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In what ratio should a 20% methyl alcohol solution be mixed with a 50%

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In what ratio should a 20% methyl alcohol solution be mixed with a 50% [#permalink]

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In what ratio should a 20% methyl alcohol solution be mixed with a 50% methyl alcohol solution so that the resultant solution has 40% methyl alcohol in it?

A. 1 : 3
B. 1 : 2
C. 2 : 3
D. 2 : 1
E. 3 : 1
[Reveal] Spoiler: OA

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Re: In what ratio should a 20% methyl alcohol solution be mixed with a 50% [#permalink]

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New post 15 Mar 2016, 18:17
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Bunuel wrote:
In what ratio should a 20% methyl alcohol solution be mixed with a 50% methyl alcohol solution so that the resultant solution has 40% methyl alcohol in it?

A. 1 : 3
B. 1 : 2
C. 2 : 3
D. 2 : 1
E. 3 : 1



Hi,

TWO things to remember:-
1) the ratio of quantity mixed to get an average is related to the ratio of each qty's distance from average.
2) More the Qty of x from that of y, More CLOSER will be average to x..


Qty of 20%/Qty of 50%= \(\frac{(50-40)}{(40-20)}=\frac{1}{2}= 1:2\)
B
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Re: In what ratio should a 20% methyl alcohol solution be mixed with a 50% [#permalink]

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New post 15 Jun 2016, 09:04
Hi chetan, i did not understand that as the explanation was too technical. Can you or someother expert please simplify?

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In what ratio should a 20% methyl alcohol solution be mixed with a 50% [#permalink]

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New post 15 Jun 2016, 11:08
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.2x+.5y=.4(x+y)
x/y=1/2=1:2

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Re: In what ratio should a 20% methyl alcohol solution be mixed with a 50% [#permalink]

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Senthil7 wrote:
Hi chetan, i did not understand that as the explanation was too technical. Can you or someother expert please simplify?


Hi,
the Q is-
In what ratio should a 20% methyl alcohol solution be mixed with a 50% methyl alcohol solution so that the resultant solution has 40% methyl alcohol in it?

Now one solution is 20% and other is 50%, and the mix has an average of 40%.....
Now had both 20% and 50% been equal, the average would be\(\frac{20+50}{2}= 35\)%. But the average is 40%, which is even further closer to 50%, so the quantity of 50% should be MORE than 20%...


HOW much? it is equal to the RATIO of distance from the average..
This is weighted average method...
QTY of 20%/QTY of 50% = Distance of average from 50%/ distance of average from 20% = \(\frac{50-40}{40-20} = \frac{1}{2}\)..

so Qty of 20%/Qty of 50% = 1/2......
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Re: In what ratio should a 20% methyl alcohol solution be mixed with a 50% [#permalink]

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In what ratio should a 20% methyl alcohol solution be mixed with a 50% [#permalink]

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New post 09 Sep 2017, 13:43
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Bunuel wrote:
In what ratio should a 20% methyl alcohol solution be mixed with a 50% methyl alcohol solution so that the resultant solution has 40% methyl alcohol in it?

A. 1 : 3
B. 1 : 2
C. 2 : 3
D. 2 : 1
E. 3 : 1

To find the ratio of two solutions in a resultant mixture, weighted average works well. I use this formula:
\((Concen_{A})(Vol_{A}) + (Concen_{B})(Vol_{B} = (Concen_{A+B})(Vol_{A+B})\)

Let A = volume of the solution with 20 percent alcohol

Let B = volume the solution with 50 percent alcohol

A + B = resultant mixture

(.20)(A) + (.50)(B) = .40(A + B) ==>

(20)(A) + (50)(B) = 40(A + B)

20A + 50B = 40A + 40B

10B = 20A

Ratio is either \(\frac{A}{B}\), or \(\frac{B}{A}\). Be careful about order. The prompt asks for the ratio of the 20 percent solution to the 50 percent solution, that is, for the ratio of A to B.

10B = 20A

\(\frac{10}{20}\) = \(\frac{A}{B}\)

A:B is 1:2

ANSWER B

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In what ratio should a 20% methyl alcohol solution be mixed with a 50%   [#permalink] 09 Sep 2017, 13:43
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