## MP Board Class 8th Maths Solutions Chapter 8 Comparing Quantities Ex 8.3

An online step-by-step percentage difference calculator.

**Class 8 Maths Chapter 8 Exercise 8.3 MP Board Question 1.**

Calculate the amount and compound interest on

(a) ₹ 10,800 for 3 years at 12\(\frac{1}{2}\)% per annum compounded annually.

(b) ₹ 18,000 for 2\(\frac{1}{2}\) years at 10% per annum compounded annually.

(c) ₹ 62,500 for 1\(\frac{1}{2}\) years at 8% per annum compounded half yearly.

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.

(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.

Solution:

(a) We have,

(b) We have,

P = ₹ 18000

R = 10 % per annum

n = 2\(\frac{1}{2}\) years or 2.5 years

Now, we calculate S.I. on this amount for \(\frac{1}{2}\) year at 10 % per annum.

∴ Amount after 2.5 years

∴ Interest = A – P = 22869 -18000 = ₹ 4869

(c) We have,

P = ₹ 62500

R = 8 % per annum = 4 % per half year

n = 1\(\frac{1}{2}\) years = 3 half years

∴ Interest = A – P = 70304 – 62500 = ₹ 7804

(d) We have,

P = ₹ 8000

R = 9 % per annum

= 4 % per half year

n = 1 year = 2 half years

∴ Interest = A – P = 8736.20 – 8000 = ₹ 736.20

(e) We have,

P = ₹ 10000

R = 8% per annum = 4 % per half year

n = 1 year = 2 half years

∴ Interest = A – P = 10816 -10000 = ₹ 816

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**MP Board Class 8 Maths Chapter 8 Exercise 8.3 Question 2.**

Kamla borrowed ₹ 26,400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

(Hint : Find 4 for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\frac{4}{12}\) years).

Solution:

We have,

P = ₹ 26400

R = 15 % per annum

n = 2 years 4 months

At the end of 2 years,

Now, P = ₹ 34914

R = 15 % per annum

n = 4 months = \(\frac{1}{3}\) years

At the end of 2 years and 4 months,

**Class 8 Maths Exercise 8.3 Question 3.**

Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Solution:

For Fabina, P = ₹ 12500

R = 12 % per annum

n = 3 years

For simple interest,

Interest = 17000 – 12500 = ₹ 4500

For Radha, P = ₹ 12500

R = 10 % per annum

n = 3 years

As this is compound interest

Interest = 16637.50 -12500 = ₹ 4137.50

Hence, Fabina pays more interest by 4500 – 4137.50 = ₹ 362.50

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**Exercise 8.3 Class 8 Question 4.**

I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Solution:

We have, P = ₹ 12000

R = 6 % per annum

n = 2 years

So, the extra amount he would have to pay = 1483.20 -1440 = ₹ 43.20

**MP Board Class 8 Maths Chapter 8 Question 5.**

Vasudevan invested X 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

(i) after 6 months?

(ii) after 1 year?

Solution:

We have,

P = ₹ 60000

R = 12 % per annum = 6 % per half year

(i) n = 6 months = 1 half year

**Class 8th Maths Chapter 8 Exercise 8.3 Solutions Question 6.**

Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1\(\frac{1}{2}\) years if the interest is

(i) compounded annually.

(ii) compounded half yearly.

Solution:

We have,

P = ₹ 80000

R = 10 % per annum = 5 % per half year

(i) If the interest is compounded annually

(ii) If interest is compounded half yearly

∴ Difference in amounts = 92610 – 92400

= ₹ 210

**Class 8 Maths Comparing Quantities Exercise 8.3 Question 7.**

Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

(i) The amount credited against her name at the end of the second year.

(ii) The interest for the 3rd year.

Solution:

We have,

P = ₹ 8000

R = 5 % per annum

(i) n = 2 years

(ii) n = 3 years

Hence, interest for 3^{rd} year = 9261 – 8820 = ₹ 441

**Class 8th Maths Chapter 8 Exercise 8.3 Question 8.**

Find the amount and the compound interest on ₹ 10,000 for 1\(\frac{1}{2}\) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually? Solution:

We have,

P = ₹ 10000

R = 10 % per annum = 5 % per half year

n = 1\(\frac{1}{2}\) years = 3 half years

(i) If interest is compounded half yearly

Interest = 11576.25 – 10000 = ₹ 1576.25

(ii) If interest is compounded annually Amount after 1 year,

Thus, more interest would be generated if interest is calculated half yearly.

**Class 8 Maths Chapter 8 8.3 Question 9.**

Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12\(\frac{1}{2}\) % per annum, interest being compounded half yearly.

Solution:

We have,

P = ₹ 4096

R = 12.5 % per annum = 6.25 % per half year

n = 18 months = 3 half years

**8th Class Maths 8th Chapter Exercise 8.3 Question 10.**

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) find the population in 2001.

(ii) what would be its population in 2005?

Solution:

We have, population in 2003 = 54000

Rate = 5% per annum

(i) Let population in 2001 be x.

The population in 2003

∴ Population = 48980 in 2001

(ii) For population in 2005

n = 2 years

**Class 8th Maths 8.3 Question 11.**

In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Solution:

Initial count of bacteria = 506000

Rate = 2.5 % per hour

n = 2 hours

∴ Count after 2 hours

**Class 8 Math Ex 8.3 Question 12.**

A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution:

Initial price = ₹ 42000

Rate of depreciation = 8% per annum

n = 1 year