# Class 8 RD Sharma Solutions – Chapter 14 Compound Interest – Exercise 14.3 | Set 2

### Chapter 14 Compound Interest – Exercise 14.3 | Set 1

**Question 15. Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in a year and a half, interest being compounded six monthly.**

**Solution:**

We have,

Principal = Rs 2000

Amount = Rs 2315.25

Time = 1 ½ years = 3/2 years

Let rate be = R % per annum

By using the formula,

A = P (1 + )

^{n}Substituting the values, we have

2315.25 = 2000 (1 + )

^{3/2}(1 + )

^{3/2}= 2315.25/2000(1 + )

^{3/2}= (1.1576)(1 + ) = 1.1025

= 1.1025 – 1

= 0.1025 × 100

= 10.25

Therefore,

Required Rate is 10.25% per annum.

**Question 16. Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.**

**Solution:**

We have,

Time = 3 years

Let rate be = R %

Also principal be = P

So, amount becomes = 2P

By using the formula,

A = P (1 + )

^{n}Substituting the values, we have

2P = P (1 + )

^{3}(1 + )

^{3}= 2(1 + ) = 2

^{1/3}1 + = 1.2599

= 1.2599-1

= 0.2599

R = 0.2599 × 100

= 25.99

Therefore,

Required Rate is 25.99% per annum.

**Question 17. Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half-yearly**

**Solution:**

We have,

Time = 2 years = 2×2 = 4 half years

Let rate = R % per annum = R/2% half years

Let principal be = P

So, Amount becomes = 4P

By using the formula,

A = P (1 + )

^{n}Substituting the values, we have

4P = P (1 + )

^{4}(1 + )

^{4}= 4(1 + ) = 4

^{1/4}1 + = 1.4142

= 1.4142-1

= 0.4142

R = 0.4142 × 200

= 82.84%

Therefore,

Required Rate is 82.84% per annum.

**Question 18. A certain sum amounts to Rs. 5832 in 2 years at 8% compounded interest. Find the sum.**

**Solution:**

We have,

Amount = Rs 5832

Time = 2 years

Rate = 8%

Let principal be = P

By using the formula,

A = P (1 + )

^{n}Substituting the values, we have

5832 = P (1 + )

^{2}5832 = P (1.1664)

P = 5832/1.1664

= 5000

Therefore,

Required sum is Rs 5000.

**Question 19. The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs. 360. Find the sum.**

**Solution:**

We have,

Time = 2 years

Rate = 7.5 % per annum

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 360

C.I – S.I = Rs 360

By using the formula,

P [(1 + )

^{n}– 1] – (PTR)/100 = 360Substituting the values, we have

P [(1 + )

^{2}– 1] – (P(2)(7.5))/100 = 360P[249/1600] – (3P)/20 = 360

249/1600P – 3/20P = 360

(249P-240P)/1600 = 360

9P = 360 × 1600

P = 576000/9

= 64000

Therefore,

The sum is Rs 64000.

**Question 20. The difference in simple interest and compound interest on a certain sum of money at 623 % per annum for 3 years in Rs. 46. Determine the sum.**

**Solution:**

We have,

Time = 3 years

Rate = 6 % per annum = 20/3%

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 46

C.I – S.I = Rs 46

By using the formula,

P [(1 + )

^{n}– 1] – (PTR)/100 = 46Substituting the values, we have

P [(1 + )

^{3}– 1] – (P(3)(20/3))/100 = 46P[(1 + )

^{3}– 1] – P/5 = 46P[721/3375] – P/5 = 46

721/3375P – 1/5P = 46

(721P-675P)/3375 = 46

46P = 46 × 3375

46P = 46 × 3375/46

= 3375

Therefore,

The sum is Rs 3375.

**Question 21. Ishita invested a sum of Rs. 12000 at 5% per annum compound interest. She received an amount of Rs. 13230 after n years. Find the value of n.**

**Solution:**

We have,

Principal = Rs 12000

Amount = Rs 13230

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + )

^{n}Substituting the values, we have

13230 = 12000 (1 + )

^{T}13230 = 12000 ()

^{T}(21/20)

^{T}= 13230/12000(21/20)

^{T}= 441/400(21/20)

^{T}= (21/20)^{2}So on comparing both the sides, n = T = 2

Therefore,

Time required is 2 years.

**Question 22. At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?**

**Solution:**

We have,

Principal = Rs 4000

Time = 2 years

CI = Rs 410

Rate be = R% per annum

By using the formula,

CI = P [(1 + )

^{n}– 1]Substituting the values, we have

410 = 4000 [(1 + )

^{2}– 1]410 = 4000 (1 + )

^{2}– 4000410 + 4000 = 4000 (1 + )

^{2}(1 + )

^{2}= 4410/4000(1 + )

^{2}= 441/400(1 + )

^{2}= (21/20)^{2}By canceling the powers on both the sides,

1 + = 21/20

= 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5% per annum.

**Question 23. A sum of money deposited at 2% per annum compounded annually becomes Rs. 10404 at the end of 2 years. Find the sum deposited.**

**Solution:**

We have,

Time = 2years

Amount = Rs 10404

Rate be = 2% per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + )

^{n}Substituting the values, we have

10404 = P [(1 + )

^{2}]10404 = P [1.0404]

P = 10404/1.0404

= 10000

Therefore,

Required sum is Rs 10000.

**Question 24. In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5% per annum compound interest?**

**Solution:**

We have,

Principal = Rs 1600

Amount = Rs 1852.20

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + )

^{n}Substituting the values, we have

1852.20 = 1600 (1 + )

^{T}1852.20 = 1600 ()

^{T}(21/20)

^{T}= 1852.20/1600(21/20)

^{T}= 9261/8000(21/20)

^{T}= (21/20)^{3}So on comparing both the sides, n = T = 3

Therefore,

Time required is 3 years.

**Question 25. At what rate percent will a sum of Rs. 1000 amount to Rs. 1102.50 in 2 years at compound interest?**

**Solution:**

We have,

Principal = Rs 1000

Amount = Rs 1102.50

Rate = R% per annum

Let time = 2 years

By using the formula,

A = P (1 + )

^{n}Substituting the values, we have

1102.50 = 1000 (1 + )

^{2}(1 + )

^{2}= 1102.50/1000(1 + )

^{2}= 4410/4000(1 +)

^{2}= (21/20)^{2}1 + = 21/20

= 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5%.

**Question 26. The compound interest on Rs. 1800 at 10% per annum for a certain period of time is Rs. 378. Find the time in years.**

**Solution:**

We have,

Principal = Rs 1800

CI = Rs 378

Rate = 10% per annum

Let time = T years

By using the formula,

CI = P [(1 + )

^{n}– 1]Substituting the values, we have

378 = 1800 [(1 + )

^{T}– 1]378 = 1800 [()

^{T}– 1]378 = 1800 [()

^{T}– 1800378 + 1800 = 1800 [()

^{T}(11/10)

^{T}= 2178/1800(11/10)

^{T}= 726/600(11/10)

^{T}= 121/100(11/10)

^{T }= (11/10)^{2}So on comparing both the sides, n = T = 2

Therefore,

Time required is 2 years.

**Question 27. What sum of money will amount to Rs. 45582.25 at 6 ¾ % per annum in two years, interest being compounded annually**

**Solution:**

We have,

Time = 2years

Amount = Rs 45582.25

Rate be = 6 ¾ % per annum = 27/4%

Let principal be = Rs P

By using the formula,

A = P [(1 + )

^{n}Substituting the values, we have

45582.25 = P [(1 + 27/4×100)

^{2}]45582.25 = P (1 + )

^{2}45582.25 = P ()

^{2}45582.25 = P × 427/400 × 427/400

P = (45582.25 × 400 × 400) / (427×427)

P = 7293160000/182329

= 40000

Therefore,

Required sum is Rs 40000.

**Question 28. Sum of money amounts to Rs. 453690 in 2 years at 6.5% per annum compounded annually. Find the sum.**

**Solution:**

We have,

Time = 2years

Amount = Rs 453690

Rate be = 6.5 % per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + )

^{n}Substituting the values, we have

453690 = P [(1 + )

^{2}]453690 = P ()

^{2}453690 = P × 106.5/100 × 106.5/100

P = (453690 × 100 × 100) / (106.5×106.5)

P = 4536900000/11342.25

= 400000

Therefore,

Required sum is Rs 400000.