X is an alloy of A and B.
Y is an alloy containing
80% of A, 4 % of B and 16% of C.
A fused mass of X and Y is found to contain 74% of A, 16% of B, and
10% of C.
The ratio of A to B in X is:
(a) 9:16
(b) 12:17
(c) 16:9
(d) None of these
Answers
Brass contains copper and zinc
100 gm bronze contains 80 gm copper, 4 gm zin and 16 gm tin
100 gm fused mass of brass and bronze contains 74 gm copper, 16 gm zinc and 10 gm tin
This entire tin must have come from bronze because brass does not have tin component
therefore, quantity of bronze in 100 gm fused mass =
100
×
10
16
= 62.5 gm
quantity of brass in 100 gm fused mass = 100 - 62.5 = 37.5 gm
62.5 gm bronze consists of
62.5 × 80/100 = 50 gm copper,
62.5 × 4/100 = 2.5 gm zinc
Amount of copper in 37.5 gm brass = 74 - 50= 24 gm
Amount of zinc in 37.5 gm zinc = 16 - 2.5 = 13.5 gm
The ratio of Copper to Zinc in Brass =
24
13.5
=
48/
27
Answer:
16:9
Step-by-step explanation:
Since X has 0% of C and Y has 16% of C the ratio of mixing in the fused mass must be 3:5 as per alligation rule:
0 6
\ /
10
/ \
6 10 = 3:5 = Ratio of fused mass of X and Y
The percentage of B in X can be got as: let b be as percentage of B in X
b 4
\ /
16
/ \
12 b-16 i.e 12 : b- 16 = 3:5.
By solving we get b = 36 ( % of B in X)
Therefore A = 100-36 =64
therefore ratio of A and B in X alloy is 64:36 = 16:9