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06 Apr 2020, 09:50
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
64% (03:09) correct
36%
(03:12)
wrong
based on 304
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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. 220 : 149
B. 229 : 141
C. 239 : 161
D. 251 : 163
E. 253 : 163
Are You Up For the Challenge: 700 Level Questions
A. 220 : 149
B. 229 : 141
C. 239 : 161
D. 251 : 163
E. 253 : 163
Are You Up For the Challenge: 700 Level Questions
06 Apr 2020, 21:43
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Bunuel
Lets solve by the following method:
A in container 1 = 18/25
A in container 2= X
Since the containers are mixed in the ratio of 3:4
hence A in container 1 = 18/25 x 3/7 and A in container 2= X x 4/7
but final ratio of A in the final mixture = 13/20
considering all the above:
18/25 x 3/7 + X x 4/7 = 13/20
solving gives X= 239/400
this means that the total in container 2 = 400 out of which X is 239.
hence answer C).
06 Apr 2020, 17:14
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The amount of paint B in drum 1 = 28gm in 100gm of mixture
The amount of paint B in drum 2 = x gm in 100gm of mixture
The amount of paint B in final mixture = 35 gm in 100gm of mixture
28.......... x
.....35.......
3.............4
\(\frac{(35-28)}{(x-35)} =\frac{ 4}{3}\)
21= 4x-140
x=\( \frac{161}{4}\)
{We can mark our answer here, as there is only 1 option in which Quantity of B is multiple of 161}
Amount of paint A in drum 2= \(100-(\frac{161}{4}) = \frac{239}{4}\)
In drum 2, then A and B were in the ratio of 239 : 161
The amount of paint B in drum 2 = x gm in 100gm of mixture
The amount of paint B in final mixture = 35 gm in 100gm of mixture
28.......... x
.....35.......
3.............4
\(\frac{(35-28)}{(x-35)} =\frac{ 4}{3}\)
21= 4x-140
x=\( \frac{161}{4}\)
{We can mark our answer here, as there is only 1 option in which Quantity of B is multiple of 161}
Amount of paint A in drum 2= \(100-(\frac{161}{4}) = \frac{239}{4}\)
In drum 2, then A and B were in the ratio of 239 : 161
Bunuel
