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# A tank holds x gallons of a saltwater solution that is 20%

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Manager
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A tank holds x gallons of a saltwater solution that is 20% [#permalink]

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22 Feb 2008, 16:33
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A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5
B. 75
C. 100
D. 150
E. 175

Can you show me how to solve this without backsolving, ie the real algabraic way?
[Reveal] Spoiler: OA
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22 Feb 2008, 18:22
neelesh wrote:
Is it e. 175 ?

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.
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22 Feb 2008, 19:13
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Manbehindthecurtain wrote:
neelesh wrote:
Is it e. 175 ?

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water.
Total = x
Out of which Salt = 0.2x and water = 0.8x
Now, 1/4 water evaporates and all salt remains.
So what remains is 0.2x salt and 0.6x water.
Now 20 gallons salt is added and 10 gallons of water is added.
So salt now becomes -> (0.2x + 20) and water --> (0.6x+10)
Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water.
So (0.2x+20) = (0.6x+10)/2,
Solving, x = 150.
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22 Feb 2008, 19:38
rgajare14 wrote:
So (0.2x+20) = (0.6x+10)/2, Solving, x = 150.

This is what I could not establish and started back solving from the answers and still got it wrong...

Good one.
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22 Feb 2008, 20:15
good explanation gajare.
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22 Feb 2008, 20:21
rgajare14 wrote:
Manbehindthecurtain wrote:
neelesh wrote:
Is it e. 175 ?

Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water.
Total = x
Out of which Salt = 0.2x and water = 0.8x
Now, 1/4 water evaporates and all salt remains.
So what remains is 0.2x salt and 0.6x water.
Now 20 gallons salt is added and 10 gallons of water is added.
So salt now becomes -> (0.2x + 20) and water --> (0.6x+10)
Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water.
So (0.2x+20) = (0.6x+10)/2,
Solving, x = 150.

Outstanding response. thank you. Can you describe how you started unpeeling this? I found the question wording really confusing.

Also, what level difficulty would you say this is?
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22 Feb 2008, 20:53
Quote:
Nope, 150. I can only get it by following PR's backsolving explanation. I hate that.

Original mixture has 20% salt and 80% water.
Total = x
Out of which Salt = 0.2x and water = 0.8x
Now, 1/4 water evaporates and all salt remains.
So what remains is 0.2x salt and 0.6x water.
Now 20 gallons salt is added and 10 gallons of water is added.
So salt now becomes -> (0.2x + 20) and water --> (0.6x+10)
Amount of salt is 33.33% of total. So amount of water is 66.66%. So salt is half of the volume of water.
So (0.2x+20) = (0.6x+10)/2,
Solving, x = 150.

Quote:
Outstanding response. thank you. Can you describe how you started unpeeling this? I found the question wording really confusing.

Also, what level difficulty would you say this is?

Thanks. Actually , I did not solve it backwards. I solved it as per the flow given in the problem question. Honestly, I dont think this was more than a 650 level question.
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Re: A tank holds x gallons of a saltwater solution that is 20% [#permalink]

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15 Feb 2014, 17:40
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Re: A tank holds x gallons of a saltwater solution that is 20% [#permalink]

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16 Feb 2014, 23:26
X gallons @20% Salt (S) & 80% Water (W)

By Gallon:
x/5 = S
4x/5 = W

x/5 + 20= S
3x/5 + 10= W

Using Salt @33.3% to solve for x:
(4x/5 + 30)/3 = x/5 + 20
x=150
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Re: A tank holds x gallons of a saltwater solution that is 20% [#permalink]

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18 Apr 2014, 14:01
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Understand the logic of the flow of question.

Consider X initial solution.

0.2X = Salt
0.8X = Water

1/4(0.8X)= Evaporated Leaving 0.6X=water

0.6X+10=Water
0.2X+20=Salt

Salt concentration now becomes= 1/3 of the Solution

(0.2X+20)/(0.8X+30) = 1/3

X=150
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A tank holds x gallons of a saltwater solution that is 20% [#permalink]

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07 Nov 2015, 09:57
x is reduced by water evaporation by (1/4)(4/5)=1/5
x/5+20=(1/3)(4x/5+30)
x/5+20=4x/15+10
3x+300=4x+150
x=150 gallons
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Re: A tank holds x gallons of a saltwater solution that is 20% [#permalink]

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19 Nov 2015, 19:17
I did it a different way but still got the right answer:
Salt + Water = Xgallons
(1/5) + (4/5) = (5/5)

water left after evaporation: (4/5)*(3/4) = (3/5) then (3/5)(5/5) = (3/4) of total is water
Amount of salt in new mix = (1/4)

New mix:
10 gals of Water plus 20 Gals of salt added = 33% Salt to 66% Water = 1:2

Amount of water plus 10 gals / amount of salt plus 20 gals = 1/2
(3/4)X + 10 gals
------------- = (1/2)
(1/4)X+20 Gals

Cross multiply:
(6/4)X +20 = (1/4)X+20
X=4/5

So X (the old mix) is (4/5) of the combined old plus new mix so...

30gals = 1/5 of X
150gals = X
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Re: A tank holds x gallons of a saltwater solution that is 20% [#permalink]

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05 Dec 2015, 12:07
oh yeah...
0.2x+20 = 1/3*(0.8x+30)

solving this, we get x = 150.
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Re: A tank holds x gallons of a saltwater solution that is 20% [#permalink]

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27 Jun 2016, 07:12
Manbehindthecurtain wrote:
A tank holds x gallons of a saltwater solution that is 20% salt by volume. One Fourth of the water is evaporated, leaving all of the salt. When 10 Gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3 % salt by volume. What is the value of x?

A. 37.5
B. 75
C. 100
D. 150
E. 175

Can you show me how to solve this without backsolving, ie the real algabraic way?

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Re: A tank holds x gallons of a saltwater solution that is 20%   [#permalink] 27 Jun 2016, 07:12
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