Chapter 3.5, Problem 3CYP

To determine

To find: the amount that she invested in three different accounts.

Answer to Problem 3CYP

  $\$18000,\$6000,\$24000$ .

Explanation of Solution

Given:

Total amount invested by Ms. Garza in different account is $\$50000$ .

The invest on 1^st account is 3 times as much as 2^nd account.

The rate of interest on different accounts are:

  $\begin{array}{l}1\text{st account}=8\%\\ 2\text{nd account}=10\%\\ 3\text{rd account}=12\%\end{array}$

Total interest earned at the end of the first year is $\$5160$ .

Calculation:

According to the question, the invest on 1^st account is 3 times as much as 2^nd account.

Let the amount invested on 2^nd account be x.

Therefore, the amount invested on 1st account is 3x.

Thus, the amount invested on 3^rd account is $\$50000-\left(x+3x\right)=\$50000-4x$ .

Now, interest on both the accounts are:

  $\begin{array}{l}1\text{st account}:\\ I=\frac{PRT}{100}\\ =\frac{3x\left(8\right)\times 1}{100}\{P=3x\}\\ =\frac{24x}{100}\end{array}$

  $\begin{array}{l}2\text{nd account}:\\ I=\frac{PRT}{100}\\ =\frac{x\left(10\right)\times 1}{100}\{P=x\}\\ =\frac{10x}{100}\end{array}$

  $\begin{array}{l}3\text{rd account}:\\ I=\frac{PRT}{100}\\ =\frac{\left(50000-4x\right)\left(12\right)\times 1}{100}\{P=50000-4x\}\\ =\frac{600000-48x}{100}\end{array}$

Now, the sim of the total interest is $\$5160$ . Therefore,

  $\begin{array}{l}\frac{24x}{100}+\frac{10x}{100}+\frac{600000-48x}{100}=\$5160\\ \frac{24x+10x+600000-48x}{100}=5160\\ \Rightarrow 34x+600000-48x=516000\\ \Rightarrow 600000-14x=516000\\ \Rightarrow 14x=600000-516000\\ 14x=84000\\ \therefore x=\frac{84000}{14}=6000\end{array}$

Thus, the amount invested on both the accounts are:

  $\begin{array}{l}1\text{st account}=3x=3\times 6000=\$18000\\ 2\text{nd account}=x=\$6000\\ 3\text{rd account}=\$50000-4x\\ =\$50000-4\times 6000\\ =\$50000-24000=\$26000\end{array}$

Hence, the amount that she invested in three different accounts is $\$18000,\$6000,\$24000$ .