Question

Gold is 19 times as heavy as water and copper is 9 times as heavy as water. In what ratio should these metals be mixed so that the mixture may be 15 times as heavy as water?
A. 2:3
B. 2:4
C. 3:2
D. 4:2

Answer 1

Hint – In order to solve this problem we have to relate gold and copper with the help of their relation with water then solve to find the ratio according to the condition provided. Doing this will solve your problem.

Complete step-by-step answer:
We know that Gold is 19 times heavier than water and copper is 9 times heavier than water. We have to find the ratio such that these metals mixed so that the mixture may be 15 times as heavy as water………(1)
Let the ratio of gold and copper that taken be G:C

Then from (1) we can write
$\dfrac{{{\text{19G}}}}{{{\text{G + C}}}}{\text{ + }}\dfrac{{{\text{9C}}}}{{{\text{G + C}}}}{\text{ = }}\dfrac{{{\text{15}}}}{{\text{1}}}$
$\dfrac{{{\text{19G + 9C}}}}{{{\text{G + C}}}} = \dfrac{{15}}{1}$
On cross multiplying we get the equation as:
19G+9C=15G+15C
4G=6C
$\dfrac{{\text{G}}}{{\text{C}}}{\text{ = }}\dfrac{{\text{6}}}{{\text{4}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{2}}}$
$\therefore $Gold and copper are taken in the ratio 3:2
So, the correct option is C.

Note – Whenever you are struck with this type of problem you have to assume the ratio Gold is to Copper then you have to make the equations as their relation has been provided with water so they can be related with the help of that. Then solve the equation to the ratio of gold to copper.