Chapter 7.1, Problem 31E

To determine

To evaluate the amount invested at the rate of 5.6% and 6.8% respectively.

Answer to Problem 31E

Amount invested at the rate of 5.6% = $3,500

Amount invested at the rate of 6.8% = $14,500

Explanation of Solution

Given information:

Total amount invested = $18000

Rates of simple interest = 5.6% and 6.8%

Total yearly interest = $1182

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 5.6% and 6.8% respectively.

Therefore , $x+y=18000\mathrm{.....}\text{(}i)$

Yearly interest at 5.6% $=\frac{x\times 5.6\times 1}{100}$ {simple interest formula used}

Yearly interest at 6.8 % $=\frac{y\times 6.8\times 1}{100}$ {simple interest formula used}

Total yearly interest $=\frac{x\times 5.6\times 1}{100}+\frac{y\times 6.8\times 1}{100}$

Therefore, $\frac{x\times 5.6\times 1}{100}+\frac{y\times 6.8\times 1}{100}=1182\mathrm{.....}(ii)$

After separating variables of equation $(i)$

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

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

  $\begin{array}{l}\frac{(18000-y)\times 5.6\times 1}{100}+\frac{y\times 6.8\times 1}{100}=1182\\ (18000-y)\times 5.6+6.8y=118200\text{ (multiplying both sides by 100)}\\ \text{100800}-5.6y+6.8y=118200\\ 100800+1.2y=118200\\ 1.2y=118200-100800\\ y=\frac{17400}{1.2}\\ y=14500\end{array}$

Now substituting the value of y in equation $(i)$

  $\begin{array}{l}x+14500=18000\\ x=3500\end{array}$

Amount invested at the rate of 5.6% and 6.8% is $3,500 and $14,500 respectively.