Question

A and B are two alloys of gold and copper prepared by mixing in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy c, the ratio of gold and copper in the C will be
A. 5:7
B. 5:9
C. 7:5
D. 9:5

Answer 1

Hint: To solve such questions we need to let them in the ratio in terms of x and then get their individual sum and get their ratio in order to obtain the desired answer for this question.

Complete Step-by-Step solution:
Given,
Gold and copper in alloy A present in ratio =7:2
Gold and copper in alloy B present in ratio=7:11
In alloy A,
Let the gold be 7x and copper be 2x …(in order to calculate we take the original mass of alloy to be x.)
Then fraction of gold in alloy A = $\dfrac{{{\text{gold}}}}{{{\text{total mass}}}}$
$ = \dfrac{{7x}}{{7x + 2x}}$ $ = \dfrac{7}{9}$
Similarly, copper in alloy A=$\dfrac{{{\text{copper}}}}{{{\text{total mass}}}}$
$ = \dfrac{{2x}}{{7x + 2x}} = \dfrac{2}{9}$
In alloy B,
Let the gold be 7x and copper be 11x,
Then fraction of gold in alloy B = $\dfrac{{{\text{gold}}}}{{{\text{total mass}}}}$
$ = \dfrac{{7x}}{{11x + 7x}}$ $ = \dfrac{7}{{18}}$
Similarly, copper in alloy B=$\dfrac{{{\text{copper}}}}{{{\text{total mass}}}}$
$ = \dfrac{{11x}}{{11x + 7x}} = \dfrac{{11}}{{18}}$
According to question,
Alloy C=Alloy A +Alloy B
Therefore, in terms of gold in alloy C=gold in alloy A + gold in alloy B
= $\dfrac{7}{9} + \dfrac{7}{{18}}$
$ = \dfrac{{14 + 7}}{{18}} = \dfrac{{21}}{{18}}$
Similarly in terms copper in alloy C= copper in alloy A+ copper in alloy B
=$\dfrac{2}{9} + \dfrac{{11}}{{18}}$
$ = \dfrac{{4 + 11}}{{18}} = \dfrac{{15}}{{18}}$
Therefore, ratio of gold and copper in alloy C $ = \dfrac{{{\text{gold in alloy C}}}}{{{\text{copper in alloy C}}}}$
=$\dfrac{{\dfrac{{21}}{{18}}}}{{\dfrac{{15}}{{18}}}}$ $ = \dfrac{{21}}{{15}} = \dfrac{7}{5}$=7:5
Hence, the answer to this question is 7:5.

Note: In these questions we need to take care of ratios and their sums and the method to calculate their sum and care to be taken to take the initial mass of alloy to be the same in order to deal in a single variable. Doing this will easily solve these types of questions.