A and B are two alloys of copper and tin prepared by mixing the respec

A and B are two alloys of copper and tin prepared by mixing the respective metals in ratio of 5:3 and 5:11 respectively. If the alloy A and B are mixed to form a third alloy C with an equal proportions of copper and tin what is the ration of alloys A and B In the new alloy C?

A 3:5
B 4:5
C 3:2
D 2:3
E 15:17

Alloy A ratio = 5:3 Copper = 5/8 = 10/16
Alloy B ratio = 5:11 Copper = 5/16 = 5/16
Final Alloy C = 1:1 Copper = 1/2 = 8/16

Make denominator common
Copper average in alloy c = 8
Using alligation
POV of copper
A = 10 B = 5

C = 8
(Average)
8 - 5 10 - 8
= 3 : 2
C

\(\frac{5}{8}\)*x + \(\frac{5}{16}\)*y = \(\frac{3}{8}\)*x + \(\frac{11}{16}\)*y

\(\frac{2}{8}\)*x = \(\frac{6}{16}\)*y

\(\frac{x}{y}\)= \(\frac{3}{2}\)

I got it this way, and I found it very intuitive:

Copper is in mix A in the ratio 5/8
Copper is in mix B in the ratio 5/16

So the question is how many of A and B I need to get a ratio 1/2 which is the same as:

(5A/8 + 5B/16) / (A+B) = 1/2

It is divided by (A+B) because is the total amount mixed

10A + 5B = 8A+8B
2A = 3B
A/B = 3/2

Hope it helps!

need help
can anyone solve this with allegation diagram ??

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