GATE2018 EC: GA-7

Question

Two alloys $A$ and $B$ contain gold and copper in the ratios of $2:3$ and $3:7$ by mass, respectively. Equal masses of alloys $A$ and $B$ are melted to make an alloy $C$. The ratio of gold to copper in alloy $C$ is ______.

• A. $ 5:10$
• B. $7:13$
• C. $6:11$
• D. $9:13$

Answer 1

Alloy $A$ is having Gold and Copper in the ratio $2:3$

Alloy $B$ is having Gold and Copper in the ratio $3:7$

Let's Assume we're taking $1 \: kg$ of alloy $A$ & $1 \: kg$ of alloy $B$

∴ in alloy $A$

• Gold will be $\dfrac{2}{5}$ kg
• Copper will be $\dfrac{3}{5}$ kg

   In alloy $B$

• Gold will be $\dfrac{3}{10}$ kg
• Copper will be $\dfrac{7}{10}$ kg

∴ In alloy $C$ Gold to Copper ratio will be  

$\quad={\dfrac{\dfrac{2}{5} + \dfrac{3}{10}}{\dfrac{3}{5} + \dfrac{7}{10}}} = {\dfrac{\dfrac{7}{10} }{\dfrac{13}{10}}} =\dfrac{7}{13} =7:13$   

Option  (B)

Answer 2

we can use weighted average concept here

$\text{the fraction of gold in alloy }A=\frac{2}{2+3} = \frac{2}{5}$

$\text{the fraction of gold in alloy }B=\frac{3}{3+7} = \frac{3}{10}$

$\text{When alloys A and B are melted in the ratio 1:1 to make an alloy C. The fraction of  gold in alloy C is}$

$${ {1*\left(\frac 25\right) + 1*\left(\frac 3{10}\right)} \over {1+1}   } = \frac 7{20}$$

$\text{the fraction of copper = }\frac{13}{20}$

$\therefore \text{ratio = 7:13 }$