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i can't understand why i can't solve following problem using numerical substitution,let us consider following problem

One gallon of fuel mixture contains antifreeze in the ratio of $5$ parts fuel to one part antifreeze. To this is added half a gallon of mixture which is one third antifreeze and two thirds fuel. What is the ratio of fuel to antifreeze in the final mixture? (Grid your answer as a fraction: fuel/antifreeze)

i have tried to solve this problem using substitution,but ones again i think it is language problem ,or understanding problem and not math itself,what i have tried is,because fuel is $5$ part and antifreeze is one part,let us take fuel as $300$ and antifreeze as $60$,then $1/3$ of antifreeze is $20$,because $60*1/3=20$ and $2/3$ of fuel is $300*2/3=200$,so in total we have $300+200=500$ fuel and $60+20=80$ antifreeze,so ratio is $500/80=25/4$,but answer is $7/2$,so i think that something did not understand,please help me figure out what is wrong in my solution

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It looks as if you started calculating before forming a "real-world" picture of what was going on. By the way, for the real problem, numerical substitution using $6$ for the initial gallon works very nicely. –  André Nicolas Aug 12 '13 at 16:16

2 Answers 2

up vote 2 down vote accepted

You have one gallon that is $5$ parts fuel to $1$ part antifreeze, so this gallon of liquid contains $\frac56$ of a gallon of fuel and $\frac16$ of a gallon of antifreeze. You also have half a gallon of a mixture that is two-thirds fuel and one-third antifreeze. Two-thirds of half a gallon is $\frac23\cdot\frac12=\frac13$ of a gallon of fuel, and the remaining $\frac12-\frac13=\frac16$ of a gallon of the mixture is antifreeze. Altogether, therefore, you have $\frac56+\frac13=\frac76$ gallons of fuel and $\frac16+\frac16=\frac13$ gallons of antifreeze. (As a quick check, $\frac76+\frac13=\frac32$, which is the correct total: you do have one and a half gallons.)

The final ratio of fuel to antifreeze is therefore $\frac76:\frac13$ or, after multiplying through by $6$, $7:2$.

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so it means that this $2/3$ is independent of the first mixture? –  giorgi Aug 12 '13 at 14:41
    
@giorgi: You mean the composition of the half gallon that is being added? Yes, $\frac23$ of it is fuel independent of the composition of the original gallon. –  Brian M. Scott Aug 12 '13 at 14:43
    
aaa,it does not matter if i take concrete numbers,just i have took $2/3$ of first fuel,i took now $300$ fuel and $60$ antifreeze and half of them is $180$, and answer is now $7:2$,it works,just only i think question should say that $2/3$ of half gallon,because it may cause confusion –  giorgi Aug 12 '13 at 14:46
    
@giorgi: This appears to be a difficulty with the English: to a native speaker it really does say that $2/3$ of the half gallon is fuel. // Yes, you can use concrete numbers, provided that the total for the added mixture is only half the total for the original gallon. –  Brian M. Scott Aug 12 '13 at 14:50
    
but i am getting experience,and i hope it would not be problem for any GRE problem –  giorgi Aug 12 '13 at 14:51

at first you have 1/6 gallon of antifreeze and 5/6 gallon of fuel, and then you add 1/2 gallon of mixture (1/6 gallon antifreeze and 2/6 gallon fuel) so in all you have 2/6 gallon of antifreeze and 7/6 gallon of fuel :) 7/6 : 2/6 = 7 : 2

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