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Solution Y is 40 percent sugar by volume, and solution X is 20 percent [#permalink]
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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? (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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28 Sep 2014, 23:58
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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23 Oct 2014, 07:19
Differential method:
402520
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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23 Oct 2014, 10:09
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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23 Oct 2014, 22:25
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% sugarSol Mix = 25 % sugarSol X = 20% sugar Sol Y = +0.15 than AvgAvg = 0.25Sol 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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07 Nov 2015, 13:04
(.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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19 Nov 2015, 16:47
I did a slightly different differential method.
402520 155
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, 12:05
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 Final Answer: GMAT assassins aren't born, they're made, Rich
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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, 18:57
Attachment:
2504 sol.jpeg [ 24.41 KiB  Viewed 727 times ]
Bunuel niks18 chetan2u gmatbusters VeritasPrepKarishmaQuote: 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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24 Apr 2018, 19:45
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 VeritasPrepKarishmaQuote: 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.) Posted from my mobile devicePosted 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, 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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