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Two mixtures A and B contain milk and water in the ratios

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bmwhype2
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Two mixtures A and B contain milk and water in the ratios [#permalink] New post   02 Jan 2008, 07:16
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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
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Re: rates - milk and water [#permalink] New post   03 Jan 2008, 12:07
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bmwhype2 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?


How can I do this without using the ratio approach?


As you may have seen in a previous post, I like to shy away from algebra if there is some other framework I can use instead. For mixtures and solutions, I simply draw this box, fill in what is given, and then make everything work out to the bottom right corner.

The rows represent the solutions being mixed. The first solution first, the added solution second, and the resulting solution third. The columns represent the parts of the solution. The first (left) column is the total solution, the middle represents the percentage of the "stuff" (in this case, milk, but could be acid, salt, alcohol, etc), and the third column is the total "stuff" in the solution.

We multiply the first column by the second column to get the third column. When we mix them together, the total amounts mix, as do the amounts of the "stuff", so we add down. Note that the middle column doesn't add - it's just there to get us from the left to the right.

So here is how I applied it to this problem. We know how much of B there is, and we are asked how much of A is mixed with it to get 40% milk all together. Follow the chart, and you see that the bottom row multiplies to the right box, and the right column adds to the right box, so we have the right box from two directions. That's how we solve.

50 + 2/7x = .4(90 + x)
x = 122.5

Check out some earlier posts I made on this topic.

https://www.gmatclub.com/forum/7-t8140
https://www.gmatclub.com/forum/7-t10946
https://www.gmatclub.com/forum/7-t40699
https://www.gmatclub.com/forum/7-t48296


Ian
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Re: rates - milk and water [#permalink] New post   02 Jan 2008, 09:22
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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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Two mixtures A and B contain milk and water in the ratios [#permalink] New post   Updated on: 04 Dec 2020, 08:03
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bmwhype2 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


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

RELATED VIDEO

Originally posted by BrentGMATPrepNow on 06 Sep 2016, 15:10.
Last edited by BrentGMATPrepNow on 04 Dec 2020, 08:03, edited 2 times in total.
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Re: Two mixtures A and B contain milk and water in the ratios [#permalink] New post   09 Nov 2017, 02:28
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bmwhype2 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



Responding to a pm:

Here is the weighted average method of solving it:
Concentration of milk in the first mixture = 2/7 = 18/63 = 90/315
Concentration of milk in the second mixture = 5/9 = 35/63 = 175/315
Concentration of milk in the resultant mixture = 2/5 = 126/315

w1/w2 = (A2 - Aavg)/(Aavg - A1)

w1/w2 = (175/315 - 126/315) / (126/315 - 90/315) = 49 / 36

So 36 gallons of mixture B needs 49 gallons of A
90 gallons of B will need (49/36)*90 = 122.5 gallons

The numbers in the question are hard to work with. In most GMAT questions, the numbers fall easily in place. It is the concept that you have to focus on.
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Two mixtures A and B contain milk and water in the ratios [#permalink] New post   09 Nov 2017, 11:23
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bmwhype2 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

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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Re: Two mixtures A and B contain milk and water in the ratios [#permalink] New post   12 Nov 2017, 08:45
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bmwhype2 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


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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Two mixtures A and B contain milk and water in the ratios [#permalink] New post   12 Jun 2018, 02:44
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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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Re: Two mixtures A and B contain milk and water in the ratios [#permalink] New post   12 Jun 2018, 05:57
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bmwhype2 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

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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Re: Two mixtures A and B contain milk and water in the ratios [#permalink] New post   12 Jun 2018, 13:05
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bmwhype2 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


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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Re: Two mixtures A and B contain milk and water in the ratios [#permalink] New post   24 Jun 2019, 03:25
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pkloeti wrote:
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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Re: Two mixtures A and B contain milk and water in the ratios [#permalink] New post   22 Nov 2023, 09:08
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