• EMPOWERgmat Slider
    1 Hour Free
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    EMPOWERgmat Slider
  • Veritas Prep
    Free Veritas GMAT Class
    Experience Lesson 1 Live Free

    Available with Beat the GMAT members only code

    MORE DETAILS
    Veritas Prep
  • PrepScholar GMAT
    5 Day FREE Trial
    Study Smarter, Not Harder

    Available with Beat the GMAT members only code

    MORE DETAILS
    PrepScholar GMAT
  • Kaplan Test Prep
    Free Practice Test & Review
    How would you score if you took the GMAT

    Available with Beat the GMAT members only code

    MORE DETAILS
    Kaplan Test Prep
  • Varsity Tutors
    Award-winning private GMAT tutoring
    Register now and save up to $200

    Available with Beat the GMAT members only code

    MORE DETAILS
    Varsity Tutors
  • e-gmat Exclusive Offer
    Get 300+ Practice Questions
    25 Video lessons and 6 Webinars for FREE

    Available with Beat the GMAT members only code

    MORE DETAILS
    e-gmat Exclusive Offer
  • Magoosh
    Magoosh
    Study with Magoosh GMAT prep

    Available with Beat the GMAT members only code

    MORE DETAILS
    Magoosh
  • Target Test Prep
    5-Day Free Trial
    5-day free, full-access trial TTP Quant

    Available with Beat the GMAT members only code

    MORE DETAILS
    Target Test Prep
  • Economist Test Prep
    Free Trial & Practice Exam
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    Economist Test Prep

ratios / fraction problem

This topic has 7 expert replies and 13 member replies
Goto page
  • 1,
  • 2
Next
sachindia Master | Next Rank: 500 Posts Default Avatar
Joined
22 Jun 2012
Posted:
258 messages
Followed by:
3 members
Thanked:
6 times
GMAT Score:
640

ratios / fraction problem

Post Thu Jul 05, 2012 11:33 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
    the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
    form a mixture containing 5 kg of wheat and 11 kg of rice?

    _________________
    Regards,
    Sach

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    Post Fri Jul 06, 2012 1:44 am
    sachindia wrote:
    Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
    the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
    form a mixture containing 5 kg of wheat and 11 kg of rice?

    1. 3kgs from A, 13 kg from B
    2. 8kgs from A, 8 kg from B
    3. 2kgs from A, 14 kg from B
    4. 6kgs from A, 10 kg from B
    Percentage of wheat in A = 2/5 = 40%
    Percentage of wheat in B = 3/10 = 30%.
    Percentage of wheat in the mixture = 5/16 = 31.25%.

    The following approach is called alligation.
    It's a very good way to handle MIXTURE PROBLEMS.

    Step 1: Plot the 3 percentages on a number line, with the two starting percentages (40% and 30%) on the ends and the goal percentage (31.25%) in the middle.
    A(40%)--------------------31.25%--------B(30%)

    Step 2: Calculate the distances between the percentages.
    A(40%)----------8.75---------31.25%----1.25----B(30%)

    Step 3: Determine the ratio in the mixture.
    The ratio of A to B in the mixture is the RECIPROCAL of the distances in red.
    A : B = 1.25 : 8.75 = 1:7.

    Only answer choice C has the same ratio:
    2:14 = 1:7.

    The correct answer is C.

    Here's another:

    Quote:
    Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ?

    (A) 10%
    (B) 33.1/3%
    (C) 40%
    (D) 50%
    (E) 66.2/3%
    X = 40% ryegrass, Y = 25% ryegrass, the mixture = 30% ryegrass.

    X(40%)--------10--------30%-----5-----Y(25%)

    X:Y = 5:10 = 1:2.
    Since 1+2 = 3, of every 3 liters, 1 liter must be X and 2 liters must be Y, implying that the fraction of X in the mixture = 1/3 = 33 1/3%.

    The correct answer is B.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

    Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.
    jaiswalamrita Senior | Next Rank: 100 Posts Default Avatar
    Joined
    24 Apr 2012
    Posted:
    34 messages
    Post Mon Jul 09, 2012 12:21 pm
    wonderful method Mitch..many thanks

    kjallow Junior | Next Rank: 30 Posts Default Avatar
    Joined
    17 Nov 2011
    Posted:
    18 messages
    Target GMAT Score:
    0
    Post Wed Sep 19, 2012 5:24 am
    GMATGuruNY - You are truly a GMAT Guru!! I have been struggling with mixture problems for a while. many thanks for this simple, easy to follow approach.

    Post Sun Nov 04, 2012 9:42 am
    Thanks Mitch for showing such a wonderful approach...

    pemdas Legendary Member Default Avatar
    Joined
    15 Apr 2011
    Posted:
    1085 messages
    Followed by:
    21 members
    Thanked:
    158 times
    Post Sun Nov 04, 2012 12:55 pm
    sachindia wrote:
    Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to form a mixture containing 5 kg of wheat and 11 kg of rice?
    no need to mystify the weighted averages as invented alligation method

    A*(2/5)+B*(3/10)=5
    A*(3/5)+B*(7/10)=11

    solve for A and B to find the required,

    4A+3B=50 (multiply by 2, 8a+6b=100)
    6A+7B=110 (subtract the expression in parentheses above, 6A+7B=110 less 8A+6B=100)

    B-2A=10 and B=2A+10
    4A+3B=50 <> 4A+3(2A+10)=50 <> 4A+6A+30=50, A=20/10=2

    Now, either solve for B or find B from (5+11-A)=16-2=14

    enjoy pure math & algebra rather invented rules

    _________________
    Success doesn't come overnight!

    Thanked by: godhuli_golpo, BKN, musejoy
    TG_GMAT Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    22 Aug 2008
    Posted:
    6 messages
    Post Mon Mar 18, 2013 7:57 am
    Hi Mitch,

    I have a question on the alligation approach. Isnt the question asking for the ratio of wheat:rice. From the alligation method in your solution, are we not getting the ratio of wheat from A to wheat from B? What about rice?

    Please help!
    Thanks

    TG_GMAT Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    22 Aug 2008
    Posted:
    6 messages
    Post Mon Mar 18, 2013 7:58 am
    Hi Mitch,

    I have a question on the alligation approach. Isnt the question asking for the ratio of wheat:rice. From the alligation method in your solution, are we not getting the ratio of wheat from A to wheat from B? What about rice? The 1:7 is for A(wheat):B(wheat), correct?

    Please help!
    Thanks

    vonsumit Junior | Next Rank: 30 Posts
    Joined
    24 Feb 2013
    Posted:
    10 messages
    Post Tue Mar 19, 2013 7:46 am
    unique but effective method

    aaggar7 Master | Next Rank: 500 Posts Default Avatar
    Joined
    11 Mar 2012
    Posted:
    124 messages
    Followed by:
    1 members
    Thanked:
    9 times
    Post Thu Apr 18, 2013 1:44 am
    Hi Mitch,

    Can u pls explain the steps for the below question using the same approach that u explained earlier

    A mixture contains wine and water in the ratio 3:2.Another contains in the ratio 4:5.
    How many gallons of latter to be mixed with 3 gallons of first so that resulting mix contains
    equal qty of wine and water

    Post Thu Apr 18, 2013 11:45 am
    aaggar7 wrote:
    Hi Mitch,

    Can u pls explain the steps for the below question using the same approach that u explained earlier

    A mixture contains wine and water in the ratio 3:2.Another contains in the ratio 4:5.
    How many gallons of latter to be mixed with 3 gallons of first so that resulting mix contains
    equal qty of wine and water
    Let F = the first solution, L = the latter solution, and M = the resulting mixture of the two solutions.
    Alligation can be performed only with percentages or fractions.

    Step 1: Convert the ratios to FRACTIONS.
    F:
    Since wine:water = 3:2, and 3+2=5, wine/total = 3/5.
    L:
    Since wine:water = 4:5, and 4+5=9, wine/total = 4/9.
    M:
    Since wine:water = 1:1, and 1+1=2, wine/total= 1/2.

    Step 2: Put the fractions over a COMMON DENOMINATOR.

    F = 3/5 = 54/90.
    L = 4/9 = 40/90.
    M = 1/2 = 45/90.

    Step 3: Plot the 3 numerators on a number line, with the two starting numerators (54 and 40) on the ends and the goal numerator (45) in the middle.
    F 54-----------45-----------40 L

    Step 4: Calculate the distances between the numerators.
    F 54-----9-----45-----5-----40 L

    Step 5: Determine the ratio in the mixture.
    The ratio of F to L in the mixture is the RECIPROCAL of the distances in red.
    F/L = 5/9.

    Since F/L = 5/9, and the actual volume of F = 3 liters, we get:
    5/9 = 3/L
    5L = 27
    L = 27/5 = 5.4 liters.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

    Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.
    ygdrasil24 Senior | Next Rank: 100 Posts
    Joined
    06 Dec 2011
    Posted:
    42 messages
    Followed by:
    1 members
    Thanked:
    3 times
    Post Thu May 23, 2013 5:28 am
    We should get the same answer when solving for rice ?

    mevicks Master | Next Rank: 500 Posts Default Avatar
    Joined
    19 Sep 2013
    Posted:
    269 messages
    Followed by:
    6 members
    Thanked:
    94 times
    Post Thu Sep 19, 2013 7:16 am
    GMATGuruNY wrote:
    sachindia wrote:
    Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
    the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
    form a mixture containing 5 kg of wheat and 11 kg of rice?

    1. 3kgs from A, 13 kg from B
    2. 8kgs from A, 8 kg from B
    3. 2kgs from A, 14 kg from B
    4. 6kgs from A, 10 kg from B
    Percentage of wheat in A = 2/5 = 40%
    Percentage of wheat in B = 3/10 = 30%.
    Percentage of wheat in the mixture = 5/16 = 31.25%.

    The following approach is called alligation.
    It's a very good way to handle MIXTURE PROBLEMS.

    Step 1: Plot the 3 percentages on a number line, with the two starting percentages (40% and 30%) on the ends and the goal percentage (31.25%) in the middle.
    A(40%)--------------------31.25%--------B(30%)

    Step 2: Calculate the distances between the percentages.
    A(40%)----------8.75---------31.25%----1.25----B(30%)

    Step 3: Determine the ratio in the mixture.
    The ratio of A to B in the mixture is the RECIPROCAL of the distances in red.
    A : B = 1.25 : 8.75 = 1:7.

    Only answer choice C has the same ratio:
    2:14 = 1:7.

    The correct answer is C.

    @Mitch : Thanks a ton for demystifying mixtures!

    @ygdrasil24 : Yes, one can work with the rice ratios and still end up with the final A to B ratio as 1 : 7

    Consider the following:


    Mixture A
    (W)40%__________________60%(R)


    Mixture B
    (W)30%__________________70%(R)


    Final Mixture
    (5KGS)31.25%_____________68.75%(11KGS)



    (60%)Ra----------8.75-----------68.75%----1.25-----Rb(70%)

    A/B = 1.25/8.75 = 1/7"

    sahilchaudhary Master | Next Rank: 500 Posts
    Joined
    11 Apr 2011
    Posted:
    153 messages
    Followed by:
    7 members
    Thanked:
    22 times
    Test Date:
    27 Dec 2013
    Target GMAT Score:
    700
    GMAT Score:
    540
    Post Mon Dec 09, 2013 5:02 am
    GMATGuruNY wrote:
    sachindia wrote:
    Two tins A and B contain mixtures of wheat and rice. In A, the weights of wheat and rice are in
    the ratio 2 : 3 and in B they are in the ratio 3 : 7. What quantities must be taken from A and B to
    form a mixture containing 5 kg of wheat and 11 kg of rice?

    1. 3kgs from A, 13 kg from B
    2. 8kgs from A, 8 kg from B
    3. 2kgs from A, 14 kg from B
    4. 6kgs from A, 10 kg from B
    Percentage of wheat in A = 2/5 = 40%
    Percentage of wheat in B = 3/10 = 30%.
    Percentage of wheat in the mixture = 5/16 = 31.25%.

    The following approach is called alligation.
    It's a very good way to handle MIXTURE PROBLEMS.

    Step 1: Plot the 3 percentages on a number line, with the two starting percentages (40% and 30%) on the ends and the goal percentage (31.25%) in the middle.
    A(40%)--------------------31.25%--------B(30%)

    Step 2: Calculate the distances between the percentages.
    A(40%)----------8.75---------31.25%----1.25----B(30%)

    Step 3: Determine the ratio in the mixture.
    The ratio of A to B in the mixture is the RECIPROCAL of the distances in red.
    A : B = 1.25 : 8.75 = 1:7.

    Only answer choice C has the same ratio:
    2:14 = 1:7.

    The correct answer is C.

    Here's another:

    Quote:
    Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ?

    (A) 10%
    (B) 33.1/3%
    (C) 40%
    (D) 50%
    (E) 66.2/3%
    X = 40% ryegrass, Y = 25% ryegrass, the mixture = 30% ryegrass.

    X(40%)--------10--------30%-----5-----Y(25%)

    X:Y = 5:10 = 1:2.
    Since 1+2 = 3, of every 3 liters, 1 liter must be X and 2 liters must be Y, implying that the fraction of X in the mixture = 1/3 = 33 1/3%.

    The correct answer is B.
    Excellent new method Mitch.

    _________________
    Sahil Chaudhary
    If you find this post helpful, please take a moment to click on the "Thank" icon.
    http://www.sahilchaudhary007.blogspot.com

    Poisson Senior | Next Rank: 100 Posts Default Avatar
    Joined
    11 Aug 2012
    Posted:
    31 messages
    Post Mon Aug 29, 2016 5:31 pm
    GMATGuruNY wrote:
    aaggar7 wrote:
    Hi Mitch,

    Can u pls explain the steps for the below question using the same approach that u explained earlier

    A mixture contains wine and water in the ratio 3:2.Another contains in the ratio 4:5.
    How many gallons of latter to be mixed with 3 gallons of first so that resulting mix contains
    equal qty of wine and water
    Let F = the first solution, L = the latter solution, and M = the resulting mixture of the two solutions.
    Alligation can be performed only with percentages or fractions.

    Step 1: Convert the ratios to FRACTIONS.
    F:
    Since wine:water = 3:2, and 3+2=5, wine/total = 3/5.
    L:
    Since wine:water = 4:5, and 4+5=9, wine/total = 4/9.
    M:
    Since wine:water = 1:1, and 1+1=2, wine/total= 1/2.

    Step 2: Put the fractions over a COMMON DENOMINATOR.

    F = 3/5 = 54/90.
    L = 4/9 = 40/90.
    M = 1/2 = 45/90.

    Step 3: Plot the 3 numerators on a number line, with the two starting numerators (54 and 40) on the ends and the goal numerator (45) in the middle.
    F 54-----------45-----------40 L

    Step 4: Calculate the distances between the numerators.
    F 54-----9-----45-----5-----40 L

    Step 5: Determine the ratio in the mixture.
    The ratio of F to L in the mixture is the RECIPROCAL of the distances in red.
    F/L = 5/9.

    Since F/L = 5/9, and the actual volume of F = 3 liters, we get:
    5/9 = 3/L
    5L = 27
    L = 27/5 = 5.4 liters.
    Hi Mitch,

    Could you please explain how to apply this approach to the following question? Each time I try it, I get an incorrect answer.

    Mix A is 35% red and 65 % blue.
    Mix B is 50% red and 50% blue.
    If you begin with 20 pounds of mix A, how many pounds of mix B do you need to end up with a combined mixture that is 40% red and 60% blue?

    A. 5
    B. 10
    C. 15
    D. 20
    E. 25

    I keep ending up with an answer of 5 once I take the reciprocal of the distances.

    Thanks very much.

    Best Conversation Starters

    1 Vincen 180 topics
    2 lheiannie07 65 topics
    3 Roland2rule 49 topics
    4 ardz24 40 topics
    5 LUANDATO 16 topics
    See More Top Beat The GMAT Members...

    Most Active Experts

    1 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    146 posts
    2 image description Rich.C@EMPOWERgma...

    EMPOWERgmat

    103 posts
    3 image description GMATGuruNY

    The Princeton Review Teacher

    100 posts
    4 image description EconomistGMATTutor

    The Economist GMAT Tutor

    92 posts
    5 image description Jay@ManhattanReview

    Manhattan Review

    79 posts
    See More Top Beat The GMAT Experts