6 …(ii) çæ 300, 80÷ö 380 = 54 2 (Maximum)
è 7 7ø 77
x£2 …(iii)

x ³ 0, y ³ 0 …(iv)

We see that, the feasible region as shaded determined by (0, 40) 40
the system of constraints (ii) to (iv) is OABC and is bounded.
So, now we shall use corner point method to determine the

maximum value of Z. Hence, the maximum value of Z is 54 2.
7
Y
11. Minimise Z = 5x - 7y,
(0, 6) C
Subject to the constraints are
x + y £ 7, 2x - 3y + 6 ³ 0, x ³ 0, y ³ 0

Y

B (2, 2) (0, 7) B
C (3, 4)
(0, 2)
(3, 0)
X¢ O
(0, 0) A (2, 0) X X¢ A
Y¢ x=2 (–3, 0) (0, 0) (7, 0)
X

2x + y = 6 2x – 3y+6=0 Y¢ x+y=7

162 CBSE Term II Applied Mathematics XII

Shaded region shown as OABC is bounded and coordinates The value of Z at the corner points are given below
of its corner points are (0, 0), (7, 0), (3, 4) and (0, 2),
respectively. Corner points Z =100x + 300y

O( 0,0) Z = 100 ´ 0 + 300 ´ 0 = 0

Corner points Corresponding value of Z A( 24,0) Z = 100 ´ 24 + 300 ´ 0 = 2400

(0, 0) 0 B(16, 8) Z = 100 ´16 + 300 ´ 8 = 4000

(7, 0) 35 (Maximum) C( 0,16) Z = 100 ´ 0 + 300 ´16 = 4800

(3, 4) -6 Hence, maximum value of Z is 4800.
13. We have, maximize Z = 0.7x + y
(0, 2) -14
Subject to the constraints are,
Hence, the maximum value of Z is 35 at (7, 0). ...(i) 2x + 3y £ 120 …(i)
12. We have, Z = 100 x + 300y. ...(ii) 2x + y £ 80 …(ii)
...(iii) …(iii)
Consider the constraints as linear equations and x, y ³ 0
x + y = 24 Now, let us draw the graph of inequalities (i) to (iii).

x + 2y = 32 Y

and x = 0, y = 0 120
Table for line x + y = 24 is

x 0 24 100

y 24 0 80

So, line passes through the points ( 0, 24) and ( 24, 0). 60
On putting (0, 0) in the inequality x + y £ 24, we get
0 + 0 £ 24Þ 0 £ 24, which is true. 40 C
So, the half plane is towards the origin.
Table for line x + 2y = 32 is 20 B
X¢ O
x 0 32 20 A 80 100 120 X
y 16 0 40 60 2x+3y=120
A
So, line passes through the points ( 0, 16) and ( 32, 0).
On putting (0, 0) in the inequality x + 2y £ 32, we get Y¢
( 0) + 2( 0) £ 32 Þ 0 £ 32, which is true.
So, the half plane is towards the origin. 2x+y=80
Also, x ³ 0, y ³ 0, which shows that the region lies in
I quadrant. Clearly, the feasible region is the shaded region, whose
The, intersection point of lines x + y = 24 and corner points are O, A, B and C.
1 x + y = 16 is B(16, 8).
2 The coordinates of O, A and C are (0, 0), (40,0) and (0, 40)
respectively.
Y
Let us find the coordinates of B which is the intersection
x + y = 24 point of 2x + y = 80 and 2x + 3y = 120.

(0, 16) C B (16, 8) 1 On solving these two equations, we get
X¢ 2 x + y = 16 x = 30

(0, 0) O (24, 0) A and y = 20

Thus, the coordinates of B are (30, 20).

Now, let us find the value of Z at corner points, as shown in
the following table.

X Corner points Value of Z = 0.7 x + y

O (0, 0) 0

Y¢ A (40,0) 28

On plotting the graph of Eqs. (i), (ii) and (iii), we get the B (30, 20) 41 (Maximum)
feasible region OABCO, whose corner points are O ( 0, 0),
A ( 24, 0), B (16, 8) and C ( 0,16). C (0, 40) 40

Hence, the maximum value of Z is 41.

CBSE Term II Applied Mathematics XII 163

14. Our problem is to minimise, Z = x + 2y ...(i) 15. We have the following LPP

Subject to constraints are, ...(ii) Maximise Z = 34x + 45y,
2x + y ³ 3 ...(iii)
x + 2y ³ 6 ...(iv) Subject to the constraints are
x + y£ 300, 2x + 3y£ 70; x, y³ 0
and x ³ 0 , y ³ 0
Table for line 2x + y = 3 is Now, considering the inequations as equations, we get

x + y= 300 …(i)

x 0 3/2 and 2x + 3y= 70 …(ii)

Table for line x + y= 300 is

y30 x 0 300
y 300 0
So, line passes through the points ( 0, 3) and æçè 3, 0øö÷.
2
So, the line passes through the points ( 0, 300) and ( 300, 0).
On putting (0, 0) in the inequality 2x + y ³ 3, we get On putting ( 0, 0) in the inequality x + y £ 300, we get
2 ´ 0 + 0 ³ 3 Þ 0 ³ 3, which is not true.
0+ 0 £ 300, which is true.
So, the half plane is away from the origin. So, the half plane is towards the origin.
Table for line x + 2y = 6 is Table for line 2x + 3y= 70 is

x06 x 35 0
y30
y 0 70/3

So, line passes through the points ( 0, 3) and ( 6, 0). So, the line passes through the points (35, 0) and (0, 70/3).
On putting (0, 0) in the inequality x + 2y ³ 6, we get On putting (0, 0) in the inequality 2x + 3y £ 70, we get

0+ 2´0³6 0 + 0 £ 70, which is true.
Þ 0 ³ 6 which is not true.
So, the half plane is towards the origin.
So, the half plane is away from the origin. Also, x, y ³ 0, so the region lies in the 1st quadrant.
Also, x ³ 0 and y ³ 0, so the region lies in the I quadrant.
The intersection point of the lines x + 2y = 6 and 2x + y = 3 The graphical representation of the system of inequations is
is B( 0, 3). as given below

Y Y
(0, 300)

6 B(0, 3) (0, 70/3) B x+y=300
5 x+2y=6 X¢
4 A(6, 0) (0, 0) O Y¢ (35, 0) (300, 0)
3 A
2 12345678 X
1 2x+3y=70
X¢
(0, 0)O X Clearly, the feasible region is OABO, where the corner
points are O( 0, 0), A( 35, 0) and B( 0, 70/ 3).
Y¢ 3 , 0 2x+y=3
2 Now, the values of Z at corner points are as follow

The corner points of the feasible region are A( 6, 0) and Corner points Z = 34 x + 45y
B( 0, 3). O( 0, 0) Z=0+ 0=0
A( 35, 0) Z = 34 ´ 35 + 45 ´ 0 = 1190
The values of Z at the corner points are given below Z = 34 ´ 0 + 45 ´ 70 = 1050
B( 0, 70/ 3)
Corner points Z = x + 2y 3
A( 6, 0) Z = 6+ 2´0 = 6
B( 0, 3) Z = 0+ 2´3 = 6

As the feasible region is unbounded, therefore 6 may or may Hence, the maximum value of Z is 1190 at ( 35, 0). ...(i)
not be the minimum value of Z. For this we draw a dotted 16. Given, minimise Z = x + y, ...(ii)
graph of the inequality x + 2y < 6 and check whether the ...(iii)
resulting half plane has points in common with a feasible Subject to constraints are
region or not. It can be seen that the feasible region has no 3x + 2y ³ 12
common point with x + 2y < 6. x + 3y ³ 11

Therefore, the minimum value of Z is 6. and x ³ 0, y ³ 0

164 CBSE Term II Applied Mathematics XII

Table for line 3x + 2y = 12 is 3x + y ³ 8
x, y ³ 0
x04
y60 The graph of the given LPP is

Y

So, the line 3x + 2y = 12 is passing through the points ( 0, 6) 8 A (0,8)
and ( 4, 0).
On putting ( 0, 0) in the inequality 3x + 2y ³ 12, we get 6

0 + 0 ³ 12, which is not true. B (1,5)
4 C (2,4)
So, the half plane is away from the origin.
Table for line x + 3y = 11 is

x 2 11
y30

So, the line x + 3y = 11 is passing through the points ( 2, 3) 2 2 3x + 5y = 26 D(10,0) X
and (11, 0). 4 6 8 10 12 x + 2y = 10
On putting ( 0, 0) in the inequality x + 3y ³ 11, we get X¢ O 3x + y = 8
Y¢ x+y=6
0 + 0 ³ 11, which is not true.
We see that the shaded region ABCD is the feasible region
So, the half plane is away from the origin. which is unbounded.
Also, x ³ 0 and y ³ 0, so the region lies in I quadrant. \ The minimum of Z may or may not occur.

The intersection point of lines corresponding to Eqs. (i) and If it occurs, it will be on the corner point.
(ii) is B( 2, 3).
Now, the values of Z on the corner point are given by
The shaded region ABD represents the graph of the given
constraints. The corner points of the boundary of the graph
ABD are A( 0, 6), B( 2, 3) and D(11, 0).

Y

6 A (0, 6) Corner point Value of Z

5 A( 0, 8) 40
=114
x + 3y B(1, 5) 28

3 B C( 2, 4) 26 (Minimum)

2 (2, 3) D( 10, 0) 30

1 x +y = 5 D(11, 0)

X¢ 0)O 1 2 3 4 5 6 7 8 9 10 11 X From the above table, we see that 26 is the smallest value
(0, of Z.
Now, draw the line 3x + 5y = 26
Y¢ 3x + 2y =12

The values of Z at corner points are given below Since, the open half plane 3x + 5y < 26 do not intersect the
feasible region at any point.
Corner points Z=x+ y \ The minimum value of Z is 26.

A(0, 6) Z = 0+ 6= 6 18. Our problem is to minimise …(i)
B(2, 3) Z = 2+ 3=5 Z = - 3x + 4y
D(11, 0) Z = 11 + 0 = 11
Subject to the constraints are
Thus, the minimum value of Z is 5 at the point B (2, 3). x + 2y £ 8 …(ii)

As the feasible region is unbounded, therefore 5 may or may 3x + 2y £ 12 …(iii)
not be the minimum value of Z. For this, we draw a dotted x ³ 0, y ³ 0 …(iv)
graph of the inequality x + y < 5 and check whether the Draw the graph of the line x + 2y = 8.
resulting half plane has points in common with a feasible
region or not. It can be seen that the feasible region has no x08
common point with x + y < 5. y40

Therefore, the minimum value of Z is 5 at B ( 2, 3). Putting (0, 0) in the inequality x + 2y £ 8, we have

17. The given LPP is 0+ 0£8 Þ 0£8 (which is true)
Minimise Z = 3x + 5y,
So, the half plane is towards the origin.
Subject to the constraints are Since, x, y ³ 0
x + 2y ³ 10
x+ y³6 So, the feasible region lies in the first quadrant.

CBSE Term II Applied Mathematics XII 165

Draw the graph of the line 3x + 2y = 12. and 10x + 20y £ 120 [transistor constraint]

x04 Þ x + 2y £ 12 ...(iii)

and 10x + 30y £ 150 [capacitor constraint]

y60 Þ x + 3y £ 15 ...(iv)

Putting (0, 0) in the inequality 3x + 2y £ 12, we have and x ³ 0, y ³ 0 [non-negative constraint] ...(v)

3 ´ 0 + 2 ´ 0 £ 12 Þ 0 £ 12 (which is true) So, maximise Z = 50x + 60y,

Y Subject to constraints
2x + y £ 20, x + 2y £ 12, x + 3y £ 15, x ³ 0, y ³ 0.

6 20. Let the firm has x number of large vans and y number of

5 small vans. From the given information, we have following
corresponding constraint table.
C(0,4) 4
3 B(2,3) Large vans ( x ) Small vans ( y) Maximum /
2 x + 2y =8 Minimum
200 80
1 A(4,0) Packages 400 200 1200
X¢ Cost
O X 3000

(0,0) 12345678 Thus, we see that objective function for minimum cost is
Z = 400 x + 200y.
Y¢ 3x + 2y = 12

So, the half plane is towards the origin. Subject to constraints are
200x + 80y ³ 1200
\ Feasible region is OABCO. [package constraint]
On solving equations x + 2y = 8 and 3x + 2y = 12 , we get Þ 5x + 2y ³ 30 ...(i)
and 400x + 200y £ 3000
x = 2 and y = 3 Þ 2x + y £ 15 [cost constraint]
...(ii)
\ Intersection point B is ( 2, 3).

The corner points of the feasible region are O(0, 0), A(4, 0), and x£y [van constraint] ...(iii)
B(2, 3) and C(0, 4). The values of Z at these points are as
follows and x ³ 0, y ³ 0 [non-negative constraints] ...(iv)

Corner point Z = - 3x + 4y Thus, required LPP to minimise cost is minimise
Z = 400x + 200y

O(0, 0) 0 Subject to constraints are
5x + 2y ³ 30, 2x + y £ 15, x £ y, x ³ 0, y ³ 0
A(4, 0) -12 (Minimum)
21. Let the company manufactures x boxes of type A screws and
B(2, 3) 6 y boxes of type B screws. From the given information, we
have following corresponding constraint table
C(0, 4) 16

Therefore, the minimum value of Z is –12 at the point Time required for Type A Type B Maximum time
A(4, 0). screws on threading (x) (y) available on each
machine machine in a week
19. Let the manufacturer produces x units of type A circuits and 28
y units of type B circuits. Time required for 60 ´ 60 (Minimum)
screws on slotting 32
Form the given information, we have following machine ` 100 ` 170 60 ´ 60 (Minimum)
corresponding constraint table.
Profit
Resistors Type A ( x ) Type B (y) Maximum stock
Transistors 20 10 200 Thus, we see that objective function for maximum profit is
Capacitors 10 20 120 Z = 100x + 170y.
10 30 150
Profit ` 50 ` 60
Subject to constraints are
Thus, we see that total profit Z = 50x + 60y (in `). 2x + 8y £ 60 ´ 60

Now, we have the following mathematical model for the [time constraint for threading machine]

given problem. Þ x + 4y £ 1800 ...(i)

Maximise Z = 50x + 60y ...(i) and 3x + 2y £ 60 ´ 60

Subject to the constraints are [resistors constraint] [time constraint for slotting machine]
20x + 10y £ 200 ...(ii)
Þ 3x + 2y £ 3600 ... (ii)
Þ 2x + y £ 20

166 CBSE Term II Applied Mathematics XII

Also, x ³ 0, y ³ 0 [non-negative constraints] ...(iii) Let us evaluate the objective function Z = 10500x + 9000y at
these vertices to find which one gives the maximum profit.
\Required LPP is,

Maximise Z = 100x + 170y Corner point Z = 10500x + 9000y

Subject to constraints O (0, 0) 0
A (40, 0) 420000
x + 4y £ 1800, 3x + 2y £ 3600, x ³ 0, y ³ 0. B (30, 20) 495000 (Maximum)
C (0, 50) 450000
22. Let the company manufactures x number of type A sweaters
and y number of type B sweaters.

From the given information we see that cost to make a type
A sweater is `360 and cost to make a type B sweater is

` 120. Y (0,50)

Also, the company spend atmost ` 72000 a day. 50 C
40
\ 360x + 120y £ 72000 2x + y = 80

Þ 3x + y £ 600 …(i)

Also, company can make atmost 300 sweaters. …(ii) 30 B (30,20)
\ x + y £ 300
20
Further, the number of sweaters of type B cannot exceed the A (40,0)
10 X
number of sweaters of type A by more than 100 …(iii) X¢ 10 20 30
i.e. x + 100 ³ y 50 60
Þ x - y ³ - 100 O x + y = 50
Y¢
Also, we have non-negative constraints for x and y

i.e. x ³ 0, y ³ 0 …(iv) Hence, the society will get the maximum profit of ` 495000
by allocating 30 hec for crop X and 20 hec for crop Y.
Hence, the company makes a profit of `200 for each sweater
24. Let the investment in bond A and B are x and y respectively.
of type A and `120 for each sweater of type B
i.e. Profit ( Z) = 200x + 120y Then, our problem is to
Maximise Z = 0.08x + 0.10y
Thus, the required LPP to maximise the profit is
Subject to constraints are
Maximise Z = 200x + 120y is subject to constraints. x + y £ 12000 x ³ 2000
3x + y £ 600 y ³ 4000 , x, y ³ 0
x + y £ 300
x - y ³ - 100 The graph of the above inequality is given by
x ³ 0, y ³ 0 Y

23. Let x hec of land be allocated to crop X and y hec to crop Y. C(2000,10000)
Obviously, x ³ 0, y ³ 0.
Profit per hectare on crop X = ` 10500
Profit per hectare on crop Y = ` 9000
Therefore, total profit = ` (10500x + 9000y)

The mathematical formulation of the problem is as follows
Maximise, Z = 10500x + 9000y

Subject to the constraints are

x + y £ 50 (constraint related to land) …(i) B(8000, 4000)
y = 4000
20x + 10y £ 800 (constraint related to use of herbicide) A (2000, 4000)

i.e. 2x + y £ 80 …(ii)

x ³ 0, y ³ 0 (non-negative constraint) …(iii)

Let us draw the graph of the system of inequalities (i) to (iii). X¢ O x = 2000 X
The feasible region OABC is shown (shaded) in the figure. Y¢ x + y = 12000
Observe that the feasible region is bounded.
In the above graph shaded region is the feasible bounded
The coordinates of the corner points O, A, B and C are region ABC.
( 0, 0), (40, 0), (30, 20) and (0, 50) respectively.

CBSE Term II Applied Mathematics XII 167

Now, the values of Z at the corner points are given by So, the line x = y is passing through ( 0, 0) and (5, 5).

Corner point Value of Z On putting ( 0, 4) is the inequality x £ y, we get
(2000, 4000) 560 0 £ 4, which is true.

So, the half plane is towards the Y-axis.

(8000, 4000) 1040 Also, x ³ 0 and y ³ 0, so the region lies in the I quadrant.

(2000, 10000) 1160 On drawing these lines on graph paper, we get the following

graph Y

From the above table, we see that the maximum value of Z is y=x
1160 which occurs at the point (2000, 10000).
A(0, 20)
Hence, to maximise the interest ` 2000 and ` 10000 must be
invested in bond A and B, respectively. D(15, 15)

25. Let x be the number of packets of screw ‘A’ and y be the (0, 10)B C(5, 5) (60,0)
X¢ O
number of packets of screw ‘B’. Then, we have the following Y¢ (10, 0) x+3y = 60 X
table from the given data. x+y = 10

Machines Packet of screws The intersection point of lines corresponding to Eqs. (i) and
Availability (iii) is D (15, 15) corresponding to Eqs. (ii) and (iii) is C(5, 5).

A(x) B(y)

Automatic machine 4 min 6 min 240 min Now, shading the common region, we get the feasible region
ABCDA.

Hand-operated machine 6 min 3 min 240 min Hence, the region is bounded and corner points of feasible
region are A( 0, 20), B( 0, 10), C(5, 5) and D(15, 15).
Now, the mathematical model of the given problem is
Maximize Z = 0.7x + y The values of Z at corner points are given below
Subject to the constraints are
Corner points Z = 3x + 9y

4x + 6y £ 240 or 2x + 3y £ 120 …(i) A( 0, 20) Z = 3( 0) + 9( 20) = 180
B( 0, 10) Z = 3( 0) + 9(10) = 90
6x + 3y £ 240 or 2x + y £ 80 …(ii) C(5, 5) Z = 3(5) + 9(5) = 60
and x, y ³ 0 …(iii) D(15, 15) Z = 3(15) + 9(15) = 180

26. Given objective function is maximise and minimise ...(i) Here, minimum value of Z is 60 which occur at point C(5, 5)
Z = 3x + 9y ...(ii) and maximum value of Z is 180 which occur at two points
...(iii) A( 0, 20) and D(15, 15). So, each point on the line segment
Subject to constraints are ...(iv) AD will give the maximum value of Z.
x + 3y £ 60
x + y ³ 10 27. Our problem is to minimise …(i)
x£y Z = 5x + 10y
…(ii)
and x ³ 0, y ³ 0 Subject to constraints are …(iii)
Table for line x + 3y = 60 is x + 2y £ 120 …(iv)
x + y ³ 60
x 60 0 x - 2y ³ 0

y 0 20 and x ³ 0, y ³ 0

So, the line x + 3y = 60 is passing through ( 60, 0) and ( 0, 20). Firstly, draw the graph of the line x + 2y = 120.
On putting ( 0, 0) in the inequality x + 3y £ 60, we get
x 0 120
0 + 0 £ 60, which is true. y 60 0
So, the half plane is towards the origin.
Table for line x + y = 10 is Put (0, 0) in the inequality x + 2y £ 120, we get
0 + 2 ´ 0 £ 120 Þ 0 £ 120, which is true.
x 10 0
So, the half plane is towards the origin. Secondly, draw the
y 0 10 graph of the line x + y = 60.

So, the line x + y = 10 is passing through (10, 0) and ( 0, 10). x0 60
On putting ( 0, 0) in the inequality x + y ³ 10, we get y 60 0

0 + 0 ³ 10, which is not true. Put (0, 0) in the inequality x + y ³ 60, we get
So, the half plane is away from the origin. 0 + 0 ³ 60Þ 0 ³ 60, which is false.
Table for line x = y is
So, the half plane is away from the origin.
x05
y05

168 CBSE Term II Applied Mathematics XII

Thirdly, draw the graph of the line x - 2y = 0. So, the half plane is towards the origin.
Table for line x + y = 4 is
x0 10
y0 5 x4 0
y0 4

Put (5, 0) in the inequality x - 2y ³ 0, we get So, the line passes through the points ( 4, 0) and ( 0, 4).
5 - 2 ´ 0 ³ 0 Þ 5 ³ 0, which is true. On putting ( 0, 0) in the inequality x + y £ 4, we get

So, the half plane is towards the X-axis. Since, x, y ³ 0 0 + 0 £ 4, which is true.

the feasible region lies in the first quadrant. So, the half plane is towards the origin.
Also, x ³ 0 and y ³ 0, so the region lies in the I quadrant.
Y

D (40, 20) The intersection point of lines corresponding to Eqs. (i) and
(0, 60)
(ii) is B çèæ 8 , 6 ø÷ö.
C (60, 30) 5 5

B (120, 0) The above inequations (i), (ii), (iii) and (iv) can be
X represented in the graph as shown below

O (60, 0) A x+2y=120 Y

x–2y=0 x+y=60 7

On solving equations x - 2y = 0 and x + y = 60, we get 6 (0, 6)
D( 40, 20) and solving equations x - 2y = 0 and x + 2y = 120,
we get C (60, 30). 5 (0, 4)
4
So, the feasible region is ABCDA. The corner points of the
feasible region are A (60, 0), B (120, 0), C (60, 30) and D (40, 3 ( (B8,6
20). The values of Z at these points are as follows 5 5
C 2
Corner point Z = 5x + 10y (0, 2)

X¢ 1 A(2, 0) (4, 0)

A (60, 0) 300 (minimum) (0, 0)O 1 2 3 4 5 6 X
B (120, 0) 600 Y¢ 7 x+2y=4
C (60, 30) 600 x+y=4
D (40, 20) 400 3x+y=6

So, the minimum value of Z is 300 at the point (60, 0). Clearly, the feasible region is OABCO, where corner points
are O( 0, 0), A( 2, 0), B çæ 8, 6÷ö and C( 0, 2).
28. We have the following LPP
è5 5ø

Maximise Z = 2x + 5y The values of Z at corner points are given below

Subject to constraints are ...(i) Corner points Z = 2x + 5y
2x + 4y £ 8 or x + 2y £ 4 ...(ii)
3x + y £ 6 ...(iii) O( 0, 0) Z = 2 ´0 + 5 ´0 = 0
x+ y£4 ...(iv) Z =2´2+5 ´0=4
A( 2, 0) Z = 2 ´ 8 + 5 ´ 6 = 46 = 9. 2
and x ³ 0, y ³ 0
B æçè 8 , 6 öø÷ 5 55
Table for line x + 2y = 4 is 5 5
Z = 2 ´ 0 + 5 ´ 2 = 10
x 40 C( 0, 2)
y 02
Hence, the maximum value of Z is 10 at C( 0, 2).
So, the line passes through the points ( 4, 0) and ( 0, 2). 29. Our problem is to minimise and maximise …(i)
On putting ( 0, 0) in the inequality x + 2y £ 4, we get
Z = x + 2y …(ii)
0 + 0 £ 4, which is true. Subject to constraints are …(iii)
So, the half plane is towards the origin. …(iv)
Table for line 3x + y= 6 is x + 2y ³ 100 …(v)
2x - y £ 0
x20 2x + y £ 200
and x ³ 0, y ³ 0
y06 Table for line x + 2y = 100 is

So, the line passes through the points ( 2, 0) and ( 0, 6). x 0 100
On putting ( 0, 0) in the inequality 3x + y £ 6, we get
y 50 0
0 + 0 £ 6, which is true.

CBSE Term II Applied Mathematics XII 169

So, the line x + 2y = 100 is passing through the points ( 0, 50) The values of Z at corner points are given below
and (100, 0).
On putting (0, 0) in the inequality x + 2y ³ 100, we get Corner points Z = x + 2y

0 + 2 ´ 0 ³ 100 A(0, 50) Z = 0 + 2 ´50 = 100
Þ 0 ³ 100, which is not true. B(20, 40) Z = 20 + 2 ´ 40 = 100
So, the half plane is away from the origin. C(50, 100) Z = 50 + 2 ´100 = 250
Table for line 2x - y = 0 is D(0, 200) Z = 0 + 2 ´ 200 = 400

x 0 10 The maximum value of Z is 400 at D(0, 200) and the

y 0 20 minimum value of Z is 100 at all the points on the line

So, the line 2x - y = 0 is passing through the points segment joining A(0, 50) and B(20, 40).
(0, 0) and (10, 20).
On putting (5, 0) in the inequality 2x - y £ 0, we get 30. Given objective function is

2 ´5 - 0 £ 0 Minimise Z = - 50x + 20y
Þ 10 £ 0, which is not true.
So, the half plane is towards Y-axis. Subject to constraints are ...(i)
Table for line 2x + y = 200 is 2x - y ³ – 5

x 0 100 3x + y ³ 3 ...(ii)

y 200 0 2x - 3y £ 12 ...(iii)

So, the line 2x + y = 200 is passing through the points and x ³ 0, y ³ 0 ...(iv)
(0, 200) and (100, 0). Table for line 2x - y = -5 is
On putting (0, 0) in the inequality 2x + y £ 200, we get
x -5 / 2 0
2 ´ 0 + 0 £ 200 y0 5
Þ 0 £ 200, which is true.
So, the line passes through the points æçè -5 , 0øö÷ and ( 0, 5 ).
So, the half plane is towards the origin. 2
Also, x, y ³ 0. So, the region lies in the I quadrant.
On solving equations 2x - y = 0 and x + 2y = 100, we get On putting ( 0, 0) in the inequality 2x - y ³ -5, we get
B(20, 40).
Again, solving the equations 2x - y = 0 and 2x + y = 200, we 0 - 0 ³ -5
get C(50, 100).
Clearly, the feasible region is ABCDA. Þ 0 ³ -5, which is true.

Y So, the half plane is towards the origin.
200 D(0, 200)
Table for line 3x + y = 3 is

x0 1
y3 0

100 C(50, 100) So, the line passes through the points ( 0, 3) and (1, 0).
On putting ( 0, 0) in the inequality 3x + y ³ 3, we get
80
0+ 0³ 3
60 B(20, 40) Þ 0 ³ 3, which is not true.

A(0, 50) So, the half plane is away from the origin.
40
Table for line 2x - 3y = 12 is

20 (10, 20) (100, 0) x0 6
X¢(0, 0) O y -4 0
X
20 40 60 80 100 120 140 160

Y¢ 2 x + y = 200 x + 2y = 100 On putting ( 0, 0) in the inequality 2x - 3y £ 12, we get
2x–y=0 0 - 0 £ 12

The corner points of the feasible region are A(0, 50), Þ 0 £ 12, which is true.

B(20, 40), C(50, 100) and D(0, 200). So, the half plane is towards the origin.

170 CBSE Term II Applied Mathematics XII

Also, x ³ 0 and y ³ 0, so the region lies in the I quadrant. So, the half plane is away from the origin.
Table for line x + 2y = 6 is
On drawing the graph of each linear equation, we get the
following graph. In first quadrant, these equations has no x06
intersection point.
y30
Y 2x – y = – 5
So, the line passes through the points ( 0, 3) and ( 6, 0).
8 On putting (0, 0) in the inequality x + 2y ³ 6, we get

7 – 5x+2y = – 30 0 + 2 ´ 0 ³ 6 Þ 0 ³ 6, which is not true.
6 So, the half plane is away from the origin.
Also, x ³ 3, so the region is away from the orgin.
5 A(0, 5) Since, x ³ 0, y ³ 0.

( (4 3 B(0, 3) So, the region lies in the I quadrant.
-5 ,0 2 (6, 0) 2x – 3y=12 The points of intersection of lines x = 3 and x + y = 5 is
2 1 C (1, 0) D C(3, 2) and lines x + 2y = 6 and x + y = 5 is B( 4, 1 ). It can be
seen that the feasible region is unbounded.
X¢ –6 –5 –4 –3 –2 –1O 1 2 3 4 5 6 7 8 X
Y
–2

–3 Y¢ 3x+y = 3
(0, -4) –4

Thus, we get the common shaded region ABCD, which gives 6 –x+2y = 1
the feasible region and it is unbounded. 5 (0, 5)
C(3, 2)
The corner points of feasible region are A( 0, 5), B( 0, 3), 4
C(1, 0) and D( 6, 0). 3 (0, 3)
The value of Z at corner points are given below
2 B(4, 1)
1 A(6, 0)
X¢ O 12345 67 8 X

Corner points Z = - 50x + 20y (0, 0) x = (5, 0) = x+2y = 6
Y¢ 3 x+y 5

A( 0, 5) Z = - 50( 0) + 20(5) = 100 The corner points of the feasible region are A(6, 0), B(4, 1)
B( 0, 3) Z = - 50( 0) + 20( 3) = 60 and C(3, 2).
C(1, 0) Z = - 50(1) + 20( 0) = - 50 The values of Z at corner points are given below
D( 6, 0) Z = - 50( 6) + 20( 0) = - 300
Corner points Z = - x + 2y

Here, feasible region is unbounded so the minimum and A( 6, 0) Z = –6+ 2´0 = - 6
maximum value may or may not be exist. B( 4, 1) Z = – 4 + 2 ´1 = - 2
C( 3, 2) Z = –3+ 2´2 =1

Now, we draw a dotted line of inequation As the feasible region is unbounded, therefore Z = 1 may or
may not be the maximum value. For this, we draw the graph
- 50x + 20y < -300 or - 5x + 2y < -30 of the inequality - x + 2y > 1 and check whether the

Here, we see that half plane determined by - 5x + 2y < - 30 resulting half plane has points in common with the feasible
has a point in common with the feasible region. region or not. The feasible region have points in common
with the - x + 2y > 1. Therefore, Z = 1 is not the maximum
Hence, no minimum value exists. value.

31. Our problem is to maximise Z = - x + 2y ...(i)

Subject to constraints are ...(ii) Hence, Z has no maximum value.
x³3
32. Our problem is to maximise Z = 3x + 2y
x + y³5 ...(iii) ...(i)

x + 2y ³ 6 ...(iv) Subject to constraints are ...(ii)
...(iii)
and x ³ 0, y ³ 0 ...(v) x + 2y £ 10 ...(iv)
3x + y £ 15
Table for line x + y = 5 is
and x ³ 0, y ³ 0
x05 Table for line x + 2y = 10 is

y50 x0 4
y5 3

So, the line passes through the points ( 0, 5) and (5, 0). So, the line passes through the points (0, 5) and (4, 3).
On putting (0, 0) in the inequality x + y ³ 5, we get On putting ( 0, 0) in the inequality x + 2y £ 10, we get

0+ 0³5 0 + 2 ´ 0 £ 10 Þ 0 £ 10, which is true.
Þ 0 ³ 5, which is not true.
So, the half plane is towards the origin.

CBSE Term II Applied Mathematics XII 171

Table for line 3x + y= 15 is Shade the region to the right of Y-axis to show x ³ 0 and
above X-axis to show y ³ 0.
x45 Table for line x + 2y = 10 is

y30 x 0 6 10
y520
So, the line passes through the points (4, 3) and (5, 0).
On putting ( 0, 0) in the inequality 3x + y £ 15, we get So, the line is passing through the points ( 0, 5), ( 4, 3) and
(10, 0).
3 ´ 0 + 0 £ 15 On putting ( 0, 0) in the inequality x + 2y £ 10, we get
Þ 0 £ 15, which is true.
So, the half plane is towards the origin. 0 + 0 £ 10, which is true.
Also, x, y ³ 0, so the region lies in the I quadrant. So, the half plane is towards the origin.
On solving equations x + 2y = 10 and 3x + y = 15, we get Table for line 2x + y = 14 is

x = 4 and y = 3 x467
So, the intersection point is B ( 4, 3).
\Feasible region is OABCO. y620

Y

6 3x + y = 15 So, the line is passing through the points ( 4, 6), ( 6, 2) and
( 7, 0).
C(0, 5) On putting ( 0, 0) in the inequality 2x + y £ 14, we get
5
4 0 + 0 £ 14, which is true.

D(0, 3) B(4, 3) So, the half plane is towards the origin.
3 x + 2y =10
The intersection point of lines corresponding to Eqs. (i) and
2 (ii) is B( 6, 2).

1 E(4, 0) A(5, 0) On shading the common region, we get the feasible region
X¢ (0, 0)O OABD.

Y¢ 12 3 45 6 X x + Y
=10
2y x=0
5
D(0, 5) 6) 2x + y =14
C(4,
4
The corner points of the feasible region are O( 0, 0), A(5, 0),
B( 4, 3) and C( 0, 5). 3 B(6, 2)
2
The values of Z at the corner points are given below
1 E(10, 0)
X¢
Corner points Z = 3x + 2y (0, 0)O 1 2 3 4 5 6 7 8 9 10 X
O( 0, 0) Z = 3´0+ 2´0 = 0
A(5, 0) Z = 3 ´5 + 2 ´ 0 = 15 Y¢ A(7, 0)
B( 4, 3) Z = 3 ´ 4 + 2 ´ 3 = 18
C( 0, 5) Z = 3 ´ 0 + 2 ´5 = 10 The values of Z at corner points are given below

Corner points Z = 2x + 3y

Therefore, the maximum value of Z is 18 at the point B( 4, 3). O(0, 0) Z = 2´0+ 3´0 = 0
A(7, 0) Z = 2 ´ 7 + 3 ´ 0 = 14
\ Area of feasible region B(6, 2) Z = 2 ´ 6 + 3 ´ 2 = 18
= Area of DBOC + Area of DOAB D(0, 5) Z = 2 ´ 0 + 3 ´5 = 15
= 1 ´ OC ´ BD + 1 ´ OA ´ BE
22 Hence, the maximum value of Z is 18 at the point B(6, 2).
= 1 ´5 ´4 + 1 ´5 ´3 34. Let the mixture contains food f1 = x kg and food f2 = y kg.
22
= 10 + 15 = 10 + 7.5 = 17.5 sq units Here, cost of food f1 = ` 50 per kg and cost of food
2 f2 = ` 75 per kg and we have to minimise the cost of
mixture.
33. Given, maximise Z = 2x + 3y
So, the objective function is
Subject to constraints are ...(i) Minimise Z = 50x + 75 y
x + 2y £ 10 ...(ii)
2x + y £ 14 ...(iii) Here, food f1 contains 2 units/kg and food f2
contains 3 units/kg of vitamin A. Also, mixture contains
and x ³ 0, y ³ 0 atleast 6 units of vitamin A.
So, first constraint is 2x + 3y ³ 6.

172 CBSE Term II Applied Mathematics XII

Similarly, food f1 contains 3 units/kg and food f2 contains As the feasible region is unbounded, therefore 150 may or
2 units/kg of vitamin B. Also, mixture contains atleast 8 units may not be the minimum value of Z. For this, we draw a
of vitamin B. So, second constraint is 3x + 2y ³ 8. graph of inequality 50x + 75y < 150 or 2x + 3y < 6 and check
whether the resulting half plane has points in common with
Also, foods f1 and f2 cannot be negative, so x ³ 0, y ³ 0. the feasible region or not. It can be seen that the feasible
Thus, linear constraints are 2x + 3y ³ 6, 3x + 2y ³ 8 and region has no common point with 2x + 3y < 6.
x, y ³ 0.

Consider, the constraints as linear equations Hence, minimum cost of the mixture will be ` 150 at each

2x + 3y = 6 ...(i) point of the line segment joining the points A( 3, 0) and
èçæ øö÷.
3x + 2y = 8 ...(ii) B 12 , 2
5 5
and x = 0, y = 0 ...(iii)
35. Let the manufacturer produces the products A and B be x
Table for line 2x + 3y = 6 is and y units, respectively.

x03 We construct the following table

y20 Products Produce Time on Time on Profit
(in units) Machine I Machine II (in `)
So, line passes through the points ( 0, 2) and ( 3, 0).
On putting ( 0, 0) in the inequality 2x + 3y ³ 6, we get (in hours) (in hours)

2( 0) + 3( 0) ³ 6 Þ 0 ³ 6 , which is not true. A x 3x 3x 7x
So, the half plane is away from the origin.
Table for line 3x + 2y = 8 is B y 2y 1y 4y
Total x+y 3x + 2y 3x + y 7x + 4y
x 0 8/3
Availability 12 9
y40
Here, total profit Z = 7x + 4y
So, line passes through the points ( 0, 4) and çæ 8, 0÷ö.
è3 ø i.e. maximise Z = 7x + 4y ...(i)

On putting ( 0, 0) in the inequality 3x + 2y ³ 8, we get Subject to the constraints are ...(ii)
3( 0) + 2( 0) ³ 8 Þ 0 ³ 8, which is not true. ...(iii)
3x + 2y £ 12
So, the half plane is away from the origin.
Also, x ³ 0, y ³ 0, so the region lies in I quadrant. 3x + y £ 9 and x ³ 0, y ³ 0
On plotting these lines, we get the following graph.
Now, consider the given inequations as equations
Y
3x + 2y = 12

Table for line 3x + 3x + y=9 = 12 - 3x is
2y = 12 or y
2
(0, 4) C

(0, 2) x 04
y 60

( )B A(3, 0) It passes through the points (0, 6) and (4, 0).
X
X¢ 1—52,—52 On putting (0, 0) in the inequality 3x + 2y £ 12, we get
O 2x+3y = 6
(8/3, 0) 0 + 0 £ 12 Þ 0 £ 12 [true]
(0, 0)
Y¢ 3x+2y = 8
So, the half plane is towards the origin.

Table for line 3x + y = 9 or y = 9 - 3x is

The intersection point of lines (i) and (ii) is B æèç 12 , 2 ø÷ö. x 03
5 5 y 90

Hence, the shaded region is a feasible region, which is

unbounded. æçè öø÷ It passes through the points (0, 9) and (3, 0).
On putting (0, 0) in the inequality 3x + y £ 9, we get
Now, the corner points are A (3, 0), B 12 , 2 and C (0, 4).
5 5 0+ 0 £ 9Þ 0 £ 9
[true]

The values of Z at the corner points are given below So, the half plane is towards the origin.
Also, x ³ 0 and y ³ 0, so the region lies in Ist quadrant.
Corner points Z = 50x + 75y
Now, the intersection point of lines (i) and (ii) is
A( 3, 0) Z = 50 ( 3) + 75( 0) = 150 ( 3x + 2y) - ( 3x + y) = 12 - 9

B çæè 12 , 2 ÷øö Z = 50 èçæ 12 ø÷ö + 75 æçè 2 öø÷ = 150 Þ y=3
5 5 5 5
and 3x = 12 - 2 ´ 3
C( 0, 4) Z = 50( 0) + 75( 4) = 300 Þ 3x = 12 - 6 Þ x = 2

Thus, the point of intersection is B (2, 3).

CBSE Term II Applied Mathematics XII 173

The graph of inequations is shown below Table for line x + y = 8 is

Y x08
12 y80

10 So, line passes through the points ( 0, 8) and ( 8, 0).
(0, 9) On putting (0, 0) in the inequality x + y £ 8, we get

8 0 + 0 £ 8 Þ 0 £ 8, which is true.

(0, 6)C 6 So, the half plane is towards the origin.
Table for line x + y = 4 is
4 B (2, 3)

2

X¢ –4 –2 O (4, 0) 8 X x04
2A 4 6 10 y40
(0, 0) –2(3,0)
3x+2y=12
–4 3x+y=9 So, line passes through the points ( 0, 4) and ( 4, 0).
Y¢ On putting (0, 0) in the inequality x + y ³ 4, we get

Here, we see that OABC is a required feasible region, whose 0 + 0 ³ 4Þ 0 ³ 4, which is not true.
corner points are O( 0, 0), A( 3, 0) , B( 2, 3) and C( 0, 6).
So, the half plane is away from the origin.
The values of Z at these corner points are as follows Draw the graph of the lines x = 5 and y = 5, which is
perpendicular to X and Y-axes. Clearly the half plane x £ 5
Corner points Z = 7x + 4y and y £ 5 is towards Y-axis and X-axis, respectively. Also, x,
y ³ 0, so the region lies in the I quadrant.
O( 0, 0) Z = 0+ 0= 0
A( 3, 0) Z = 7 ´ 3 + 0 = 21 The points of intersection of lines (i), (ii) and (iii) are C (5, 3)
B( 2, 3) Z = 7 ´ 2 + 4 ´ 3 = 26 (maximum) and D ( 3, 5).
C( 0, 6) Z = 7 ´ 0 + 4 ´ 6 = 24 The graphical representation of these lines is given below

Hence, for maximum profit, manufacturer produce 2 units Y x=5
of product A and 3 units of product B. 8 (0,8)

36. Let the unit supplied from P to A be x and the unit supplied 7 D(3,5) y=5
from P to B be y.
Then, the unit supplied from P to C is 8 - ( x + y), the unit 6
supplied from Q to A is 5 - x, the unit supplied from Q to B is 5 E(0,5)
5 - y and the unit supplied from Q to C is x + y - 4.
F(0,4)4 C(5,3)
3
2 x+y = 8
x+y = 4
P 1 B(5,0) (8,0) X
8 units 56 7 8
X¢
x y 8 – (x+y) (0,0) O 1 2 34

Y¢ A(4,0)

A B C The shaded region in the graph represents the feasible
5 units 5 units 4 units region ABCDEFA and its corner points are A ( 4, 0), B (5, 0),
C (5, 3), D ( 3, 5), E ( 0, 5) and F ( 0, 4).
5 –y
5 –x x+y – 4 The values of Z at the corner points are given below

Q Corner points Z = x - 7y + 190
6 units

\The objective function is to A ( 4, 0) Z = 4 - 7( 0) + 190 = 194
B (5, 0) Z = 5 - 7( 0) + 190 = 195
Minimise Z = 16x + 10y + 15 ( 8 - x - y) + 10 (5 - x) C (5, 3) Z = 5 - 7( 3) + 190 = 174
+ 12 (5 - y) + 10 ( x + y - 4) D ( 3, 5) Z = 3 - 7(5) + 190 = 158
E ( 0, 5) Z = 0 - 7(5) + 190 = 155
= 16x + 10y + 120 - 15x - 15y + 50 - 10x F ( 0, 4) Z = 0 - 7( 4) + 190 = 162
+ 60 - 12y + 10x + 10y - 40

= x - 7y + 190

subject to constraints are Thus, Z is minimum at x = 0 and y = 5 and minimum value of
x + y £ 8, x + y ³ 4, x £ 5, y £ 5 and x, y ³ 0 Z is 155.

Consider the constraints as linear equations ...(i) Hence, the optimal transportation strategy will be deliver to
x+y=8 ...(ii) 0, 5 and 3 units from the factory at P and 5, 0 and 1 units
x+y=4 ...(iii) from the factory at Q to the depots at A, B and C,
...(iv) respectively. Also, the minimum transportation cost will be
x = 5, y = 5 ` 155.
and x = 0, y = 0

174 CBSE Term II Applied Mathematics XII

37. Let the number of pieces of two types of teaching aids A and \ Feasible region is OABCO.
B be x and y, respectively.
The corner points of the feasible region are O( 0, 0),
We construct the following table A( 20, 0), B(12, 6) and C( 0, 10).

Items Number Time on Time on Profit Corner points Z = 80x + 120y
of pieces fabricating finishing (in `)

(in hours) (in hours) O(0, 0) Z = 80( 0) + 120( 0) = 0
A(20, 0) Z = 80( 20) + 120( 0) = 1600
A x 9x x 80x B(12, 6) Z = 80(12) + 120( 6) = 1680
C(0, 10) Z = 80( 0) + 120(10) = 1200
B y 12y 3y 120y
Total x+y 9x + 12y x + 3y 80x + 120y

Availability 180 30

The profit on type A is ` 80 and on type B is ` 120. ...(i) Hence, the maximum value of Z is ` 1680.
Our objective is to maximise,
...(ii) The manufacturer should produce 12 pieces of type A and 6
Z = 80x + 120y ...(iii) pieces of type B to earn maximum profit of ` 1680.
...(iv) 38. Suppose the farmer uses x kg of F1 and y kg of F2.
Subject to constraints are ...(v) We have to construct the following table
9x + 12y £ 180 or 3x + 4y £ 60
x + 3y £ 30 Type Quantity Nitrogen Phosphoric Cost

and x ³ 0, y ³ 0 F1 (in kg) acid (in `)
Table for line 9x + 12y = 180 is F2
Total
x 0 20 x 10 x = 1 x 6x 6x
y 15 0 100 10 100

So, line passes through the points ( 0, 15) and ( 20, 0). y 5 y= 1 y 10 y 5y
On putting (0, 0) in the inequality 9x + 12y £ 180, we get 100 20 100

9( 0) + 12( 0) £ 180 Þ 0 £ 180, which is true. x+y x+ y 6x + 10y 6x + 5y
10 20 100 100
So, the half plane is towards the origin.
Table for line x + 3y = 30 is Requirement
(in kg)
x 0 30 14 14

y 10 0 Total cost of fertilisers, Z = 6x + 5y ...(i)
So, our problem is to minimise Z = 6x + 5y .
So, line passes through the points ( 0, 10) and ( 30, 0). Subject to constraints are ...(ii)
...(iii)
On putting (0, 0) in the inequality x + 3y £ 30, we get x + y ³ 14 ...(iv)
10 20
0 + 3( 0) £ 30 Þ 0 £ 30, which is true. Þ 2x + y ³ 280

So, the half plane is towards the origin. 6x + 10y ³ 14
100 100
Also, x ³ 0 and y ³ 0, so the region lies in the I quadrant.
Þ 3x + 5y ³ 700
On solving 3x + 4y = 60 and x + 3y = 30, we get the point of and x ³ 0, y ³ 0
intersection is B(12, 6). Table for line 2x + y = 280 is

The graphical representation of the lines is given below

Y

20

x+3y = 15 D (0,15)
30 C(0,10)

10 3x+4y = 60 x 0 140
y 280 0
5 B(12, 6)
A(20, 0)

X¢ O 5 E(30, 0)
X
(0, 0) Y ¢
10 15 20 25 30

CBSE Term II Applied Mathematics XII 175

So, line passes through the points ( 0, 280) and (140, 0). graph of the inequality 6x + 5y < 1000 and check whether
Y the resulting half plane has points in common with the
feasible region or not.

320 It can be seen that the feasible region has no common point
280 C(0, 280) with 6x + 5y < 1000.

240 Therefore, 100 kg of fertiliser F1 and 80 kg of fertiliser F2
should be used to minimise the cost and the minimum cost
200
is ` 1000.
160 (0,140) B(100, 80) 39. (i) Total cost (in `) = 12000x + 15000y
120
Then, minimise Z = 12000x + 15000y,
80
subject to constraints are
40 A(700/3, 0) 50x + 40y ³ 6400 or 5x + 4y ³ 640 ...(i)
50x + 20y ³ 4000 or 5x + 2y ³ 400 ...(ii)
X¢ O 40 80 120 160 200 240 280 320 X 30x + 40y ³ 4800 or 3x + 4y ³ 480 ...(iii)
…(iv)
(0, 0) 2x+y = 280 3 x+5y = 700 and x ³ 0, y ³ 0
Y¢ (ii) Table for line 5x + 4y = 640 is

6x+ 5y = 1000

On putting (0, 0) in the inequality 2x + y ³ 280, we get x 128 0
2 ´ 0 + 0 ³ 280 y0 160

Þ 0 ³ 280 , which is not true. So, the line passes through the points (128, 0) and
( 0, 160).
So, the half plane is away from the origin. On putting ( 0, 0) in the inequality 5x + 4y ³ 640, we get
Table for line 3x + 5y = 700 is

x 0 700/3 5( 0) + ( 0) ³ 640, which is not true.
y 140 0 So, the half plane is away from the origin.
Table for line 5x + 2y = 400 is
èæç 700 0ö÷ø.
So, line passes through the points (0, 140) and 3 , x 80 0

On putting (0, 0) in the inequality 3x + 5y ³ 700, we get y0 200
3 ´ 0 + 5 ´ 0 ³ 700
So, the line passes through the points ( 80, 0) and
Þ 0 ³ 700, which is not true. ( 0, 200).
On putting ( 0, 0) in the inequality 5x + 2y ³ 400, we get
So, the half plane is away from the origin.
5( 0) + 2( 0) ³ 400, which is not true.
Also, x, y ³ 0, so the region lies in the I quadrant.
So, the half plane is away from the origin.
On solving the equations Table for line 3x + 4y = 480 is

2x + y = 280 and 3x + 5y = 700, we get B(100, 80).

It can be seen that the feasible region is unbounded. x 160 0

The corner points of the feasible region are Açèæ 700 , 0÷øö , y 0 120
3

B(100, 80) and C( 0, 280). So, the line passes through the points (160, 0) and
(0, 120).
The values of Z at the corner points are given below On putting ( 0, 0) in the inequality 3x + 4y ³ 480, we get

Corner points Z = 6x + 5y 0 + 0 ³ 480, which is not true.
Z = 6 ´ 700 + 5 ´ 0 = 1400
A çèæ 700 , 0÷øö So, the half plane is away from the origin.
3 3
Z = 6 ´100 + 5 ´ 80 =1000 Also, x ³ 0 and y ³ 0, so the feasible region lies in the
B(100, 80) Z = 6 ´ 0 + 5 ´ 280 = 1400 I quadrant.
C( 0, 280)
The point of intersection of lines corresponding to (i)
As the feasible region is unbounded, therefore 1000 may or and (iii) is B ( 80, 60) and corresponding to (i) and (ii) is
may not be the minimum value of Z. For this, we draw a C ( 32, 120).

176 CBSE Term II Applied Mathematics XII

The above Eqs. (i), (ii), (iii) and (iv) can be represented (iii) The values of Z at corner points are given below
graphically as
Z = 12000x + 15000y
Y Corner
points Z = 12000 ´ 160 + 15000 ´ 0 = 1920000
220 A(160, 0) Z = 12000 ´ 80 + 15000 ´ 60 = 1860000
B( 80, 60) Z = 12000 ´ 32 + 15000 ´ 120 = 2184000
200 D(0, 200) C( 32, 120) Z = 12000 ´ 0 + 15000 ´ 200 = 3000000
D( 0, 200)
180

160 (0, 160)

140 C(32, 120) In the above table, we find that minimum value of Z is
120 (0, 120) 1860000 occur at the point B ( 80, 60).
100
But we cannot say that it is a minimum value of Z as
80 B(80, 60) region is unbounded.
60
Therefore, we have to draw the graph of the inequality
40 12000x + 15000y < 1860000

20 (80, 0) (128, 0) A(160, 0) or 12x + 15y < 1860
X¢(0, 0)O or x + y < 1
20 40 60 80 100 120 140 160 180 200 220 X
Y¢ 155 124
64012x +153yx=+1846y0= 480
4y = Now, check whether the resulting open half plane has
5x + any point common with feasible region. From figure,
5x + 2y = 400 we see that it has no point in common.

Clearly, the feasible region ABCD (shaded region), Thus, the minimum value of Z is ` 1860000 attained at
where corner points are A(160, 0), B( 80, 60), C( 32, 120) the point B ( 80, 60).
and D( 0, 200) is unbounded.
Hence, factory I should run for 80 days.

Chapter Test (ii) Constraints related to vitamin C is given by

(a) x + y ³ 10 (b) x + 2 y ³ 10

(c) x - y ³ 8 (d) None of these

Multiple Choice Questions (iii) Objective function of the problem is
(a) Minimise Z = 50 x + 70 y (b) Maximise Z = 50 x + 70 y
(c) Minimise Z = 70 x + 50 y (d) Maximise Z = 70 x + 50 y

1. The linear inequalities or equations or restrictions on (iv) The minimum value of objective function is
the variables of a linear programming problem are
called (a) 380 (b) 560 (c) 180 (d) None of these

(v) How much kg of food I should be used by the

(a) linear relations (b) constraints dietician for the optimal mixing strategy
(c) functions (d) objective functions
(a) 1 (b) 2 (c) 3 (d) 4

2. The feasible region (shaded) for a LPP is shown in Short Answer Type Questions
following figure.
5. Find the maximum value of Z for the problem
B (3, 4) maximise Z = x + y, subject to constraints
x - y £ - 1, - x + y £ 0, x, y ³ 0.
C (0, 2)
6. A factory makes tennis rackets and cricket bats. A tennis
O (0, 0) A (7, 0) racket takes 1.5 h of machine time and 3 h of craftman’s
time in its making, while a cricket bat takes 3 h of
Then,the maximum value of Z = 5x + 7 y is machine time and 1 h of craftman’s time in a day, the
factory has the availability of not more than 42 h of
(a) 14 (b) 43 (c) 35 (d) 0 machine time and 24 h of craftsman’s time.Formulate the
given problem as LPP. For maximising the profit, if profit
3. The feasible region for a LPP is shown in following on one racket is ` 20 and profit on one bat is ` 10.
figure.
7. A small firm manufactures necklaces and bracelets.
Y The total number of necklaces and bracelets that it
can handle per day is at most 24. It takes one hour to
(0, 5) make a bracelet and half an hour to make a necklace.
The maximum number of hours available per day is 16.
(0, 3) If the profit on a necklace is ` 100 and that on a
bracelet is ` 300. Formulate linear programming
(3, 2) problem for finding how many, each should be
produced daily to maximise the profit, if it is being
O X given that atleast one of each must be produced?
x+3y=9
x+y=5 Long Answer Type Questions

Then, the minimum value of Z = 11x + 7 y is 8. Find the maximum value of Z, for the problem
maximise Z = 1000x + 600 y, subject to constraints
(a) 21 (b) 47 (c) 35 (d) 18 x + y £ 200, x ³ 20, y - 4x ³ 0, x, y ³ 0.

Case Based MCQs 9. A manufacturer makes two types of toys A and B. Three
machines are needed for this purpose and the time (in
4. A dietician wishes to mix two types of foods in such a min) required for each toy on the machines is given
way the vitamin contents of the mixture contains below
atleast 8 units of vitamin A and 10 units of vitamin C.
Food ‘I’ contains 2 units/kg of vitamins A and 1 unit/kg Types Machines
of vitamin C. Food ‘II’ contains 1 unit/kg of vitamin A I II III
and 2 units/kg of vitamin C. It costs ` 50 per kg to
purchase Food ‘I’ and ` 70 per kg to purchase Food ‘II’. A 12 18 6

B6 0 9

Resources Food Requirement Each machine is available for a maximum of 6 h per
I ( x) II ( y) day. If the profit on each toy of type A is ` 7.50 and
Vitamin A (units/kg) 8 that the each toy of type B is ` 5, then show that 15
Vitamin C (units/kg) 21 10 toys of type A and 30 of type B should be
12 manufactured in a day to get maximum profit.
Cost (Rs/kg) 50 70

Let the mixture contain x kg of food I and y kg of food II. Answers

On the basis of above information, answer the 1. (b) 2. (b) 3. (a) 4. (i) (a) (ii) (b) (iii) (a) (iv) (a) (v) (b)
following questions. 5. No feasible region 8. 136000

(i) Constraints related to vitamin A is given by For Detailed Solutions
Scan the code
(a) 2x + y ³ 8 (b) x + 2 y ³ 8

(c) x + y ³ 1 (d) None of these







CBSEATeprmpIIliAepdplieMd MaatthhemeamticsaXtII ics 181
Class 12th (Term II)

Practice Paper 1*

(Solved)

Ins tructions Tim e : 2 Hr
M a x. M a rks : 4 0
1. The ques tion paper contains three s ections A, B and C.
2. Section A has 5 ques tions w ith 3 internal choices .
3. Section B has 4 ques tions w ith 3 internal choices .
4 . Section C has 1 Cas e Bas ed MCQs com pris es of 5 MCQs .
5. There is no neg ative m arking .

As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this
paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised
not to consider the pattern of this paper as official.

Section A (3 Marks Each)

This section consists of 5 questions of Short Answer Type.

1. Evaluate ò dx .

x ax - x 2

Or Evaluate ò 1 + 3x - x 2 dx.

2. The marginal cost function of producing x units of a product is given by MC = x . Find the total
x 2 + 2500

cost function and the average cost function, if the fixed cost is ` 1000.

3. Two batches of the same product are tested for their mean life. Assuming that the lives of the product follow

a normal distribution with an unknown variance, test the hypothesis that the mean life is the same for both
the branches, given the following information. (given t16(0.05) = 2.2120)

Batch Sample size Mean life in hours Standard deviation

Batch I 10 750 12

Batch II 8 820 14

Or Consider the following hypothesis test

H0 : m1 - m 2 = 0
Ha : m1 - m 2 ¹ 0