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

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

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New post 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?

(A) 37.5
(B) 75
(C) 150
(D) 240
(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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New post 28 Sep 2014, 23:58
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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



Answer = 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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New post 23 Oct 2014, 07:19
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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
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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

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New post 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 -> \(60 + \frac{20}{100}(x) = \frac{25}{100}(150+x)\)

This is solving equation with one variable.

\(60 + 0.2x = 37.5 + .25x\)

\(.05x = 22.5\)

\(x = 450\)

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

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New post 23 Oct 2014, 22:25
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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Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

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New post 07 Nov 2015, 13:04
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(.4)(150)+.2x=.25(150+x)
x=450 gallons
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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

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New post 19 Nov 2015, 16:47
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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New post 10 Mar 2018, 12:05
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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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New post 24 Apr 2018, 18:57
Attachment:
2504 sol.jpeg
2504 sol.jpeg [ 24.41 KiB | Viewed 727 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.

Please help if my approach is correct (with help taken from this blog.)
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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]

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New post 24 Apr 2018, 19:45
1
Your method is correct, and infact I find weighted average/mixture alligation method to be the best approach to solve such questions.


adkikani wrote:
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.

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


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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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New post 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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Re: Solution Y is 40 percent sugar by volume, and solution X is 20 percent   [#permalink] 28 Apr 2018, 04:30
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Solution Y is 40 percent sugar by volume, and solution X is 20 percent

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