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Two vessels A and B contain milk and water mixed in the ratio 8:5 and

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Two vessels A and B contain milk and water mixed in the ratio 8:5 and [#permalink]

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Two vessels A and B contain milk and water mixed in the ratio 8:5 and 5:2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing milk and water in the ratio 9:4

A) 2:7
B) 5:2
C) 3:5
D) 5:7
[Reveal] Spoiler: OA

Last edited by ruchi857 on 14 May 2016, 08:41, edited 1 time in total.

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ruchi857 wrote:
Two vessels A and B contain milk and water mixed in the ratio 8:5 and 5:2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing milk and water in the ratio 9:4

A) 2:7
B) 5:2
C) 3:5
D) 5:7


Hi,
your OA is wrong ..
say A qty of vessel A and B qty of vessel B is added together..
Lets concentrate on quantity of milk


\(\frac{8A}{13}+ \frac{5B}{7}= \frac{9}{13}(A+B)\) ...

\(\frac{8A}{13}+ \frac{5B}{7}= \frac{9A}{13}+\frac{9B}{13})\) ..

\(\frac{5B}{7}-\frac{9B}{13}=\frac{9A}{13}-8A/13\).....

\(\frac{65B-63B}{7*13}=\frac{A}{13}\)...

\(\frac{2B}{7*13}=\frac{A}{13}\)....

\(\frac{A}{B}=\frac{2}{7}\)

A
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Re: Two vessels A and B contain milk and water mixed in the ratio 8:5 and [#permalink]

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Another method:

Required Ratio = \(\frac{\frac{5}{7} - \frac{9}{13}}{\frac{9}{13} - \frac{8}{13}}\) = \(\frac{2}{7}\)

Answer: A

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Re: Two vessels A and B contain milk and water mixed in the ratio 8:5 and [#permalink]

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New post 06 Jun 2016, 03:42
Not sure what’s wrong with the following approach:

In vessel A: milk = 8x, water = 5x
In vessel B: milk = 5y, water = 2y
According to the question stem 8x + 5y : 5x + 2y should equal to 9:4

Hence, 8x + 5y / 5x + 2y = 9/4
=> 2y = 13x
=> x/y = 2/13 :|

Could someone please clarify?
Regards.
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ranaazad wrote:
Not sure what’s wrong with the following approach:

In vessel A: milk = 8x, water = 5x
In vessel B: milk = 5y, water = 2y
According to the question stem 8x + 5y : 5x + 2y should equal to 9:4

Hence, 8x + 5y / 5x + 2y = 9/4
=> 2y = 13x
=> x/y = 2/13 :|

Could someone please clarify?
Regards.



Hi,

you have started well but left it in between....

it is fine till
\(\frac{x}{y} = \frac{2}{13}\).....
But we are not looking for x/y..
what is the total of FIRST mixture = 8x+5x = 13x....
and SECOND mix = 5y+2y = 7y...

so we are looking for \(\frac{13x}{7y}\)......
\(\frac{x}{y}= \frac{2}{13}\), so\(\frac{x}{y}*\frac{13}{7} = \frac{2}{13}*\frac{13}{7}.......... \frac{13x}{7y} = \frac{2}{7}\)....
ans\(\frac{2}{7}\)
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Two vessels A and B contain milk and water mixed in the ratio 8:5 and [#permalink]

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New post 17 Jul 2016, 10:08
8A/13+5B/7=9/13(A+B)
A/B=2/7
A:B=2:7

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Two vessels A and B contain milk and water mixed in the ratio 8:5 and [#permalink]

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New post 16 Oct 2017, 08:16
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There are two equations here, either one of which, when solved, will get you the final answer.

1) \((\frac{8}{13})*X + (\frac{5}{7})*Y = (\frac{9}{13})*(X+Y)\)

2) \((\frac{5}{13})*X + (\frac{2}{7})*Y = (\frac{4}{13})*(X+Y)\)

Both equations lead to\(\frac{X}{Y} = \frac{2}{7}\)
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Two vessels A and B contain milk and water mixed in the ratio 8:5 and [#permalink]

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New post 14 Nov 2017, 06:08
In questions like this concentrate only on one of the ingredients of the mixture.

Let us take milk.

what is the percentage of milk in A=8/13. (When we have a ratio like given in this question 8:5 add them to get the total)

% of milk in B=5/7

% of milk after mixing=9/13=average after mixing 2 solutions or mixtures.

use weighted average formula

w1/w2=(A2-avg)/(avg-A1). Here A1=A and A2=B

A1=8/13, A2=5/7 and Avg=9/13. (You can take A1 and A2 as any of the given value and you will get the same answer)

w1/w2=(5/7-9/13)÷(9/13-8/13)=[(65-63)/91]÷[(9-8)/13]=2/91÷1/13=2/7 or 2:7 answer is =A

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Two vessels A and B contain milk and water mixed in the ratio 8:5 and   [#permalink] 14 Nov 2017, 06:08
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