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

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

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New post 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?
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Re: Salt Storage Tank  [#permalink]

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New post 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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Re: Salt Storage Tank  [#permalink]

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New post 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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Re: Salt Storage Tank  [#permalink]

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New post 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. :wall
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Re: Salt Storage Tank  [#permalink]

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New post 22 Feb 2008, 20:15
good explanation gajare.
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Re: Salt Storage Tank  [#permalink]

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New post 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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Re: Salt Storage Tank  [#permalink]

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

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

Mixture composition adjusted for 1/4 W evaporation and additional S/W gallons:
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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New post 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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New post Updated on: 23 Sep 2017, 20:25
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

after evaporation, x-(.25)(.8)x=.8x gallons
salt now=25% of .8x gallons
water now=75% of .8x gallons
(.25)(.8x)+20=(.8x+30)/3
x=150 gallons
D

Originally posted by gracie on 07 Nov 2015, 09:57.
Last edited by gracie on 23 Sep 2017, 20:25, edited 1 time in total.
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Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

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New post 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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New post 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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New post 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?


Got the right answer but the wording is super bad!
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Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

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New post 06 Sep 2017, 01:15
1
x gallon was solution = 0.2x salt and 0.8x water, 0.25% evaporated = 0.6x water remains
new solution is 0.2x salt and 0.6x water
20 gallon salt is added, new salt = 20+0.2x
10 gallon water added , new water = 10+0.6x
total solution = 30+0.8x

new solution is 33.33% salt by vol = 1/3

\(\frac{(20+0.2x)}{(30+0.8x)}\) = \(\frac{1}{3}\)
x=150 Answer

D is the answer
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Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

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New post 06 Sep 2017, 02:46
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?


It can be easily solved by front solving.

Salt Water Total
0.2x 0.8x x
After Evaporation
0.2x 0.6x 0.8x
After adding 10 gallons of water and 20 gallons of salt
20+0.2x 10+0.6x 30+0.8x

20+0.2x )/ (30+0.8x = 1/3
x = 150

Very Simple . Will take around 1 minute 40 seconds
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Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

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New post 11 Sep 2017, 11:09
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


We are given that a tank holds x gallons of a saltwater solution that is 20% salt by volume. Thus, the solution contains 0.2x salt and 0.8x water. After 1/4 of the water is evaporated, there is 0.6x water left.

We are also given that when 10 gallons of water and 20 gallons of salt are added, the resulting mixture is 33 1/3% (or 1/3) salt by volume. Thus:

(0.2x + 20)/(0.2x + 0.6x + 10 + 20) = 1/3

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

Multiplying 10/10 on the left side of the equation, we have:

(2x + 200)/(8x + 300) = 1/3

8x + 300 = 6x + 600

2x = 300

x = 150

Answer: D
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Re: A tank holds x gallons of a saltwater solution that is 20%  [#permalink]

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New post 17 Sep 2017, 06:32
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?

Initially the solution contains 0.8x of water and 0.2x of salt. 1/4th of the water is evaporated which is (1/4*0.8x) = x/5. Hence, 1-x/5= 4x/5 is the remaining content of solution.
20 gallons of salt is added , hence, new salt content in solution= 0.2x+20
Total content = 4x/5+30

0.2x+20/4x/5+30 = 1/3
Solving for x, we get x=150. (option D)
Re: A tank holds x gallons of a saltwater solution that is 20% &nbs [#permalink] 17 Sep 2017, 06:32
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