Slot -2 – Quantitative Aptitude – Arithmetic – Mixtures – There are two drums, each containing a mixture
There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio?
a) 251 : 163
b) 239 : 161
c) 229 : 141
d) 220 : 149
Answer: b) 239 : 161
Let in drum 2, A and B are in the ratio m : n.
Given, In drum 1, A and B were in the ratio 18 : 7 = 18x : 7x .
Final mixtures ratio of contribution of drum 1 and drum 2 = 3: 4 = 75x : 100x
So A in final mixture = 18x × 3+100x × m/(m+n) = ( 154m+54n)/(m+n) x
B in final mixture = 7x × 3+100x × n/(m+n) = ( 21m+121n)/(m+n) x
Given A : B in final mixture = 13:7
Or (154m+54n )/(21m+121n )=13/7
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