Chapter 8.4, Problem 3E

To determine

To calculate : the total annual yield if equal amount of money was invested at each given rate.

To infer the account which has more money, considering the actual yield was $\$144$ .

Answer to Problem 3E

The total annual yield is $\$126.5$ , if equal amount of money is invested at each rate.

The account earning $7\%$ interest annually has more money.

Explanation of Solution

Given information :

Consider the Problem $11$ on page $246$

Total money invested is $\$2200$

The two rates given: $4.5\%$ and $7\%$

The actual annual yield $=\$144$

Formula used :

The interest I on a sum of money P at rate r % in time t years is

  $I=\frac{P\times R\times T}{100}$

Calculation :

As per the problem,

If equal amount of money is invested at rates $4.5\%$ and $7\%$

Then money invested at each rate, $P=\frac{\$2200}{2}=\$1100$

Annual interest earned at rate $4.5\%$

  $I_1=\frac{\$1100\times 4.5\times 1}{100}=\$49.5$

Annual interest earned at rate $7\%$

  $I_2=\frac{\$1100\times 7\times 1}{100}=\$77$

Therefore, total annual yield is

  $I=I_1+I_2=\$49.5+\$77=\$126.5$

Now, in Problem $11$ on page $246$

The money is invested partly at $4.5\%$ and rest at $7\%$

Assuming money invested at $4.5\%$ is $\$x$ .

So, money invested at $7\%$ will be $\$(2200-x)$ .

Calculate the interest earned at each rate.

For $R_1=4.5\%$

  $I_1=\frac{x\times 4.5\times 1}{100}=\$\frac{4.5x}{100}$

For $R_2=7\%$

  $I_2=\frac{(\$2200-x)\times 7\times 1}{100}=\$\frac{7(2200-x)}{100}$

The actual annual yield is $=\$144$

Therefore,

  $\begin{array}{l}\frac{4.5x}{100}+\frac{7(2200-x)}{100}=144\\ 4.5x+7(2200-x)=14400\text{ [Multiplying each side by 100]}\\ \text{4}\text{.5}x+15400-7x=14400\text{ [Distibutive Property]}\\ -\text{2}\text{.5}x=-1000\\ x=400\end{array}$

So, $\$400$ is invested at $4.5\%$ .

Therefore, the money invested at $7\%$ is $(\$2200-\$400)=\$1800$

Hence, the account earning $7\%$ interest annually has more money.