Two mixtures A and B have the following compositions.
Mixture A has copper and tin in the ratio 1:2. Mixture B has copper and tin in the ratio 1:3 , If equal quantities of mixtures A and B are used for producing mixture C , then find the ratio of copper and tin in mixture c.

Correct option is A

Given:

Mixture A has copper and tin in the ratio 1 : 2
Let the quantity of Mixture A used be x.
Hence, copper in A = $$\frac{1}{3}x,$$​ tin in A = $$\frac{2}{3}x.$$​​

Mixture B has copper and tin in the ratio 1:3.
Let the quantity of Mixture B used be x.
Hence, copper in B = $$\frac{1}{4}x$$​, tin in B = $$\frac{3}{4}x.$$​​

Equal quantities of mixtures A and B are used to produce mixture C.

Concept Used:

The ratio of copper to tin in Mixture C is:

Ratio of copper to tin = $$\frac{\text{Total copper in C}}{\text{Total tin in C}}$$​​

Solution:

Copper in Mixture C:
Copper from Mixture A: $$\frac{1}{3}x$$ ​

Copper from Mixture B: $$\frac{1}{4}x$$​​
Total copper in C:

​$$\frac{1}{3}x + \frac{1}{4}x = \frac{4}{12}x + \frac{3}{12}x = \frac{7}{12}x$$​​

Tin in Mixture C:
Tin from Mixture A: $$\frac{2}{3}x$$​

Tin from Mixture B: $$\frac{3}{4}x$$​​
Total tin in C:

​$$\frac{2}{3}x + \frac{3}{4}x = \frac{8}{12}x + \frac{9}{12}x = \frac{17}{12}x$$​​

Ratio of copper to tin in Mixture C:

Ratio = $$\frac{\frac{7}{12}x}{\frac{17}{12}x} = \frac{7}{17}$$​​

Thus, The ratio of copper to tin in Mixture C is 7 : 17