Chapter 7.1, Problem 30E

To determine

To evaluate the amount invested at the rate of 6% and 3% respectively.

Answer to Problem 30E

Amount invested at the rate of 6% = $10,000

Amount invested at the rate of 3% = $8,000

Explanation of Solution

Given information:

Total amount invested = $18000

Rates of simple interest = 6% and 3%

Total yearly interest = $840

Formula used:

Simple interest formula:

  $i=\frac{p\times t\times r}{100}$

Here, i = interest, t = time(years), r = rate of interest, p = principal amount.

Calculation:

Considering x and y be the amount invested at 6% and 3% respectively, the equation can be obtained as,

  $x+y=18000\mathrm{.....}\text{(}i)$

When yearly interest at 6%,

  $\frac{x\times 6\times 1}{100}$ {simple interest formula used}

When yearly interest at 3%,

  $\frac{y\times 3\times 1}{100}$ {simple interest formula used}

Total yearly interest $\frac{x\times 6\times 1}{100}+\frac{y\times 3\times 1}{100}$

Therefore,

  $\frac{x\times 6\times 1}{100}+\frac{y\times 3\times 1}{100}=840\mathrm{.....}\text{(}ii)$

After separating variables of equation $(i)$

  $x=18000-y\mathrm{.....}\text{(}iii)$

Substituting the expression obtained for x from equation ( iii ) into equation ( ii )

  $\begin{array}{l}\frac{(18000-y)\times 6\times 1}{100}+\frac{y\times 3\times 1}{100}=840\\ (18000-y)\times 6+3y=84000\text{ (multiplying both sides by 100)}\\ \text{108000}-6y+3y=84000\\ 108000-3y=84000\\ 3y=108000-84000\\ y=\frac{24000}{3}\\ y=8000\end{array}$

Now substituting the value of y in equation

  $\begin{array}{l}x+8000=18000\\ x=10000\end{array}$

Amount invested at the rate of 6% and 3% is $10,000 and $8,000 respectively.