RD Sharma Solutions Class 8 Compound Interest Exercise 14.2

RD Sharma Solutions Class 8 Chapter 14 Exercise 14.2

RD Sharma Class 8 Solutions Chapter 14 Ex 14.2 PDF Free Download

Exercise 14.2

Q1) Compute the amount and the compound interest in each of the following by using the formulae when:

(i) Principal = Rs 3000, Rate = 5%, Time = 2 years

(ii) Principal = Rs 3000, Rate =18%, Time = 2 years

(iii) Principal = Rs 5000, Rate = 10 paise per rupee per annum, Time = 2 years

(iv) Principal = Rs 2000, Rate = 4 paise per rupee per annum, Time = 3 years

(v) Principal = Rs 12800, Rate = \(7\frac{1}{2}\)%, Time = 3 years

(vi) Principal =Rs 10000, Rate 20% per annum compounded half-yearly, Time = 2 years

(vii) Principal = Rs 160000, Rate =10 paise per rupee per annum compounded half-yearly, Time = 2 years.

Solution:

Applying the rule A = \(P(1+\frac{R}{100})^{n}\) on the given situations, we get:

(i) A = \(3000(1+\frac{5}{100})^{2}\) = \(3000(1.05)^{2}\) = Rs 3307.5

Now, CI = A – P = Rs 3307.50 – Rs 3000 = Rs. 307.50

(ii) A = \(3000(1+\frac{18}{100})^{2}\) = \(3000(1.18)^{2}\) = Rs 4177.2

Now, CI = A – P = Rs 4177.20 – Rs 3000 = Rs. 1177.20

(iii) A = \(5000(1+\frac{10}{100})^{2}\) = \(5000(1.10)^{2}\) = Rs 6050

Now, CI = A – P = Rs 6050 – Rs 5000 = Rs. 1050

(iv) A = \(2000(1+\frac{4}{100})^{3}\) = \(2000(1.04)^{3}\) = Rs 2249.68

Now, CI = A – P = Rs 2249.68 – Rs 2000 = Rs. 249.68

(v) A = \(12800(1+\frac{7.5}{100})^{3}\) = \(12800(1.075)^{3}\) = Rs 15901.40

Now, CI = A – P = Rs 15901.40 – Rs 12800 = Rs. 3101.40

(vi) A = \(10000(1+\frac{20}{200})^{4}\) = \(10000(1.1)^{4}\) = Rs 14641

Now, CI = A – P = Rs 14641 – Rs 10000 = Rs. 4641

(vii) A = \(160000(1+\frac{10}{200})^{4}\) = \(160000(1.05)^{4}\) = Rs 194481

Now, CI = A – P = Rs 194481 – Rs 160000 = Rs. 34481

Q2) Find the amount of Rs 2400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.

Solution:

Given:

P = Rs 2400

R = 20 % p.a

n = 3 years

We know that amount A at the end of n years at the rate R% per annum when the interest is compounded annually is given by

A = \(P(1+\frac{R}{100})^{n}\)

A = \(2400(1+\frac{20}{100})^{3}\)

A = \(2400(1.2)^{3}\)

A = 4147.20

Thus, the required amount is Rs 4147.20.

 

Q3) Rahman lent Rs 16000 to Rasheed at the rate of \(12\frac{1}{2}\)% per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.

Solution:

Given:

P = Rs 16000

R = 12.5 % p.a

n = 3 years

We know that amount A at the end of n years at the rate R% per annum when the interest is compounded annually is given by

A = \(P(1+\frac{R}{100})^{n}\)

A = \(16000(1+\frac{12.5}{100})^{3}\)

A = \(16000(1.125)^{3}\)

A = 22781.25

Thus, the required amount is Rs 22781.25.

Q4) Meera borrowed a sum of Rs 1000 from Sita for two years. If the rate of interest is 10% compounded annually, find the amount that Meera has to pay back.

Solution:

Given:

P = Rs 1000

R = 10 % p.a

n = 2 years

We know that amount A at the end of n years at the rate R% per annum when the interest is compounded annually is given by

A = \(P(1+\frac{R}{100})^{n}\)

A = \(1000(1+\frac{10}{100})^{2}\)

A = \(1000(1.1)^{2}\)

A = 1210

Thus, the required amount is Rs 1210.

Q5) Find the difference between the compound interest and simple interest. On a sum of       Rs 50,000 at 10% per annum for 2 years.

Solution:

Given:

P = Rs 50000

R = 10 % p.a

n = 2 years

We know that amount A at the end of n years at the rate R% per annum when the interest is compounded annually is given by

A = \(P(1+\frac{R}{100})^{n}\)

A = \(50000(1+\frac{10}{100})^{2}\)

A = \(50000(1.1)^{2}\)

A = Rs 60500

Also,

CI = A – P = Rs 60500 – Rs 50000 = Rs 10500

We know that SI = \(\frac{PRT}{100}=\frac{50000\times 10\times 2}{100}\) = Rs 10000

Therefore, Difference between CI and SI = Rs 10500 – Rs 10000 = Rs 500

Q6) Amit borrowed Rs 16000 at \(17\frac{1}{2}\)% per annum simple interest. On the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years?

Solution:

Amount to be paid by Amit:

SI = \(\frac{PRT}{100}=\frac{16000\times 17.5\times 2}{100}\) = Rs 5600

Amount gained by Amit:

A = \(P(1+\frac{R}{100})^{n}\)

A = \(16000(1+\frac{17.5}{100})^{2}\)

A = \(50000(1.175)^{2}\)

A = Rs 22090

We know that:

CI = A – P = Rs 22090 – Rs 16000 = Rs 6090

Amit’s gain in the whole transaction = Rs 6090 – Rs 5600 = Rs 490

Q7) Find the amount of Rs 4096 for 18 months at \(12\frac{1}{2}\)% per annum, the interest being compounded semi-annually.

Solution:

Given:

P = Rs 4096

R = 12.5 % p.a

n = 1.5 years

We have:

A = \(P(1+\frac{R}{100})^{n}\)

When the interest is compounded annually, we have:

A = \(P(1+\frac{R}{200})^{2n}\)

A = \(4096(1+\frac{12.5}{200})^{3}\)

A = \(4096(1.0625)^{3}\)

A = Rs 4913

Thus, the required amount is Rs 4913.

 

Q8) Find the amount and the compound interest on Rs 8000 for \(1\frac{1}{2}\) years at 10% per annum, compounded half-yearly.

Solution:

Given:

P = Rs 8000

R = 10% p.a

n = 1.5 years

When compounded half-yearly,

We have:

A = \(P(1+\frac{R}{200})^{2n}\)

A = \(8000(1+\frac{10}{200})^{3}\)

A = \(8000(1.05)^{3}\)

A = Rs 9261

Also, CI = A – P = Rs 9261 – Rs 8000 = Rs 1261

 Q9) Kama] borrowed Rs 57600 from LIC against her policy at \(12\frac{1}{2}\)% per annum to build a house. Find the amount that she pays to the LIC after \(1\frac{1}{2}\) years if the interest is calculated half-yearly.

Solution:

Given:

P = Rs 57600

R = 12.5% p.a

n = 1.5 years

When the interest is compounded half-yearly,

We have:

A = \(P(1+\frac{R}{200})^{2n}\)

A = \(57600(1+\frac{12.5}{200})^{3}\)

A = \(57600(1.0625)^{3}\)

A = Rs 69089.06

Thus, the required amount is Rs 69089.06

 

Q10) Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.

Solution:

Given:

P = Rs 64000

R = 5% p.a

n = 1.5 years

When the interest is compounded half-yearly,

We have:

A = \(P(1+\frac{R}{200})^{2n}\)

A = \(64000(1+\frac{5}{200})^{3}\)

A = \(64000(1.025)^{3}\)

A = Rs 68921

Also, CI = A – P = Rs 68921 – Rs 64000 = Rs 4921

Thus, the required interest is Rs 4921.

 

Q11) Rakesh lent out Rs 10000 for 2 years at 20% per annum, compounded annually. How much more he could earn if the interest be compounded half-yearly?

Solution:

Given:

P = Rs 10000

R = 20% p.a

n = 2 years

A = \(P(1+\frac{R}{100})^{n}\)

A = \(10000(1+\frac{20}{100})^{2}\)

A = \(10000(1.2)^{2}\)

A = 14400

When the interest is compounded half-yearly,

We have:

A = \(P(1+\frac{R}{200})^{2n}\)

A = \(10000(1+\frac{20}{200})^{4}\)

A = \(10000(1.1)^{4}\)

A = Rs 14641

Difference = Rs 14641 – Rs 14400 = Rs 241

 

Q12) Romesh borrowed a sum of Rs 245760 at 12.5% per annum, compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest, but compounded semi-annually. Find his gain after 2 years.

Solution:

Given:

P = Rs 245760

R = 12.5% p.a

n = 2 years

When compounded annually,

We have: A = \(P(1+\frac{R}{100})^{n}\)

A = \(245760(1+\frac{12.5}{100})^{2}\)

A = \(245760(1.125)^{2}\)

A = Rs 311040

When compounded semi-annually,

We have:

A = \(P(1+\frac{R}{200})^{2n}\)

A = \(245760(1+\frac{12.5}{200})^{4}\)

A = \(245760(1.0625)^{4}\)

A = Rs 313203.75

Romesh’s gain = Rs 313203.75 – Rs 311040 = Rs 2163.75

  

Q13) Find the amount that David would receive if he invests Rs 8192 for 18 months at \(12\frac{1}{2}\)% per annum, the interest being compounded half-yearly.

Solution:

Given:

P = Rs 8192

R = 12.5% p.a

n = 1.5 years

When the interest is compounded half-yearly, we have:

A = \(P(1+\frac{R}{200})^{2n}\)

A = \(8192(1+\frac{12.5}{200})^{3}\)

A = \(8192(1.0625)^{3}\)

A = Rs 9826

Thus, the required amount is Rs 9826

 

Q14) Find the compound interest on Rs 15625 for 9 months, at 16% per annum, compounded quarterly.

Solution:

Given:

P = Rs 15625

R = 16 % = \(\frac{16}{4}\) = 4 % quarterly

n = 9 months = 3 quarters

We know that:

A = \(P(1+\frac{R}{100})^{n}\)

A = \(15625(1+\frac{4}{100})^{3}\)

A = \(15625(1.043)^{3}\)

A = Rs 17576

Also, CI = A – P = Rs 17576 – Rs 15625 = Rs 1951

Thus, the required compound interest is Rs 1951.

Q15) Rekha deposited Rs 16000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.

Solution:

Given:

P = Rs 16000

R = 20 % p.a

n = 1 year

We know that:

A = \(P(1+\frac{R}{100})^{n}\)

When compounded quarterly, we have:

A = \(P(1+\frac{R}{400})^{4n}\)

A = \(16000(1+\frac{20}{400})^{4}\)

A = \(16000(1.054)^{4}\)

A = Rs 19448.10

Also, CI = A – P = Rs 194448.10 – Rs 16000 = Rs 3448.10

Thus, the interest received by Rekha after one year is Rs 3448.10.

Q16) Find the amount of Rs 12500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.

Solution:

Given:

P = Rs 12500

\(R_{1}\) = 15 % p.a

\(R_{2}\) = 16 % p.a

Therefore, the amount after two years = \(P(1+\frac{R_{1}}{100}\times 1+\frac{R_{2}}{100})\) = \(12500(1+\frac{15}{100}\times 1+\frac{16}{100})\) = \(12500(1.15\times 1.16)\) = Rs 16675

Thus, the required amount is Rs 16675.

Q17) Ramu borrowed Rs 15625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment will he have to make after \(2\frac{1}{4}\) years?

Solution:

Given:

P = Rs 15625

R = 16 % p.a

n = \(2\frac{1}{4}\)

Therefore, Amount after \(2\frac{1}{4}\) years = \(P(1+\frac{R}{100})^{2}\times (1+\frac{\frac{1}{4}\times R}{100})\)

= \(15625(1+\frac{16}{100})^{2}\times (1+\frac{\frac{16}{4}}{100})\)

= \(15625(1+\frac{16}{100})^{2}\times (1+\frac{4}{100})\)

= 15625 (1.16)2 (1.04)

= Rs 21866

Thus, the required amount is Rs 21866.

Q18) What will Rs 125000 amount to at the rate of 6%, if the interest is calculated after every four months?

Solution:

Because interest is calculated after every 3 months, it is compounded quarterly.

Given:

P = Rs 125000

R = 6 % p.a = \(\frac{6}{4}\) % quarterly = 1.5 % quarterly

n = 3

So, A = \(P(1+\frac{R}{100})^{n}\)

A = \(125000(1+\frac{1.5}{100})^{3}\)

A = \(125000(1.015)^{3}\)

A = Rs 132670 approx

Thus, the required amount is Rs 132670.

Q19) Find the compound interest at the rate of 5% for three years on that principle which in three years at the rate of 5% per annum gives Rs 12000 as simple interest.

Solution:

\(P=\frac{SI\times 100}{RT}\)

According to the given values, we have:

P = \(\frac{12000\times 100}{5\times 3}\) = 80000

The principal is to be compounded annually.

So, A = \(P(1+\frac{R}{100})^{n}\)

A = \(80000(1+\frac{5}{100})^{3}\)

A = \(80000(1.05)^{3}\) => A = Rs 92610

Now, CI = A – P = 92610 – 80000 = Rs 12610

Thus, the required compound interest is Rs 12610.

Q20) A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.

Solution:

A = A = \(P(1+\frac{R}{100})^{n}\)

Also, P = A – CI

Let the sum of money be Rs x.

If the compound annually, then: \(A_{1}=x(1+\frac{20}{100})^{2}\) = 1.44x

Therefore, CI = 1.44x – x = 0.44x …..(1)

If the interest is compounded half-yearly, then : \(A_{2}=x(1+\frac{10}{100})^{4}\) = 1.4641x

Therefore, CI = 1.4641x – x = 0.4641x …..(2)

It is given that if interest is compounded half-yearly, then it will be Rs 482 more.

∴ 0.4641x = 0.44x + 482         [From (1) and (2)]

0.4641x – 0.44x = 482

0.0241x = 482

x = \(\frac{482}{0.0241}\) = 20000

Thus, the required sum is Rs 20000.

 

Q21) Simple interest on a sum of money for 2 years at \(6\frac{1}{2}\)% per annum is Rs 5200. What will be the compound interest on the sum at the same rate for the same period?

Solution:

\(P=\frac{SI\times 100}{RT}\)

According to the given values, we have:

P = \(\frac{5200\times 100}{6.5\times 2}\) = 40000

Now, A = \(P(1+\frac{R}{100})^{n}\)

A = \(40000(1+\frac{6.5}{100})^{2}\)

A = \(40000(1.065)^{2}\)

A = Rs 45369

Also, CI = A – P = Rs 45369 – Rs 40000 = Rs 5369

Thus, the required compound interest is Rs 5369.

 Q22) Find the compound interest at the rate of 5% per annum for 3 years on that principle which in 3 years at the rate of 5% per annum gives Rs 1200 as simple interest.

Solution:

\(P=\frac{SI\times 100}{RT}\)

According to the given values, we have:

P = \(\frac{1200\times 100}{5\times 3}\) = 8000

Now, A = \(P(1+\frac{R}{100})^{n}\)

A = \(8000(1+\frac{5}{100})^{3}\)

A = \(8000(1.05)^{3}\)

A = Rs 9261

Also, CI = A – P = Rs 9261 – Rs 8000 = Rs 1261

Thus, the required compound interest is Rs 1261.


Practise This Question

The graph of the equation y = a is a straight line parallel to