BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Sun Jul 08, 2012 10:10 pm
gmatter2012 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?
Algebraic Approach:
Say, m volume of first mixture and n volumes of second mixture were mixed. We need to find m:n.

Amount of X in the new mixture = (3m/5 + 3n/7) = (21m + 15n)/35
Amount of Y in the new mixture = (2m/5 + 4n/7) = (14m + 20n)/35

So, [(21m + 15n)/35]/[(14m + 20n)/35] = 5/4
--> 4(21m + 15n) = 5(14m + 20n)
--> (84m + 60n) = (70m + 100n)
--> 14m = 40n
--> m/n = 40/14 = 20/7

The correct answer is C.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
User avatar
Legendary Member
Posts: 520
Joined: Sat Apr 28, 2012 9:12 pm
Thanked: 339 times
Followed by:49 members
GMAT Score:770

by eagleeye » Sun Jul 08, 2012 11:41 pm
gmatter2012 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?
Let's assume that we mix amount a of 3:2 mixture and amount b of 3:4 mixture to get (a+b) of 5:4 mixture. We need to find a/b. Let's equate the amount of X in all of them.
Amount of X in 3:2 mixture = 3*a/5. Amount of X in b = 3*b/7. Amount in overall mixture = 5*(a+b)/9
Then 3a/5+3b/7 = 5(a+b)/9
=> 9* (21a+15b) = 35*5*(a+b)
=> 189a + 135b = 175a+175b
=> a(189-175) = b(175-135)
=> a*14 = b*40
=> a/b = 40/14 = 20/7
So we need to mix them in 20:7 ratio :)
Top
Master | Next Rank: 500 Posts
Posts: 141
Joined: Tue Oct 04, 2011 5:17 am
Thanked: 25 times

by coolhabhi » Sun Jul 08, 2012 11:45 pm
gmatter2012 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
In first mixture, X = 3/5
In second mixture, X = 3/7
resultant mixture, X = 5/9
Use allegations rule.

3/7 3/5
5/9
(3/5)-(5/9) (5/9)-(3/7)
2/45 8/63

Mixture2 : Mixture1 = (2/45) : (8/63)
=>Mixture2 7
--------- = ---
Mixture1 20
=>M1 : M2 = 20 : 7
Top
Master | Next Rank: 500 Posts
Posts: 141
Joined: Tue Oct 04, 2011 5:17 am
Thanked: 25 times

by coolhabhi » Mon Jul 09, 2012 12:10 am
gmatter2012 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
In first mixture, X = 3/5
In second mixture, X = 3/7
resultant mixture, X = 5/9
Use allegations rule.

3/7 3/5
5/9
(3/5)-(5/9) (5/9)-(3/7)
2/45 8/63

Mixture2 : Mixture1 = (2/45) : (8/63)
=>Mixture2 7
--------- = ---
Mixture1 20
=>M1 : M2 = 20 : 7
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 138
Joined: Sat May 05, 2012 7:03 pm
Thanked: 1 times

by gmatter2012 » Mon Jul 09, 2012 12:33 am
Anurag@Gurome wrote:
gmatter2012 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?
Algebraic Approach:
Say, m volume of first mixture and n volumes of second mixture were mixed. We need to find m:n.

Amount of X in the new mixture = (3m/5 + 3n/7) = (21m + 15n)/35
Amount of Y in the new mixture = (2m/5 + 4n/7) = (14m + 20n)/35

So, [(21m + 15n)/35]/[(14m + 20n)/35] = 5/4
--> 4(21m + 15n) = 5(14m + 20n)
--> (84m + 60n) = (70m + 100n)
--> 14m = 40n
--> m/n = 40/14 = 20/7

The correct answer is C.

Anurag this method was awesome , using this I have solved another similar question using this same method

https://www.beatthegmat.com/two-alloys-a ... tml#485112

Thank you
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Jul 09, 2012 3:01 am
gmatter2012 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
The following approach is called alligation.
It's a very good way to handle MIXTURE PROBLEMS.

Let F = the first mixture and S = the second mixture.

Step 1: Convert the ratios to FRACTIONS.
F:
Since X:Y = 3:2, and 3+2=5, X/total = 3/5.
S:
Since X:Y = 3:4, and 3+4=7, X/total = 3/7.
New mixture:
Since X:Y = 5:4, and 5+4=9, X/total = 5/9.

Step 2: Put the fractions over a COMMON DENOMINATOR.
F = 3/5 = (3*7*9)/(5*7*9) = 189/(5*7*9).
S = 3/7 = (3*5*9)/(5*7*9) = 135/(5*7*9).
New = 5/9 = (5*5*7)/(5*7*9) = 175/(5*7*9).

Step 3: Plot the 3 numerators on a number line, with the two starting numerators (189 and 135) on the ends and the goal numerator (175) in the middle.
F(189)--------------------(175)--------S(135)

Step 4: Calculate the distances between the numerators.
F(189)----------14---------(175)----40----S(135)

Step 5: Determine the ratio in the mixture.
The ratio of F to S in the mixture is the RECIPROCAL of the distances in red.
F : S = 40 : 14 = 20:7.

The correct answer is C.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Post new topic Post Reply
• Page 1 of 1