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??
mixtures problem!
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- chaitanya.bhansali
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Hi Chaitanya the mistake ischaitanya.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??
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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- chaitanya.bhansali
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Aah! Thankyou, sir! That was quite a stupid mistake there! I didn't read the question carefully!
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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Since Y is 40% sugar, the amount of sugar in 150 gallons of Y = 40% of 150 = 60 gallons.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
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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When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.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
Start with 150 gallons of solution that is 40% sugar:
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:
At this point, we can ADD the two solutions (PART BY PART) to get the following volumes:
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
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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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