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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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Intern
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Joined: 14 Nov 2011
Posts: 39

Kudos [?]: 8 [0], given: 245

Location: Kenya
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 29 Jan 2016, 09:37
Let me try to explain and clarify the doubts. This is actually a weighted average problem, which is really easy to solve once we understand and apply the "number line" logic.

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

Focus on any one of the two elements i.e. either mixture X or mixture Y (in this question). Forget about the other one. One reason why such questions seem to be tricky is because of the ratios / volume / weight is provided for every element. But, the most important thing is to FOCUS ON ANY ONE OF THE ELEMENTS.

Now, back to solving the question with the number line method:-

1) Focus on element X in both mixtures X and Y respectively and base all your calculations on element X.

2) Draw a simple number line and mark mixture X and mixture Y at the ends with depiction of ratios element X (or Y) in X and Y respectively. Mark average of the mixtures somewhere in the middle and depict the ratio of element X in the mixture too.. It will look something like this:-


2/45 (3/5 - 5/9) 8/63 (5/9 - 3/7)
!!----------------------!!------------------------------!!
X Mixture (X + Y) Y
(3/3+2) (5/5+4) (3/3+4)

3) Find the difference in the ratios of the element in focus from the average - 2/45 between X and average and 8/63 between Y and average in this case, as shown in the line diagram above.

4) Finally, answer the question stem - "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?" - Divide 8/63 by 2/45 => 8/63 * 45/2 = 20:7. Bingo!

Please refer to this extremely helpful link for further clarifications. Karishma's explanations are super!

the-final-exam-of-a-particular-class-makes-up-40-of-the-105651.html#p828579

Cheers!

Kudos [?]: 8 [0], given: 245

Director
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Joined: 13 Mar 2017
Posts: 561

Kudos [?]: 145 [0], given: 64

Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
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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Kudos [?]: 145 [0], given: 64

Intern
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Joined: 10 Dec 2015
Posts: 22

Kudos [?]: 6 [0], given: 49

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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Kudos [?]: 6 [0], given: 49

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

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