**NCERT Solutions for Class 8 Maths Chapter 8
Comparing Quantities Ex 8.3**

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities
Ex 8.3 are the part of NCERT Solutions for Class 8 Maths. Here you can find the
NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3**.**

**Ex 8.3 Class 8 Maths Question 1.**

Calculate
the amount and compound interest on**(a)**₹ 10,800 for 3 years at 12

^{1}/

_{2}% per annum compounded annually.

**(b)**₹ 18,000 for 2

^{1}/

_{2}years at 10% per annum compounded annually.

**(c)**₹ 62,500 for 1

^{1}/

_{2}years at 8% per annum compounded half yearly.

**(d)**₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

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

**Solution:**

**Ex
8.3 Class 8 Maths Question 2.**

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

**Hint:**Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for 4/12 years)

**Solution:**

Here, P = ₹ 26,400, R = 15% p.a. and n = 2 years 4 months =2^{1}/_{3} years.

Hence, Kamala will pay ₹ 36659.70 to the bank to clear the loan.

**Ex 8.3 Class 8
Maths 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:**

In case of Fabina:

P = ₹ 12,500, R = 12% p.a. and T = 3 years. Then,

**Ex 8.3 Class 8
Maths 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:**

Here, P = ₹ 12,000, R = 6% p.a. and T = 2 years.

**Ex 8.3 Class 8
Maths Question 5.**

Vasudevan invested ₹ 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:**

Here, P = ₹ 60,000, R = 12% p.a. = 6% per half-year.

**(i)** Time
= 6 months = 1 half-year

**Ex
8.3 Class 8 Maths 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^{1}/

_{2}years if the interest is

**(i)**compounded annually

**(ii)**compounded half-yearly.

**Solution:**

Here, P = ₹ 80,000

R = 10% p.a. = 5% per half-year,

Time = 1^{1}/_{2} years = 3 half-years.

**Ex 8.3 Class 8
Maths 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:**

**Ex
8.3 Class 8 Maths Question 8. **

Find the amount and the compound interest on ₹ 10,000 for 1^{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:
**

Here, Principal = ₹ 10,000

Time = 1^{1}/_{2} years = 3 half years,

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

Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12^{1}/

_{2}% per annum, interest being compounded half yearly.

**Solution:**

Here, Principal = ₹ 4096,

Time = 18 months = 3 half years

**Ex 8.3 Class 8
Maths 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?

**Ex 8.3 Class 8
Maths 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:**

We have, P = Original count of bacteria = 5,06,000

Rate of increase = R = 2.5% per hour, Time = 2 hours.

**Ex 8.3 Class 8
Maths 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:**

We have, V_{0} = Initial value = ₹ 42,000

R = Rate of depreciation = 8% p.a.