mixtures problem!

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mixtures problem!

by chaitanya.bhansali » Fri Jul 18, 2014 8:50 am
Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

1.) 37.5
2.) 50
3) 62.5
4) 300
5) 450

Whats wrong with the method I used, attached herewith??
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by GMATinsight » Fri Jul 18, 2014 9:01 am
chaitanya.bhansali wrote:Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

1.) 37.5
2.) 50
3) 62.5
4) 300
5) 450

Whats wrong with the method I used, attached herewith??
Image
Hi Chaitanya the mistake is
Question's phrase How many gallons of solution X must be added to 150 gallons of solution Y

Therefore we should take X = 150

whereas you have taken 4X=150

if X = 150 then 3X = 450
Answer Option E
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by chaitanya.bhansali » Fri Jul 18, 2014 10:55 am
Aah! Thankyou, sir! That was quite a stupid mistake there! I didn't read the question carefully!

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by [email protected] » Fri Jul 18, 2014 11:00 am
Hi chaitanya.bhansali,

Mixture questions can also be answered with the Part/Whole formula:

In this case, we're dealing with Sugar/Total

Solution X = 20% sugar
Solution Y = 40% sugar

We're told to mix a certain amount of Solution X with 150 gallons of Solution Y to get a 25% sugar solution:

X = number of gallons of Solution X

[(.2)(X) + (.4)(150)] / (X + 150) = .25

Now, it's just algebra steps:

.2X + 60 = .25X + 37.5

22.5 = .05X

450 = X

Final Answer: E

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by GMATGuruNY » Fri Jul 18, 2014 1:51 pm
chaitanya.bhansali wrote:Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

1.) 37.5
2.) 50
3) 62.5
4) 300
5) 450
Since Y is 40% sugar, the amount of sugar in 150 gallons of Y = 40% of 150 = 60 gallons.
An alternate approach is to PLUG IN THE ANSWERS, which represent the amount of X that must be added.
When the correct answer is plugged in, sugar/total = 25% = 1/4.

Answer choice D: 300 gallons
Since X is 20% sugar, the amount of sugar in 300 gallons of X = 20% of 300 = 60 gallons.
(sugar in X + sugar in Y)/(total X + total Y) = (60+60)/(300+150) = 120/450 = 4/15.
Since 4/15 is greater than 1/4, the proportion of sugar must DECREASE.
X has a lower sugar percentage than does Y.
Thus, to decrease the percentage of sugar in the resulting mixture, MORE X must be added.

The correct answer is E.
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by Brent@GMATPrepNow » Fri Jul 18, 2014 2:12 pm
chaitanya.bhansali wrote:Solution Y is 40 percent sugar by volume, and solution X is 20 percent sugar by volume. How many gallons of solution X must be added to 150 gallons of solution Y to create a solution that is 25 percent sugar by volume?

A) 37.5
B) 50
C) 62.5
D) 300
E) 450
When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.

Start with 150 gallons of solution that is 40% sugar:
Image
When we draw this with the ingredients separated, we see we have 60 gallons of sugar in the mixture.

Next, we'll let x = the number of gallons of solution X we need to add.
Since 20% of the solution X is sugar, we know that 0.2x = the volume of sugar in this solution:
Image
At this point, we can ADD the two solutions (PART BY PART) to get the following volumes:
Image
Since the resulting solution is 25% sugar (i.e., 25/100 of the solution is sugar), we can write the following equation:
(60 + 0.2x)/(150 + x) = 25/100
Simplify to get: (60 + 0.2x)/(150 + x) = 1/4
Cross multiply to get: 4(60 + 0.2x) = 1(150 + x)
Expand: 240 + 0.8x = 150 + x
Rearrange: 90 = 0.2x
Solve: x = 450

Answer: E

Cheers,
Brent

Here are some additional mixture questions to practice with:
https://www.beatthegmat.com/liters-of-mi ... 71387.html
https://www.beatthegmat.com/percentage-m ... 68631.html
https://www.beatthegmat.com/rodrick-mixe ... 70387.html
https://www.beatthegmat.com/mixure-probl ... 61767.html
https://www.beatthegmat.com/mixture-rati ... 91643.html
Brent Hanneson - Creator of GMATPrepNow.com
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by mcdesty » Thu Jul 24, 2014 7:56 pm
This is how I would do this just by looking at it. You have to know the methods above before trying this.
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