Two alloys contain tin and iron in the ratio | vedclass

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Question

(D) Let the weight of the first alloy be $3x$ and the second alloy be $4x$.In the first alloy $(A)$,the ratio of tin to iron is $1:2$. Total parts = $

Vedclass · Accepted Answer

$13:22$

Vedclass · Answer

(D) Let the weight of the first alloy be $3x$ and the second alloy be $4x$.In the first alloy $(A)$,the ratio of tin to iron is $1:2$. Total parts = $1+2=3$.Amount of tin in $A = \frac{1}{3} 	imes 3x = x$.Amount of iron in $A = \frac{2}{3} 	imes 3x = 2x$.In the second alloy $(B)$,the ratio of tin to iron is $2:3$. Total parts = $2+3=5$.Amount of tin in $B = \frac{2}{5} 	imes 4x = \frac{8x}{5}$.Amount of iron in $B = \frac{3}{5} 	imes 4x = \frac{12x}{5}$.Total tin in the new mixture = $x + \frac{8x}{5} = \frac{13x}{5}$.Total iron in the new mixture = $2x + \frac{12x}{5} = \frac{22x}{5}$.The ratio of tin to iron in the new alloy = $\frac{13x}{5} : \frac{22x}{5} = 13:22$.

Two alloys contain tin and iron in the ratio of $1:2$ and $2:3$. If the two alloys are mixed in the proportion of $3:4$ respectively (by weight),the ratio of tin and iron in the newly formed alloy is:

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