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

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Re: Two mixtures A and B contain milk and water in the ratios  [#permalink]

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New post 02 Aug 2016, 02:37
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



I tried two approaches.one in which i convereted 40% to fraction form and the second in which i converted 2/7 to % form.The later method yielded the correct OA.why???
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Re: Two mixtures A and B contain milk and water in the ratios  [#permalink]

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New post 02 Aug 2016, 05:34
How abt this approach.

of all the options provided only 122.5 is divisible by 7 ie. ratio of A is 2:5, so the mixture need to be in multiple of 7.
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Two mixtures A and B contain milk and water in the ratios  [#permalink]

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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:
Image
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:
Image

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

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:

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Last edited by GMATPrepNow on 16 Apr 2018, 11:57, edited 1 time in total.
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Re: Two mixtures A and B contain milk and water in the ratios  [#permalink]

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New post 23 Dec 2016, 08:03
2x+50/5x+40=4/6, find x, then don't get into decimals, approx 17.something then 2(17)+5(17)= approx 122
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Re: Two mixtures A and B contain milk and water in the ratios  [#permalink]

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New post 09 Nov 2017, 01:28
3
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]

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New post 09 Nov 2017, 10:23
1
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. Formula:

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


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 A, \(\frac{M}{W}=\frac{2}{5}\)
2 parts milk, 5 parts water, total parts = 7
So milk is \(\frac{2parts}{7parts}=\frac{2}{7}\)

B: \(\frac{M}{W}=\frac{5}{4}.\) Milk \(=\frac{5}{4+5}=\frac{5}{9}\)

Resultant mixture: 40% milk \(=\frac{2}{5}\)

We have 90 gallons for the volume of B. How much A?

2) Weighted average to find 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]

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New post 12 Nov 2017, 07:45
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]

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New post 12 Jun 2018, 01:44
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]

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New post 12 Jun 2018, 04:57
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]

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New post 12 Jun 2018, 12:05
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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 &nbs [#permalink] 12 Jun 2018, 12:05

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