Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 21 Aug 2010
Posts: 176
Location: United States

Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
29 Sep 2011, 12:49
Question Stats:
68% (02:02) correct 32% (02:27) wrong based on 574 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________





Current Student
Joined: 08 Jan 2009
Posts: 309

Re: Ratio of Two Mixtures
[#permalink]
Show Tags
29 Sep 2011, 17:01
3/5x + 3/7(1x) = 5/9
x = 20 / 27 (1x) = 7 / 27
(20 / 27) / (7 / 27) = 20 / 7




Senior Manager
Status: MBAing!!!!
Joined: 24 Jun 2011
Posts: 255
Location: United States (FL)
Concentration: Finance, Real Estate
Schools: Wharton '14 (D), CBS '14 (WL), Ross '14 (WL), Haas '14 (D), Johnson '14 (A), McCombs '14 (II), KenanFlagler '14 (M), Madison (A), Consortium (D), Consortium (A), Consortium (M), Consortium (II), Consortium (A), Consortium (WL)
GPA: 3.65
WE: Project Management (Real Estate)

Re: Ratio of Two Mixtures
[#permalink]
Show Tags
Updated on: 30 Sep 2011, 07:39
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, 16:00.
Last edited by arzad on 30 Sep 2011, 07:39, edited 1 time in total.



Intern
Joined: 24 Sep 2011
Posts: 7

Re: Ratio of Two Mixtures
[#permalink]
Show Tags
29 Sep 2011, 16:38
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 rewrite 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.



Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 230

Re: Ratio of Two Mixtures
[#permalink]
Show Tags
30 Sep 2011, 05:05
mixture1 = 3/5 or 189/315 mixture2 = 3/7 or 135/315 final mix = 5/9 or 175/315 final ratio = \(\frac{175135}{189175} = \frac{20}{7}\)
_________________
press +1 Kudos to appreciate posts Download Valuable Collection of Percentage Questions (PS/DS)



Current Student
Status: :)
Joined: 29 Jun 2010
Posts: 108
WE: Information Technology (Consulting)

Re: Ratio of Two Mixtures
[#permalink]
Show Tags
30 Sep 2011, 06:41
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 (3/55/9) (5/93/7) hence, the proportion in which both should be mixed is 8/63 * 45/2 = 20/7 (C)[code][/code]
_________________
Thanks, GC24
Please click Kudos ,if my post helped you



Manager
Joined: 12 Oct 2011
Posts: 118
GMAT 1: 700 Q48 V37 GMAT 2: 720 Q48 V40

Re: Ratio of Two Mixtures
[#permalink]
Show Tags
25 Apr 2012, 03:06
Why can't I apply this formula here?
w1/w2=(A2Avg.)/(AvgA1)
I get 1/2 if I use it, which seems to be incorrect.



Senior Manager
Joined: 23 Oct 2010
Posts: 362
Location: Azerbaijan
Concentration: Finance

Re: Ratio of Two Mixtures
[#permalink]
Show Tags
28 Apr 2012, 11:04
BN1989 wrote: Why can't I apply this formula here?
w1/w2=(A2Avg.)/(AvgA1)
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=(3/75/9) /(5/93/5)=20/7
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth



Manager
Joined: 07 Apr 2012
Posts: 105
Location: United States
Concentration: Entrepreneurship, Operations
GPA: 3.9
WE: Operations (Manufacturing)

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
04 Sep 2013, 07:29
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



Intern
Joined: 04 Dec 2012
Posts: 11
GMAT Date: 12262013

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
16 Dec 2013, 07:36
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, (3/55/9)X  (5/93/7)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.



SVP
Joined: 06 Sep 2013
Posts: 1852
Concentration: Finance

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
03 Jan 2014, 07:22
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 J



Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 871
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
29 Mar 2014, 08:32
Mixture A  Mixture B x:y  x:y 3:2  3:4 M  N \(\frac{\frac{3}{5}M + \frac{3}{7}N} {\frac{2}{5}M + \frac{4}{7}N} = \frac{5}{4}\) \(\frac{M}{N} = \frac{20}{7}\)
_________________
Piyush K
 Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press> Kudos My Articles: 1. WOULD: when to use?  2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".



Manager
Joined: 10 Jun 2015
Posts: 120

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
14 Aug 2015, 00:27
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 (175135):(189175) 40:14 or 20:7



VP
Joined: 07 Dec 2014
Posts: 1064

Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
Updated on: 03 Feb 2016, 15:06
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, 15:33.
Last edited by gracie on 03 Feb 2016, 15:06, edited 1 time in total.



Director
Joined: 23 Jan 2013
Posts: 598

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
02 Oct 2015, 23:12
Differential approach:
3/55/93/7
(3/55/9)x=(5/93/7)y
(2/45)x=(8/63)y
x/y=(8/63)/(2/45) => 20/7



Retired Moderator
Joined: 23 Sep 2015
Posts: 386
Location: France
GMAT 1: 690 Q47 V38 GMAT 2: 700 Q48 V38
WE: Real Estate (Mutual Funds and Brokerage)

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
09 Nov 2015, 02:55
I always do it as a weighted average: \(\frac{3x}{5}\)+\(\frac{3(1x)}{7}\)=\(\frac{5}{9}\) x = \(\frac{20}{27}\) So y = 7
_________________
New Application Tracker : update your school profiles instantly!



Manager
Joined: 05 Jul 2015
Posts: 104
Concentration: Real Estate, International Business
GPA: 3.3

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
19 Nov 2015, 16:30
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)



Manager
Joined: 05 Jul 2015
Posts: 104
Concentration: Real Estate, International Business
GPA: 3.3

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
22 Nov 2015, 09:39
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?



Retired Moderator
Joined: 23 Sep 2015
Posts: 386
Location: France
GMAT 1: 690 Q47 V38 GMAT 2: 700 Q48 V38
WE: Real Estate (Mutual Funds and Brokerage)

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
23 Nov 2015, 14:59
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.
_________________
New Application Tracker : update your school profiles instantly!



Intern
Joined: 15 Nov 2011
Posts: 34
Location: Kenya

Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3
[#permalink]
Show Tags
29 Jan 2016, 10:37
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! thefinalexamofaparticularclassmakesup40ofthe105651.html#p828579Cheers!




Re: Two mixtures of X and Y have X and Y in the ratio 3:2 and 3 &nbs
[#permalink]
29 Jan 2016, 10:37



Go to page
1 2
Next
[ 22 posts ]



