Question

Taylor deposits and withdraws money in her bank account every alternate month. She started by depositing $\$900$. If every successive amount withdrawn/deposited is one-third of the previous amount deposited/withdrawn, then find her account balance after the second deposit?

• A. $\$700$
• B. $\$1800$
• C. $\$900$
• D. $\$2100$

Answer 1

The correct option is A $\$700$

Amount deposited the first time
$=\$900$

Amount withdrawn the first time $=\frac{1}{3}\times 900$

Amount deposited the second time $=\frac{1}{3}\times \frac{1}{3}\times 900=\frac{1}{9}\times 900$

Current account balance $=900+(-\frac{1}{3}\times 900)+\frac{1}{9}\times 900$

$=900+900\times (-\frac{1}{3}+\frac{1}{9})\text{[Distributive property]}$

$=900+900\times (-\frac{1}{9}+\frac{1}{3})\text{[Commutative property]}$

$=900+900\times \left(\frac{1-3}{9}\right)$

$=900+900\times (-\frac{2}{9})$

$=900-900\times \frac{2}{9}$

$=900-200$
$=700$
After the second deposit, the account balance will be $\$700$.