question_answer

A sum of money was invested for 14 yr in scheme A which offers simple interest at a rate of 8% per annum. The amount received from scheme A after 14 yr was then invested for 2 yr in scheme B which offers compound interest (compounded annually) at a rate of 10% per annum. If the interest received from scheme B was Rs. 6678, what was the sum invested in scheme A?                                                         [IBPS RRB (Office Assistant) 2015]

A) Rs. 15500                     

B) Rs. 14500

C) Rs. 16000                     

D) Rs. 12500

E) Rs. 15000

Correct Answer: E

Solution :

Let the principal invested in scheme A.

\[SI=\frac{P\times R\times T}{100}\]\[\Rightarrow \]\[SI=\frac{P\times 14\times 8}{100}\]

\[SI=\frac{112\,P}{100}\]

\[A=P+SI=P+\frac{112\,\,P}{100}=\frac{212}{100}P\]

On compound interest in scheme B.

\[A=\frac{212\,\,P}{100}{{\left( 1+\frac{10}{100} \right)}^{2}}=\frac{212\,\,P}{100}\times {{\left( \frac{110}{100} \right)}^{2}}\]

\[=\,\,\frac{212\,\,P}{100}\times \frac{121}{100}\,\,=\,\,\frac{25652\,\,P}{10000}\]

Interest received from scheme B

\[=\,\,\frac{25652\,\,P}{10000}-\frac{212\,\,P}{100A}\]

\[=\,\,\frac{25652\,\,P-21200\,P}{10000}\,\,=\,\,\frac{4452\,\,P}{10000}\]

But given, \[\frac{4452\,\,P}{10000}=6678\]

\[P=\,\,\frac{6678\times 10000}{4452}=Rs.\,\,15000\]