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

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29 Sep 2011, 11:49
6
35
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Difficulty:

65% (hard)

Question Stats:

66% (02:38) correct 34% (02:54) wrong based on 586 sessions

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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

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Re: Ratio of Two Mixtures  [#permalink]

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29 Sep 2011, 16:01
7
2
3/5x + 3/7(1-x) = 5/9

x = 20 / 27
(1-x) = 7 / 27

(20 / 27) / (7 / 27) = 20 / 7
##### General Discussion
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Re: Ratio of Two Mixtures  [#permalink]

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Updated on: 30 Sep 2011, 06:39
2
2
I picked C.

(3x+3y)/(2x+4y)=5/4....results in x = 4y
and
mixt. 1/mixt. 2= 5x/7y
therefore
Substitute...mixt. 1/mixt. 2 = 5(4y)/7y = 20/7

Originally posted by arzad on 29 Sep 2011, 15:00.
Last edited by arzad on 30 Sep 2011, 06:39, edited 1 time in total.
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Re: Ratio of Two Mixtures  [#permalink]

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29 Sep 2011, 15:38
4
2
Props to Arzad.
I got C but with a different, yet slightly longer approach, which I constantly use for proportion problems such as this one. Usually, this approach is very efficient when used with slightly difficult questions, but this problem is quite hard so I had to slightly tweak it. I will show you how I usually use the approach and later show you how to use it with this specific problem.

Example.
1 Liter of Solution A contains 45% alcohol, while 1 Liter of Solution B contains 20% alcohol. In what ratio must the two solutions be used to get a solution with 30% alcohol

Solution:
1. 45/100*[A/(A+B)]+20/100*[B/(A+B)] = 30/100
2. Multiply 100 to both sides to arrive at 45A/(A+B) + 20B/(A+B) = 30
3. Multiply (A+B) to both sides to arrive at 45A + 20B = 30A + 30B
4. Distribute to arrive at 15A = 10B
5. Thus the ratio is A/B = 10/15 = 2/3

Now using this same approach, we tackle Gopu106’s question. It is important to first think of X in the mixture as the alcohol in the problem above; hence, a mixture of X and Y in the ratio of 3:2 translates to X is 3/5 of the solution. Applying this concept to all three equations, we write:
1. 3/5*[A/(A+B)]+3/7*[B/(A+B)] = 5/9
2. Now here is the tweak that must be made to continue with this approach. You must find the common denominator for all three numbers and organize the fractions accordingly. By finding the common denominator of 5,7,9 (or 315) we re-write the equations as follows
3. 189/315*[A/(A+B)]+135/315*[B/(A+B)] = 175/315
4. Multiply 315 to both sides to arrive at 189A/(A+B) + 135B/(A+B) = 175
5. Multiply (A+B) to both sides to arrive at 189A + 135B = 175A + 175B
6. Distribute to arrive at 14A = 40B
7. Thus the ratio is A/B = 40/14 = 20/7 or answer C

Finding the common denominator and adjusting the numerator is time consuming, but knowing some number property rules would speed the process. For example, if you know that your common denominator is (5)(7)(9), and you want to apply this to 3/5, then you just multiply 3*(7)(9) and omit the (5) because that is already present in the denominator and arrive at 189/315.
Again, this process is much longer than that of Arzad’s, but it is always good to know how to solve a problem multiple ways.
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Re: Ratio of Two Mixtures  [#permalink]

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30 Sep 2011, 04:05
3
mixture1 = 3/5 or 189/315
mixture2 = 3/7 or 135/315
final mix = 5/9 or 175/315

final ratio = $$\frac{175-135}{189-175} = \frac{20}{7}$$
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Re: Ratio of Two Mixtures  [#permalink]

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30 Sep 2011, 05:41
2
2
gopu106 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

Just use alligation method.

Mixture A - 3:2 ;Mixture B - 3:4 .

Now , we know each mixture consists of two components.For using allegations just take
one component.

Mixture A - 3/5 of component 1
Mixture B - 3/7 of component 1

We need a mixture which has 5/9 of component 1

Alligations :

3/5 3/7

5/9

(3/5-5/9) (5/9-3/7)

hence, the proportion in which both should be mixed is

8/63 * 45/2 = 20/7 (C)[code][/code]
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Re: Ratio of Two Mixtures  [#permalink]

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25 Apr 2012, 02:06
Why can't I apply this formula here?

w1/w2=(A2-Avg.)/(Avg-A1)

I get 1/2 if I use it, which seems to be incorrect.
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Re: Ratio of Two Mixtures  [#permalink]

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28 Apr 2012, 10:04
1
BN1989 wrote:
Why can't I apply this formula here?

w1/w2=(A2-Avg.)/(Avg-A1)

I get 1/2 if I use it, which seems to be incorrect.

you certainly can apply the formula above

w1=3/5
w2=3/7
a=5/9
w1/w2=(3/7-5/9) /(5/9-3/5)=20/7
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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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04 Sep 2013, 06:29
1
For mixtures problem, use allegation rule to arrive faster.

If we take ratio of X

3/5(X) ------5/9(final mixture)---------3/7(Y)

Find diff b/w 3/5 5/9 and 5/9 3/7

8/63 and 2/45

The reqd ratio will be 2/45 of X to 8/63 of Y.
or 20/7
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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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16 Dec 2013, 06:36
1
2
Shortest Method(based on Manhanttan approach):

According to the shortcut, " Suppose that A and B are averaged together. If they are in a ratio of a: b, then you can multiply the differential of A by a, and it will cancel out with the differential of B times b."

For ex:
suppose there is a group of men and women in a ratio of 2:3. If the men have an average age of 50, and the average age of the group is 56, you can easily figure out the average age of the women in the group. Men have a - 6 differential, and there are 2 of them for every 3 women. If the average age of women is w , then:

2 x (-6) + 3 x (w) = 0
-12 + 3w = 0
w = 4
Women have a +4 differential. The average age of the women in the group is 56 + 4 = 60 years old.

Applying the idea in our problem,
(3/5-5/9)X - (5/9-3/7)Y = 0[Here negative sign is applied for 'Y' as its value is lower than the average(3/7<5/9)], which gives us:
X/Y=20/7

This technique works on almost all of the problems involving Mixtures.
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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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03 Jan 2014, 06:22
1
1
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

Alligations is preferred approach here

3/7 - 5/9 / 5/9 - 3/5 = 20/7

Answer is C

Hope it helps!
Cheers

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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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29 Mar 2014, 07:32
5
3
Mixture A -------------------------- Mixture B
x:y ------------------------------------ x:y
3:2 ------------------------------------ 3:4
M --------------------------------------- N

$$\frac{\frac{3}{5}M + \frac{3}{7}N} {\frac{2}{5}M + \frac{4}{7}N} = \frac{5}{4}$$

$$\frac{M}{N} = \frac{20}{7}$$
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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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13 Aug 2015, 23:27
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

x in the first mixture is 3/5 or 189/315
x in the second is 3/7 or 135/315
x in the resultant mix is 5/9 or 175/315
(Go for a convenient number which is 315)
The required ratio is (175-135):(189-175)
40:14 or 20:7
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Two mixtures of X and Y have X and Y in the ratio 3:2 and 3  [#permalink]

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Updated on: 03 Feb 2016, 14:06
let volume of two mixtures=a and b respectively
3a/5+3b/7=5(a+b)/9
14a=40b
a/b=20/7

Originally posted by gracie on 15 Sep 2015, 14:33.
Last edited by gracie on 03 Feb 2016, 14:06, edited 1 time in total.
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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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02 Oct 2015, 22:12
1
Differential approach:

3/5------5/9-------3/7

(3/5-5/9)x=(5/9-3/7)y

(2/45)x=(8/63)y

x/y=(8/63)/(2/45) => 20/7
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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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09 Nov 2015, 01:55
I always do it as a weighted average:

$$\frac{3x}{5}$$+$$\frac{3(1-x)}{7}$$=$$\frac{5}{9}$$

x = $$\frac{20}{27}$$

So y = 7
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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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19 Nov 2015, 15:30
I'm not sure if my way is correct but I got the right answer.

3/2=1.5
3/4=3.5

(1.5/3.5)*x=5/4
x is about 3:1

Answer C) 20:7 is closest (21/7 = 3)
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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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22 Nov 2015, 08:39
When I use the allegation method with Fractions I get the correct answer but when I use decimals or percents it's wrong. Not sure if I'm messing up or if it's because of rounding.

3/5= 60%
3/7= 43%

.6-.56 = 4%
.56-.43=13%
13/4 = 3.25/1 ??????? Not exactly right

From here, I can't seem to get an answer that is 20:7 (2.86) although I do get 3.25:1 which is close. Did anyone have success using percents and allegation on this problem?
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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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23 Nov 2015, 13:59
DJ1986 wrote:
When I use the allegation method with Fractions I get the correct answer but when I use decimals or percents it's wrong. Not sure if I'm messing up or if it's because of rounding.

3/5= 60%
3/7= 43%

.6-.56 = 4%
.56-.43=13%
13/4 = 3.25/1 ??????? Not exactly right

From here, I can't seem to get an answer that is 20:7 (2.86) although I do get 3.25:1 which is close. Did anyone have success using percents and allegation on this problem?

Could you please explain what you did in more detail? I just don't understand your method at all.
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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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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!
Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3 &nbs [#permalink] 29 Jan 2016, 09:37

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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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