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

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Re: rates - milk and water  [#permalink]

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New post 18 Aug 2009, 10:38
I found Ian7777's all posts very clear, detail and useful. More than that his attitude is humble and highly positive. I always enjoy reading his posts and learned a lot from him.

Thanks Ian.


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ian7777 wrote:
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.

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


Ian

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Re: rates - milk and water  [#permalink]

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New post 25 Aug 2009, 02:57
Milk Water Total

A 2 5 x

B 5 4 90

----------------------------------------------
Total 2 3 90+x
-----------------------------------------------

Equation 2/5(90+x) = 50 + 2/7x...sove for x....
x = 122.5

sorry...tried but unable to edit....tab is not working.. :evil:
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Two mixtures A and B contain milk and water in the ratios  [#permalink]

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New post Updated on: 12 Jun 2018, 20:33
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

let x=gallons of A to be mixed
(2x/7)+(5*90)/9=.4(x+90)
x=122.5 gallons
B

Originally posted by gracie on 15 Jun 2016, 12:44.
Last edited by gracie on 12 Jun 2018, 20:33, edited 2 times 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 02 Aug 2016, 03:37
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, 06:34
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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New post 23 Dec 2016, 09:03
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 12 Nov 2017, 08:45
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, 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]

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New post 12 Jun 2018, 05:57
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, 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,
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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 Jun 2019, 00:37
GMATPrepNow
I'm not able to figure this question out. I'm not sure how to use a ratio of solution A to find an answer. Could you help me solve this one?

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

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New post 24 Jun 2019, 03:25
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]

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New post 04 Aug 2019, 10:54
i tried like this:

Out of all the given options for amount of A, only one option is completely divisible by 7 i.e., B) 122.5
verified the answer by considering A=122.5 gallons.
So total mixture =90+122.5=212.5
(2/7)*(122.5)+(5/9)*(90)=(4/10)*(212.5)
35+50=4*21.25
85=85

so option B) is the answer
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Re: Two mixtures A and B contain milk and water in the ratios   [#permalink] 04 Aug 2019, 10:54

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