Question

Divide $900$ into two parts such that $60\%$ of one part exceeds $30\%$ of the other by $270$.

• A. $800$ and $900$
• B. $300$ and $500$
• C. $600$ and $300$
• D. $200$ and $700$

Answer 1

The correct option is C $600$ and $300$

Let one part of $900$ be $x$
$\therefore$ other part is $900-x$
Given,
$60\%$ of $x=30\%$ of $(900-x)+270$
$\Rightarrow \frac{60x}{100}=\frac{30(900-x)}{100}+270$
$\Rightarrow \frac{60x}{100}=\frac{30(900-x)+27000}{100}$
$\Rightarrow 60x=27000-30x+27000$
$\Rightarrow 90x=54000$
$\Rightarrow x=600$
One part is $600$ and
other part is $900-600=300$