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# Solution Y is 40 percent sugar by volume, and solution X is 20 percent

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Manager
Joined: 17 Oct 2012
Posts: 65
Location: India
Concentration: Strategy, Finance
WE: Information Technology (Computer Software)
Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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28 Sep 2014, 04:01
10
00:00

Difficulty:

35% (medium)

Question Stats:

78% (01:25) correct 22% (01:59) wrong based on 514 sessions

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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?

(A) 37.5
(B) 75
(C) 150
(D) 240
(E) 450
Manager
Joined: 21 Jul 2014
Posts: 125
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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23 Oct 2014, 09:09
2
6
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 -> $$60 + \frac{20}{100}(x) = \frac{25}{100}(150+x)$$

This is solving equation with one variable.

$$60 + 0.2x = 37.5 + .25x$$

$$.05x = 22.5$$

$$x = 450$$

##### General Discussion
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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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28 Sep 2014, 22:58
2
2
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

Solution Y

Sugar ................ Water .............. Total

40 ............................ 60 .................. 100

60 .......................... 90 .................... 150

Say x litres of Solution x added; New equation would be

$$60+\frac{20x}{100}$$ .............. $$90+\frac{80x}{100}$$ ............ 150+x

New sugar concentration is 25%; setting up the equation

$$\frac{25}{100} (150+x) = 60 + \frac{20x}{100}$$

x = 450
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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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23 Oct 2014, 06:19
2
Differential method:

40----------25----20

15x=5y
x/y=5/15=1/3, so 1 part of 40% sol should be added 3 part of 20%.
So 150*3=450

E
Manager
Status: I am not a product of my circumstances. I am a product of my decisions
Joined: 20 Jan 2013
Posts: 117
Location: India
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GPA: 3.92
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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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23 Oct 2014, 21:25
1
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

0.15 Y - 0.05 X = 0
0.15 Y = 0.05 X

X/Y = 0.15/0.05 = 3:1

X:Y:Total = 3:1:4

If Y = 150 Gallons then by ratio X:Y = 3:1

X/150= 3/1
X = 150 * 3 = 450 Gallons

Ans E
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Joined: 07 Dec 2014
Posts: 1128
Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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07 Nov 2015, 12:04
1
(.4)(150)+.2x=.25(150+x)
x=450 gallons
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GPA: 3.3
Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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19 Nov 2015, 15:47
1
I did a slightly different differential method.

40----25----20
--15------5---

So Y has to be 5/20 of the new solution. Given Y = 150 gallons

150 = 5/20(total)
Total = 600
X = 600 - 150

Ans E) 450
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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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10 Mar 2018, 11:05
1
Hi All,

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

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Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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24 Apr 2018, 17:57
Attachment:

2504 sol.jpeg [ 24.41 KiB | Viewed 3198 times ]
Bunuel niks18 chetan2u gmatbusters VeritasPrepKarishma

Quote:
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.

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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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24 Apr 2018, 18:45
1
Your method is correct, and infact I find weighted average/mixture alligation method to be the best approach to solve such questions.

Attachment:
2504 sol.jpeg
Bunuel niks18 chetan2u gmatbusters VeritasPrepKarishma

Quote:
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.

Posted from my mobile device

Posted from my mobile device
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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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28 Apr 2018, 03:30
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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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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03 Aug 2018, 11:15
2
Top Contributor
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.

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

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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent  [#permalink]

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23 Nov 2018, 06:36
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.

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

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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent &nbs [#permalink] 23 Nov 2018, 06:36
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