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Solution Y is 40 percent sugar by volume, and solution X is 20 percent
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28 Sep 2014, 05:01

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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?

Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent
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23 Oct 2014, 10:09

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In order to solve this problem you must be able to translate English into Algebra equations. Once you have written one or more algebraic equations it becomes a simple math problem to solve. You should look out for some keywords such as added, percent and so on while creating the equations. In this case, lets go step by step:

1. The problem states that there is 150 gallons of Y solution and X will be added to create a solution - this shows that the total number of gallons in the new solution will be \(150+x\). 2. The new solution is 25 percent sugar by volume. So, that becomes \(\frac{25}{100}* (150+x)\). 3. Solution Y is 40 percent sugar and we know that there are 150 gallons of Y solution. Solving for that gives \(\frac{40}{100} * 150\). This gives us 60 gallons of sugar by volume in Solution Y out of the 150 total gallons 4. Solution X is 20 percent sugar and this will be added to 60 gallons of Y. So the translation here is \(60 + \frac{20}{100}(x)\)

Now we can form the equation -> \(60 + \frac{20}{100}(x) = \frac{25}{100}(150+x)\)

Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent
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28 Sep 2014, 23:58

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chetan86 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?

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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent
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23 Oct 2014, 22:25

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chetan86 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) 75 (C) 150 (D) 240 (E) 450

Sol Y = 40% sugar---------------------Sol Mix = 25 % sugar-----------------Sol X = 20% sugar

Sol Y = +0.15 than Avg---------------Avg = 0.25------------------------------Sol X = -0.05 than avg

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?

I am trying to master concept of weighted average by scaling since I find it very time efficient and applicable in multiple scenarios.

Please help if my approach is correct (with help taken from this blog.)
_________________

It's the journey that brings us happiness not the destination.

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?

I am trying to master concept of weighted average by scaling since I find it very time efficient and applicable in multiple scenarios.

Please help if my approach is correct (with help taken from this blog.)

Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent
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28 Apr 2018, 04:30

chetan86 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) 75 (C) 150 (D) 240 (E) 450

Weighted Average Method :

Let's add G gallons of solution X

So we have, \(\frac{G}{150}\) = \(\frac{40 - 25}{25 - 20}\)

\(G = 150 * \frac{15}{5}\)

\(G = 450\)

Algebraic Method:

\(0.20X + 0.40Y = 0.25(X +Y)\)

\(0.20X + 0.40Y = 0.25X + 0.25Y\)

\(0.5X = 0.15Y\)

\(X = \frac{0.15*150}{0.5}\)

\(X = 450\)
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"Do not watch clock; Do what it does. KEEP GOING."

Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent
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03 Aug 2018, 12:15

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chetan86 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) 75 (C) 150 (D) 240 (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: 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

Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent
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23 Nov 2018, 07:36

GMATPrepNow wrote:

chetan86 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) 75 (C) 150 (D) 240 (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: 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

RELATED VIDEO

great explanation! this post deserves to be the most helpful expert reply in this thread

gmatclubot

Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent
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23 Nov 2018, 07:36