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-by-Step Explanation:

1. Define the Variables:
Let the total sum available for investment be Rs. 40,000.
Let the sum invested in Account B be x.
Therefore, the remaining sum invested in Account A is (40000-x).

2. Calculate Simple Interest from Account A:
Account A offers simple interest at a rate of 30% per annum for 3 years.
The formula for Simple Interest (SI) is:

SI=P×R×T100

Substituting the values for Account A:
Principal (P) = 40000-x
Rate (R) = 30% per annum
Time (T) = 3 years

SI=(40000-x)×30×3100

SI=(40000-x)×90100

SI=0.9×(40000-x)

SI=36000-0.9x

3. Calculate Compound Interest from Account B:
Account B yields interest at 20% per annum compounded annually for 3 years.
The formula for Compound Interest (CI) is:

CI=P×1+R100T-1

Substituting the values for Account B:
Principal (P) = x
Rate (R) = 20%
Time (T) = 3 years

CI=x×1+201003-1

CI=x×1.23-1

CI=x×1.728-1

CI=0.728x

4. Set up and Solve the Equation:
The problem states that the interest accrued from Account A is Rs. 9952 more than the interest accrued from Account B:

SI-CI=9952

Substituting the expressions derived for SI and CI:

(36000-0.9x)-0.728x=9952

36000-1.628x=9952

1.628x=36000-9952

1.628x=26048

x=260481.628

x=16000

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

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