Compound Interest Class IX( Revision)

 Important types of Question and Answer

1). Find the compound interest on ₹6250 at 8% p.a. for 2 years 

Ans. P = ₹6250, R = 8% p.a. and T = 2 years

Amount = P(1+R/100) ^t

                = {6250(1+8/100) ^2}

               = {6250×27/25×27/25} = ₹7290

CI = (7290 - 6250) = ₹1040

2). A deposited ₹6000 in a bank at 5% per annum simple interest. B deposited ₹5000 at 8% p.a. compound interest. After 2 years, the difference between their interests will be. 

Ans. SI = ( P×R×T/100) = (6000×5×2/100) 

             = ₹600

       CI = [5000×(1+8/100) ^2 - 5000]

           = [ (5000×27/25×27/25) - 5000]

           = (5832-5000) = ₹832

Difference = CI - SI = (832-600) = ₹232

3). Calculate the compound interest accrued on ₹6,000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.

Ans. P = ₹ 6,000; n = 3 years; r1 = 5%; r2 = 8% and r3 =10%

Amount = P(1 + r1/100)(1 + r2/100)

= 6000(1 + 5/100)(1 + 8/100)(1 + 10/100)

= 6000(21/20)(27/25)(11/10) = ₹7,484.40

Therefore,

C.I. = ₹7,484.40 – ₹6,000 = ₹1,484.40

4).In how many years will ₹7,000 amount to ₹9,317 at 10% per annum compound interest?

Ans. P = ₹7,000; A = ₹9,317 and r = 10%

A = P(1 + r/100)n

9317 = 7000(1 + 10/100)n

9317/7000 = (11/10)n

1331/1000 = (11/10)n

(11/10)3 = (11/10)n

On comparing, we have, n = 3

Therefore, the number of years is 3

5). Divide ₹28,730 between A and B so that when their shares are lent out at 10% compound interest compounded per year, the amount that A receives in 3 years is the same as what B receives in 5 years.

Ans. Let’s assume the share of A as ₹y

Share of B = ₹(28,730 – y) & Rate of interest = 10%

Then, according to question

Amount of A in 3 years = Amount of B in 5 years

y(1 + 10/100)3 = (28730 – y)(1 + 10/100)5

y = (28730 – y)(1 + 10/100)2

y = (28730 – y)(11/10)2

y = (121/100)(28730 – y)

100y = 121 (28730 – y)

100y = 121 x 28730 – 121y

221y = (121 x 28730)

y = (121 x 28730)/ 221 = 15730

Therefore, the share of A = ₹15,730 and share of B = ₹28,730 – ₹15,730 = ₹13,000

6). Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹12,600 at the end of the first year and ₹17,640 at the end of the second year. Find the sum borrowed.

Ans. Let’s assume the principal sum to be P

Now, For the payment of ₹12,600 at the end of the first year, we have

A = ₹12,600; R = 5% and n = 1 year

So, A = P(1 + R/100)n

12600 = P(1 + 5/100)1

12600 = P(21/20)

P = (12600 x 20)/21 = ₹12,000

Now, for the payment of ₹17,640 at the end of second year

A = ₹17,640; R = 5% and n = 2 years

A = P(1 + R/100)n

17640 = P(1 + 5/100)2

17640 = P(21/20)2

P = (20/21)2 x 17640 = 16000

Thus, the sum borrowed = ₹(12,000 + 16,000) = ₹28,000

7). A sum on compound interest amounts to ₹2809 in 2 years and ₹2977.54 in 3 years. Find the sum and rate per cent per annum. 

Ans. Interest on ₹2809 for 1 years = ( 2977.54 - 2809) = ₹168.54

Let's assume the sum be ₹x

Amount in 2 years = x × (1+6/100) ^2

 = (x × 53/50 × 53/50) 

Therefore, 2809 = (x × 53/50 × 53/50) 

 x = (2809×50/53×50/53) = 2500

Thus, the sum is ₹2500

8). A sum of money is invested at 10% per annum compounded half yearly. If the difference of amounts at the end of 6 months and 12 months is ₹189, find the sum of money invested.

Ans. Let’s assume the sum of money to be ₹y

And, given rate = 10% p.a. compounded half yearly

Now, for first 6 months

A = P[1 + r/(2 x 100)]n x 2

= y[1 + 10/(2 x 100]1/2 x 2

= y(1 + 10/200)1

= (21/20)y

And, For first 12 months

A = P[1 + r/(2 x 100)]n x 2

= y[1 + 10/(2 x 100]1 x 2

= y(1 + 10/200)2

= (441/400)y

Also given, the difference between the above amounts = ₹189

So, (441/400)y – (21/20)y = 189

(21/400)y = 189

y = (189 x 400)/21

y = 3600

Thus, the sum of money invested is ₹3,600

9). The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years?

Ans. Let’s assume that ₹x and ₹y to be the money invested by Pramod and Rohit respectively such that they will get the same sum on attaining the age of 25 years.

Now, Pramod will attain the age of 25 years after (25 – 16) = 9 years

Rohit will attain the age of 25 years after (25 -18) = 7 years

So, we have

x (1 + 5/100)9 = y(1 + 5/100)7

x/y = 1/(1 + 5/100)2

x/y = 400/441

Therefore, Pramod and Rohit should invest in the ratio 400:441 respectively such that they will get the same sum on attaining the age of 25.

10). A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money becomes twenty-seven times of itself at the same rate of interest p.a.

Ans.  Let’s assume the principal as x

Amount (A) = 3x, n = 8 years, R =?

A = P(1 + R/100)n

Now,

Case I: 

3x = x(1 + R/100)8

Taking the 8th root on both sides, we have

31/8 = (1 + R/100) … (1)

Case II: P = x, A = 27x, T =?

27x = x(1 + R/100)T

271/T = 1 + R/100 … (2)

From (1) and (2), we have

31/8 = 271/T

31/8 = (33)1/T

31/8 = 33/T

On comparing the exponents,

1/8 = 3/T

T = 3 x 8 = 24

Thus, it will take 24 years for the money to become twenty-seven times of itself at the same rate of interest p.a.

---------------------------------

Contacts of PathMaker

Numbers - 8240042490, 9875616291

Social media -

https://www.facebook.com/PathMaker-112826467702117/

https://www.instagram.com/pathmaker06/