1. A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?

1. 10 years

2. 8 years

3. 6 years

4. 12 years

Answer & Explanation

Answer:- 4

Explanation :-

Let P = Rs.100 Rate = R% A = Rs. 200 Time = 4 years.

Then , A = P*[1+ (r/100)]^{n}

200 = 100*[1+(R/100)]^{4}

2 = [1+(R/100)]^{4 }————- (1)

If sum become 8 times in the time T years,

Then,8 = (1+(R/100))^{T}

2^{3} = (1+(R/100))^{T} ————- (2)

By putting value of 2 from equation (1) in equation (2)

([1+(R/100)]^{4})^{3} = (1+(R/100))^{T}

[1+(R/100)]^{12} = (1+(R/100))^{T}

Thus, T = 12 years.

2. On what sum will the compound interest for 2 1⁄2 years at 10% amount to Rs. 31762.5?

1. Rs.5000

2. Rs.20000

3. Rs.25000

4. Rs.30000

Answer & Explanation

Answer:- 3

Explanation :-

= [ P(1+10/100)^2 ] [ (1+10/200)^T ] = Rs. 31762.5

P [(11/10)^2] × (21/20) = 31762.5

P = (31762.5×20×10×10) / (21×11×11) = (1512.5×20×10×10)/(11×11)

P = (137.5×20×10×10)/11 =12.5×20×10×10 = Rs. 25000

3. The compound interest on Rs. 20,000 at 8% per annum is Rs. 3,328. What is the period (in year)?

1. 1 year

2. 4 years

3. 3 years

4. 2 years

Answer & Explanation

Answer:- 4

Explanation :-

Let the period be t years.

Amount after t years = Rs.20000 + Rs.3328 = Rs. 23328

P(1+R/100)^T = 23328

20000(1+8100)^T = 23328/20000

(108/100)^T = 23328/20000

(11664/10000) = (108/100)^2

(108/100)^T = (108/100)^2

Therefore, T = 2 years

4. A man borrows Rs. 20,000 at 10% compound interest. At the end of every year he pays Rs. 2000 as part repayment. How much does he still owe after three such installments?

1. Rs.24000

2. Rs.15000

3. Rs.20000

4. Rs.10000

Answer:- 3

Explanation :-

Principal (For 1st year)= Rs. 12500

Therefore Interest (for 1st year) = 20% of 12500 = Rs. 2500

Amount paid At the end of 1st year = Rs. 2000

Therefore , Principal (for 2nd year) = 12500 +2500 – 2000 = 13000

Therefore Interest (for 2nd year) = 20% of 13000 = Rs. 2600

Amount paid At the end of 2nd year = Rs. 2000

Therefore , Principal (for 3rd year) = 13000 +2600 – 2000 = 13600

Therefore Interest (for 3rd year) = 20% of 13600 = Rs. 2720

Amount paid At the end of 3rd year = Rs. 2000

Therefore , Amount after three installments = 13600 +2720 – 2000 = Rs.14320

5. The present worth of Rs. 242 due in 2 years at 10% per annum compound interest is:

1. Rs. 180

2. Rs. 240

3. Rs. 220

4. Rs. 200

Answer & Explanation

Answer:- 4

Explanation :-

Present worth of Rs. x due T years hence is given by

Present Wort = x/ [ (1+R/100)^T]

Present Worth = 242/ [(1+10/100)^2] = 242 / [(11/10)^2] = Rs. 200