Solve Study Textbooks Guides

Join / Login

Question

An alloy of gold and copper weights 50 g. It contains 80% gold. How much gold should be added to the alloy so that percentage of gold is increased to 90?

Easy

Open in App

Solution

Verified by Toppr

Inital amount of gold = $\frac{8}{1 0} \times 50$ $=$ $40$ gm
Let amount of gold required to me mixed = $X$ gm
According to question , $40 + x = \frac{9}{1 0} \times$ ( $50 + x$ )
Solving we get $x = 50$ gm

Was this answer helpful?

0

0

Similar questions

A and B are two alloys of gold and copper prepared by mixing metals in the ratio of 5:3 and 5:11 respectively. Equal quantities of these alloys are melted to form a third alloy C. The ratio of gold and copper in alloy C is:

A and B are two alloys of gold and copper prepared by mixing metals in the ration 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 C will be

A and B are two alloys of gold and copper prepared by mixing metals 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 C will be

$50$ $g$ of an alloy of gold and silver contains $80$ $%$ gold(by weight). The quantity of gold, that is to be mixed up with this alloy, so that it may contain $95$ $%$ gold, is:

$A$ and $B$ are two alloys of gold and copper in the ratio $7 : 2$ and $7 : 11$ respectively. If equal quantities of these two alloys are melted to from a new alloy $C$ , then the ratio of gold and copper in $C$ is $:$

View more