Night Class High Level CI Questions

1. What is the difference between the compound interests on Rs. 5000 for 11⁄2 years at 4% per annum compounded yearly and half-yearly?

A. Rs. 2.04

B. Rs. 4.80

C. Rs. 3.06

D. Rs. 8.30

Here is the answer and explanation

Answer: Option A

Explanation:

Amount after 11⁄2 years when interest is compounded yearly

=5000×(1+4/100)×(1+(1/2×4)/100)=5000×104/100×(1+2/100)=5000×104100×102100 =50×104×51/50=104×51=Rs. 5304

Compound Interest for 1 1⁄2 years when interest is compounded yearly = Rs.(5304 - 5000)

Amount after 11⁄2 years when interest is compounded half-yearly =P(1+(R/2)100)2T=5000(1+(4/2)/100)2×32= 5000(1+2/100)^3=5000(102/100)^3=5000(102/100)(102/100)(102/100)=50×102×51/50×51/50=102×51×51/50=51×51×5125=Rs. 5306.04

Compound Interest for 1 1⁄2 years when interest is compounded half-yearly

= Rs.(5306.04 - 5000)

Difference in the compound interests = (5306.04 - 5000) - (5304 - 5000) = 5306.04 - 5304 = Rs. 2.04

2. A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is:

A. Rs. 120

B. Rs. 121

C. Rs. 123

D. Rs. 122

Here is the answer and explanation

Answer: Option B

Explanation:

Amount after 1 year on Rs. 1600 (deposited on 1st Jan) at 5% when interest calculated half-yearly

=P(1+(R/2)/100)^2T=1600(1+(5/2)/100)^2×1=1600(1+1/40)^2

Amount after 1/2 year on Rs. 1600 (deposited on 1st Jul) at 5% when interest calculated half-yearly

=P(1+(R/2)/100)^2T=1600(1+(5/2)/100)^(2×12)=1600(1+1/40)

Total Amount after 1 year

=1600(1+1/40)^2+1600(1+1/40)=1600(41/40)2+1600(41/40)=1600(41/40)[1+41/40]=1600(41/40)(81/40)=41×81=Rs. 3321

Compound Interest = Rs.3321 - Rs.3200 = Rs.121

3. There is 80% increase in an amount in 8 years at simple interest. What will be the compound interest of Rs. 14,000 after 3 years at the same rate?

A. Rs.3794

B. Rs.3714

C. Rs.4612

D. Rs.4634

4. The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:

A. 1

B. 2

C. 3

D. 3.5

Here is the answer and explanation

Answer: Option B

Explanation:

Let the period be n years

Then, amount after n years = Rs.(30000 + 4347) = Rs. 34347

P(1+R/100)^T=34347

30000(1+7/100)^n=34347

30000(107/100)^n

34347(107/100)^n

34347/30000

11449/10000

(107/100)^2 

n=2 years

5. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum is:

A. Rs.600

B. Rs.645

C. Rs.525

D. Rs.625

Here is the answer and explanation

Answer: Option D

Explanation :

Solution 1

Let the sum be Rs.x

Amount after 2 years on Rs.x at 4% per annum when interest is compounded annually 

x(1+4/100)^2

x(104/100)^2

Compound Interest = x(104/100)^2−x

x[(104/100)2−1]

x[(26/25)2−1]

x[676/625−1]

x[51/625]=51x/625

Simple Interest = PRT/100

=(x×4×2)/100

=2x/25

Given that difference between compound interest and simple interest is Rs.1

⇒51x/625−2x/25=1

⇒(51x−50x)/625=1

⇒x=625

Solution 2

The difference between compound interest and simple interest on Rs. P for 2 years at R% per annum

=P(R100)2

P(R100)2=1P(4100)2=1P(125)2=1P(1625)=1P=625

 required sum is Rs.625