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A salt water solution initially contains 20% salt and 80% water. Water 15 Jan 2017, 05:28
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E
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95% (hard)

Question Stats:

37% (02:47) correct 63% (02:51) wrong based on 19 sessions

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A salt water solution initially contains 20% salt and 80% water. Water evaporates at a constant rate, changing the solution to 40% salt after one hour. What was the approximate percentage decrease in water per minute?

A. 0.1%
B. 0.2%
C. 0.3%
D. 0.4%
E. 0.5%

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E



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E
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A salt water solution initially contains 20% salt and 80% water. Water 15 Jan 2017, 15:46
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winionhi
A salt water solution initially contains 20% salt and 80% water. Water evaporates at a constant rate, changing the solution to 40% salt after one hour. What was the approximate percentage decrease in water per minute?

A. 0.1%
B. 0.2%
C. 0.3%
D. 0.4%
E. 0.5%

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E
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(.8-.6)/.8=.25
.25/60=.004
.4%
D
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A salt water solution initially contains 20% salt and 80% water. Water 11 Feb 2017, 06:04
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Shouldn't it be D
How is the answer E? Some please explain
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A salt water solution initially contains 20% salt and 80% water. Water 11 Feb 2017, 09:34
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winionhi
A salt water solution initially contains 20% salt and 80% water. Water evaporates at a constant rate, changing the solution to 40% salt after one hour. What was the approximate percentage decrease in water per minute?

A. 0.1%
B. 0.2%
C. 0.3%
D. 0.4%
E. 0.5%

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E
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D is ofcourse not the answer. The solution above doesn't take into consideration that the total value does not remain same...

Let salt be 20 ml and water 80 ml..
Now 20 ml has become 40%, so 0.4*total =20..... Total=50 ml..
This means water is 50-20=30....
Decrease % in water = \(\frac{80-30}{80}*100\)=5000/80=62.5% in one hour, so 62.5/60 in 1 min.., nearly 1%

So none of the answers match...
If I take the answer as 0.5%, total in an hour will be 0.5*60=30%
30% of 80 is 24, so remaining water is 80-24=56..
So salt % = 20/(56+20)=20/76, nearly 25%..
So not 40%..
Q is flawed..

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A salt water solution initially contains 20% salt and 80% water. Water 27 Jun 2023, 09:44
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