Question 1 |

An amount of Rs.9000 is invested for 2 years at interest rate of 15% per annual and compunded annually. At the end of 2nd year how much amount will be obtained as interest?

Rs. 2902.50 | |

Rs. 2900.50 | |

Rs. 2899.50 | |

Rs. 2899 |

**Compound Interest**

**Discuss it**

Question 1 Explanation:

Amount = P[1+ (R/100)]

where P = principal, R = rate of interest and n = time(years)

Amount= Rs.9000 x [1 + (15/100)]

The amount obtained by the way of interest in compound interest = Amount - principal = Rs.(11902.5 - 9000) = Rs.2902.50

^{n}where P = principal, R = rate of interest and n = time(years)

Amount= Rs.9000 x [1 + (15/100)]

^{n}= 9000 x (23/20)^{n}= 23805/2 = Rs.11902.5The amount obtained by the way of interest in compound interest = Amount - principal = Rs.(11902.5 - 9000) = Rs.2902.50

Question 2 |

Ram deposits Rs.2000 each on 1

^{st}January and 1^{st}July of a year at the rate of 8% compound interest calculated on half-yearly basis. At the end of the year how much amount he would have?Rs.2150.50 | |

Rs.3182.40 | |

4243.20 | |

280.40 |

**Compound Interest**

**Discuss it**

Question 2 Explanation:

We can break this problem into two parts: Rs. 1500 invested for 1 year (Jan to Dec) and Rs. 2000 invested for 6 months (Jul to Dec)

When interest is compounded Half-yearly:

Amount = P[1+ (R/2)/100 ]

The total amount for the investment on 1

When interest is compounded Half-yearly:

Amount = P[1+ (R/2)/100 ]

^{2n}The total amount for the investment on 1

^{st}Jan is: Amount1 = Rs. 2000 x [1+ (8/2)/100]^{2x1}= Rs. 2000 x [1 + (4/100)]^{2}= Rs. 2000 x [26/25]^{2}The total amount for investment on 1^{st}july is: Amount2 = Rs. 2000 x [1+ (8/2)/100][2 x(1/2)] = Rs. 2000 x [1+ 4/100 ] = Rs. 2000 x [26/25] The total amount at the end of the year = amount1 + amount2 = 2000 x [26/25]^{2}+ 2000 x [26/25] = 2000 x [26/25] x [(26/25) + 1] = 2000 x 26/25 x 51/25 = 4243.20Question 3 |

Ram invests Rs. 10,000 for 1 year at a rate of 10% per annum compounded yearly and Sita invests the same amount for same time at same rate per annum compounded half yearly. What is the difference between the interests earned by both?

Rs. 25.50 | |

Rs. 25 | |

Rs.20.50 | |

Rs.23.75 |

**Compound Interest**

**Discuss it**

Question 3 Explanation:

For Ram:

Amount = 10000.[1 + 10/100]

For Sita:

Amount= Rs. 10000.[1 +(10/2)/100]

Amount = 10000.[1 + 10/100]

^{1}= Rs.10000 x 11/10 = Rs.11000For Sita:

Amount= Rs. 10000.[1 +(10/2)/100]

^{2}= 10000 x (21/20)^{2}= Rs.11025 Difference = Rs.(11000-11025) = Rs.25Question 4 |

Ram has a sum of Rs.8000. He lends it for 20% per annum at compound interest. In how much time the sum of the amount will be Rs.13824?

2 | |

1 | |

3 | |

4 |

**Compound Interest**

**Discuss it**

Question 4 Explanation:

3824 = 8000 x (1 + 20/100)

13824/8000 = (120/100)

(24/20)

(12/10)

=> n = 3 years

^{n}13824/8000 = (120/100)

^{n}(24/20)

^{3}= (12/10)^{n}(12/10)

^{3}= (12/10)^{n}=> n = 3 years

Question 5 |

Find the amount returned for an investment of Rs.5,000 after 2 years, if the rate of interest for the 1st year is 5% and for the 2nd year is 10%.

772 | |

775 | |

622 | |

820 |

**Compound Interest**

**Discuss it**

Question 5 Explanation:

Amount = (Principal + Compound interest) = P(1 + R1/100)(1 + R2/100)(1 + R3/100)

Amount = 5000(1 + 5/100)(1 + 10/100)

= 5000 x (21/20)(11/10)

= 5000 x (231/200)

=5775

Interest = 5775 - 5000 = 775

Amount = 5000(1 + 5/100)(1 + 10/100)

= 5000 x (21/20)(11/10)

= 5000 x (231/200)

=5775

Interest = 5775 - 5000 = 775

Question 6 |

What sum will be amount to Rs.30000 at CI in 3 years, if the rate of interest for 1

^{st}, 2^{nd}and 3^{rd}year being 10%, 20% and 30% respectively?17482.5 | |

20145 | |

16524 | |

17000 |

**Compound Interest**

**Discuss it**

Question 6 Explanation:

Let Rs.P be the required sum.

30000= p(1 + 10/100)(1 + 20/100)(1 + 30/100)

= p (110/100) x (120/100) x (130/100)

p = 30000x 100 x 100 x 100 / (110 x 120 x 130)

p = 17482.5

30000= p(1 + 10/100)(1 + 20/100)(1 + 30/100)

= p (110/100) x (120/100) x (130/100)

p = 30000x 100 x 100 x 100 / (110 x 120 x 130)

p = 17482.5

Question 7 |

What will be the amount if Rs.10,00,000 is invested at CI for 3 years at a rate of interest 10%, 11% and 12% respectively?

1367520 | |

1367602 | |

1367420 | |

1365520 |

**Compound Interest**

**Discuss it**

Question 7 Explanation:

= p(1 + R1/100)(1 + R2/100)(1 + R3/100)

= 10,00,000 x(1 + 10/100)x(1 + 11/100)x(1 + 12/100)

= 10,00,000 x (110/100) x (111/100) x (112/100)

= 110 x 111 x 112

= 10,00,000 x(1 + 10/100)x(1 + 11/100)x(1 + 12/100)

= 10,00,000 x (110/100) x (111/100) x (112/100)

= 110 x 111 x 112

Question 8 |

An amount of Rs.9600 lent out at a rate of 4.5 % per annum for a 1 year and 9 months. At the end of the period, the amount he earned as C.I was:

10450.69 | |

10455.69 | |

10465.69 | |

10555.69 |

**Compound Interest**

**Discuss it**

Question 8 Explanation:

Amount = 9600 * (1+4.5/100)

= 9600*(104.5/100)

=10455.69

^{7/4}= 9600*(104.5/100)

^{7/4}=10455.69

Question 9 |

An amount of Rs.8000 is landed out for 2 years at compound interest rate 5 % per annum. How much interest will be incurred on maturity of the FD?

850 | |

832 | |

800 | |

820 |

**Compound Interest**

**Discuss it**

Question 9 Explanation:

= 8000(1+5/100)

= 8000(105/100)

= 8000 [(105/100)

= 8000 [1.1025-1]

= 8000 X 0.1025

= 820

^{2}– 8000= 8000(105/100)

^{2}– 8000= 8000 [(105/100)

^{2}-1]= 8000 [1.1025-1]

= 8000 X 0.1025

= 820

Question 10 |

An amount of Rs.15,000 is invested at 10% per annum for one year. If the interest is compounded half-yearly, then what amount will be received at the end of the year?

16527.50 | |

16427.50 | |

16527.50 | |

16537.50 |

**Compound Interest**

**Discuss it**

Question 10 Explanation:

A= 15000 [1+(10/2)/100]

= 15000 [105/100]

= 15000 X 1.1025

= 16537.50

^{2}= 15000 [105/100]

^{2}= 15000 X 1.1025

= 16537.50

Question 11 |

Find the compound interest on Rs. 15,000 for 9 months at 16% per annum compounded quarterly.

1872.96 | |

1972.96 | |

2072.96 | |

2172.96 |

**Compound Interest**

**Discuss it**

Question 11 Explanation:

CI = 15000[1+(16/4)/100]

= 15000[104/100]

= 15000 [1.125-1]

= 15000 X 0.125

= 1872.96

^{4X(9/12)}-15000= 15000[104/100]

^{3}- 15000= 15000 [1.125-1]

= 15000 X 0.125

= 1872.96

Question 12 |

An amount of 200000 is deposited in bank under a scheme which provides 10% rate 1st years, 20% rate 2nd year and 5% rate 3rd year. After 3rd year what amount will be returned?

277400 | |

277220 | |

277200 | |

277320 |

**Compound Interest**

**Discuss it**

Question 12 Explanation:

A = 200000 (1+10/100)(1+20/100)(1+5/100)

A= 200000(110/100)(120/100)(105/100)

A= 277200

A= 200000(110/100)(120/100)(105/100)

A= 277200

Question 13 |

A sum of money is borrowed and paid back in two annual installments of Rs. 882 each allowing 5% compound interest. The sum borrowed was:

1540 | |

1640 | |

1580 | |

1680 |

**Compound Interest**

**Discuss it**

Question 13 Explanation:

Principal = Present worth of the installment at the end of first year + Present worth of the installment at the end of second year
= 882/(1 + 5%) + 882/(1 + 5%)^2
= 882/(1 + 5/100) + 882/(1 + 5/100)^2
= 882/(105/100) + 882/(105/100)^2
= 882/(21/20) + 882/(21/20)^2
= 882*20/21 + 882*400/441
= 42*20 + 2*400
= 840 + 800
= 1640.

Question 14 |

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

6 | |

4 | |

5 | |

3 |

**Compound Interest**

**Discuss it**

Question 14 Explanation:

Let the Principal be P. Then,
P(1 + 20%)^n > 2P
Or, P(1 + 1/5)^n > 2P
Or, P(6/5)^n > 2P
Or (6/5)^n > 2.
Now,
(6/5)^3 = 1.728
(6/5)^4 = 2.074
Therefore, in 4 years.

Question 15 |

The difference between compound interest and simple interest on an amount of Rs 15,000 for 2 years is Rs 96. What is the rate of interest per annum?

8 | |

9 | |

10 | |

None of these |

**Compound Interest**

**Discuss it**

Question 15 Explanation:

Simple Interest = 15000 * R/100 * 2
Compound Interest = 15000 * (1 + R/100)^2 - 15000
We have,
[ 15000 * (1 + R/100)^2 - 15000 ] - [15000 * R/100 * 2] = 96
15000 [ (1 + R/100)^2 - 1 - 2R/100 ] = 96
15000 [ 1 + R^2/10000 + 2R/100 - 1 - 2R/100 ] = 96
15000 [ R^2/10000 ] = 96
R^2 = 96 * 10000/15000 = 96 * 2/3 = 64
R = 8.

There are 15 questions to complete.