Question

Gold and copper are as heavy as water by $19$ and $9$ times respectively. The ratio in which these two metals be mixed so that the mixture is $17$ times as heavy as water is

• A. $2:3$
• B. $3:4$
• C. $3:2$
• D. $4:1$

Answer 1

The correct option is D $4:1$

Let the mass of $1$ volume unit of water be $1$.
Then,
$1$ volume unit of gold has mass $19$ and $1$ volume unit of copper has mass $9$.
Now let $g$ be the volume of gold used and $c$ be the volume of copper.
Then,
$(19g+9c)/(g+c)=17$
$19g+9c=17g+17c$
$2g=8c$
$\frac{g}{c}=\frac{8}{2}=\frac{4}{1}$
$\therefore$ Required ratio $=4:1$.