Question

Kevin invested some money at 25% per annum simple interest in bank A. At the end of one year, he withdrew the amount and invested in bank B at the rate of R% per annum under compound interest for two years. He received $\$$75,000 as the total interest from bank B. The first year’s interest at bank B was $\$$36,000. How much money did Kevin invest in bank A? (Take R to nearest whole number.)

• A. $\$157,890$
• B. $\$153,400$
• C. $\$15,789$
• D. $\$1579$

Answer 1

The correct option is A $\$157,890$

Let the money he invested in A = 100x

Then, at the end of one year,

$\text{Amount}=\frac{PTR}{100}+P$

$=\frac{100x\times 1\times 20}{100}+100x$

$=120x$

Given:
Compound interest for 2 years
$=\$75,000$

$\text{C.I. for 1 year}=\$36,000$

$Difference=\$75,000–\$36,000=\$29,000$

Now, $\$36,000–\$29,000=\$7,000$

At R% per annum, $\$$36,000 gives an interest of $\$$7,000 in 1 year.
So,
$\frac{36,000\times R}{100}=7000$

$36,000\times R=7000\times 100$

$R=\frac{7000\times 100}{36,000}$

$R=19.44\%$

$R=19$% per annum (Approx.)

At 19% per annum, 120x gives a compound interest of $\$$75,000 in two years or $\$$36,000 in one year.

$\text{C.I.}=P(1+\frac{R}{100}{)}^n–P$

$36,000=120x(1+\frac{19}{100})–120x$

$36,000=120x\left(\frac{100+19}{100}\right)–120x$

$36,000=120x\left(\frac{119}{100}\right)–120x$

$36,000=120x\left(1.19\right)–120x$

$36,000=142.8x–120x$

$36,000=22.8x$

$x=\$1578.9$

Money invested in A
$=100x=100\times 1578.9=\$157,890$