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

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.

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

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

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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good explanation gajare.

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.

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

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

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

oh yeah...
0.2x+20 = 1/3*(0.8x+30)

solving this, we get x = 150.

Got the right answer but the wording is super bad!

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

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

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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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)

Hi, please help..
I understand the approach and I was trying to solve it using the same approach. However,
When I calculated for 20% evaporated water I did something like this
0.8 * 0.8x = 0.64x
instead of
1/4 * 0.8x, which leads to 0.6 x

Why am I wrong ?

HeySagarV,

The highlighted portion is wrong in your solution. Remember, 1/4th of the water is being evaporated. Which means, you're left with 75% water. So either you multiple it by 0.75 or by 3/4.

Your approach is correct but you made the silly mistake of confusing 1/4th with 20% as we all do at some point . Hope this helps!

x gallons of water --> 0.2x salt & 0.8x water

\(\frac{1}{4}\) Only Water evaporates ... then new solution has....
0.2x salt & 0.8 - \(\frac{1}{4}\)*0.8 = 0.6x water

Hence new solution volume is 0.2x+0.6x = 0.8x = W1

Ci.Vi = Cf.Vf
0.2*x = Cf*0.8x
Cf=\(\frac{1}{4}\) = C1

Now to this new solution 20 gallon salt and 10 gallon water is added to make the new solution : 33.3% saltwater. This means C(Avg)=\(\frac{1}{3}\) in 30 gallon saltwater (W2)

This means added solution has \(\frac{20}{30}\) = \(\frac{2}{3}\) salt = C2

\(\frac{W1}{W2}\) = \(\frac{C2-C(Avg)}{C(Avg)-C1}\)
=> \(\frac{0.8x}{30}\) =\((\frac{1}{4}-\frac{1}{3})\)/\((\frac{1}{3}-\frac{2}{3})\)
\(\\
x=150\)

Hence D.
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