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A tank holds x gallons of a saltwater solution that is 20% 03 Mar 2021, 12:19
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SagarV
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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

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


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 ?
Show more

Response:

The question tells us that one fourth of the water evaporated. If you prefer to work with decimals (or percents), 1/4 is equivalent to 0.25 (or 25%). So, you should have calculated 75% of 0.8x (or multiplied 0.8x by 0.75). With the correct numbers, it will work.
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A tank holds x gallons of a saltwater solution that is 20% 03 Mar 2021, 15:53
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Tank holds x gallon of water out of which
Salt - 20X / 100
Water - 80x/ 100 ; After evaporation - 60x/100

After 20 gallon salt ad 30 gallon water added, total -
Salt - 20x/100 + 20
Water - 60x/100 + 30

Now 33 1/3 i.e. 100/3 = (salt / total ) * 100
1/3 = (20x/100 + 20) / (60x/100 + 30)

Solve for x and we can easily get x =150
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A tank holds x gallons of a saltwater solution that is 20% 09 Apr 2021, 09:48
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SagarV
ScottTargetTestPrep
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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

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


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 ?
Show more

Response:

Simply because “one fourth of the water” is not 20% of the water, but instead 25% of the water. If you perform the same operations with the correct percent of evaporated water, you’ll get the same expression as I did, which is 0.75 * 0.8x = 0.6x.
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A tank holds x gallons of a saltwater solution that is 20% 18 Feb 2025, 07:18
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Initial volume of salt in the saltwater solution = \(\frac{20}{100}x\) = \(0.2x\)

Initial volume of water in the saltwater solution = \(\frac{80}{100}x\) = \(0.8x\)

If \(1/4\) of the water is evaporated, then \(1 - 1/4 = 3/4\) is remaining.

Remaining water = \(\frac{3}{4}(\frac{80}{100}x)\) = \(\frac{3}{5}x\) = \(0.6x\)

After adding 10 gallons of water and 20 gallons of salt to the solution ->

New volume of salt = \(0.2x + 20\)
New volume of water = \(0.6x + 10\)

Total volume of the new salt water solution = \((0.2x + 20) + (0.6x + 10) = 0.8x + 30\)

We know that the new solution has a salt concentration of \(33 \frac{1}{3}\)% = \(33.3333......\)%
\(33.3333......\)% = \(\frac{33.33333.....}{100} = 0.33333...... = \frac{1}{3}\)

So, \(1/3\) volume of the new saltwater solution is equal to the volume of salt we calculated after adding 20 gallons of salt to the initial solution.

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

\(3(0.2x + 20) = 0.8x + 30\)

\(0.6x + 60 = 0.8x + 30\)

\(0.6x - 0.8x = 30 - 60\)

\(-0.2x = -30\)

\(x = \frac{30}{0.2}\)

\(x = \frac{300}{2}\)

\(x = 150\)

Correct option : D
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