vipulgoyal wrote:Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?
A. 144
B. 122.5
C. 105.10
D. 72
E. 134
The following approach is called ALLIGATION -- a great way to handle mixture problems.
For an explanation of alligation, check my 2 posts here:
https://www.beatthegmat.com/ratios-fract ... 15365.html
Note:
Alligation can be performed only with percentages or fractions.
In the problem above, the given ratios must be converted.
Step 1: Convert the ratios to FRACTIONS.
A:
Since milk:water = 2:5, and 2+5=7, milk/total = 2/7.
B:
Since milk:water = 5:4, and 5+4=9, milk/total = 5/9.
Mixture:
milk/total = 40% = 2/5.
Step 2: Put the fractions over a COMMON DENOMINATOR.
A = 2/7 = (2*5*9)/(5*7*9) = 90/(5*7*9).
B = 5/9 = (5*5*7)/(5*7*9) = 175/(5*7*9).
Mixture = 2/5 = (2*7*9)/(5*7*9) = 126/(5*7*9).
Step 3: Plot the 3 numerators on a number line, with the numerators for A and B on the ends and the numerator for the mixture in the middle.
A 90------------126------------175 B
Step 4: Calculate the distances between the numerators.
A 90-----
36-----126-----
49-----175 B
Step 5: Determine the ratio in the mixture.
The required ratio of A to B is the RECIPROCAL of the distances in red.
A/B = 49/36.
Since A/B = 49/36, and B=90 gallons, we get:
49/36 = A/90
49/2 = A/5
2A = 5*49
2A = 245
A = 122.5.
The correct answer is
B.
Last edited by
GMATGuruNY on Mon Oct 20, 2014 4:32 am, edited 1 time in total.
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