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Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective

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Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]

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New post 03 Nov 2017, 08:56
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

79% (01:05) correct 21% (01:33) wrong based on 85 sessions

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Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respectively. The four solutions are mixed in a proportion of 1:2:3:4 respectively. What will be the percentage of alcohol in the resultant solution?

A. 45%
B. 50%
C. 55%
D. 65%
E. 75%

Source: Experts Global
[Reveal] Spoiler: OA

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Re: Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]

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New post 03 Nov 2017, 09:25
Just take out the weighted average. (20*1+40*2+50*3+100*4)/10
65%

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Re: Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]

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New post 03 Nov 2017, 09:31
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option D.

Concentration=(1*20+2*40+3*50+4*100)/(1+2+3+4)%
= 650/10%
=65%
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Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]

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Ratio=1:2:3:4 ie the quantity is in this ratio
if A=100 ml than B=200ml, C=300ml, and D=400ml as per above ratio
In A % of alcohol is 20% ie in 100ml=20ml, similarly % of alcohol in B is 40%=80ml, C is 50%=150ml and D is 100%=400ml
Total alcohol is =A+B+C+D=20+80+150+400=650ml
Total solution=100+200+300+400=1000ml
% of alcohol=(650/1000)*100=65% Ans D.

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Re: Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]

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New post 02 Dec 2017, 18:40
why people in this post receive so many kudos even though the math problem is not hard at all?

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Re: Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]

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pushpitkc wrote:
Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respectively. The four solutions are mixed in a proportion of 1:2:3:4 respectively. What will be the percentage of alcohol in the resultant solution?

A. 45%
B. 50%
C. 55%
D. 65%
E. 75%

Source: Experts Global



Ofcourse the straight method is given..
Merely looking at the concentration and ratio should get you to the answer.
1) 75% can be ruled out as then 50% and 100% should have equal proportions. But 50% and below is 1+2+3=6 parts and 100% is 4 parts so answer < 75%
2) 20:40:50 are in ratio 1:2:3 so the ratio is more towards 50%..
The combined 6 parts of 20,40 and 50% has to be more than 40.
So 6 parts of 40 and 4 part of 100 will get you 100-6*60/10=100-36=64
Our answer has to be>64%

Only 65% is left
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Re: Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]

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New post 02 Dec 2017, 19:29
chesstitans wrote:
why people in this post receive so many kudos even though the math problem is not hard at all?


The answer to your question is "not hard at all" :-)
the post helped me so i will complement it with kudos :thumbup:
Not everybody is at the same level and to assume that something is easy for you will be easy for others is a "flaw in the reasoning" ;) ;)

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Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]

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New post 06 Dec 2017, 13:23
pushpitkc wrote:
Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respectively. The four solutions are mixed in a proportion of 1:2:3:4 respectively. What will be the percentage of alcohol in the resultant solution?

A. 45%
B. 50%
C. 55%
D. 65%
E. 75%

Source: Experts Global

Four solutions. Different concentrations of alcohol. Different amounts (volumes) of each mixed. Result: final solution of \(x\) percent alcohol.

Known quantities (do not get an "x" in the equation*):
--The 4 solutions' alcohol concentrations and volume
--total resultant mix's volume (amount)

Unknown quantity: resultant mix's concentration of alcohol

Decimal percentages = alcohol concentrations (100% = 100/100 = 1)
\(.20(1) + .40(2) + .50(3) + 1(4) =\frac{x}{100}(1+2+3+4)\)

\(.20 + .80 + 1.5 + 4 = \frac{x}{100}(10)\)
\(6.5 = \frac{x}{100}(10)\)
\(\frac{6.5}{10}=\frac{x}{100}\)
\(.65(100) = x\)
\(x = 65\)
percent

Answer D

*Let different solutions = A, B, C, D, etc. There can be as many as needed in the equation below

1) multiply A's concentration by A's volume, and
2) do the same for each solution; then
3) add all of them (LHS); to yield
3) both resultant volume and resultant concentration (RHS)

\((Conc_{A})(Vol_{A}) + (Conc_{B})(Vol_{B})=(Conc_{A+B})(Vol_{A+B})\)

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Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective   [#permalink] 06 Dec 2017, 13:23
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