Question : If $60 \% \text{ of} (x-y)= 45\%(x+y)$ and $y= k \% \text{ of} \ x$, then $21\%$ of $k$ is equal to:
Option 1: 7
Option 2: 1
Option 3: 6
Option 4: 3
Correct Answer: 3
Solution :
Given: $60 \% \text{ of} (x-y)= 45\%(x+y)$ and $y= k \% \text{ of} \ x$
So, $\frac{60}{100} \times (x-y)= \frac{45}{100}(x+y)$
⇒ $60x-60y=45x+45y$
⇒ $15x=105y$
⇒ $\frac{x}{y}=\frac{7}{1}$
⇒ $x:y=7:1$
Given, $y= \frac{k}{100}\times x$
Putting the values, we get:
$1= \frac{k}{100}\times 7$
⇒ $k= \frac{100}{7}$
$\therefore 21\%$ of $k$ = $\frac{21}{100}\times\frac{100}{7}$ = 3
Hence, the correct answer is 3.
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