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

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Math Expert
Joined: 02 Sep 2009
Posts: 47112
In what ratio should a 20% methyl alcohol solution be mixed with a 50%  [#permalink]

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15 Mar 2016, 12:34
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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

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Posts: 6257
Re: In what ratio should a 20% methyl alcohol solution be mixed with a 50%  [#permalink]

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15 Mar 2016, 19:17
1
2
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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: In what ratio should a 20% methyl alcohol solution be mixed with a 50%  [#permalink]

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15 Jun 2016, 10:04
Hi chetan, i did not understand that as the explanation was too technical. Can you or someother expert please simplify?
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Joined: 07 Dec 2014
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In what ratio should a 20% methyl alcohol solution be mixed with a 50%  [#permalink]

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15 Jun 2016, 12:08
1
.2x+.5y=.4(x+y)
x/y=1/2=1:2
Math Expert
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Posts: 6257
Re: In what ratio should a 20% methyl alcohol solution be mixed with a 50%  [#permalink]

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15 Jun 2016, 20:00
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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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

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09 Sep 2017, 14:43
3
1
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%

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