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Parin Sir > Quantitative Aptitude (Maths) > Compound Interest > Examples 9 to 11 > Examples 9 to 11

Examples 9 to 11

June 15, 2020
Category: Examples 9 to 11

(9) The least number of complete yrs. in which a sum of money put out at 10 % C. I. will more than double is

Soln. Here, N = $\frac{72}{R}$ = $\frac{72}{10}$ = 7.2 years.

$\therefore$ The least number of complete yrs. in which a sum of money will

more than double = 8 years.

(10) Divide Rs. 841 between A and B, so that the amount of A after 10 years is equal to the amount of B after 12 years, the interest being compounded at 5 % per annum.

Soln. Let the two parts be Rs. x and Rs. ( 841 – x ).

x $\left (\frac{105}{100} \right )^{7}$ = ( 841 – x ) $\left (\frac{105}{100} \right )^{9}$ $\large \Leftrightarrow$ $\frac{x}{\left (841-x \right )}$ = $\left (\frac{105}{100} \right )^{2}$ = $\left (\frac{21}{20} \times \frac{21}{20} \right )$ = $\frac{441}{400}$ $\large \Leftrightarrow$ 400 x = 441 ( 841 – x ) $\large \Leftrightarrow$ 841 x = 441 $\times$ 841 $\large \Leftrightarrow$ x = 441.

So, the two parts are Rs. 441 and Rs.(841 – 441 ) i.e. Rs. 841 and Rs. 400.

Present worth of Rs. x due n years hence is given by : P. W. = $\frac{x}{\left (1+\frac{R}{100} \right ) ^{n}}$

(11) What annual payment will discharge a debt of Rs. 6620 due in 3 years at 10 % per annum compound interest ?

Soln.

Let, each installment be Rs. x. Then,

( P.W. of Rs. x due 1 year hence ) + ( P.W. of Rs. x due 2 years ) + ( P.W. of Rs. x due 3 years ) = 6620

$\therefore$ $\frac{x}{\frac{110}{100}}+\frac{x}{\frac{110}{100}\times \frac{110}{100}}+\frac{x}{\frac{110}{100}\times \frac{110}{100}\times \frac{110}{100}}$ = 6620