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Mixture B is a 5:4 ratio mixture of Milk:Water. In other words, for every 5 gallons of milk in B there are 4 gallons of water. To find out the actual number of gallons of milk in X gallons of mixture B there are a couple ways to go about it:

If we have 5 gallons of milk there are 4 gallons of water. So milk = 5/(5+4) or 5/9 is milk. You add 4 and 5 together to get the total volume of the mixture (milk + water)
90 gallons of mixture * 5/9 milk = 50 gallons of milk.

OR

You can see that 5:4 add up to 9 and that 90 is a multiple of 9. If 5:4 = 9 and 9*10 = 90 and we have 90 gallons of the mixture just multiply the ratio by 10. 5*10:4*10 = 50:40 = 50 gallons of milk

Let's say we have milk:water in a 1:9 ratio. We have 15 gallons of the mixture, how many gallons of milk?

1+9=10 so 1 gallon of milk for every 10 gallons of mixture
15/10 = 1.5
1*1.5:9*1.5 multiply each side by (15/10 or 1.5)
1.5:13.5
1.5+13.5 = 15 and 1.5 gallons of it is milk.
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bmwhype2
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

When solving some mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.

First recognize that if mixture A has a milk to water ratio of 2:5, then the mixture is 2/7 milk.
Also recognize that if mixture B has a milk to water ratio of 5:4, then the mixture is 5/9 milk.

Start with 90 gallons of mixture B, which is 5/9 milk:

When we draw this with the ingredients separated, we see we have 50 gallons of milk in the mixture.

Next, we'll let x = the number of gallons of mixture A we need to add.
Since 2/7 of mixture A is milk, we know that (2/7)x = the volume of MILK in this mixture:


At this point, we can ADD the two solutions (PART BY PART) to get the following volumes:


Since the RESULTING mixture is 40% milk (i.e., 40/100 of the mixture is milk), we can write the following equation:
[50 + (2/7)x]/(90 + x) = 40/100
Simplify to get: [50 + (2/7)x]/(90 + x) = 2/5
Cross multiply to get: 5[50 + (2/7)x] = 2(90 + x)
Expand: 250 + (10/7)x = 180 + 2x
Subtract 180 from both sides to get: 70 + (10/7)x = 2x
Multiply both sides by 7 to get: 490 + 10x = 14x
Rearrange: 490 = 4x
Solve: x = 490/4 = 245/2 = 122.5

Answer: B

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bmwhype2
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
Another weighted average approach, expressed a bit differently.

Track on milk. We know the desired concentration of milk in the resultant mixture.

Milk is a fraction (or percentage or concentration) of all three mixtures of milk and water. The weighted average formula accounts for water by way of volume.

This formula is easy (concentration can be a percentage or a fraction):

\((Concentration_{A})(Vol_{A}) + (Concentration_{B})(Vol_{B})\)
\(=(Concentration_{A+B})(Vol_{A+B})\)


Let A = # of gallons of A (volume)

1) Use ratios and desired percentage to find the concentration of milk in A, B, and end mixture
(With ratios, remember to find \(\frac{part}{whole}\))

In mixture A, \(\frac{Milk}{Water}=\frac{2}{5}\)

2 parts milk, 5 parts water, total parts = 7

So in A, milk is \(\frac{2parts}{7parts}=\frac{2}{7}\)

In the second mixture, B, milk is what fraction?
\(\frac{M}{W}=\frac{5}{4}\)

B, concentration of milk \(=\frac{5}{4+5}=\frac{5}{9}\)

Resultant mixture, desired concentration =
40% milk \(=\frac{40}{100}=\frac{2}{5}\)

The volume of B is 90 gallons.
The volume of the resultant mixture is (A + B).
What is the volume of A?

2) Use weighted average to find the unknown volume of A (steps can be combined)

\(\frac{2}{7}A + \frac{5}{9}(90)=\frac{2}{5}(A+90)\)

\(\frac{2}{7}A + 50=\frac{2}{5}A +\\
\frac{2}{5}(90)\)

\(\frac{2}{7}A + 50=\frac{2}{5}A + 36\)

\(14=\frac{2}{5}A-\frac{2}{7}A\)

\(14 = \frac{4}{35}A\)

\(A = (14*\frac{35}{4})=122.5\) gallons of A

Answer B
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bmwhype2
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

Mixture A has a ratio of milk : water = 2x : 5x.

Mixture B has a ratio of milk : water = 5y : 4y.

Since there are 90 gallons of mixture B, we have:

milk : water = 50 : 40

We can now create the following equation to determine how many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk:

(2x + 50)/(7x + 90) = 40/100

(2x + 50)/(7x + 90) = 2/5

5(2x + 50) = 2(7x + 90)

10x + 250 = 14x + 180

70 = 4x

x = 17.5

So, we need 7(17.5) = 122.5 gallons of A.

Answer: B
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Hi,
I solved this in 15 seconds by just seeing that 122,5 is the only number that yields a "comfortable" terminating decimal (17,5) when divided by 7 (taking the ratio of 2:5). The other answers are also terminating decimals but in these type of GMAT questions they usually do not make you calculate with numbers that have more then 3 decimals.

Does that approach hold up in general? Bunuel VeritasPrepKarishma

Thanks a lot for the feedback!
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bmwhype2
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
An alternate approach is to use ALLIGATION.
Alligation can be performed only with percentages or fractions.

Step 1: Convert the ratios to FRACTIONS.
A:
Since M:W = 2:5, and 2+5=7, \(\frac{Milk}{Total}\) = \(\frac{2}{7}\).
B:
Since M:W = 5:4, and 5+4=9, \(\frac{Milk}{Total}\) = \(\frac{5}{9}\).
Mixture:
\(\frac{Milk}{Total}\)= \(\frac{2}{5}\).


Step 2: Put the fractions over a COMMON DENOMINATOR.

A = \(\frac{2}{7}\) = \(\frac{(2*9*5)}{(7*9*5)}\) = \(\frac{90}{315}\).
B = \(\frac{5}{9}\) = \(\frac{(5*7*5)}{(9*7*5)}\) = \(\frac{175}{315}\).
Mixture = \(\frac{2}{5}\) = \(\frac{(2*7*9)}{(5*7*9)}\) = \(\frac{126}{315}\).


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 ratio of A to B is equal to the RECIPROCAL of the distances in red.
A:B = 49:36.


Since \(\frac{A}{B}\) = \(\frac{49}{36}\), and the actual volume of B=90, we get:
\(\frac{A}{90}\) = \(\frac{49}{36}\)
36A = 49*90
2A = 49*5
2A = 245
A = 122.5.

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bmwhype2
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

Given, Mixture A with Milk: water = 2 : 5 & Mixture B with Milk : water = 5 : 4

Let X be the Quantity of Mixture A , we have

Quantity of Milk in Mixture A = 2X/7

Given Quantity of Mixture B = 90 gallons

Quantity of Milk in Mixture B = 5*90/9 = 50 gallons

When Mixture A & B are mixed we get 40% milk.

hence we have, 2X/7 + 50 = 4/10* (X + 90)

Solving we get X = 122.5 gallons

Answer B.


Thanks,
GyM
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pkloeti
Hi,
I solved this in 15 seconds by just seeing that 122,5 is the only number that yields a "comfortable" terminating decimal (17,5) when divided by 7 (taking the ratio of 2:5). The other answers are also terminating decimals but in these type of GMAT questions they usually do not make you calculate with numbers that have more then 3 decimals.

Does that approach hold up in general? Bunuel VeritasPrepKarishma

Thanks a lot for the feedback!

I understand what you are saying and that is a valid point. Though these numbers are not very GMAT-like. If they have given 122.5 as the answer (presumably the calculations would involve decimals), I would worry about some other option being the answer with the intermediate steps having decimals.
Hence, with 15 secs on hand to make a quick guess and move on, your logic is great - but given 2 mins, I would actually solve the question.
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