KnowledgeBoat Logo
|
OPEN IN APP

Chapter 2

Compound Interest

Class - 9 RS Aggarwal Mathematics Solutions

Exercise 2(A)

Question 1

Calculate the amount and the compound interest on ₹ 25,000 for 2 years at 8% per annum, compounded annually.

Answer

For first year :

P = ₹ 25,000

T = 1 year

R = 8%

I = P×R×T100\dfrac{P \times R \times T}{100}

=25000×8×1100= \dfrac{25000 \times 8 \times 1}{100} = ₹ 2,000.

Amount = P + I = ₹ 25,000 + ₹ 2,000 = ₹ 27,000.

For second year :

P = ₹ 27,000

T = 1 year

R = 8%

I = P×R×T100\dfrac{P \times R \times T}{100}

=27000×8×1100= \dfrac{27000 \times 8 \times 1}{100} = ₹ 2160.

Amount = P + I = ₹ 27,000 + ₹ 2,160 = ₹ 29,160.

Compound interest = Final amount - Initial principal

= ₹ 29,160 - ₹ 25,000 = ₹ 4,160.

Hence, compound interest = ₹ 4,160 and amount = ₹ 29,160.

Question 2

Rohit borrows ₹ 62,500 from Arun for 2 years at 10% per annum, simple interest. He immediately lends out this sum to Kunal at 10% per annum for the same period, compounded annually. Calculate Rohit's profit in the transaction at the end of two years.

Answer

For Rohit,

P = ₹ 62,500

T = 2 year

R = 10% per annum simple interest

Interest Rohit pays to Arun:

I = P×R×T100\dfrac{P \times R \times T}{100}

=62500×10×2100= \dfrac{62500 \times 10 \times 2}{100} = ₹ 12,500.

For Kunal,

For first year :

P = ₹ 62,500

T = 1 year

R = 10% per annum compounded annually

I = P×R×T100\dfrac{P \times R \times T}{100}

=62500×10×1100= \dfrac{62500 \times 10 \times 1}{100} = ₹ 6,250.

Amount = P + I = ₹ 62,500 + ₹ 6,250 = ₹ 68,750.

For second year :

P = ₹ 68,750

T = 1 year

R = 10% per annum compounded annually

I = P×R×T100\dfrac{P \times R \times T}{100}

=68750×10×1100= \dfrac{68750 \times 10 \times 1}{100} = ₹ 6,875.

Amount = P + I = ₹ 68,750 + ₹ 6,875 = ₹ 75,625.

Compound interest = Final amount - Initial principal

= ₹ 75,625 - ₹ 62,500 = ₹ 13,125.

∴ Interest Kunal pays to Rohit = ₹ 13,125

Rohit's profit = Compound interest received from Kunal - Simple interest paid to Arun = ₹ 13,125 - ₹ 12,500 = ₹ 625.

Hence, Rohit's profit in the transaction at the end of two years = ₹ 625.

Question 3

A man invests ₹ 10,000 for 3 years at a certain rate of interest, compounded annually. At the end of one year, it amounts to ₹ 11,200. Calculate :

(i) the rate of interest per annum;

(ii) the interest accrued in the second year;

(iii) the amount at the end of the third year.

Answer

(i) Given,

P = ₹ 10,000

T = 3 year

Amount at the end of first year = ₹ 11,200

Interest in the first year = Amount - Principal

= ₹ 11,200 - ₹ 10,000 = ₹ 1,200.

So, for 1 year interest equals to ₹ 1,200 on ₹ 10,000. Let rate of interest be R%. Substituting values we get :

⇒I=P×R×T100⇒1200=10000×R×1100⇒1200=100×R⇒R=1200100⇒R=12\Rightarrow I = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 1200 = \dfrac{10000 \times R \times 1}{100} \\[1em] \Rightarrow 1200 = 100 \times R \\[1em] \Rightarrow R = \dfrac{1200}{100} \\[1em] \Rightarrow R = 12%.

Hence, the rate of interest per annum = 12% p.a.

(ii) Given,

For second year :

P = ₹ 11,200

T = 1 year

R = 12%

Interest accrued in the second year,

I = P×R×T100\dfrac{P \times R \times T}{100}

=11200×12×1100= \dfrac{11200 \times 12 \times 1}{100}

= ₹ 1,344.

Hence, the interest accrued in the second year = ₹ 1,344.

(iii) For third year,

P = ₹ 11,200 + ₹ 1,344 = ₹ 12,544

I = P×R×T100\dfrac{P \times R \times T}{100}

=12544×12×1100= \dfrac{12544 \times 12 \times 1}{100}

= ₹ 1,505.28

Amount at the end of the third year = P + I = ₹ 12,544 + ₹ 1,505.28 = ₹ 14,049.28

Hence, the amount at the end of the third year = ₹ 14,049.28.

Question 4

Sudhakar borrows ₹ 22,500 at 10% per annum, compounded annually. If he repays ₹ 11,250 at the end of first year and ₹ 12,550 at the end of the second year, find the amount of loan outstanding against him at the end of the third year.

Answer

For first year :

P = ₹ 22,500

T = 1 year

R = 10 %

I = P×R×T100\dfrac{P \times R \times T}{100}

=22500×10×1100= \dfrac{22500 \times 10 \times 1}{100} = ₹ 2,250.

Amount = P + I = ₹ 22,500 + ₹ 2,250 = ₹ 24,750.

Amount payed at end of first year = ₹ 11,250.

Amount left at beginning of second year = ₹ 24,750 - ₹ 11,250 = ₹ 13,500.

For second year :

P = ₹ 13,500

R = 10%

T = 1 year

I = P×R×T100\dfrac{P \times R \times T}{100}

=13500×10×1100= \dfrac{13500 \times 10 \times 1}{100} = ₹ 1,350.

Amount = P + I = ₹ 13,500 + ₹ 1,350 = ₹ 14,850.

Amount payed at end of second year = ₹ 12,550.

Amount left at beginning of third year = ₹ 14,850 - ₹ 12,550 = ₹ 2,300

For third year :

P = ₹ 2,300

R = 10%

T = 1 year

I = P×R×T100\dfrac{P \times R \times T}{100}

=2300×10×1100= \dfrac{2300 \times 10 \times 1}{100} = ₹ 230.

Amount due at the end of third year = P + I = ₹ 2,300 + ₹ 230 = ₹ 2,530.

Hence, the amount outstanding at the end of the third year = ₹ 2,530.

Question 5

A man borrows ₹ 15,000 at 12% per annum, compounded annually. If he repays ₹ 4,400 at end of each year, find the amount outstanding against him at the beginning of third year.

Answer

For first year :

P = ₹ 15,000

T = 1 year

R = 12%

I = P×R×T100\dfrac{P \times R \times T}{100}

=15000×12×1100= \dfrac{15000 \times 12 \times 1}{100} = ₹ 1,800.

Amount = P + I = ₹ 15,000 + ₹ 1,800 = ₹ 16,800.

Amount payed at end of first year = ₹ 4,400.

Amount left at beginning of second year = ₹ 16,800 - ₹ 4,400 = ₹ 12,400.

For second year :

P = ₹ 12,400

R = 12%

T = 1 year

I = P×R×T100\dfrac{P \times R \times T}{100}

=12400×12×1100= \dfrac{12400 \times 12 \times 1}{100} = ₹ 1,488.

Amount = P + I = ₹ 12,400 + ₹ 1,488 = ₹ 13,888.

Amount payed at end of second year = ₹ 4,400.

Amount left at beginning of third year = ₹ 13,888 - ₹ 4,400 = ₹ 9,488.

Hence, amount left at beginning of third year = ₹ 9,488.

Question 6

Mr. Ravi borrows ₹ 16,000 for 2 years. The rate of interest for the two successive years are 10% and 12% respectively. If he repays ₹ 5,600 at the end of first year, find the amount outstanding at the end of the second year.

Answer

For first year :

P = ₹ 16,000

T = 1 year

R = 10 %

I = P×R×T100\dfrac{P \times R \times T}{100}

=16000×10×1100= \dfrac{16000 \times 10 \times 1}{100} = ₹ 1,600.

Amount = P + I = ₹ 16,000 + ₹ 1,600 = ₹ 17,600.

Amount payed at end of first year = ₹ 5,600.

Amount left at beginning of second year = ₹ 17,600 - ₹ 5,600 = ₹ 12,000.

For second year :

P = ₹ 12,000

R = 12%

T = 1 year

I = P×R×T100\dfrac{P \times R \times T}{100}

=12000×12×1100= \dfrac{12000 \times 12 \times 1}{100} = ₹ 1,440.

Amount = P + I = ₹ 12,000 + ₹ 1,440 = ₹ 13,440.

Hence, amount outstanding at end of second year = ₹ 13,440.

Question 7

Calculate the amount of ₹ 30,000 at the end of 2 years 4 months, compounded annually at 10% per annum.

Answer

For first year :

P = ₹ 30,000

T = 1 year

R = 10%

I = P×R×T100\dfrac{P \times R \times T}{100}

=30000×10×1100= \dfrac{30000 \times 10 \times 1}{100} = ₹ 3,000.

Amount = P + I = ₹ 30,000 + ₹ 3,000 = ₹ 33,000.

For second year :

P = ₹ 33,000

T = 1 year

R = 10%

I = P×R×T100\dfrac{P \times R \times T}{100}

=33000×10×1100= \dfrac{33000 \times 10 \times 1}{100} = ₹ 3,300

Amount = P + I = ₹ 33,000 + ₹ 3,300 = ₹ 36,300

For next 4 months :

P = ₹ 36,300

T = 4 months = 412\dfrac{4}{12} year = 13\dfrac{1}{3} year

R = 10%

I = P×R×T100\dfrac{P \times R \times T}{100}

=36300×10×13100= \dfrac{36300 \times 10 \times \dfrac{1}{3}}{100}

=363000300= \dfrac{363000}{300} = ₹ 1,210.

Amount = P + I = ₹ 36,300 + ₹ 1,210 = ₹ 37,510.

Hence, final amount = ₹ 37,510.

Question 8

Calculate the amount of ₹ 31,250 at the end of 2122\dfrac{1}{2} years, compounded annually at 8% per annum.

Answer

For first year :

P = ₹ 31,250

T = 1 year

R = 8%

I = P×R×T100\dfrac{P \times R \times T}{100}

=31250×8×1100= \dfrac{31250 \times 8 \times 1}{100} = ₹ 2,500.

Amount = P + I = ₹ 31,250 + ₹ 2,500 = ₹ 33,750.

For second year :

P = ₹ 33,750

T = 1 year

R = 8%

I = P×R×T100\dfrac{P \times R \times T}{100}

=33750×8×1100= \dfrac{33750 \times 8 \times 1}{100} = ₹ 2,700.

Amount = P + I = ₹ 33,750 + ₹ 2,700 = ₹ 36,450.

For next 12\dfrac{1}{2} year :

P = ₹ 36,450

T = 12\dfrac{1}{2} year

R = 8%

I = P×R×T100\dfrac{P \times R \times T}{100}

=36450×8×12100= \dfrac{36450 \times 8 \times \dfrac{1}{2}}{100}

=291600200= \dfrac{291600}{200} = ₹ 1,458.

Amount = P + I = ₹ 36,450 + ₹ 1,458 = ₹ 37,908.

Hence, final amount = ₹ 37,908.

Question 9

Calculate the amount and the compound interest on ₹ 15,000 for 2 years compounded annually, the rates of interest for successive years being 8% and 9% per annum respectively.

Answer

For first year :

P = ₹ 15,000

T = 1 year

R = 8%

I = P×R×T100\dfrac{P \times R \times T}{100}

=15000×8×1100= \dfrac{15000 \times 8 \times 1}{100} = ₹ 1,200.

Amount = P + I = ₹ 15,000 + ₹ 1,200 = ₹ 16,200.

For second year :

P = ₹ 16,200

T = 1 year

R = 9%

I = P×R×T100\dfrac{P \times R \times T}{100}

=16200×9×1100= \dfrac{16200 \times 9 \times 1}{100} = ₹ 1,458.

Amount = P + I = ₹ 16,200 + ₹ 1,458 = ₹ 17,658.

Compound interest = Final amount - Initial principal

= ₹ 17,658 - ₹ 15,000 = ₹ 2,658.

Hence, final amount = ₹ 17,658 and compound interest = ₹ 2,658.

Question 10

Calculate the amount and the compound interest on ₹ 25,000 for 3 years compounded annually, the rates of interest for successive years being 8%, 9% and 10% respectively.

Answer

For first year :

P = ₹ 25,000

T = 1 year

R = 8%

I = P×R×T100\dfrac{P \times R \times T}{100}

=25000×8×1100= \dfrac{25000 \times 8 \times 1}{100} = ₹ 2,000.

Amount = P + I = ₹ 25,000 + ₹ 2,000 = ₹ 27,000.

For second year :

P = ₹ 27,000

T = 1 year

R = 9%

I = P×R×T100\dfrac{P \times R \times T}{100}

=27000×9×1100= \dfrac{27000 \times 9 \times 1}{100} = ₹ 2,430.

Amount = P + I = ₹ 27,000 + ₹ 2,430 = ₹ 29,430.

For third year :

P = ₹ 29,430

T = 1 year

R = 10%

I = P×R×T100\dfrac{P \times R \times T}{100}

=29430×10×1100= \dfrac{29430 \times 10 \times 1}{100} = ₹ 2,943.

Amount = P + I = ₹ 29,430 + ₹ 2,943 = ₹ 32,373.

Compound interest = Final amount - Initial principal

= ₹ 32,373 - ₹ 25,000 = ₹ 7,373.

Hence, final amount = ₹ 32,373 and compound interest = ₹ 7,373.

Question 11

Peter invested ₹ 2,40,000 for 2 years at 10% per annum compounded annually. If 20% of the accrued interest at the end of each year is deducted as income tax, find the amount he received at the end of 2 years.

Answer

For first year :

P = ₹ 2,40,000

T = 1 year

R = 10%

I = P×R×T100\dfrac{P \times R \times T}{100}

=240000×10×1100= \dfrac{240000 \times 10 \times 1}{100} = ₹ 24,000.

Income tax deducted = 20% of Interest

= 20100×24000\dfrac{20}{100} \times 24000 = ₹ 4,800

Interest after deduction = ₹ 24,000 - ₹ 4,800 = ₹ 19,200.

Amount = P + I = ₹ 2,40,000 + ₹ 19,200 = ₹ 2,59,200.

For second year :

P = ₹ 2,59,200

T = 1 year

R = 10%

I = P×R×T100\dfrac{P \times R \times T}{100}

=259200×10×1100= \dfrac{259200 \times 10 \times 1}{100} = ₹ 25,920.

Income tax deducted = 20% of Interest

= 20100×25920\dfrac{20}{100} \times 25920 = ₹ 5,184.

Interest after deduction = ₹ 25,920 - ₹ 5,184 = ₹ 20,736.

Amount = P + I = ₹ 2,59,200 + ₹ 20,736 = ₹ 2,79,936.

Hence, final amount received at the end of 2 years = ₹ 2,79,936.

Question 12

Find the amount and the compound interest on ₹ 10,000 for 1 year at 12% per annum, compounded half-yearly.

Answer

Given,

Rate = 12%

Half yearly rate (R) = Rate2=122\dfrac{\text{Rate}}{2} = \dfrac{12}{2} = 6%

For first half year :

P = ₹ 10,000

T = 1 half year

I = P×R×T100\dfrac{P \times R \times T}{100}

=10000×6×1100= \dfrac{10000 \times 6 \times 1}{100} = ₹ 600.

Amount = P + I = ₹ 10,000 + ₹ 600 = ₹ 10,600.

For second half year :

P = ₹ 10,600

T = 1 half year

Half yearly rate = 6%

I = P×R×T100\dfrac{P \times R \times T}{100}

=10600×6×1100= \dfrac{10600 \times 6 \times 1}{100} = ₹ 636.

Amount = P + I = ₹ 10,600 + ₹ 636 = ₹ 11,236.

Compound interest = Final amount - Initial principal

= ₹ 11,236 - ₹ 10,000 = ₹ 1,236.

Hence, final amount = ₹ 11,236 and compound interest = ₹ 1,236.

Question 13

Find the amount and the compound interest on ₹ 64,000 for 1121\dfrac{1}{2} year at 15% per annum, compounded half-yearly.

Answer

Given,

Rate = 15%

Half yearly rate (R) = Rate2=152\dfrac{\text{Rate}}{2} = \dfrac{15}{2} % = 7.5%

Time = 1121\dfrac{1}{2} year = 32×2\dfrac{3}{2} \times 2 = 3 half-year.

For first half year :

P = ₹ 64,000

T = 1 half year

I = P×R×T100\dfrac{P \times R \times T}{100}

=64,000×7.5×1100= \dfrac{64,000 \times 7.5 \times 1}{100} = ₹ 4,800

Amount = P + I = ₹ 64,000 + ₹ 4,800 = ₹ 68,800

For second half year :

P = ₹ 68,800

Half yearly rate (R) = 7.5%

T = 1 half year

I = P×R×T100\dfrac{P \times R \times T}{100}

=68800×7.5×1100= \dfrac{68800 \times 7.5 \times 1}{100} = ₹ 5,160.

Amount = P + I = ₹ 68,800 + ₹ 5,160 = ₹ 73,960.

For third half year :

P = ₹ 73,960

Half yearly rate (R) = 7.5%

T = 1 year

I = P×R×T100\dfrac{P \times R \times T}{100}

=73960×7.5×1100= \dfrac{73960 \times 7.5 \times 1}{100} = ₹ 5,547.

Amount = P + I = ₹ 73,960 + ₹ 5,547 = ₹ 79,507.

Compound interest = Final amount - Initial principal

= ₹ 79,507 - ₹ 64,000 = ₹ 15,507.

Hence, final amount = ₹ 79,507 and compound interest = ₹ 15,507.

Question 14

The simple interest on a sum of money for 2 years at 10% p.a. is ₹ 1,700. Find:

(i) the sum of money,

(ii) the compound interest on this sum for 1 year, payable half yearly at the same rate.

Answer

(i) Given,

The simple interest on a sum of money for 2 years at 10% p.a. is ₹ 1700.

I = ₹ 1,700

T = 2 year

R = 10%

Let sum of money be ₹ P.

I = P×R×T100\dfrac{P \times R \times T}{100}

Substituting values we get :

⇒1700=P×10×2100⇒1700=P×20100⇒1700=P5⇒P=1700×5=8,500.\Rightarrow 1700 = \dfrac{P \times 10 \times 2}{100} \\[1em] \Rightarrow 1700 = \dfrac{P \times 20}{100}\\[1em] \Rightarrow 1700 = \dfrac{P}{5} \\[1em] \Rightarrow P = 1700 \times 5 = 8,500.

Hence, the sum of money = ₹ 8,500

(ii) Given,

For first half year :

P = ₹ 8,500

R = 10%

Half yearly rate = Rate2=102\dfrac{Rate}{2} = \dfrac{10}{2} = 5%

T = 1 half year

I = P×R×T100\dfrac{P \times R \times T}{100}

=8500×5×1100=₹425= \dfrac{8500 \times 5 \times 1}{100} = ₹ 425

Amount = P + I = ₹ 8,500 + ₹ 425 = ₹ 8,925.

For second half year :

P = ₹ 8,925

Half yearly rate = 5%

T = 1 half year

I = P×R×T100\dfrac{P \times R \times T}{100}

=8925×5×1100=₹446.25= \dfrac{8925 \times 5 \times 1}{100} = ₹ 446.25

Amount = P + I = ₹ 8,925 + ₹ 446.25 = ₹ 9,371.25

Compound interest = Final Amount - Initial Pincipal

= ₹ 9,371.25 - ₹ 8,500

= ₹ 871.25

Hence, compound interest = ₹ 871.25

Exercise 2(B)

Question 1

Calculate the amount and the compound interest on ₹ 10,000 for 2 years at 8% p.a., compounded annually.

Answer

Given,

P = ₹ 10,000

n = 2 years

r = 8%

By formula,

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒A=10000(1+8100)2⇒A=10000(100+8100)2⇒A=10000(108100)2⇒A=10000(2725)2⇒A=10000×729625⇒A=₹11,664\Rightarrow A = 10000\Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\dfrac{100+8}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\dfrac{108}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow A = 10000 \times \dfrac{729}{625} \\[1em] \Rightarrow A = ₹ 11,664

Compound interest = Final amount - Initial principal

= ₹ 11,664 - ₹ 10,000 = ₹ 1,664

Hence, amount = ₹ 11,664 and compound interest = ₹ 1,664.

Question 2

Calculate the amount and the compound interest on ₹ 64,000 for 3 years at 7127\dfrac{1}{2}% per annum, compounded annually.

Answer

Given,

P = ₹ 64,000

n = 3 years

r = 7127\dfrac{1}{2}% = 7.5%

By formula,

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒A=64000(1+7.5100)3⇒A=64000(100+7.5100)3⇒A=64000(107.5100)3⇒A=64000×(21.520)3⇒A=64000×9938.3758000⇒A=₹79,507.\Rightarrow A = 64000\Big(1 + \dfrac{7.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000\Big(\dfrac{100+7.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000\Big(\dfrac{107.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000 \times \Big(\dfrac{21.5}{20}\Big)^3 \\[1em] \Rightarrow A = 64000 \times \dfrac{9938.375}{8000} \\[1em] \Rightarrow A = ₹ 79,507.

Compound interest = Final amount - Initial principal = ₹ 79,507 - ₹ 64,000 = ₹ 15,507

Hence, amount = ₹ 79,507 and compound interest = ₹ 15,507.

Question 3

How much will ₹ 12,000 amount to in 2 years at compound interest, the rates of interest for successive years being 10% and 11% respectively ?

Answer

Given,

P = ₹ 12,000

r1 = 10%

r2 = 11%

n = 2 years

By formula,

A = P(1+r1100)(1+r2100)P\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)

Substituting values we get :

⇒A=12000×(1+10100)×(1+11100)=12000×110100×111100=12000×1110×111100=12000×11×11110×100=₹14,652.\Rightarrow A = 12000 \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 + \dfrac{11}{100}\Big) \\[1em] = 12000 \times \dfrac{110}{100} \times \dfrac{111}{100} \\[1em] = 12000 \times \dfrac{11}{10} \times \dfrac{111}{100} \\[1em] = \dfrac{12000 \times 11 \times 111}{10 \times 100} \\[1em] = ₹ 14,652.

Hence, final amount = ₹ 14,652.

Question 4

Calculate the amount and the compound interest on ₹ 25,000 for 3 years, the rates of interest for the successive years being 8%, 9% and 10%, compounded annually.

Answer

Given,

P = ₹ 25,000

r1 = 8%

r2 = 9%

r3 = 10%

n = 3 years

By formula,

A = P(1+r1100)(1+r2100)(1+r3100)P\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)\Big(1 + \dfrac{r_3}{100}\Big)

Substituting values we get :

⇒A=25000×(1+8100)×(1+9100)×(1+10100)=25000×108100×109100×110100=25000×2725×109100×1110=25000×27×109×1125×100×10=27×109×11=₹32,373.\Rightarrow A = 25000 \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{9}{100}\Big) \times \Big(1 + \dfrac{10}{100}\Big) \\[1em] = 25000 \times \dfrac{108}{100} \times \dfrac{109}{100} \times \dfrac{110}{100} \\[1em] = 25000 \times \dfrac{27}{25} \times \dfrac{109}{100} \times \dfrac{11}{10}\\[1em] = \dfrac{25000 \times 27 \times 109 \times 11}{25 \times 100 \times 10} \\[1em] = 27 \times 109 \times 11 \\[1em] = ₹ 32,373.

Compound interest = Final amount - Initial principal

= ₹ 32,373 - ₹ 25,000

= ₹ 7,373.

Hence, amount = ₹ 32,373 and compound interest = ₹ 7,373.

Question 5

Find the amount and the compound interest on ₹ 7,500 for 2 years 8 months at 10% p.a., compounded annually.

Answer

Given,

P = ₹ 7,500

n = 2 years 8 months

= 2 812\dfrac{8}{12} years = 2 23\dfrac{2}{3} years

r = 10%

By formula,

A=P(1+r100)n(1+23r100)A = P\Big(1 + \dfrac{r}{100}\Big)^n \Big(1 + \dfrac{\dfrac{2}{3}r}{100}\Big)

Substituting values we get :

⇒A=7500(1+10100)2(1+23×10100)⇒A=7500(100+10100)2(1+20300)⇒A=7500(110100)2(300+20300)⇒A=7500(1110)2(320300)⇒A=7500×121100×3230⇒A=₹9,680\Rightarrow A = 7500\Big(1 + \dfrac{10}{100}\Big)^2 \Big(1 + \dfrac{\dfrac{2}{3} \times 10}{100}\Big) \\[1em] \Rightarrow A = 7500\Big(\dfrac{100+10}{100}\Big)^2 \Big(1 + \dfrac{20}{300}\Big) \\[1em] \Rightarrow A = 7500\Big(\dfrac{110}{100}\Big)^2 \Big(\dfrac{300 + 20}{300}\Big) \\[1em] \Rightarrow A = 7500\Big(\dfrac{11}{10}\Big)^2 \Big(\dfrac{320}{300}\Big)\\[1em] \Rightarrow A = 7500 \times \dfrac{121}{100} \times \dfrac{32}{30}\\[1em] \Rightarrow A = ₹ 9,680

C.I. = A - P = ₹ 9,680 - ₹ 7,500 = ₹ 2,180

Hence, amount = ₹ 9,680 and compound interest = ₹ 2,180.

Question 6

If simple interest on sum of money for 3 years at 8% per annum is ₹ 7,500, find the compound interest on the same sum for the same period at same rate.

Answer

Given,

I = ₹ 7,500

T = 3 years

R = 8% p.a. simple interest

I = P×R×T100\dfrac{P \times R \times T}{100}

⇒7500=P×8×3100⇒7500=P×24100⇒P=7500×10024⇒P=₹31,250.\Rightarrow 7500 = \dfrac{P \times 8 \times 3}{100} \\[1em] \Rightarrow 7500 = \dfrac{P \times 24}{100} \\[1em] \Rightarrow P = \dfrac{7500 \times 100}{24} \\[1em] \Rightarrow P = ₹ 31,250.

Let's calculate compound interest for this principal, rate of interest and time.

By formula,

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

⇒A=31250(1+8100)3⇒A=31250(100+8100)3⇒A=31250(108100)3⇒A=31250(2725)3⇒A=31250×1968315625⇒A=₹39366.\Rightarrow A = 31250\Big(1 + \dfrac{8}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\dfrac{100+8}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\dfrac{108}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\dfrac{27}{25}\Big)^3 \\[1em] \Rightarrow A = 31250 \times \dfrac{19683}{15625} \\[1em] \Rightarrow A = ₹ 39366.

Compound interest = Final amount - Initial principal

= ₹ 39,366 - ₹ 31,250 = ₹ 8,116.

Hence, compound interest = ₹ 8,116.

Question 7

Calculate the amount and compound interest on ₹ 16,000 for 1 year at 15% per annum, compounded half yearly.

Answer

Given,

Principal (P) = ₹ 16,000

Time (n) = 1 year

Rate (r) = 15% compounded half-yearly

When rate of interest is compounded half-yearly :

By formula,

A = P(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

⇒A=16000×(1+152×100)1×2⇒A=16000×(200+15200)2⇒A=16000×(215200)2⇒A=16000×(4340)2⇒A=16000×(18491600)⇒A=16000×18491600=₹18,490.\Rightarrow A = 16000 \times \Big(1 + \dfrac{15}{2 \times 100}\Big)^{1 \times 2} \\[1em] \Rightarrow A = 16000 \times \Big(\dfrac{200+15}{200}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\dfrac{215}{200}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\dfrac{43}{40}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\dfrac{1849}{1600}\Big) \\[1em] \Rightarrow A = \dfrac{16000 \times 1849}{1600} = ₹ 18,490.

Compound interest = Amount - Principal = ₹ 18,490 - ₹ 16,000 = ₹ 2,490

Hence, amount = ₹ 18,490 and compound interest = ₹ 2,490.

Question 8

Find the amount and compound interest on ₹ 1,25,000 for 1121\dfrac{1}{2} years at 12% per annum, compounded half yearly.

Answer

Given,

Principal (P) = ₹ 1,25,000

Time (n) = 1 12\dfrac{1}{2} years = 1.5 years

Rate (r) = 12% compounded half-yearly

When rate of interest is compounded half-yearly :

By formula,

A = P(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

⇒A=125000×(1+122×100)1.5×2⇒A=125000×(1+350)3⇒A=125000×(50+350)3⇒A=125000×(5350)3⇒A=125000×(148877125000)⇒A=125000×148877125000=₹1,48,877\Rightarrow A = 125000 \times \Big(1 + \dfrac{12}{2 \times 100}\Big)^{1.5 \times 2} \\[1em] \Rightarrow A = 125000 \times \Big(1 + \dfrac{3}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\dfrac{50 + 3}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\dfrac{53}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\dfrac{148877}{125000}\Big) \\[1em] \Rightarrow A = \dfrac{125000 \times 148877}{125000} = ₹ 1,48,877

Compound interest = Amount - Principal

= ₹ 1,48,877 - ₹ 1,25,000 = ₹ 23,877

Hence, amount = ₹ 1,48,877 and compound interest = ₹ 23,877.

Question 9

A sum of ₹ 12,500 is deposited for 1121\dfrac{1}{2} years, compounded half yearly. It amounts to ₹ 13,000 at the end of first half year. Find:

(i) The rate of interest

(ii) The final amount. Give your answer correct to the nearest rupee.

Answer

(i) Given,

P = ₹ 12,500

n = 1 half year

Amount = ₹ 13,000

When rate of interest is compounded half-yearly :

By formula,

A=P(1+r2×100)2×nA = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2 \times n}

For first half year:

⇒13000=12500(1+R2×100)0.5×2⇒1300012500=(1+R200)1⇒1.04−1=(R200)⇒0.04=(R200)⇒R=0.04×200⇒R=8\Rightarrow 13000 = 12500\Big(1 + \dfrac{R}{2 \times 100}\Big)^{0.5 \times 2} \\[1em] \Rightarrow \dfrac{13000}{12500}= \Big(1 + \dfrac{R}{200}\Big)^1 \\[1em] \Rightarrow 1.04 - 1 = \Big(\dfrac{R}{200}\Big) \\[1em] \Rightarrow 0.04 = \Big(\dfrac{R}{200}\Big) \\[1em] \Rightarrow R = 0.04 \times 200 \\[1em] \Rightarrow R = 8%

Hence, Rate of interest = 8% p.a .

(ii) Given,

P = ₹ 12,500

n = 1.5 years

R = 8%

Let's calculate compound interest:

When rate of interest is compounded half-yearly :

By formula,

A=P(1+r2×100)2×nA = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2 \times n}

⇒A=12500(1+82×100)2×1.5⇒A=12500(200+8200)3⇒A=12500(208200)3⇒A=12500(2625)3⇒A=12500×1757615625⇒A=₹14,060.80≈₹14,061\Rightarrow A = 12500 \Big(1 + \dfrac{8}{2 \times 100}\Big)^{2 \times 1.5} \\[1em] \Rightarrow A = 12500 \Big(\dfrac{200 + 8}{200}\Big)^3 \\[1em] \Rightarrow A = 12500 \Big(\dfrac{208}{200}\Big)^3 \\[1em] \Rightarrow A = 12500 \Big(\dfrac{26}{25}\Big)^3 \\[1em] \Rightarrow A = 12500 \times \dfrac{17576}{15625} \\[1em] \Rightarrow A = ₹ 14,060.80 \approx ₹ 14,061

Hence, compound interest = ₹ 14,061.

Question 10

The simple interest on a sum of money at 12% per annum for 1 year is ₹ 900. Find :

(i) the sum of money and

(ii) the compound interest on this sum for 1 year, payable half-yearly at the same rate.

Answer

(i) Given,

I = ₹ 900

R = 12%

T = 1 year

I = P×R×T100\dfrac{P \times R \times T}{100}

⇒900=P×12×1100⇒900=P×12100⇒P=900×10012⇒P=₹7,500.\Rightarrow 900 = \dfrac{P \times 12 \times 1}{100}\\[1em] \Rightarrow 900 = \dfrac{P \times 12}{100}\\[1em] \Rightarrow P = \dfrac{900 \times 100}{12} \\[1em] \Rightarrow P = ₹ 7,500.

Hence, principal = ₹ 7,500.

(ii) When rate of interest is compounded half-yearly :

By formula,

A=P(1+r2×100)2×nA = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2 \times n}

Substituting values we get :

⇒A=7500(1+122×100)2×1⇒A=7500(200+12200)2⇒A=7500(212200)2⇒A=7500(1.06)2⇒A=7500×1.1236⇒A=₹8,427\Rightarrow A = 7500 \Big(1 + \dfrac{12}{2 \times 100}\Big)^{2 \times 1} \\[1em] \Rightarrow A = 7500 \Big(\dfrac{200 + 12}{200}\Big)^2 \\[1em] \Rightarrow A = 7500 \Big(\dfrac{212}{200}\Big)^2 \\[1em] \Rightarrow A = 7500 \Big(1.06\Big)^2 \\[1em] \Rightarrow A = 7500 \times 1.1236 \\[1em] \Rightarrow A = ₹ 8,427

By formula,

Compound interest = Amount - Principal = ₹ 8,427 - ₹ 7500 = ₹ 927.

Hence, compound interest = ₹ 927.

Question 11

What sum of money will amount to ₹ 18,150 in 2 years at 10% per annum, compounded annually?

Answer

Let sum of money be ₹ x.

Given,

P = ₹ x

r = 10%

n = 2 years

A = ₹ 18,150

By formula,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒18150=x×(1+10100)2⇒18150=x×(110100)2⇒18150=x×(1110)2⇒18150=x×121100⇒x=18150×100121⇒x=₹15,000.\Rightarrow 18150 = x \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \dfrac{121}{100} \\[1em] \Rightarrow x = \dfrac{18150 \times 100}{121} \\[1em] \Rightarrow x = ₹ 15,000.

Hence, sum of money = ₹ 15,000.

Question 12

What sum of money will amount to ₹ 93,170 in 3 years at 10% per annum, compounded annually?

Answer

Let sum of money be ₹ x.

Given,

P = ₹ x

r = 10%

n = 3 years

A = ₹ 93,170

By formula,

A = P(1+r100)nP\Big(1+ \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒93170=x×(1+10100)3⇒93170=x×(110100)3⇒93170=x×(1110)3⇒93170=x×13311000⇒x=93170×10001331⇒x=₹70,000.\Rightarrow 93170 = x \times \Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \Big(\dfrac{110}{100}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \Big(\dfrac{11}{10}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \dfrac{1331}{1000} \\[1em] \Rightarrow x = \dfrac{93170 \times 1000}{1331} \\[1em] \Rightarrow x = ₹ 70,000.

Hence, sum of money = ₹ 70,000.

Question 13

On what sum of money will the compound interest for 2 years at 8% per annum be ₹ 7,488?

Answer

Let sum of money be ₹ x.

Given,

P = ₹ x

n = 2 years

r = 8%

C.I. = ₹ 7,488

A = P + I = ₹ x + ₹ 7,488

By formula,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒x+7488=x×(1+8100)2⇒x+7488=x×(108100)2⇒x+7488=x×(2725)2⇒x+7488=x×729625⇒625(x+7488)=729x⇒625x+4680000=729x⇒729x−625x=4680000⇒104x=4680000⇒x=4680000104⇒x=₹45,000.\Rightarrow x + 7488 = x \times \Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \Big(\dfrac{108}{100}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \dfrac{729}{625} \\[1em] \Rightarrow 625(x + 7488) = 729x \\[1em] \Rightarrow 625x + 4680000 = 729x \\[1em] \Rightarrow 729x - 625x = 4680000 \\[1em] \Rightarrow 104x = 4680000 \\[1em] \Rightarrow x = \dfrac{4680000}{104} \\[1em] \Rightarrow x = ₹ 45,000.

Hence, sum of money = ₹ 45,000.

Question 14

The difference between the simple interest and the compound interest on a sum of money for 2 years at 12% per annum is ₹ 216. Find the sum.

Answer

Given,

n = 2 years

r = 12%

Let sum of money be ₹ P.

C.I. = A - P

C.I.=P(1+r100)n−P=P(1+12100)2−P=P×(112100)2−P=P×(2825)2−P=P×784625−P=784P625−P=784P−625P625=159P625.C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \dfrac{12}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{112}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{28}{25}\Big)^2 - P\\[1em] = P \times \dfrac{784}{625} - P \\[1em] = \dfrac{784P}{625} - P \\[1em] = \dfrac{784P - 625P}{625} \\[1em] = \dfrac{159P}{625}.

By formula,

T = 2 years

S.I.=P×R×T100=P×12×2100=6P25.S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 12 \times 2}{100} \\[1em] = \dfrac{6P}{25}.

Given,

Difference between S.I. and C.I. = ₹ 216

⇒159P625−6P25=216⇒159P−150P625=216⇒9P625=216⇒P=216×6259⇒P=24×625⇒P=₹15,000.\Rightarrow \dfrac{159P}{625} - \dfrac{6P}{25} = 216 \\[1em] \Rightarrow \dfrac{159P - 150P}{625} = 216 \\[1em] \Rightarrow \dfrac{9P}{625} = 216 \\[1em] \Rightarrow P = \dfrac{216 \times 625}{9} \\[1em] \Rightarrow P = 24 \times 625 \\[1em] \Rightarrow P = ₹ 15,000.

Hence, sum = ₹ 15,000.

Question 15

The difference between the simple interest and the compound interest on a sum of money for 3 years at 10% per annum is ₹ 558. Find the sum.

Answer

Given,

n = 3 years

r = 10 %

Let sum of money be ₹ P.

C.I. = A - P

C.I.=P(1+r100)n−P=P(1+10100)3−P=P×(110100)3−P=P×(1110)3−P=P×13311000−P=1331P1000−P=1331P−1000P1000=331P1000.C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \dfrac{10}{100}\Big)^3 - P \\[1em] = P \times \Big(\dfrac{110}{100}\Big)^3 - P \\[1em] = P \times \Big(\dfrac{11}{10}\Big)^3 - P\\[1em] = P \times \dfrac{1331}{1000} - P \\[1em] = \dfrac{1331P}{1000} - P \\[1em] = \dfrac{1331P - 1000P}{1000} \\[1em] = \dfrac{331P}{1000}.

Calculating S.I.,

R = 10%

T = 3 years

S.I.=P×R×T100=P×10×3100=3P10.S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 10 \times 3}{100} \\[1em] = \dfrac{3P}{10}.

Given,

Difference between S.I. and C.I. = ₹ 558

⇒331P1000−3P10=558⇒331P−300P1000=558⇒31P1000=558⇒P=558×100031⇒P=18×1000⇒P=₹18,000.\Rightarrow \dfrac{331P}{1000} - \dfrac{3P}{10} = 558 \\[1em] \Rightarrow \dfrac{331P - 300P}{1000} = 558 \\[1em] \Rightarrow \dfrac{31P}{1000} = 558 \\[1em] \Rightarrow P = \dfrac{558 \times 1000}{31} \\[1em] \Rightarrow P = 18 \times 1000 \\[1em] \Rightarrow P = ₹ 18,000.

Hence, the sum = ₹ 18,000.

Question 16

The difference between the compound interest for 1 year, compounded half-yearly and the simple interest for 1 year on a certain sum of money at 10% per annum is ₹ 360. Find the sum.

Answer

Let sum of money lent out be ₹ x.

Calculating C.I. payable half-yearly :

P = ₹ x

r = 10%

n = 1 year

C.I. = A - P

When rate of interest is compounded half-yearly :

By formula,

A = P(1+r2×100)2nP\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2n}

C.I. = A - P

Substituting values we get :

C.I.=P(1+r2×100)n×2−P=x×(1+10200)1×2−x=x×(210200)2−x=x×(2120)2−x=441x400−x=441x−400x400=₹41x400.C.I. = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] = x \times \Big(1 + \dfrac{10}{200}\Big)^{1 \times 2} - x \\[1em] = x \times \Big(\dfrac{210}{200}\Big)^2 - x \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^2 - x \\[1em] = \dfrac{441x}{400} - x \\[1em] = \dfrac{441x - 400x}{400} \\[1em] = ₹ \dfrac{41x}{400}.

Calculating S.I. :

T = 1 year

R = 10%

S.I.=P×R×T100=x×10×1100=₹x10.S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{x \times 10 \times 1}{100} \\[1em] = ₹ \dfrac{x}{10}.

Given,

Difference between compound interest for a year payable half-yearly and simple interest on ₹ x lent out at 10% for a year is ₹ 360.

∴41x400−x10=360⇒41x−40x400=360⇒x400=360⇒x=360×400=₹1,44,000.\therefore \dfrac{41x}{400} - \dfrac{x}{10} = 360 \\[1em] \Rightarrow \dfrac{41x - 40x}{400} = 360 \\[1em] \Rightarrow \dfrac{x}{400} = 360 \\[1em] \Rightarrow x = 360 \times 400 = ₹ 1,44,000.

Hence, the sum = ₹ 1,44,000.

Question 17

At what rate per cent per annum compound interest will ₹ 6,250 amount to ₹ 7,290 in 2 years?

Answer

Given,

P = ₹ 6,250

A = ₹ 7,290

n = 2 years

Let rate of interest be r%.

By formula,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒7290=6250×(1+r100)2⇒72906250=(1+r100)2⇒729625=(1+r100)2⇒(2725)2=(1+r100)2⇒2725=1+r100⇒2725−1=r100⇒27−2525=r100⇒225=r100⇒r=20025=8\Rightarrow 7290 = 6250 \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{7290}{6250} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{729}{625} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{27}{25} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{27}{25} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{27 - 25}{25} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{2}{25} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{200}{25} = 8%.

Hence, rate of interest = 8% p.a.

Question 18

At what rate per cent per annum will ₹ 3,000 amount to ₹ 3,993 in 3 years, the interest being compounded annually?

Answer

Given,

P = ₹ 3,000

A = ₹ 3,993

n = 3 years

Let rate of interest be r%.

By formula,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒3993=3000×(1+r100)3⇒39933000=(1+r100)3⇒13311000=(1+r100)3⇒(1110)3=(1+r100)3⇒1110=1+r100⇒1110−1=r100⇒11−1010=r100⇒110=r100⇒r=10010=10\Rightarrow 3993 = 3000 \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{3993}{3000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{11}{10} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{11}{10} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{11 - 10}{10} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{10} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{10} = 10%.

Hence, rate of interest = 10% p.a.

Question 19

In what time will ₹ 5,120 amount to ₹ 7,290 at 121212\dfrac{1}{2}% per annum, compounded annually?

Answer

Given,

P = ₹ 5,120

Rate = 121212\dfrac{1}{2}% = 12.5%

A = ₹ 7,290

Let time required be n years.

⇒A=P(1+r100)n⇒7290=5120×(1+12.5100)n⇒72905120=(100+12.5100)n⇒729512=(112.5100)n⇒729512=(11251000)n⇒729512=(98)n⇒(98)3=(98)n⇒n=3\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^{n } \\[1em] \Rightarrow 7290 = 5120 \times \Big(1 + \dfrac{12.5}{100}\Big)^{n} \\[1em] \Rightarrow \dfrac{7290}{5120} = \Big(\dfrac{100 + 12.5}{100}\Big)^{n} \\[1em] \Rightarrow \dfrac{729}{512} = \Big(\dfrac{112.5}{100}\Big)^{n} \\[1em] \Rightarrow \dfrac{729}{512} = \Big(\dfrac{1125}{1000}\Big)^{n} \\[1em] \Rightarrow \dfrac{729}{512} = \Big(\dfrac{9}{8}\Big)^{n} \\[1em] \Rightarrow \Big(\dfrac{9}{8}\Big)^3 = \Big(\dfrac{9}{8}\Big)^{n} \\[1em] \Rightarrow n = 3

Hence, required time = 3 years.

Question 20

A certain sum of money amounts to ₹ 7,260 in 2 years and to ₹ 7,986 in 3 years, interest being compounded annually. Find the rate per cent per annum.

Answer

Let original sum of money invested be ₹ x and rate of percent be r%.

By formula,

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Given,

The sum of money, invested at compound interest, amounts to ₹ 7,260 in 2 years.

⇒A=P(1+r100)n⇒7260=x×(1+r100)2⇒7260=x(1+r100)2......(1)\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 7260 = x \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 7260 = x\Big(1 + \dfrac{r}{100}\Big)^2 ......(1)

The sum of money, invested at compound interest, amounts to ₹ 7,986 in 3 years.

⇒A=P(1+r100)n⇒7986=x×(1+r100)3⇒7986=x(1+r100)3......(2)\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 7986 = x \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow 7986 = x\Big(1 + \dfrac{r}{100}\Big)^3 ......(2)

Dividing equation (2) by (1), we get :

⇒79867260=x(1+r100)3x(1+r100)2⇒13311210=(1+r100)⇒(1+r100)=13311210⇒r100=13311210−1⇒r100=1331−12101210⇒r100=1211210⇒r=100×1211210=10\Rightarrow \dfrac{7986}{7260} = \dfrac{x\Big(1 + \dfrac{r}{100}\Big)^3}{x\Big(1 + \dfrac{r}{100}\Big)^2} \\[1em] \Rightarrow \dfrac{1331}{1210} = \Big(1 + \dfrac{r}{100}\Big) \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big) = \dfrac{1331}{1210} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1331}{1210} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1331 - 1210}{1210} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{121}{1210} \\[1em] \Rightarrow r = \dfrac{100 \times 121}{1210} = 10%.

Hence, rate percent = 10% p.a.

Exercise 2(C)

Question 1

A town has 15625 inhabitants. If the population of this town increases at the rate of 4% per annum, find the number of inhabitants of the town at the end of 3 years.

Answer

Given,

P = 15625

R = 4% p.a.

n = 3 years

By formula,

Population after n years = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Population after 3 years =15625×(1+4100)3=15625×(100+4100)3=15625×(104100)3=15625×(2625)3=15625×1757615625=17576.\text{Population after 3 years }= 15625 \times \Big(1 + \dfrac{4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{100 + 4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{104}{100}\Big)^3 \\[1em] = 15625 \times \Big(\dfrac{26}{25}\Big)^3 \\[1em] = 15625 \times \dfrac{17576}{15625} \\[1em] = 17576.

Hence, the number of inhabitants of the town after 3 years = 17576.

Question 2

The population of a town is increasing at the rate of 10% per annum. If its present population is 36300, find:

(i) its population after 2 years,

(ii) its population 2 years ago.

Answer

(i) Given,

P = 36300

R = 10% p.a.

n = 2 years

By formula,

Population after n years = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Population after 2 years =36300×(1+10100)2=36300×(100+10100)2=36300×(110100)2=36300×(1110)2=36300×121100=43923.\text{Population after 2 years } = 36300 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{100 + 10}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] =36300 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] =36300 \times \dfrac{121}{100} \\[1em] =43923.

Hence, population of the town after 2 years = 43923.

(ii) Given,

P = 36300

R = 10% p.a.

n = 2 years

By formula,

Population before n years = P(1+r100)n\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}

Substituting the values in formula,

Population 2 years ago =36300(1+10100)2=36300(100+10100)2=36300(110100)2=36300(1110)2=36300121100=36300×100121=30000.\text{Population 2 years ago }= \dfrac{36300}{\Big(1 + \dfrac{10}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{100+10}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{110}{100}\Big)^2} \\[1em] = \dfrac{36300}{\Big(\dfrac{11}{10}\Big)^2} \\[1em] = \dfrac{36300}{\dfrac{121}{100}} \\[1em] = \dfrac{36300 \times 100} {121} \\[1em] = 30000.

Hence, population of the town 2 years ago = 30000.

Question 3

The present population of a town is 176400. If the rate of growth in its population is 5% per annum, find:

(i) its population 2 years hence,

(ii) its population one year ago.

Answer

(i) Given,

P = 176400

R = 5% p.a.

n = 2 years

By formula,

Population after n years = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Population after 2 years=176400×(1+5100)2=176400×(100+5100)2=176400×(105100)2=176400×(2120)2=176400×441400=194481\text{Population after 2 years}=176400 \times \Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{100 + 5}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] =176400 \times \dfrac{441}{400} \\[1em] =194481

Hence, population of the town after 2 years = 194481.

(ii) Given,

P = 176400

R = 5% p.a.

n = 1 year

By formula,

Population before n years = P(1+r100)n\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}

Substituting the values in formula,

Population one year ago=176400(1+5100)=176400(105100)=1764002120=176400×2021=168000\text{Population one year ago} =\dfrac{176400} {\Big(1 + \dfrac{5}{100}\Big)} \\[1em] = \dfrac{176400} {\Big(\dfrac{105}{100}\Big)} \\[1em] = \dfrac{176400} {\dfrac{21}{20}} \\[1em] = \dfrac{176400 \times 20} {21} \\[1em] = 168000

Hence, population of the town before 1 year = 168000.

Question 4

Three years ago, the population of a city was 50000. If the annual increase during three successive years be 5%, 8% and 10% respectively, find the present population of the city.

Answer

Given,

P = 50000

r1 = 5%

r2 = 8%

r3 = 10%

By formula,

Population = P×(1+r1100)×(1+r2100)×(1+r3100)P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big)

Substituting the values in formula,

Present population=50000×(1+5100)×(1+8100)×(1+10100)=50000×(100+5100)×(100+8100)×(100+10100)=50000×(105100)×(108100)×(110100)=50000×(2120)×(2725)×(1110)=50000×21×27×115000=10×21×27×11=62370.\text{Present population} = 50000 \times \Big(1 + \dfrac{5}{100}\Big) \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{10}{100}\Big) \\[1em] = 50000 \times \Big(\dfrac{100 + 5}{100}\Big) \times \Big(\dfrac{100 + 8}{100}\Big) \times \Big(\dfrac{100 + 10}{100}\Big) \\[1em] = 50000 \times \Big(\dfrac{105}{100}\Big) \times \Big(\dfrac{108}{100}\Big) \times \Big(\dfrac{110}{100}\Big) \\[1em] = 50000 \times \Big(\dfrac{21}{20}\Big) \times \Big(\dfrac{27}{25}\Big) \times \Big(\dfrac{11}{10}\Big) \\[1em] = \dfrac{50000 \times 21 \times 27 \times 11}{5000} \\[1em] = 10 \times 21 \times 27 \times 11 \\[1em] = 62370.

Hence, present population of the city = 62370.

Question 5

A farmer has an increase of 12.5% in the output of wheat in his farm every year. This year, he produced 2,916 quintals of wheat. What was his annual production of wheat 2 years ago?

Answer

Given,

P = 2916 quintals

R = 12.5% p.a.

n = 2 year

By formula,

Wheat production before n years = P(1+r100)n\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}

Substituting the values in formula,

Wheat production before 2 years =2916(1+12.5100)2=2916(100+12.5100)2=2916(112.5100)2=291612656.2510000=2916×1000012656.25=2304\text{Wheat production before 2 years }= \dfrac{2916}{\Big(1 + \dfrac{12.5}{100}\Big)^2} \\[1em] = \dfrac{2916}{\Big(\dfrac{100 + 12.5}{100}\Big)^2} \\[1em] = \dfrac{2916}{\Big(\dfrac{112.5}{100}\Big)^2} \\[1em] = \dfrac{2916}{\dfrac{12656.25}{10000}} \\[1em] = \dfrac{2916 \times 10000}{12656.25} \\[1em] = 2304

Hence, farmer's annual production of wheat 2 years ago = 2304 quintals.

Question 6

The population of a town is 64000. If the annual birth rate is 11.7% and the annual death rate is 4.2%, calculate the population of the town after 3 years.

Answer

Given,

P = 64000

Net growth rate (R) = Birth rate - Death rate

= 11.7% - 4.2% = 7.5%

n = 3 years

By formula,

Population after n years = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Population after 3 years=64000×(1+7.5100)3=64000×(100+7.5100)3=64000×(107.5100)3=64000×(4340)3=64000×7950764000=79507\text{Population after 3 years} = 64000 \times \Big(1 + \dfrac{7.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\dfrac{100 + 7.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\dfrac{107.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\dfrac{43}{40}\Big)^3 \\[1em] = 64000 \times \dfrac{79507}{64000} \\[1em] = 79507

Hence, the population of the town after 3 years = 79,507.

Question 7

A mango tree was planted 2 years ago. The rate of its growth is 20% per annum. If at present, the height of the tree is 162 cm, what it was when the tree was planted?

Answer

Given,

Present height (P) = 162 cm

R = 20% p.a.

n = 2 years

By formula,

Height before n years = P(1+r100)n\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}

Substituting the values in formula,

Height before 2 years =162(1+20100)2=162(100+20100)2=162(120100)2=162(65)2=1623625=162×2536=112.5\text{Height before 2 years }=\dfrac{162}{\Big(1 + \dfrac{20}{100}\Big)^2} \\[1em] = \dfrac{162}{\Big(\dfrac{100 + 20}{100}\Big)^2} \\[1em] = \dfrac{162}{\Big(\dfrac{120}{100}\Big)^2} \\[1em] = \dfrac{162}{\Big(\dfrac{6}{5}\Big)^2} \\[1em] = \dfrac{162}{\dfrac{36}{25}} \\[1em] = \dfrac{162 \times 25}{36} \\[1em] = 112.5

Hence, the height of the tree when the tree was planted was 112.5 cm.

Question 8

Two years ago, the population of a village was 4000. During next year it increased by 6% but due to an epidemic, it decreased by 5% in the following year. What is its population now?

Answer

Given,

P = 4000

r1 = 6%

r2 = 5%

Given,

The population increased by 6% in first year and decreased by 5% in second year.

By formula,

Population after n years = P×(1+r100)×(1−r100)P \times \Big(1 + \dfrac{r}{100}\Big) \times \Big(1 - \dfrac{r}{100}\Big)

Substituting the values in formula,

Population after 2 years =4000×(1+6100)×(1−5100)=4000×(100+6100)×(100−5100)=4000×(106100)×(95100)=4000×(5350)×(1920)=4000×53×1950×20=4×53×19=4028.\text{Population after 2 years }= 4000 \times \Big(1 + \dfrac{6}{100}\Big) \times \Big(1 - \dfrac{5}{100}\Big) \\[1em] = 4000 \times \Big(\dfrac{100 + 6}{100}\Big) \times \Big(\dfrac{ 100 - 5}{100}\Big) \\[1em] = 4000 \times \Big(\dfrac{106}{100}\Big) \times \Big(\dfrac{95}{100}\Big) \\[1em] = 4000 \times \Big(\dfrac{53}{50}\Big) \times \Big(\dfrac{19}{20}\Big) \\[1em] = \dfrac{4000 \times 53 \times 19}{50 \times 20} \\[1em] = 4 \times 53 \times 19 \\[1em] = 4028.

Hence, the present population of the village = 4028.

Question 9

The count of bacteria in a culture grows by 10% during first hour, decreases by 8% during second hour and again increases by 12% during third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours?

Answer

Given,

P = 13125000

r1 = 10%

r2 = 8%

r3 = 12%

∴Count of bacteria after 3 years=13125000×(1+10100)×(1−8100)×(1+12100)=13125000×(100+10100)×(100−8100)×(100+12100)=13125000×(110100)×(92100)×(112100)=13125000×(1110)×(2325)×(2825)=13125000×11×23×2810×25×25=14876400\therefore \text{Count of bacteria after 3 years} = 13125000 \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 - \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \\[1em] = 13125000 \times \Big(\dfrac{100 + 10}{100}\Big) \times \Big(\dfrac{100 - 8}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \\[1em] = 13125000 \times \Big(\dfrac{110}{100}\Big) \times \Big(\dfrac{92}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \\[1em] = 13125000 \times \Big(\dfrac{11}{10}\Big) \times \Big(\dfrac{23}{25}\Big) \times \Big(\dfrac{28}{25}\Big) \\[1em] = \dfrac{13125000 \times 11 \times 23 \times 28}{10 \times 25 \times 25} \\[1em] = 14876400

Hence, the count of bacteria after 3 hours = 14876400.

Question 10

In a factory, the production of scooters was 40000 per year, which rose to 57600 in 2 years. Find the rate of growth per annum.

Answer

Given,

Initial Production = 40000

Production after 2 years = 57600

n = 2 years

Let the rate of growth per annum be r.

By formula,

Production after n years = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

⇒57600=40000×(1+r100)2⇒5760040000=(1+r100)2⇒576400=(1+r100)2⇒(2420)2=(1+r100)2⇒2420=1+r100⇒2420−1=r100⇒r100=24−2020⇒r100=420⇒r=4×10020⇒r=20\Rightarrow 57600 = 40000 \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{57600}{40000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{576}{400} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{24}{20}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{24}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{24}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{24 - 20}{20} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{4}{20} \\[1em] \Rightarrow r = \dfrac{4 \times 100}{20} \\[1em] \Rightarrow r = 20%

Hence, the rate of growth per annum = 20%.

Question 11

Amit started a shop by investing ₹ 5,00,000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.

Answer

Given,

P = ₹ 5,00,000

r1 = 5%

r2 = 10%

r3 = 12%

Given,

In the first year, Amit incurred a loss of 5%, in second year, he earned a profit of 10% which in the third year rose to 12%.

Substituting the values in formula,

Value after 3 years=500000×(1−5100)×(1+10100)×(1+12100)=500000×(100−5100)×(100+10100)×(100+12100)=500000×(95100)×(110100)×(112100)=500000×(1920)×(1110)×(2825)=500000×19×11×2820×10×25=₹5,85,200\text{Value after 3 years} = 500000 \times \Big(1 - \dfrac{5}{100}\Big) \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \\[1em] = 500000 \times \Big(\dfrac{100 - 5}{100}\Big) \times \Big(\dfrac{100 + 10}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \\[1em] = 500000 \times \Big(\dfrac{95}{100}\Big) \times \Big(\dfrac{110}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \\[1em] = 500000 \times \Big(\dfrac{19}{20}\Big) \times \Big(\dfrac{11}{10}\Big) \times \Big(\dfrac{28}{25}\Big) \\[1em] = \dfrac{500000 \times 19 \times 11 \times 28}{20 \times 10 \times 25} \\[1em] = ₹ 5,85,200

Net profit = Value after 3 years - Investment

= ₹ 5,85,200 - ₹ 5,00,000

= ₹ 85,200

Hence, the net profit for the entire period of three years = ₹ 85,200.

Question 12

The value of a machine depreciates 10% annually. Its present value is ₹ 64,800. Find :

(i) its value after 2 years,

(ii) its value 2 years ago.

Answer

(i) Given,

Present value of machine (V) = ₹ 64,800

R = 10%

n = 2 years

By formula,

Value of machine after n years = ₹ [V×(1−r100)n]\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]

Substituting the values in formula,

Value of machine after 2 years=64800×(1−10100)2=64800×(100−10100)2=64800×(90100)2=64800×(910)2=64800×81100=64800×81100=₹52,488\text{Value of machine after 2 years} = 64800 \times \Big(1 - \dfrac{10}{100}\Big)^2 \\[1em] =64800 \times \Big(\dfrac{100 - 10}{100}\Big)^2 \\[1em] =64800 \times \Big(\dfrac{90}{100}\Big)^2 \\[1em] =64800 \times \Big(\dfrac{9}{10}\Big)^2 \\[1em] =64800 \times \dfrac{81}{100} \\[1em] =\dfrac{64800 \times 81}{100} \\[1em] = ₹ 52,488

Hence, value of machine after two years = ₹ 52,488.

(ii) Given,

Present value of machine (V) = ₹ 64,800

R = 10%

n = 2 years

By formula,

Value of machine n years ago = ₹ V(1−r100)n\dfrac{V}{\Big(1 - \dfrac{r}{100}\Big)^n}

Substituting the values in formula,

Value of machine 2 years ago =64800(1−10100)2=64800(100−10100)2=64800(90100)2=64800(910)2=6480081100=64800×10081=₹80,000\text{Value of machine 2 years ago } = \dfrac{64800}{\Big(1 - \dfrac{10}{100}\Big)^2} \\[1em] =\dfrac{64800}{\Big(\dfrac{100 - 10}{100}\Big)^2} \\[1em] =\dfrac{64800}{\Big(\dfrac{90}{100}\Big)^2} \\[1em] =\dfrac{64800}{\Big(\dfrac{9}{10}\Big)^2} \\[1em] =\dfrac{64800}{\dfrac{81}{100}} \\[1em] =\dfrac{64800 \times 100}{81} \\[1em] =₹ 80,000

Hence, value of machine two years ago = ₹ 80,000.

Question 13

A refrigerator was purchased one year ago for ₹ 20,000. Its value depreciates at the rate of 15% per annum. Find:

(i) its present value,

(ii) its value after 1 year.

Answer

(i) Given,

V = ₹ 20,000

R = 15%

n = 1 year

By formula,

Value of refrigerator after n years = ₹ [V×(1−r100)n]\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]

Substituting the values in formula,

Present value =20000×(1−15100)1=20000×(100−15100)=20000×(85100)=20000×1720=20000×1720=₹17,000\text{Present value }=20000 \times \Big(1 - \dfrac{15}{100}\Big)^1 \\[1em] =20000 \times \Big(\dfrac{100 - 15}{100}\Big) \\[1em] =20000 \times \Big(\dfrac{85}{100}\Big) \\[1em] =20000 \times \dfrac{17}{20} \\[1em] =\dfrac{20000 \times 17}{20} \\[1em] = ₹ 17,000

Hence, present value of refrigerator = ₹ 17,000.

(ii) Given,

V = ₹ 17,000

R = 15%

n = 1 year

By formula,

Value of refrigerator after n years = ₹ [V×(1−r100)n]\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]

Substituting the values in formula,

Value after 1 year =17000×(1−15100)1=17000×(100−15100)=17000×(85100)=17000×1720=17000×1720=₹14,450\text{Value after 1 year }=17000 \times \Big(1 - \dfrac{15}{100}\Big)^1 \\[1em] =17000 \times \Big(\dfrac{100 - 15}{100}\Big) \\[1em] =17000 \times \Big(\dfrac{85}{100}\Big) \\[1em] =17000 \times \dfrac{17}{20} \\[1em] =\dfrac{17000 \times 17}{20} \\[1em] =₹ 14,450

Hence, value of refrigerator after 1 year = ₹ 14,450.

Question 14

A machine depreciates each year at 8% of its value in the beginning of the year. If its value be ₹ 57,500 at the end of the year 2015, find :

(i) its value at the end of the year 2014,

(ii) its value at the end of the year 2016.

Answer

(i) Given,

Value of machine at the end of the year 2015 (V) = ₹ 57,500

R = 8%

n = 1 year

By formula,

Value of machine n years ago = ₹ V(1−r100)n\dfrac{V}{\Big(1 - \dfrac{r}{100}\Big)^n}

Substituting the values in formula,

Value of machine at the end of the year 2014 =57500(1−8100)1=57500100−8100=5750092100=575002325=57500×2523=₹62,500\text{Value of machine at the end of the year 2014 }=\dfrac{57500}{\Big(1 - \dfrac{8}{100}\Big)^1} \\[1em] =\dfrac{57500}{\dfrac{100 - 8}{100}} \\[1em] =\dfrac{57500}{\dfrac{92}{100}} \\[1em] =\dfrac{57500}{\dfrac{23}{25}} \\[1em] =\dfrac{57500 \times 25}{23} \\[1em] =₹ 62,500

Hence, value of machine at the end of the year 2014 = ₹ 62,500.

(ii) Given,

Value of machine at the end of the year 2015 (V) = ₹ 57,500

R = 8%

n = 1 year

By formula,

Value of machine after n years = [V×(1−r100)n]\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]

Substituting the values in formula,

Value of machine at the end of 2016=57500×(1−8100)1=57500×(100−8100)=57500×(92100)=57500×2325=57500×2325=₹52,900.\text{Value of machine at the end of 2016}=57500 \times \Big(1 - \dfrac{8}{100}\Big)^1 \\[1em] =57500 \times \Big(\dfrac{100 - 8}{100}\Big) \\[1em] =57500 \times \Big(\dfrac{92}{100}\Big) \\[1em] =57500 \times \dfrac{23}{25} \\[1em] =\dfrac{57500 \times 23}{25} \\[1em] =₹ 52,900.

Hence, value of machine at the end of the year 2016 = ₹ 52,900.

Question 15

The value of a machine depreciates at the rate of 162316\dfrac{2}{3}% per annum. It was purchased 3 years ago. If its present value is ₹ 62,500, find its purchase price.

Answer

Given,

Present value of machine (V) = ₹ 62,500

R = 1623=48+23=50316\dfrac{2}{3} = \dfrac{48 + 2}{3} = \dfrac{50}{3}%

n = 3 years

By formula,

Value of machine n years ago = ₹ V(1−r100)n\dfrac{V}{\Big(1 - \dfrac{r}{100}\Big)^n}

Substituting the values in formula,

Value of machine 3 years ago=62500(1−503100)3=62500(1−503×100)3=62500(300−50300)3=62500(250300)3=62500(2530)3=625001562527000=62500×2700015625=₹1,08,000\text{Value of machine 3 years ago}=\dfrac{62500}{\Big(1 - \dfrac{\dfrac{50}{3}}{100}\Big)^3} \\[1em] =\dfrac{62500}{\Big(1 - \dfrac{50}{3 \times 100}\Big)^3} \\[1em] =\dfrac{62500}{\Big(\dfrac{300 - 50}{300}\Big)^3} \\[1em] =\dfrac{62500}{\Big(\dfrac{250}{300}\Big)^3} \\[1em] =\dfrac{62500}{\Big(\dfrac{25}{30}\Big)^3} \\[1em] =\dfrac{62500}{\dfrac{15625}{27000}} \\[1em] =\dfrac{62500 \times 27000}{15625} \\[1em] =₹ 1,08,000

Hence, its purchase price = ₹ 1,08,000.

Multiple Choice Questions

Question 1

The compound interest on ₹ 3,750 for 2 years at 8% p.a., compounded annually is:

  1. ₹ 604

  2. ₹ 614

  3. ₹ 624

  4. ₹ 642

Answer

Given,

P = ₹ 3,750

n = 2 years

r = 8%

By formula,

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒A=3750×(1+8100)2⇒A=3750×(100+8100)2⇒A=3750×(108100)2⇒A=3750×(2725)2⇒A=3750×729625⇒A=₹4,374\Rightarrow A = 3750 \times \Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \Big(\dfrac{100 + 8}{100}\Big)^2 \\[1em] \Rightarrow A = 3750\times \Big(\dfrac{108}{100}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \dfrac{729}{625} \\[1em] \Rightarrow A = ₹ 4,374

Compound interest = Final amount - Initial principal

= ₹ 4374 - ₹ 3750 = ₹ 624.

Hence, option 3 is correct option.

Question 2

A man invests ₹ 46,875 at 4% p.a. compound interest for 3 years. The interest for the 1st year will be:

  1. ₹ 1,785

  2. ₹ 1,587

  3. ₹ 1,875

  4. ₹ 1,758

Answer

Given,

P = ₹ 46,875

n = 1 year

r = 4%

By formula,

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒A=46875×(1+4100)1⇒A=46875×(100+4100)⇒A=46875×(104100)⇒A=46875×2625⇒A=₹48,750.\Rightarrow A = 46875 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] \Rightarrow A = 46875 \times \Big(\dfrac{100 + 4}{100}\Big) \\[1em] \Rightarrow A = 46875 \times \Big(\dfrac{104}{100}\Big) \\[1em] \Rightarrow A = 46875 \times \dfrac{26}{25} \\[1em] \Rightarrow A = ₹ 48,750.

Compound interest = Final amount - Initial principal

= ₹ 48,750 - ₹ 46,875 = ₹ 1,875.

Hence, option 3 is correct option.

Question 3

A man deposits ₹ 10,000 in a cooperative bank for 3 years at 9% p.a. If interest is compounded annually, then the amount he will get from the bank after 3 years is:

  1. ₹ 12,950.29

  2. ₹ 12,905.29

  3. ₹ 12,059.29

  4. ₹ 12,095.29

Answer

Given,

P = ₹ 10,000

n = 3 years

r = 9%

By formula,

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒A=10000×(1+9100)3⇒A=10000×(100+9100)3⇒A=10000×(109100)3⇒A=10000×12950291000000⇒A=₹12,950.29\Rightarrow A = 10000 \times \Big(1 + \dfrac{9}{100}\Big)^3 \\[1em] \Rightarrow A = 10000 \times \Big(\dfrac{100 + 9}{100}\Big)^3 \\[1em] \Rightarrow A = 10000\times \Big(\dfrac{109}{100}\Big)^3 \\[1em] \Rightarrow A = 10000 \times \dfrac{1295029}{1000000} \\[1em] \Rightarrow A = ₹ 12,950.29

Hence, option 1 is correct option.

Question 4

₹ 16,000 is deposited in a bank for three years. The rates of compound interest for first and second year are 8% and 12% respectively. At the end of third year the amount becomes ₹ 21,384. The rate of interest for the third year will be:

  1. 7%

  2. 10%

  3. 11%

  4. 12%

Answer

Given,

P = ₹ 16,000

r1 = 8%

r2 = 12%

A = ₹ 21,384

Let the interest for third year be r3.

By formula,

Amount = P×(1+r1100)×(1+r2100)×(1+r3100)P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big)

Substituting the values in formula,

⇒21384=16000×(1+8100)×(1+12100)×(1+r3100)⇒21384=16000×(100+8100)×(100+12100)×(1+r3100)⇒21384=16000×(108100)×(112100)×(1+r3100)⇒21384=16000×(2725)×(2825)×(1+r3100)⇒21384=16000×27×2825×25×(1+r3100)⇒21384×25×25=16000×27×28×(1+r3100)⇒21384×25×2516000×27×28=(1+r3100)⇒21384×62516000×756=(1+r3100)⇒1336500012096000=(1+r3100)⇒1+r3100=1.10⇒r3100=1.10−1⇒r3=0.10×100⇒r3=10\Rightarrow 21384 = 16000 \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{100 + 8}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{108}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\dfrac{27}{25}\Big) \times \Big(\dfrac{28}{25}\Big) \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = \dfrac{16000 \times 27 \times 28}{25 \times 25} \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 \times 25 \times 25= 16000 \times 27 \times 28 \times \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow \dfrac{21384 \times 25 \times 25}{16000 \times 27 \times 28} = \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow \dfrac{21384 \times 625}{16000 \times 756} = \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow \dfrac{13365000}{12096000} = \Big(1 + \dfrac{r_3}{100}\Big) \\[1em] \Rightarrow 1 + \dfrac{r_3}{100} = 1.10 \\[1em] \Rightarrow \dfrac{r_3}{100} = 1.10 - 1 \\[1em] \Rightarrow r_3 = 0.10 \times 100\\[1em] \Rightarrow r_3 = 10 %

Hence, option 2 is correct option.

Question 5

A man borrows ₹ 5,000 at 12% compound interest p.a., interest payable every six months. He pays back ₹ 1,800 at the end of every six months. The third payment he has to make at the end of 18 months in order to clear the entire loan will be:

  1. ₹ 2,024.60

  2. ₹ 2,204.60

  3. ₹ 2,240.60

  4. ₹ 2,402.60

Answer

For first six moths :

P = ₹ 5,000

T = 6 months = 0.5 year

R = 12%

I = P×R×T100\dfrac{P \times R \times T}{100}

=5000×12×0.5100= \dfrac{5000 \times 12 \times 0.5}{100} = ₹ 300.

Amount = P + I = ₹ 5,000 + ₹ 300 = ₹ 5,300.

Amount payed at end of six months = ₹ 1,800.

Amount left at beginning of second six months = ₹ 5,300 - ₹ 1,800 = ₹ 3,500.

For next six months :

P = ₹ 3,500

R = 12%

T = 6 months = 0.5 year

I = P×R×T100\dfrac{P \times R \times T}{100}

=3500×12×0.5100= \dfrac{3500 \times 12 \times 0.5}{100} = ₹ 210.

Amount = P + I = ₹ 3,500 + ₹ 210 = ₹ 3,710.

Amount payed at end of second six months = ₹ 1,800.

Amount left at beginning of third six months = ₹ 3,710 - ₹ 1,800 = ₹ 1,910

For next six months :

P = ₹ 1,910

R = 12%

T = 6 months = 0.5 year

I = P×R×T100\dfrac{P \times R \times T}{100}

=1910×12×0.5100= \dfrac{1910 \times 12 \times 0.5}{100} = ₹ 114.6

Amount due at the end of third year = P + I = ₹ 1,910 + ₹ 114.6 = ₹ 2,024.60

Hence, option 1 is correct option.

Question 6

The compound interest for the second year on ₹ 8,000 invested for 3 years at 10% p.a. is:

  1. ₹ 780

  2. ₹ 880

  3. ₹ 890

  4. ₹ 1,080

Answer

Given,

P = ₹ 8,000

T = 1 year

r = 10%

For the first year,

By formula,

I = P×R×T100\dfrac{P \times R \times T}{100}

=8000×10×1100= \dfrac{8000 \times 10 \times 1}{100} = ₹ 800.

Amount = P + I = ₹ 8,000 + ₹ 800 = ₹ 8,800.

Amount at beginning of second year = ₹ 8,800.

P = ₹ 8,800

T = 1 year

r = 10%

For the first year,

By formula,

I = P×R×T100\dfrac{P \times R \times T}{100}

=8800×10×1100= \dfrac{8800 \times 10 \times 1}{100} = ₹ 880.

The compound interest for the second year is ₹ 880

Hence, option 2 is correct option.

Question 7

A person took a loan of ₹ 6,000 from a bank and agreed to pay back the amount along with interest in 2 years. If the rate of compound interest for the first year is 10% and second year is 12%, the amount he had to pay after 2 years will be:

  1. ₹ 7,329

  2. ₹ 7,932

  3. ₹ 7,292

  4. ₹ 7,392

Answer

Given,

P = ₹ 6,000

r1 = 10%

r2 = 12%

n = 2 years

By formula,

Amount = P×(1+r1100)×(1+r2100)P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big)

Substituting the values in formula,

⇒A=6000×(1+10100)×(1+12100)⇒A=6000×(100+10100)×(100+12100)⇒A=6000×(110100)×(112100)⇒A=6000×(1110)×(2825)⇒A=6000×11×2810×25⇒A=₹7,392.\Rightarrow A = 6000 \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 + \dfrac{12}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\dfrac{100 + 10}{100}\Big) \times \Big(\dfrac{100 + 12}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\dfrac{110}{100}\Big) \times \Big(\dfrac{112}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\dfrac{11}{10}\Big) \times \Big(\dfrac{28}{25}\Big) \\[1em] \Rightarrow A = \dfrac{6000 \times 11 \times 28}{10 \times 25} \\[1em] \Rightarrow A = ₹ 7,392.

Hence, option 4 is correct option.

Question 8

Nikita invests ₹ 6,000 for two years at a certain rate of interest compounded annually. At the end of the first year, it amounts to ₹ 6,720. The rate of interest p.a. is:

  1. 8%

  2. 10%

  3. 12%

  4. 14%

Answer

Given,

P = ₹ 6,000

n = 1 year

A = ₹ 6,720

Let the rate of interest be r,

By formula,

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒6720=6000×(1+r100)1⇒67206000=1+r100⇒1+r100=1.12⇒r100=1.12−1⇒r=0.12×100⇒r=12\Rightarrow 6720 = 6000 \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{6720}{6000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = 1.12 \\[1em] \Rightarrow \dfrac{r}{100} = 1.12 - 1 \\[1em] \Rightarrow r = 0.12 \times 100 \\[1em] \Rightarrow r = 12%

Hence, option 3 is correct option.

Question 9

The compound interest on ₹ 8,640 for 3 years at 8% p.a. is:

  1. ₹ 2,345

  2. ₹ 3,245

  3. ₹ 3,425

  4. ₹ 3,452

Answer

Given,

P = ₹ 8,640

n = 3 year

r = 8%

By formula,

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

⇒A=8640×(1+8100)3⇒A=8640×(100+8100)3⇒A=8640×(108100)3⇒A=8640×(2725)3⇒A=8640×1968315625⇒A=₹10,883.9\Rightarrow A = 8640 \times \Big(1 + \dfrac{8}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\dfrac{100 + 8}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\dfrac{108}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\dfrac{27}{25}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \dfrac{19683}{15625} \\[1em] \Rightarrow A = ₹ 10,883.9

Compound interest = Final amount - Initial principal

= ₹ 10,883.9 - ₹ 8,640

= ₹ 2,243.9

Question 10

If the interest is compounded half-yearly, then, C.I. when the principal is ₹ 7,400, the rate of interest is 5% p.a. and the duration is one year, is:

  1. ₹ 373.63

  2. ₹ 374.63

  3. ₹ 373.36

  4. ₹ 373

Answer

Given,

P = ₹ 7,400

r = 5%

n = 1 year

Given,

When interest is compounded half-yearly.

By formula,

A=P×(1+r2×100)n×2A = P \times \Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

⇒A=7400×(1+52×100)1×2⇒A=7400×(200+5200)2⇒A=7400×(205200)2⇒A=7400×(4140)2⇒A=7400×16811600⇒A=7774.625\Rightarrow A = 7400 \times \Big(1 + \dfrac{5}{2 \times 100}\Big)^{1 \times 2} \\[1em] \Rightarrow A = 7400 \times \Big(\dfrac{200 + 5}{200}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \Big(\dfrac{205}{200}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \Big(\dfrac{41}{40}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \dfrac{1681}{1600} \\[1em] \Rightarrow A = 7774.625

Compound interest = Final amount - Initial principal

= ₹ 7774.625 - ₹ 7400

= ₹ 374.625 ≈ ₹ 374.63

Hence, option 2 is correct option.

Question 11

The simple interest on a sum of money for 2 years at 4% per annum is ₹ 340. The compound interest on this sum for one year payable half-yearly at the same rate is:

  1. ₹ 170.70

  2. ₹ 107.70

  3. ₹ 171.70

  4. ₹ 270.70

Answer

Given,

I = ₹ 340

R = 4%

T = 2 years

Let sum of money be ₹ P.

By formula,

I = P×R×T100\dfrac{P \times R \times T}{100}

⇒340=P×4×2100⇒340=P×8100⇒P=340×1008⇒P=₹4,250.\Rightarrow 340 = \dfrac{P \times 4 \times 2}{100} \\[1em] \Rightarrow 340 = \dfrac{P \times 8}{100} \\[1em] \Rightarrow P = \dfrac{340 \times 100}{8} \\[1em] \Rightarrow P = ₹ 4,250.

Given, interest is compounded half-yearly.

By formula,

A=P(1+r2×100)2×nA = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2 \times n}

Substituting values we get :

⇒A=4250(1+42×100)2×1⇒A=4250(200+4200)2⇒A=4250(204200)2⇒A=4250(1.02)2⇒A=4250×1.0404⇒A=₹4,421.7\Rightarrow A = 4250 \Big(1 + \dfrac{4}{2 \times 100}\Big)^{2 \times 1} \\[1em] \Rightarrow A = 4250 \Big(\dfrac{200 + 4}{200}\Big)^2 \\[1em] \Rightarrow A = 4250 \Big(\dfrac{204}{200}\Big)^2 \\[1em] \Rightarrow A = 4250 \Big(1.02\Big)^2 \\[1em] \Rightarrow A = 4250 \times 1.0404 \\[1em] \Rightarrow A = ₹ 4,421.7

Compound interest = Amount - Principal = ₹ 4,421.7 - ₹ 4,250 = ₹ 171.70.

Hence, option 3 is correct option.

Question 12

The compound interest on a certain sum of money at 5% p.a. for two years is ₹ 246. The simple interest on the same sum for three years at 6% p.a. will be:

  1. ₹ 432

  2. ₹ 430.50

  3. ₹ 432.75

  4. ₹ 431.75

Answer

Given,

I = ₹ 246

R = 5%

n = 2 years

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^{n}

Compound interest = Amount - Principal

CI=P(1+r100)n−PCI = P\Big(1 + \dfrac{r}{100}\Big)^{n} - P

⇒246=P(1+5100)2−P⇒246=P(100+5100)2−P⇒246=P(105100)2−P⇒246=P[(1.05)2−1]⇒246=P[1.1025−1]⇒246=0.1025P⇒2460.1025=P⇒P=₹2,400.\Rightarrow 246 = P \Big(1 + \dfrac{5}{100}\Big)^{2} - P \\[1em] \Rightarrow 246 = P \Big(\dfrac{100 + 5}{100}\Big)^2 - P \\[1em] \Rightarrow 246 = P \Big(\dfrac{105}{100}\Big)^2 - P \\[1em] \Rightarrow 246 = P [(1.05)^2 - 1] \\[1em] \Rightarrow 246 = P [1.1025 - 1] \\[1em] \Rightarrow 246 = 0.1025P \\[1em] \Rightarrow \dfrac{246}{0.1025} = P\\[1em] \Rightarrow P = ₹ 2,400.

For calculating Simple interest,

P = ₹ 2,400

R = 6%

T = 3 years

I = P×R×T100\dfrac{P \times R \times T}{100}

⇒I=2400×6×3100⇒I=43200100⇒I=₹432\Rightarrow I = \dfrac{2400 \times 6 \times 3}{100} \\[1em] \Rightarrow I = \dfrac{43200}{100} \\[1em] \Rightarrow I = ₹ 432

Hence, option 1 is correct option.

Question 13

Ramesh wants to get ₹ 6,050 from a bank after 2 years. If the bank gives 10% p.a. compound interest, then the amount of money he has to keep now in the bank is:

  1. ₹ 5,500

  2. ₹ 5,000

  3. ₹ 5,600

  4. ₹ 5,800

Answer

Given,

A = ₹ 6,050

R = 10%

n = 2 years

Let amount he needs to keep be ₹ P.

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^{n}

Substituting values we get :

⇒6050=P(1+10100)2⇒6050=P(100+10100)2⇒6050=P(110100)2⇒6050=P(1.1)2⇒6050=P×1.21⇒60501.21=P⇒P=₹5,000.\Rightarrow 6050 = P \Big(1 + \dfrac{10}{100}\Big)^{2} \\[1em] \Rightarrow 6050 = P \Big(\dfrac{100 + 10}{100}\Big)^2 \\[1em] \Rightarrow 6050 = P \Big(\dfrac{110}{100}\Big)^2 \\[1em] \Rightarrow 6050 = P (1.1)^2 \\[1em] \Rightarrow 6050 = P \times 1.21 \\[1em] \Rightarrow \dfrac{6050}{1.21} = P \\[1em] \Rightarrow P = ₹ 5,000. \\[1em]

Hence, option 2 is correct option.

Question 14

The difference between the compound and simple interest on a certain sum deposited for 2 years at 5% p.a. is ₹ 12. The sum will be :

  1. ₹ 4,500

  2. ₹ 4,600

  3. ₹ 4,800

  4. ₹ 5,000

Answer

By formula,

Given,

T = 2 years

r = 5%

Let sum of money be ₹ P.

By formula,

S.I.=P×R×T100=P×5×2100=P10.S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 5 \times 2}{100} \\[1em] = \dfrac{P}{10}.

By formula,

C.I. = A - P

C.I.=P(1+r100)n−P=P(1+5100)2−P=P×(105100)2−P=P×(2120)2−P=P×441400−P=441P400−P=441P−400P400=41P400.C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \dfrac{5}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{105}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{21}{20}\Big)^2 - P\\[1em] = P \times \dfrac{441}{400} - P \\[1em] = \dfrac{441P}{400} - P \\[1em] = \dfrac{441P - 400P}{400} \\[1em] = \dfrac{41P}{400}.

Given,

Difference between S.I. and C.I. = ₹ 12

⇒41P400−P10=12⇒41P−40P400=12⇒P400=12⇒P=400×12⇒P=₹4,800.\Rightarrow \dfrac{41P}{400} - \dfrac{P}{10} = 12 \\[1em] \Rightarrow \dfrac{41P - 40P}{400} = 12 \\[1em] \Rightarrow \dfrac{P}{400} = 12 \\[1em] \Rightarrow P = 400 \times 12 \\[1em] \Rightarrow P = ₹ 4,800.

Hence, option 3 is correct option.

Question 15

At what rate of compound interest p.a. will ₹ 20,000 amount to ₹ 26,620 in 3 years?

  1. 4%

  2. 6%

  3. 8%

  4. 10%

Answer

Given,

A = ₹ 26,620

P = ₹ 20,000

n = 3 years

Let rate of interest be r.

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^{n}

⇒26620=20000(1+R100)3⇒2662020000=(1+R100)3⇒13311000=(1+R100)3⇒(1110)3=(1+R100)3⇒1110=(1+R100)⇒1110−1=R100⇒11−1010=R100⇒110=R100⇒10010=R⇒R=10\Rightarrow 26620 = 20000 \Big(1 + \dfrac{R}{100}\Big)^{3} \\[1em] \Rightarrow \dfrac{26620}{20000} = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(1 + \dfrac{R}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{11}{10} = \Big(1 + \dfrac{R}{100}\Big) \\[1em] \Rightarrow \dfrac{11}{10} - 1 = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{11-10}{10} = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{1}{10} = \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{100}{10} = R \\[1em] \Rightarrow R = 10%

Hence, option 4 is correct option.

Question 16

In what time will ₹ 5,000 amount to ₹ 5,832 at 8% rate of compound interest p.a.?

  1. 2 years

  2. 4 years

  3. 6 years

  4. 8 years

Answer

Given,

Amount = ₹ 5,832

Principal = ₹ 5,000

R = 8%

Let time be n years.

A=P(1+r100)nA = P\Big(1 + \dfrac{r}{100}\Big)^{n}

⇒5832=5000(1+8100)n⇒58325000=(100+8100)n⇒1.1664=(108100)n⇒1.1664=(1.08)n⇒(1.08)2=(1.08)n⇒n=2\Rightarrow 5832 = 5000 \Big(1 + \dfrac{8}{100}\Big)^{n} \\[1em] \Rightarrow \dfrac{5832}{5000} = \Big(\dfrac{100 + 8}{100}\Big)^n \\[1em] \Rightarrow 1.1664 = \Big(\dfrac{108}{100}\Big)^n \\[1em] \Rightarrow 1.1664 = (1.08)^n \\[1em] \Rightarrow (1.08)^2 = (1.08)^n \\[1em] \Rightarrow n = 2

Hence, option 1 is correct option.

Question 17

A machine depreciates at the rate of 10% of its value at the beginning of a year. If the present value of a machine is ₹ 8,000, its value after 3 years will be:

  1. ₹ 5,382

  2. ₹ 5,832

  3. ₹ 5,238

  4. ₹ 5,638

Answer

Given,

V = ₹ 8,000

n = 3 years

R = 10%

Value of machine after n years = [V×(1−r100)n]\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]

Value of machine after 3 years =8000(1−10100)3=8000(100−10100)3=8000(90100)3=8000(0.9)3=8000×0.729=₹5,832.\text{Value of machine after 3 years }= 8000 \Big(1 - \dfrac{10}{100}\Big)^{3} \\[1em] = 8000 \Big(\dfrac{100 - 10}{100}\Big)^3 \\[1em] = 8000 \Big(\dfrac{90}{100}\Big)^3 \\[1em] = 8000 (0.9)^3 \\[1em] = 8000 \times 0.729 \\[1em] = ₹ 5,832.

Value of machine after 3 years = ₹ 5,832

Hence, option 2 is correct option.

Question 18

The present population of a town is 200000. The population will increase by 10% in the first year and 15% in the second year. The population of the town after two years will be:

  1. 253000

  2. 235000

  3. 203500

  4. 352000

Answer

Given,

P = 200000

r1 = 10% p.a.

r2 = 15% p.a.

By formula,

Population after two years = P×(1+r1100)×(1+r2100)P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big)

Substituting the values in formula,

Population after two years =200000×(1+10100)×(1+15100)=200000×(100+10100)×(100+15100)=200000×(110100)×(115100)=200000×(1110)×(2320)=200000×1.10×1.15=253000.\text{Population after two years }=200000 \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 + \dfrac{15}{100}\Big) \\[1em] =200000 \times \Big(\dfrac{100 + 10}{100}\Big) \times \Big(\dfrac{100 + 15}{100}\Big) \\[1em] =200000 \times \Big(\dfrac{110}{100}\Big) \times \Big(\dfrac{115}{100}\Big) \\[1em] =200000 \times \Big(\dfrac{11}{10}\Big) \times \Big(\dfrac{23}{20}\Big) \\[1em] =200000 \times 1.10 \times 1.15\\[1em] =253000.

Hence, option 1 is correct option.

Question 19

A machine depreciates at the rate of 12% of its value at the beginning of a year. The machine was purchased for ₹ 10,000 and is sold for ₹ 7,744. The number of years, that the machine was used is:

  1. 2

  2. 4

  3. 6

  4. 8

Answer

Given,

Initial value (P) = ₹ 10,000

Depreciated value (A) = ₹ 7,744

R = 12%

Value of machine after n years = [V×(1−r100)n]\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]

⇒7744=10000(1−12100)n⇒7744=10000(100−12100)n⇒7744=10000(88100)n⇒774410000=(88100)n⇒(88100)2=(88100)n⇒n=2.\Rightarrow 7744 = 10000 \Big(1 - \dfrac{12}{100}\Big)^{n} \\[1em] \Rightarrow 7744 = 10000 \Big(\dfrac{100 - 12}{100}\Big)^n \\[1em] \Rightarrow 7744 = 10000 \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow \dfrac{7744}{10000} = \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{88}{100}\Big)^2 = \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow n = 2.

Hence, option 1 is correct option.

Question 20

The value of a machine depreciates every year at a constant rate. If the values of the machine in 2006 and 2008 are ₹ 25,000 and ₹ 19,360 respectively, then the annual rate of depreciation is:

  1. 8%

  2. 10%

  3. 12%

  4. 14%

Answer

Given,

Initial value (P) = ₹ 25,000

Depreciated value (A) = ₹ 19,360

n = 2 years

Let rate of depreciation be R.

Value of machine after n years = [V×(1−R100)n]\Big[V \times \Big(1 - \dfrac{R}{100}\Big)^n \Big]

⇒19360=25000(1−R100)2⇒19360=25000(1−R100)2⇒1936025000=(1−R100)2⇒484625=(1−R100)2⇒(2225)2=(1−R100)2⇒2225=1−R100⇒R100=1−2225⇒R100=25−2225⇒R100=325⇒R=3×10025⇒R=3×4⇒R=12\Rightarrow 19360 = 25000 \Big(1 - \dfrac{R}{100}\Big)^{2} \\[1em] \Rightarrow 19360 = 25000 \Big(1 - \dfrac{R}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{19360}{25000} = \Big(1 - \dfrac{R}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{484}{625} = \Big(1 - \dfrac{R}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{22}{25}\Big)^2 = \Big(1 - \dfrac{R}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{22}{25} = 1 - \dfrac{R}{100} \\[1em] \Rightarrow \dfrac{R}{100} = 1 - \dfrac{22}{25} \\[1em] \Rightarrow \dfrac{R}{100} = \dfrac{25 - 22}{25} \\[1em] \Rightarrow \dfrac{R}{100} = \dfrac{3}{25} \\[1em] \Rightarrow R = \dfrac{3 \times 100}{25} \\[1em] \Rightarrow R = 3 \times 4\\[1em] \Rightarrow R = 12%

Hence, option 3 is correct option.

Case Study Based Questions

Question 1

Case Study

In 2015, the population of a town was 20000. During 2016 and 2017, it increases by 4% and 5% every year respectively. In 2018, due to an epidemic and migration to cities, the population decreases at 5%. Based on the above information answer the following questions:

  1. The population of the town in 2016 was:
    (a) 20800
    (b) 20500
    (c) 20460
    (d) 20300

  2. The difference in the population of the town at the end of the years 2017 and 2015 was:
    (a) 1500
    (b) 1700
    (c) 1800
    (d) 1840

  3. The population of the town at the end of the year 2018 was:
    (a) 20100
    (b) 20500
    (c) 20748
    (d) 20850

  4. Which of the following expressions gives the population at the end of the year 2018?

    (a) 20000×104100×105100×10510020000 \times \dfrac{104}{100} \times \dfrac{105}{100} \times \dfrac{105}{100}

    (b) 20000×104100×105100×9510020000 \times \dfrac{104}{100} \times \dfrac{105}{100} \times \dfrac{95}{100}

    (c) 20000×104100×95100×9510020000 \times \dfrac{104}{100} \times \dfrac{95}{100}\times \dfrac{95}{100}

    (d) 20000×96100×95100×9510020000 \times \dfrac{96}{100} \times \dfrac{95}{100} \times \dfrac{95}{100}

  5. If the population of the town at the end of 2019 was 21,163, then during 2019, the population increases at the rate of:
    (a) 2%
    (b) 3%
    (c) 3.5% (d) 4%

Answer

1. Given,

Population (in 2015) = 20000

r = 4% p.a.

n = 1 year

By formula,

Population in 2016 = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Population in 2016 =20000×(1+4100)1=20000×(100+4100)1=20000×(104100)1=20000×(1.04)=20800.\text{Population in 2016 } = 20000 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] =20000 \times \Big(\dfrac{100 + 4}{100}\Big)^1 \\[1em] =20000 \times \Big(\dfrac{104}{100}\Big)^1 \\[1em] =20000 \times (1.04)\\[1em] =20800.

Hence, option (a) is correct option.

2. Given,

Population (in 2016) = 20800

r = 5% p.a.

n = 1 year

By formula,

Population in 2017 = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Population in 2017=20800×(1+5100)1=20800×(100+5100)1=20800×(105100)1=20800×(1.05)=21840.\text{Population in 2017} = 20800 \times \Big(1 + \dfrac{5}{100}\Big)^1 \\[1em] =20800 \times \Big(\dfrac{100 + 5}{100}\Big)^1 \\[1em] =20800 \times \Big(\dfrac{105}{100}\Big)^1 \\[1em] =20800 \times (1.05)\\[1em] =21840.

The difference in population at the end of the years 2017 and 2015 = 21840 - 20000 = 1840.

Hence, option (d) is correct option.

3. Given,

Population (in 2017) = 21840

r (decrease) = 5% p.a.

n = 1 year

By formula,

Population in 2018 = P×(1−r100)nP \times \Big(1 - \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Population in 2018=21840×(1−5100)1=21840×(100−5100)1=21840×(95100)1=21840×(0.95)=20748.\text{Population in 2018}=21840 \times \Big(1 - \dfrac{5}{100}\Big)^1 \\[1em] =21840 \times \Big(\dfrac{100 - 5}{100}\Big)^1 \\[1em] =21840 \times \Big(\dfrac{95}{100}\Big)^1 \\[1em] =21840 \times (0.95)\\[1em] =20748.

Hence, option (c) is correct option.

4. Given,

P = ₹ 20,000

r1 = 4%

r2 = 5%

r3 = 5% (decrease)

By formula,

Population at the end = P×(1+r1100)×(1+r2100)×(1−r3100)P \times \Big(1 + \dfrac{r_1}{100}\Big) \times \Big(1 + \dfrac{r_2}{100}\Big) \times \Big(1 - \dfrac{r_3}{100}\Big)

Substituting the values in formula,

⇒Population at the end of year 2018=20000×(1+4100)×(1+5100)×(1−5100)=20000×(100+4100)×(100+5100)×(100−5100)=20000×(104100)×(105100)×(95100)\Rightarrow \text{Population at the end of year 2018} = 20000 \times \Big(1 + \dfrac{4}{100}\Big) \times \Big(1 + \dfrac{5}{100}\Big) \times \Big(1 - \dfrac{5}{100}\Big) \\[1em] = 20000 \times \Big(\dfrac{100 + 4}{100}\Big) \times \Big(\dfrac{100 + 5}{100}\Big) \times \Big(\dfrac{100 - 5}{100}\Big) \\[1em] = 20000 \times \Big(\dfrac{104}{100}\Big) \times \Big(\dfrac{105}{100}\Big) \times \Big(\dfrac{95}{100}\Big)

Hence, option (b) is correct option.

5. Given,

A (population in 2019) = 21,163

P (population in 2018) = 20,748

n = 1 year

By formula,

A = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

⇒21163=20748×(1+R100)1⇒2116320748=(1+R100)1⇒1.02=(1+R100)1⇒1.02−1=(R100)⇒0.02=(R100)⇒0.02×100=R⇒R=2\Rightarrow 21163 = 20748 \times \Big(1 + \dfrac{R}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{21163}{20748} = \Big(1 + \dfrac{R}{100}\Big)^1 \\[1em] \Rightarrow 1.02 = \Big(1 + \dfrac{R}{100}\Big)^1 \\[1em] \Rightarrow 1.02 - 1 = (\dfrac{R}{100}) \\[1em] \Rightarrow 0.02 = (\dfrac{R}{100}) \\[1em] \Rightarrow 0.02 \times 100 = R \\[1em] \Rightarrow R = 2%

Hence, option (a) is correct option.

Assertion Reasoning Questions

Question 1

Assertion (A): ₹ 1,00,000 amounts to ₹ 1,21,000 in 1 year and to ₹1,46,410 in 2 years, compounded annually.

Reason (R): When the interest is compounded annually, then, we have A = P×(1+R100)nP \times \Big(1 + \dfrac{R}{100} \Big)^n

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

Given,

P = ₹ 1,00,000

A = ₹ 1,21,000

n = 1 year

By formula,

A = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

⇒121000=100000×(1+r100)1⇒121000100000=1+r100⇒r100=1.21−1⇒r=0.21×100⇒r=21\Rightarrow 121000 = 100000 \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{121000}{100000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{r}{100} = 1.21 - 1 \\[1em] \Rightarrow r = 0.21 \times 100 \\[1em] \Rightarrow r = 21%

Lets calculate the amount for 2 years with this rate of interest.

By formula,

A = P×(1+r100)nP \times \Big(1 + \dfrac{r}{100}\Big)^n

Substituting the values in formula,

⇒A=100000×(1+21100)2⇒A=100000×(100+21100)2⇒A=100000×(121100)2⇒A=100000×(1.21)2⇒A=100000×1.4641⇒A=146410.\Rightarrow A = 100000 \times \Big(1 + \dfrac{21}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times \Big(\dfrac{100 + 21}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times \Big(\dfrac{121}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times (1.21)^2 \\[1em] \Rightarrow A = 100000 \times 1.4641 \\[1em] \Rightarrow A = 146410.

Assertion (A) is true.

Formula,

A = P×(1+R100)nP \times \Big(1 + \dfrac{R}{100} \Big)^n

Reason is true, as it is the formula to calculate the amount for compound interest when the interest is compounded annually.

Hence, option 3 is correct option.

Question 2

Assertion (A): The present population of a town decreases at the rate of 5% p.a. due to migration. Its present population is 21,600 and 2 years ago it was 24,000.

Reason (R): Let there be an increase in population at R% p.a. Then population after n years is given by P×(1−R100)nP \times \Big(1 - \dfrac{R}{100} \Big)^n, where P is the present population.

  1. A is true, R is false

  2. A is false, R is true

  3. Both A and R are true

  4. Both A and R are false

Answer

Given,

P (two years ago) = 24,000

R = 5% p.a.

n = 2 year

By formula,

Population after n years = P(1−r100)nP\Big(1 - \dfrac{r}{100}\Big)^n

Substituting the values in formula,

Present population =24000×(1−5100)2=24000×(100−5100)2=24000×(95100)2=24000×(0.95)2=24000×0.9025=21,660\text{Present population }=24000 \times \Big(1 - \dfrac{5}{100}\Big)^2 \\[1em] =24000 \times \Big(\dfrac{100 - 5}{100}\Big)^2 \\[1em] =24000 \times \Big(\dfrac{95}{100}\Big)^2 \\[1em] =24000 \times (0.95)^2\\[1em] =24000 \times 0.9025\\[1em] =21,660

Assertion is False

Reason is also false as we do not use that formula to calculate the growth in population. We use,

Population growth after n years at R% with initial Population as P is calulated using formula : P×(1+R100)nP \times \Big(1 + \dfrac{R}{100}\Big)^n

Hence, option 4 is correct option.

Competency Focused Questions

Question 1

What annual instalment will discharge a debt of ₹ 4,600 due in 4 years at 10% simple interest?

  1. ₹ 1,000

  2. ₹ 1,150

  3. ₹ 1,500

  4. ₹ 1,610

Answer

Let annual installment be ₹ x.

Annual installment 1 (Paid at the end of 1st year) : This installment will earn interest for 3 years.

⇒x+x×10100×3⇒x+3x10⇒13x10⇒1.3x\Rightarrow x + x \times \dfrac{10}{100} \times 3 \\[1em] \Rightarrow x + \dfrac{3x}{10} \\[1em] \Rightarrow \dfrac{13x}{10} \\[1em] \Rightarrow 1.3x

Annual installment 2 (Paid at the end of 2nd year) : This installment will earn interest for 2 years.

⇒x+x×10100×2⇒x+2x10⇒12x10⇒1.2x\Rightarrow x + x \times \dfrac{10}{100} \times 2 \\[1em] \Rightarrow x + \dfrac{2x}{10} \\[1em] \Rightarrow \dfrac{12x}{10} \\[1em] \Rightarrow 1.2x

Annual installment 3 (Paid at the end of 3rd year) : This installment will earn interest for 1 year.

⇒x+x×10100×1⇒x+x10⇒11x10⇒1.1x\Rightarrow x + x \times \dfrac{10}{100} \times 1 \\[1em] \Rightarrow x + \dfrac{x}{10} \\[1em] \Rightarrow \dfrac{11x}{10} \\[1em] \Rightarrow 1.1x

Annual installment 4 (Paid at the end of 4th year) : This installment will not earn any interest.

⇒ x

The sum of these values must be equal to debt value of ₹ 4,600.

⇒ 1.3x + 1.2x + 1.1x + x = 4,600

⇒ 4.6x = 4,600

⇒ x = 4,6004.6\dfrac{4,600}{4.6}

⇒ x = ₹ 1,000.

Hence, Option 1 is correct option.

Question 2

The simple interest on a sum of money at 8% per annum for 6 years is half the sum. The sum is:

  1. ₹ 4,000

  2. ₹ 6,000

  3. ₹ 10,000

  4. Data inadequate

Answer

Let the sum be ₹ P.

Given,

I = P2\dfrac{P}{2}

T = 6 years

R = 8%

By formula,

I=P×R×T100I = \dfrac{P \times R \times T}{100}

Substituting values we get :

⇒P2=P×8×6100\Rightarrow \dfrac{P}{2} = \dfrac{P \times 8 \times 6}{100}

By doing this calculation we will not obtain the sum, we need to have the interest to calculate the sum.

Hence, Option 4 is correct option.

Question 3

What is the principal amount which earns ₹ 132 as compound interest for the second year at 10% per annum?

  1. ₹ 800

  2. ₹ 1,050

  3. ₹ 1,200

  4. Data inadequate

Answer

Let the principal amount be ₹ P.

For 1 year :

R = 10%

T = 1 year

I = P×R×T100\dfrac{P \times R \times T}{100}

Substituting values we get :

⇒I=P×10×1100=P10\Rightarrow I = \dfrac{P \times 10\times 1}{100} = \dfrac{P}{10}

A = P+P10=11P10P + \dfrac{P}{10} = \dfrac{11P}{10}

For 2nd year :

Principal = 11P10\dfrac{11P}{10}

R = 10%

T = 1 year

I = P×R×T100\dfrac{P \times R \times T}{100}

I=11P10×10×1100=11P×101000=11P100.I = \dfrac{\dfrac{11P}{10} \times 10 \times 1}{100} \\[1em] = \dfrac{11P \times 10}{1000} \\[1em] = \dfrac{11P}{100}.

Given,

Interest for 2nd year = ₹ 132

∴11P100=132⇒P=132×10011⇒P=1,200.\therefore \dfrac{11P}{100} = 132 \\[1em] \Rightarrow P = \dfrac{132 \times 100}{11} \\[1em] \Rightarrow P = 1,200.

Hence, Option 3 is correct option.

Question 4

In an industrial area, the price of a plot of area 1 acre is ₹ 5 crores. If the price of land increases by 2% quarterly, then what will be the price of the plot after 1 year?

  1. 5×(1+2100)45 \times \Big(1 + \dfrac{2}{100}\Big)^4 crores

  2. 5×(1+8100)45 \times \Big(1 + \dfrac{8}{100}\Big)^4 crores

  3. 5×(1+2100)5 \times \Big(1 + \dfrac{2}{100}\Big) crores

  4. 5×(1+8100)5 \times \Big(1 + \dfrac{8}{100}\Big) crores

Answer

Given,

V = ₹ 5 crores

R = 2%

Price of land increases by 2% quarterly.

n = There are 4 Quarters in a year = 4

By Formula,

Value after n years = V×(1+R100)nV \times \Big(1 + \dfrac{R}{100}\Big)^n

By substituting values,

Value after 1 year=5×(1+2100)4\text{Value after 1 year}= 5 \times \Big(1 + \dfrac{2}{100}\Big)^4 crores.

Hence, Option 1 is correct option.

Question 5

If the compound interest on a certain sum for 2 years at 10% p.a. is ₹ 2,100, the simple interest on it at the same rate for 2 years will be:

  1. ₹ 1,500

  2. ₹ 1,800

  3. ₹ 2,000

  4. ₹ 2,050

Answer

Given,

C.I = ₹ 2,100

n = 2 years

R = 10%

By Formula,

C.I = Amount - Principal

C.I = P×(1+R100)nP \times \Big(1 + \dfrac{R}{100}\Big)^n - P

Substituting the values in formula

⇒2100=P×(1+10100)2−P⇒2100=P×(100+10100)2−P⇒2100=P×(110100)2−P⇒2100=P×(1.10)2−P⇒2100=1.21P−P⇒2100=0.21PP=21000.21P=10,000\Rightarrow 2100 = P \times \Big(1 + \dfrac{10}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times \Big(\dfrac{100 + 10}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times \Big(\dfrac{110}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times (1.10)^2 - P\\[1em] \Rightarrow 2100 = 1.21P - P \\[1em] \Rightarrow 2100 = 0.21P \\[1em] P = \dfrac{2100}{0.21} \\[1em] P = 10,000

To calculate the simple interest on the same principal.

T = 2 years

I = P×R×T100\dfrac{P \times R \times T}{100}

Substituting the Values,

I = 10000×10×2100\dfrac{10000 \times 10 \times 2}{100} = ₹ 2,000

Hence, Option 3 is correct option.

Question 6

Mr. Goyal bought a car for ₹ 6,25,000. Its value depreciates at 20% p.a. How many years it would take for the price of the car to go down by ₹ 3,69,000?

Answer

Given,

Present value of machine (V) = ₹ 6,25,000

r = 20%

Depreciated value = ₹ 6,25,000 - ₹ 3,69,000 = ₹ 2,56,000

By formula,

Value of machine after n years = ₹ V×(1−r100)nV \times \Big(1 - \dfrac{r}{100}\Big)^n

Substituting the values in formula,

⇒256000=625000×(1−20100)n⇒(1−20100)n=256000625000⇒(1−15)n=256625⇒(45)n=(45)4⇒n=4\Rightarrow 256000 = 625000 \times \Big(1 - \dfrac{20}{100}\Big)^n\\[1em] \Rightarrow \Big(1 - \dfrac{20}{100}\Big)^n=\dfrac{256000}{625000} \\[1em] \Rightarrow \Big(1 - \dfrac{1}{5}\Big)^n=\dfrac{256}{625} \\[1em] \Rightarrow \Big(\dfrac{4}{5}\Big)^n=\Big(\dfrac{4}{5}\Big)^4 \\[1em] \Rightarrow n = 4

Hence, It will take 4 years for the price of the car to depreciate.

Question 7

The population of a town was decreasing every year due to migration. The present population of the town is 6,31,680. Last year the migration was 4% and the year before last, it was 6%. What was the population 2 years ago?

Answer

Given,

P = 6,31,680

r1 = 4%

r2 = 6%

n = 1 year

By formula,

Population n years ago = P(1−r1100)×(1−r2100)\dfrac{P}{\Big(1 - \dfrac{r_1}{100}\Big) \times \Big(1 - \dfrac{r_2}{100}\Big)}

Substituting the values in formula,

Population 2 years ago=631680(1−4100)×(1−6100)=631680(100−4100)×(100−6100)=631680(96100)×(94100)=631680×100×10096×94=63168000009024=700000.\text{Population 2 years ago}= \dfrac{631680}{\Big(1 - \dfrac{4}{100}\Big) \times \Big(1 - \dfrac{6}{100}\Big)} \\[1em] = \dfrac{631680}{\Big(\dfrac{100 - 4}{100}\Big) \times \Big(\dfrac{100 - 6}{100}\Big)} \\[1em] = \dfrac{631680}{\Big(\dfrac{96}{100}\Big) \times \Big(\dfrac{94}{100}\Big)} \\[1em] = \dfrac{631680 \times 100 \times 100}{96 \times 94} \\[1em] = \dfrac{6316800000}{9024} \\[1em] = 700000.

Hence, the population of the town 2 years ago was 700000.

PrevNext