Question

Alloy A contains metals x and y only in the ratio 5 : 2 and alloy B contains these metals in the ratio 3 : 4. Alloy C is prepared by mixing A and B in the ratio 4 : 5. The percentage of x in alloy C is:

• A. 45 %
• B. 56 %
• C. 44 4/9 %
• D. 55 5/9 %

Answer 1

Answer:

D. 55 5/9 %

Solution:

Given :

The ratio of x and y in alloy A = 5 : 2

The ratio of x and y in alloy B = 3 : 4

The ratio of A and B in alloy C = 4 : 5

To find :

Percentage of metal x in alloy C

Calculation :

⇒ Quantity of metal x in alloy $A =\frac{5}{7}$

⇒ Quantity of metal y in alloy $A=\frac{2}{7}$

⇒ Quantity of metal x in alloy $B=\frac{3}{7}$

⇒ Quantity of metal y in alloy $B = \frac{4}{7}$

A.T.Q.,

⇒ The ratio of x and y in alloy C $=\frac{(\frac{5}{7})\times{4}+(\frac{3}{7})\times{5}}{(\frac{2}{7})\times{4}+(\frac{4}{7})\times{5}}$

⇒ The ratio of x and y in alloy C $=\frac{(\frac{20}{7})+(\frac{15}{7})}{(\frac{8}{7})+(\frac{20}{7})}$

⇒ The ratio of x and y in alloy $C = \frac{35}{28}$

⇒ Quantity of x in alloy $C = \frac{35}{63}$

⇒ Quantity of x in alloy $C = \frac{5}{9}$

⇒ Percentage of x in alloy $C = (\frac{5}{9})\times{100}$

∴ The percentage of x in alloy C is  $55\frac{5}{9}$%