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Two mixtures of X and Y have X and Y in the ratio 3:2 and 3

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Joined: 13 Mar 2017
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Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3  [#permalink]

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New post 16 Aug 2017, 12:00
g106 wrote:
Two mixtures of X and Y have X and Y in the ratio 3:2 and 3:4. In what proportion should these two mixtures be mixed to get a new mixture in which the ration of X to Y is 5:4?

A. 6:1
B. 5:4
C. 20:7
D. 10:9
E. 14:11

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
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Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3  [#permalink]

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New post 18 Jan 2018, 10:19
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
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Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3 &nbs [#permalink] 18 Jan 2018, 10:19

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