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16 Aug 2017, 13:00
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g106
First mixture X/Y = 3/2 So, X/T = 3/5
Second mixture X/Y = 3/4 So, X/T = 3/7
Final mixture X/Y = 5/4 So, X/T = 5/9
Let the proportion in which first mixture is mixed with second be p/q
So, by allegation formula
3/5 5/9 3/7
_____________________________
p q
p/q = (5/9 - 3/7)/(3/5-5/9) = ((35-27)/9/7)/((27-25)/9/5) = (8/9/7) *(9*5/2) = 20:7
Answer C
18 Jan 2018, 11:19
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For mixture problems, we can use the quantity of one of the components in both the mixtures for the solution.
Here, we have 3/5 is the conc. of X in soln. 1
And, we have 3/7 is the conc. of Y in soln. 2
Note: The denominators are the total parts of individual solutions(Viz. 5 in soln. 1 & 7 in soln. 2)
Our combined ratio is 5/9.
Simply, substract the individual parts from these two to arrive at the ratio.
(3/5)-(5/9)=(2/45)
(5/9)-(3/7)=(8/63)
Now, rationalize the ratios:
(8/63):(2/45) = 20:7. Ans. C
Here, we have 3/5 is the conc. of X in soln. 1
And, we have 3/7 is the conc. of Y in soln. 2
Note: The denominators are the total parts of individual solutions(Viz. 5 in soln. 1 & 7 in soln. 2)
Our combined ratio is 5/9.
Simply, substract the individual parts from these two to arrive at the ratio.
(3/5)-(5/9)=(2/45)
(5/9)-(3/7)=(8/63)
Now, rationalize the ratios:
(8/63):(2/45) = 20:7. Ans. C
Gmat2Cracker
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27 Jan 2019, 13:56
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weighted avg approach-
3/7- 5/9
-----------
5/9 - 3/5
=20/7 (c)
3/7- 5/9
-----------
5/9 - 3/5
=20/7 (c)
