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Class 8 RD Sharma Solutions – Chapter 14 Compound Interest – Exercise 14.3 | Set 1

  • Last Updated : 08 Apr, 2021

Question 1. On what sum will the compound interest at 5% per annum for 2 years compounded annually be 164.

Solution:

We have,

Rate = 5 % per annum

Compound Interest (CI) = ₹164

Time (t) = 2 years



By using the formula,

Let P be ‘x’

CI = A – P

164 = P (1 + R/100) n – P

Substituting the values, we have

= P [(1 + R/100)n – 1]

= x [(1 + 5/100)2 – 1]

= x [(105/100)2 – 1]

164 = x ((1.05)2 – 1)

x = 164 / ((1.05)2 – 1)

= 164/0.1025

= ₹1600

Therefore 

The required sum is ₹1600.

Question 2. Find the principal if the interest compounded annually at the rate of 10% for two years is 210.

Solution:

We have,

Rate = 10 % per annum

Compound Interest (CI) = ₹210



Time (t) = 2 years

By using the formula,

Let P be ‘x’

CI = A – P

210 = P (1 + R/100)n – P

Substituting the values, we have

= P [(1 + R/100)n – 1]

= x [(1 + 10/100)2 – 1]

= x [(110/100)2 – 1]

210 = x ((1.1)2 – 1)

x = 210 / ((1.1)2 – 1)

= 210/0.21

= ₹1000

Therefore,

The required sum is ₹1000.

Question 3. A sum amounts to 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.

Solution:

We have,

Rate = 10 % per annum

Amount = ₹756.25

Time (t) = 2 years



By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

756.25 = P (1 + 10/100)2

P = 756.25/(1 + 10/100)2

= 756.25/1.21

= 625

Therefore, 

The principal amount is ₹625.

Question 4.  What sum will amount to 4913 in 18 months, if the rate of interest is 12 ½ % per annum, compounded half-yearly?

Solution:

We have,

Rate = 12 ½% per annum = 25/2% = 25/2/2 = 25/4% half yearly

Amount = ₹4913

Time (t) = 18months = 18/12years = 3/2 × 2 = 3 half years

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

4913 = P (1 + 25/4 ×100)3

P = 4913 / (1 + 25/400)3

= 4913/1.19946

= 4096

Therefore, 

The principal amount is ₹4096.

Question 5. The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is 283.50. Find the sum.

Solution:

We have,

Rate = 15 % per annum

Compound Interest (CI) – Simple Interest (SI)= ₹283.50

Time (t) = 3 years

By using the formula,

CI – SI = 283.50



P [(1 + R/100)n – 1] – (PTR)/100 = 283.50

Substituting the values, we have

P [(1 + 15/100)3 – 1] – (P(3)(15))/100 = 283.50

P[1.520 – 1] – (45P)/100 = 283.50

0.52P – 0.45P = 283.50

0.07P = 283.50

P = 283.50/0.07

= 4000

Therefore,

The sum is ₹4000.

Question 6. Rachana borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years 1290 as interest compounded annually, find the sum she borrowed.

Solution:

We have,

Rate = 15 % per annum

Time = 2 years

CI = Rs 1290

By using the formula,

CI = P [(1 + R/100)n – 1]

Substituting the values, we have

1290 = P [(1 + 15/100)2 – 1]

1290 = P [0.3225]

P = 1290/0.3225

= 4000

Therefore,

The sum is ₹4000.

Question 7.  The interest on a sum of 2000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is 163.20.

Solution:

We have,

Rate = 4 % per annum

CI = ₹163.20

Principal (P) = Rs 2000

By using the formula,

CI = P [(1 + R/100)n – 1]

Substituting the values, we have

163.20 = 2000[(1 + 4/100)n – 1]

163.20 = 2000[(1.04)n -1]

163.20 = 2000 × (1.04)n – 2000

163.20 + 2000 = 2000 × (1.04)n

2163.2 = 2000 × (1.04)n

(1.04)n = 2163.2/2000

(1.04)n = 1.0816

(1.04)n = (1.04)2



So on comparing both the sides, n = 2

Therefore,

Time required is 2 years.

Question 8. In how much time would 5000 amount to 6655 at 10% per annum compound interest?

Solution:

We have,

Rate = 10% per annum

A = ₹6655

Principal (P) = ₹5000

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

6655 = 5000 (1 + 10/100)n

6655 = 5000 (11/10)n

(11/10)n = 6655/5000

(11/10)n = 1331/1000

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

So on comparing both the sides, n = 3

Therefore,

Time required is 3 years.

Question 9. In what time will 4400 become 4576 at 8% per annum interest compounded half-yearly?

Solution:

We have,

Rate = 8% per annum = 8/2 = 4% (half yearly)

A = Rs 4576

Principal (P) = ₹4400

Let n be ‘2T’

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

4576 = 4400 (1 + 4/100)2T

4576 = 4400 (104/100)2T

(104/100)2T = 4576/4400

(104/100)2T= 26/25

(26/25)2T = (26/25)1

So on comparing both the sides, n = 2T = 1

Therefore, 

Time required is 1/2 year.

Question 10. The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is 20. Find the sum.

Solution:

We have,

Rate = 4 % per annum

Time = 2 years

Compound Interest (CI) – Simple Interest (SI)= ₹20

By using the formula,

CI – SI = 20

P [(1 + R/100)n – 1] – (PTR)/100 = 20

Substituting the values, we have

P [(1 + 4/100)2 – 1] – (P(2)(4))/100 = 20

P[51/625] – (2P)/25 = 20

51/625P – 2/25P = 20

(51P-50P)/625 = 20

P = 20 × 625



P = 20/7.918

= 12500

Therefore 

The sum is ₹12500.

Question 11. In what time will Rs. 1000 amount to Rs. 1331 at 10% per annum, compound interest?

Solution:

We have,

Principal = Rs 1000

Amount = Rs 1331

Rate = 10% per annum

Let time = T years

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

1331 = 1000 (1 + 10/100)T

1331 = 1000 (110/100)T

(11/10)T = 1331/1000

(11/10)T = (11/10)3

So on comparing both the sides, n = T = 3

Therefore,

Time required is 3 years.

Question 12. At what rate percent compound interest per annum will Rs. 640 amount to Rs. 774.40 in 2 years?

Solution:

We have,

Principal = Rs 640

Amount = Rs 774.40

Time = 2 years

Let rate = R%

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

774.40 = 640 (1 + R/100)2

(1 + R/100)2 = 774.40/640

(1 + R/100)2 = 484/400

(1 + R/100)2 = (22/20)2

By canceling the powers on both sides,

(1 + R/100) = (22/20)

R/100 = 22/20 – 1

= (22-20)/20

= 2/20

= 1/10

R = 100/10

= 10%

Therefore,

Required Rate is 10% per annum.

Question 13. Find the rate percent per annum if Rs. 2000 amount to Rs. 2662 in 1 ½ years, interest being compounded half-yearly?

Solution:

We have,

Principal = Rs 2000

Amount = Rs 2662

Time = 1 ½ years = 3/2 × 2 = 3 half years

Let rate be = R% per annum = R/2 % half yearly

By using the formula,

A = P (1 + R/100)n

Substituting the values, we have

2662 = 2000 (1 + R/2×100)3

(1 + R/200)3 = 1331/1000

(1 + R/100)3 = (11/10)3

By canceling the powers on both sides,

(1 + R/200) = (11/10)

R/200 = 11/10 – 1

= (11-10)/10

= 1/10

R = 200/10

= 20%

Therefore,

Required Rate is 20% per annum.

Question 14. Kamala borrowed from Ratan a certain sum at a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years she received Rs. 210 as compound interest, but paid Rs. 200 only as simple interest. Find the sum and the rate of interest.

Solution:

We have,

C.I that Kamala receives = Rs 210

S.I that Kamala paid = Rs 200

Time = 2 years

So,



We know, SI = PTR/100

= P×2×R/100

P×R = 10000 ………….. Equation 1

CI = A – P

CI = P [(1 + R/100)n – 1]

Substituting the values, we have

210 = P [(1 + R/100)2 – 1]

210 = P (12 + R2/1002 + 2(1)(R/100) – 1) (by using the formula (a+b)2)

210 = P (1 + R2/10000 + R/50 – 1)

210 = P (R2/10000 + R/50)

210 = PR2/10000 + PR/50

We know PR = 10000 from Equation 1

210 = 10000R/10000 + 10000/50

210 = R + 200

R = 210 – 200

= 10%

In Equation 1, PR = 10000

P = 10000/R

= 10000/10

= 1000

Therefore,

Required sum is Rs 1000.

Chapter 14 Compound Interest – Exercise 14.3 | Set 2

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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