Question

In an alloy of zinc, tin and lead, quantity of zinc is ${\left(\frac{3}{4}\right)}^{th}$ that of tin and quantity of tin is ${\left(\frac{4}{5}\right)}^{th}$ that of lead. How much of each metal will be there in $12$ kg of the alloy?

• A. $5,4,3$
• B. $3,4,5$
• C. $4,5,3$
• D. $5,3,4$

Answer 1

Answer: (D)

$5,3,4$

• Let the quantity of the tin be ${}^{\prime }{x}^{\prime }$ , zinc be ${}^{\prime }{y}^{\prime }$ and lead be ${}^{\prime }{z}^{\prime }$.

Given quantity of zinc is $\frac{3}{4}th$ of tin,

$⟹y=\frac{3}{4}\left(x\right)--------\left(1\right)$

Quantity of tin is $\frac{4}{5}th$ of lead,

$⟹x=\frac{4}{5}\left(z\right)------\left(2\right)$

But given total weight $=12kg$

$⟹x+y+z=12$

From $\left(1\right)\&\left(2\right)$equations

$x+\frac{3}{4}x+\frac{5x}{4}=12$

$\frac{12x}{4}=12$

$⟹x=4$

if $x=4,y=3,z=5$

So the quantity of tin is $4kg$ , zinc is $3kg$ , lead is $5kg$