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Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]
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03 Nov 2017, 09:56
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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
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Re: Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]
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03 Nov 2017, 10:25
Just take out the weighted average. (20*1+40*2+50*3+100*4)/10 65% Sent from my Moto G (5) Plus using GMAT Club Forum mobile app



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Re: Solutions A,B,C,and D contain 20%,40%,50%, and 100% alcohol respective [#permalink]
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03 Nov 2017, 10:31
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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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14 Nov 2017, 03:20
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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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02 Dec 2017, 19:40
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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02 Dec 2017, 20:17
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 GlobalOfcourse 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 1006*60/10=10036=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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02 Dec 2017, 20:29
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 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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06 Dec 2017, 14:23
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 GlobalFour 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
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