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A 60 ml mixture of juice concentrate and water contained 40 percent ju
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05 Dec 2023, 01:41
05 Dec 2023, 01:41
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Question Stats:
47% (02:18) correct
52% (02:15) wrong based on 42 sessions
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A 60 ml mixture of juice concentrate and water contained 40 percent juice. A part of the mixture was removed and replaced with an equal quantity of water. The resulting mixture contained 10 percent juice.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Volume of the mixture that was removed |
45 ml |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: A 60 ml mixture of juice concentrate and water contained 40 percent ju
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06 Dec 2023, 10:49
06 Dec 2023, 10:49
2
We can interpret this as (the amount of juice in the original mixture) * (volume of the original mixture mixed into new mixture) = the amount of juice in the new mixture. Also, the amount of juice in either mixture = concentration*total volume.
\((60-v)(0.4) = (0.1)(60)\)
\(24-0.4v = 6\)
\(v = 45\)
\((60-v)(0.4) = (0.1)(60)\)
\(24-0.4v = 6\)
\(v = 45\)
Re: A 60 ml mixture of juice concentrate and water contained 40 percent ju
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10 Mar 2026, 08:40
10 Mar 2026, 08:40
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