Question Details

Choose the correct option to solve the problem.
An investor has a total of Rs. 40000. He allocates a portion of this sum to Account A, which yields 30% per annum simple interest, and the remainder to Account B, which earns 20% interest compounded annually. If, after three years, the total interest generated from Account A is Rs. 9952 more than the interest accrued from Account B, find the sum invested in Account B.

Options

A

Rs.18000

B

Rs.16000

C

Rs.24000

D

Rs.12000

E

Rs.21000

Show Answer

Correct Answer :

Option B

Rs.16000

Solution :

The correct option is Rs.16000.

Step 1: Understand the given information and define the variables.
Total sum invested = Rs. 40000.
Let the sum invested in Account B be x.
Therefore, the sum invested in Account A is (40000-x).

Step 2: Calculate the Simple Interest earned from Account A.
Account A offers a 30% per annum simple interest for a period of 3 years.
The formula for Simple Interest is:

Simple Interest=Principal×Rate×Time100

Substituting the values for Account A:

InterestA=(40000-x)×30×3100

InterestA=(40000-x)×90100=0.9(40000-x)

InterestA=36000-0.9x

Step 3: Calculate the Compound Interest earned from Account B.
Account B earns 20% interest compounded annually for 3 years.
The formula for the total amount under Compound Interest is:

Amount=Principal(1+Rate100)Time

Substituting the values for Account B:

AmountB=x(1+20100)3=x(1.2)3=1.728x

The Compound Interest is calculated by subtracting the principal from the total amount:

InterestB=1.728x-x=0.728x

Step 4: Set up the equation and solve for x.
We are given that the interest generated from Account A is Rs. 9952 more than the interest accrued from Account B:

InterestA-InterestB=9952

Substitute the expressions obtained in Step 2 and Step 3:

(36000-0.9x)-0.728x=9952

36000-1.628x=9952

1.628x=36000-9952

1.628x=26048

x=260481.628

x=16000

Hence, the sum invested in Account B is Rs. 16000.

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