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# The 20% acid liquid solution is produced by adding a gallons of 10% ac

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6663
GMAT 1: 760 Q51 V42
GPA: 3.82
The 20% acid liquid solution is produced by adding a gallons of 10% ac  [#permalink]

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26 Feb 2018, 01:36
1
00:00

Difficulty:

25% (medium)

Question Stats:

82% (02:05) correct 18% (02:33) wrong based on 73 sessions

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[GMAT math practice question]

The 20% acid liquid solution is produced by adding a gallons of 10% acid liquid solution to b gallons of 50% acid liquid solution. To get 10 gallons of 20% acid liquid solution, how many gallons of 50% acid liquid solution are needed?

A. 1
B. 1.5
C. 2
D. 2.5
E. 3

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Intern Joined: 20 Dec 2017 Posts: 36 Location: Singapore Re: The 20% acid liquid solution is produced by adding a gallons of 10% ac [#permalink] ### Show Tags 26 Feb 2018, 02:03 For most weighted average questions, there are usually 3 key values: A, B & C A = Highest Value B = Desired Average C = Lowest Value we can use the following formula to get the ratio of "pull": $$\frac{A - B}{B - C}$$ For the question, A = 50 B = 20 C = 10 $$\frac{A - B}{B - C} = \frac{50-20}{20-10} = \frac{30}{10} = \frac{3}{1}$$ Since B is closer to C than A, the ratio of A:C is 1:3 Gallons of A required = $$\frac{10}{4}*1 = 2.5$$ Ans: D PS Forum Moderator Joined: 25 Feb 2013 Posts: 1217 Location: India GPA: 3.82 Re: The 20% acid liquid solution is produced by adding a gallons of 10% ac [#permalink] ### Show Tags 26 Feb 2018, 08:55 1 MathRevolution wrote: [GMAT math practice question] The 20% acid liquid solution is produced by adding a gallons of 10% acid liquid solution to b gallons of 50% acid liquid solution. To get 10 gallons of 20% acid liquid solution, how many gallons of 50% acid liquid solution are needed? A. 1 B. 1.5 C. 2 D. 2.5 E. 3 Given $$0.1a+0.5b=0.2(a+b)$$. Resulting mixture's volume will be summation of volumes of individual contents $$=>3b=a => a:b=3:1$$ New mixture has volume of 10 gallons, Hence $$b= \frac{1}{(1+3)}*10=2.5$$ Option D VP Joined: 07 Dec 2014 Posts: 1130 Re: The 20% acid liquid solution is produced by adding a gallons of 10% ac [#permalink] ### Show Tags 26 Feb 2018, 17:08 MathRevolution wrote: [GMAT math practice question] The 20% acid liquid solution is produced by adding a gallons of 10% acid liquid solution to b gallons of 50% acid liquid solution. To get 10 gallons of 20% acid liquid solution, how many gallons of 50% acid liquid solution are needed? A. 1 B. 1.5 C. 2 D. 2.5 E. 3 .1(10-b)+.5b=.2*10 b=2.5 gallons D Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6663 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The 20% acid liquid solution is produced by adding a gallons of 10% ac [#permalink] ### Show Tags 28 Feb 2018, 00:29 => We have $$a + b = 10.$$ $$\frac{(0.1a + 0.5b)}{10} = 0.2$$ $$⇔ 0.1a + 0.5b = 2$$ $$⇔ 0.1(10-b) + 0.5b = 2$$ $$⇔ 1 – 0.1b + 0.5b = 2$$ $$⇔ 0.4b = 1$$ $$⇔ b = 2.5$$ Therefore, D is the answer. Answer: D _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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The 20% acid liquid solution is produced by adding a gallons of 10% ac  [#permalink]

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28 Feb 2018, 09:59
MathRevolution wrote:
[GMAT math practice question]

The 20% acid liquid solution is produced by adding a gallons of 10% acid liquid solution to b gallons of 50% acid liquid solution. To get 10 gallons of 20% acid liquid solution, how many gallons of 50% acid liquid solution are needed?

A. 1
B. 1.5
C. 2
D. 2.5
E. 3

To find the amount of one solution in a resultant mixture, I use this weighted average formula:
$$(Concen_{A})(Vol_{A}) + (Concen_{B})(Vol_{B}) = (Concen_{A+B})(Vol_{A+B})$$

Let A = volume of the solution with 10% acid
Let B = volume of the solution with 50% acid

A + B = resultant mixture
A + B = 10 (gallons in volume)
A = (10 - B)
Desired concentration of resultant solution: 20%

1) the equation without substitution
$$(.10)(A) + (.50)(B) = .20(A + B)$$

2) Substitute (10-B) for A, and 10 for (A+B):

$$.10(10-B) + .50B = .20(10)$$
$$1 - .10B + .50B = 2$$
$$.40B = 1$$
$$B =\frac{1}{.4}=\frac{10}{4}=2.5$$

The 20% acid liquid solution is produced by adding a gallons of 10% ac &nbs [#permalink] 28 Feb 2018, 09:59
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