NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Comparing Quantities |

Exercise |
Ex 8.3 |

Number of Questions Solved |
12 |

Category |
NCERT Solutions |

## 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 % per annum compounded annually.

**(b)** ₹ 18,000 for 2 years at 10% per annum compounded annually.

**(c)** ₹ 62,500 for 1 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 years)

**Solution:**

Here, P = ₹ 26400, R =15% per annum

and n = 2 years 4 months =2 years.

Hence, Kamala will pay ₹ 36659.70 to the bank.

**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 = ₹ 12500, R =12% per annum and T =3 years. Then,

Hence, Fabina pays 362.50 more as interest ₹ (4500 – 4137.50), i.e., ₹ 362.50 more as interest.

**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 = ₹ 12000, R = 6% per annum and T = 2 years.

So, I have to pay ₹ (1483.20 -1440), i.e., ₹ 43.20 in excess.

**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, Principal =₹ 60000, Rate = 12% per annum = 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 years if the interest is 2

**(i)** compounded annually

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

**Solution:**

Here, P = ₹ 80000

Rate = 10% per annum = 5% per half-year,

Time = 1 years = 3 half-years.

**Ex 8.3 Class 8 Maths Question 7.**

Maria invested ? 8,000 in a business. She would he 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 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 = ? 10000

Time = 1 years = 3 half years,

This interest is more than the interest that he would get if it was compounded annually.

**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 % 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?

**Solution:**

**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 = 506000;

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 = ₹ 42000

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

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