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09 May 2016, 20:50
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
80% (01:37) correct
20%
(02:08)
wrong
based on 177
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After an ice began to melt out from the freezer, in the first hour lost 3/4, in the second hour lost 3/4 of its remaining. If after two hours, the volume is 0.5 cubic inches, what is the original volume of the cubic ice, in cubic inches?
A. 2.5
B. 3.0
C. 4.0
D. 6.5
E. 8.0
* A solution will be posted in two days.
A. 2.5
B. 3.0
C. 4.0
D. 6.5
E. 8.0
* A solution will be posted in two days.
09 May 2016, 21:30
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Let initial volume of ice be = x
Ice Remaining after 1 hour = x - 0.75x = 0.25x
Ice Remaining after 2 hour = (1/4)x - (3/4 * 1/4 * x) = (1/16)x
(1/16)x = 0.5
x = 8
Alternate solution: Try to backsolve.
Initial volume = 8
After one hour --> (1/4)8 = 2
After two hours --> (1/4)2 = 0.5
Answer: E
Ice Remaining after 1 hour = x - 0.75x = 0.25x
Ice Remaining after 2 hour = (1/4)x - (3/4 * 1/4 * x) = (1/16)x
(1/16)x = 0.5
x = 8
Alternate solution: Try to backsolve.
Initial volume = 8
After one hour --> (1/4)8 = 2
After two hours --> (1/4)2 = 0.5
Answer: E
09 May 2016, 21:32
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Let total volume be x initially.
At the end of 1st hr
Lost: \(\frac{3}{4} x\)
Remaining: \(x - \frac{3}{4} x = \frac{1}{4} x\)
At the end of 2nd hr
Lost: \(\frac{3}{4} * \frac{x}{4} = \frac{3x}{16}\)
Remaining: \(\frac{x}{4} - \frac{3x}{16} = \frac{x}{16}\)
Now, we are givn that remaining after 2 hours is 0.5 cubic inches
Therefore, \(\frac{x}{16} = 0.5\)
=> \(x = 8\) cubic inches
Option E
At the end of 1st hr
Lost: \(\frac{3}{4} x\)
Remaining: \(x - \frac{3}{4} x = \frac{1}{4} x\)
At the end of 2nd hr
Lost: \(\frac{3}{4} * \frac{x}{4} = \frac{3x}{16}\)
Remaining: \(\frac{x}{4} - \frac{3x}{16} = \frac{x}{16}\)
Now, we are givn that remaining after 2 hours is 0.5 cubic inches
Therefore, \(\frac{x}{16} = 0.5\)
=> \(x = 8\) cubic inches
Option E
