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There are two drums, each containing a mixture of paints A and B. In d

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Bunuel
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There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   06 Apr 2020, 09:50
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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


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sachinsavailable
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Re: There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   06 Apr 2020, 21:43
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Bunuel wrote:
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


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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).
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There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   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


Bunuel wrote:
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
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sachinsavailable
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Re: There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   06 Apr 2020, 19:53
nick1816 wrote:
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


\(\frac{(35-28)}{(x-70)} =\frac{ 4}{3}\)

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 wrote:
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


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nick1816
Could you advise how did you calculate. If possible in detail.
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There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   06 Apr 2020, 20:42
Very long calculation..

3(18/25)+4(x/x+y) = 3(7/25) + 4(y/x+y)

3(18/25)+4(x/x+y)/3(7/25) + 4(y/x+y) = 13/7
7[3(18/25)+4(x/x+y)] = 13[3(7/25) + 4(y/x+y)]
7(54x+54y+100x) = 13(21x+21y+100y)
378x+378y+700x = 273x+273y+1300y
805x = 1195y
1195/805 = y/x

239/161 = y/x
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Re: There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   07 Apr 2020, 23:09
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nick1816 wrote:
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


\(\frac{(35-28)}{(x-70)} =\frac{ 4}{3}\)

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 wrote:
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


nick1816
mistakenly you wrote X-70
it should be X-35 instead of X-70
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Re: There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   08 Apr 2020, 01:37
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Let's say 75 part of drum 1 is mixed with 100 part of drum 2.
A:B in drum 1 =54:21
A:B in drum 2 =A :100-A

Now,
(54 +A)/(21+100-A)= 13/7

A =239/4
Therefore
B= 161/4
A:B =239:161
Option C is the answer

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There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   10 Jun 2021, 08:42
A+B in Drum 1 = 25
So, the total of A and B in Drum 2 should be a multiple of 25 and A+B ( =20) of the final mixture.

Only C qualifies the above-mentioned criterion.

What's your opinion on my approach?
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Re: There are two drums, each containing a mixture of paints A and B. In d [#permalink] New post   17 Dec 2023, 11:03
Hello from the GMAT Club BumpBot!

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Re: There are two drums, each containing a mixture of paints A and B. In d [#permalink]
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