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

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

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

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06 Sep 2016, 14:10
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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.

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

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

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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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09 Nov 2017, 01:28
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Expert's post
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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09 Nov 2017, 10:23
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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. 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

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

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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.

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Re: Two mixtures A and B contain milk and water in the ratios   [#permalink] 12 Nov 2017, 07:45

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