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Solution X is 15% salt and solution Y is 50% salt.

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Solution X is 15% salt and solution Y is 50% salt. [#permalink]

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Solution X is 15% salt and solution Y is 50% salt. If the two solutions are combined to form 7 gallons of a 30% solution, how many gallons of solution X must be used?

(A) 2

(B) 2.5

(C) 3

(D) 4

(E) 4.2
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Re: Solution X is 15% salt and solution Y is 50% salt. [#permalink]

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New post 14 Mar 2017, 05:03
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vikasp99 wrote:
Solution X is 15% salt and solution Y is 50% salt. If the two solutions are combined to form 7 gallons of a 30% solution, how many gallons of solution X must be used?

(A) 2

(B) 2.5

(C) 3

(D) 4

(E) 4.2



By Weighted Average method, we can find the ratio of X and Y..
\(\frac{X}{Y}=\frac{50-30}{30-15}=4/3\)..
So in 7 litres x is 4 and y is 3..

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Re: Solution X is 15% salt and solution Y is 50% salt. [#permalink]

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New post 14 Mar 2017, 17:00
vikasp99 wrote:
Solution X is 15% salt and solution Y is 50% salt. If the two solutions are combined to form 7 gallons of a 30% solution, how many gallons of solution X must be used?

(A) 2

(B) 2.5

(C) 3

(D) 4

(E) 4.2


let x=gallons of solution X to be used
.15x+.5(7-x)=.3*7
x=4
E
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Re: Solution X is 15% salt and solution Y is 50% salt. [#permalink]

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vikasp99 wrote:
Solution X is 15% salt and solution Y is 50% salt. If the two solutions are combined to form 7 gallons of a 30% solution, how many gallons of solution X must be used?

(A) 2

(B) 2.5

(C) 3

(D) 4

(E) 4.2


We can let the number of gallons of solution in solution X = x and the number of gallons of solution in solution Y = y. Thus:

0.15x + 0.5y = 0.3(x + y)

15x + 50y = 30x + 30y

20y = 15x

4y = 3x

We also know that x + y = 7, or y = 7 - x, and thus:

4(7 - x) = 3x

28 - 4x = 3x

28 = 7x

x = 4

Answer: D
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Re: Solution X is 15% salt and solution Y is 50% salt. [#permalink]

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New post 16 Mar 2017, 22:55
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Here are two posts. The first one discusses the weighted average formula
w1/w2 = (A2 - Aavg)/(Aavg - A1)
and the second one discusses its application in mixtures.

https://www.veritasprep.com/blog/2011/0 ... -averages/
https://www.veritasprep.com/blog/2011/0 ... -mixtures/
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Re: Solution X is 15% salt and solution Y is 50% salt. [#permalink]

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New post 17 Mar 2017, 01:16
[quote="vikasp99"]Solution X is 15% salt and solution Y is 50% salt. If the two solutions are combined to form 7 gallons of a 30% solution, how many gallons of solution X must be used?

(A) 2

(B) 2.5

(C) 3

(D) 4


let x amount be from soln X and y amount be from soln Y

x +y = 7
0.15x + 0.5y = 0.3(x+y)

solving 2 equation
x = 4, y = 3

Option D
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Re: Solution X is 15% salt and solution Y is 50% salt. [#permalink]

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New post 18 Mar 2017, 12:39
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Hi vikasp99,

This question can be solved rather easily by TESTing THE ANSWERS.

We're told to mix a certain amount of 15% salt solution with a certain amount of 50% salt solution to get a total solution that is 7 GALLONS and 30% salt. We're asked for the number of gallons of the 15% solution in that mixture.

To start, if we had the SAME amount of each solution, then the mixture would be (15 + 50)/2 = 65/2 = 32.5% salt. This is clearly TOO MUCH salt, so we need MORE of the 15% solution - meaning more than 3.5 of the 7 total gallons. Thus, the correct answer must be either D or E.

Let's TEST Answer D: 4 gallons

IF we have....
4 gallons of 15%
3 gallons of 50%
then the average would be....
[4(.15) + 3(.5)]/7 = (.6 + 1.5)/7 = 2.1/7 = 21/70 = 3/10 = 30%
This is an exact MATCH for what we were told, so this MUST be the answer.

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Re: Solution X is 15% salt and solution Y is 50% salt. [#permalink]

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New post 12 May 2018, 04:54
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Re: Solution X is 15% salt and solution Y is 50% salt.   [#permalink] 12 May 2018, 04:54
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