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20 May 2021, 02:15
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B
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Difficulty:
65%
(hard)
Question Stats:
64% (02:50) correct
36%
(03:29)
wrong
based on 135
sessions
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Amit and Ian paint a wall in alternating shifts. First Amit paints alone, then Ian paints alone, then Amit paints alone, etc. During each of his shifts, Amit paints 1/2 of the remaining unpainted area of the wall, while Ian paints 1/3 of the remaining unpainted area of the wall during each of his shifts. If Amit goes first, what fraction of the wall's area will remain unpainted after Amit has completed his 4th shift?
A. 1/27
B. 1/54
C. 1/81
D. 1/162
E. 1/216
A. 1/27
B. 1/54
C. 1/81
D. 1/162
E. 1/216
23 May 2021, 07:45
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Bunuel
OFFICIAL SOLUTION:
The issue of this question is (efficiently) finding out a term in a sequence. The terms in this sequence are the fractions of unpainted wall area. To find out how to calculate the terms, start writing them down and look for a pattern.
The first term - the fraction of unpainted area after Amit's first shift is 1/2.
In the next shift, Ian paints a 1/3 of the unpainted area, i.e., a 1/3 of 1/2. But how much is left unpainted? Ian leaves 2/3 of the 1/2 unpainted. Thus, the second term is (2/3)(1/2).
In the following shift, (Amit's second shift,) half of the unpainted area is painted, or, in other words, half of unpainted area remains unpainted, hence, the third term is (1/2)(2/3)(1/2). Continue the pattern from here until after Amit's 4th shift.
The final expression after Amit's 4th shift (labeled Amit (4)):
Amit(1) Ian Amit(2) Ian Amit(3) Ian Amit(4)
\(\frac{1}{2}*\frac{2}{3}*\frac{1}{2}*\frac{2}{3}*\frac{1}{2}*\frac{2}{3}\\
*\frac{1}{2}=\frac{1}{54}\).
Answer: B.
General Discussion
20 May 2021, 09:42
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Amit finishes \(\frac{1}{2}\) and leaves \(\frac{1}{2}\).
=> \(\frac{1}{2}\) of work will be left 4 times. The total fraction that remains unpainted after 4 shifts of Amit is (\(\frac{1}{2})^4\).
Ian works after Amit so he will paint 3 times. He paints \(\frac{1}{3}\) of unpainted part(\(\frac{1}{2}) = \frac{1}{3} * \frac{1}{2} = \frac{1}{6}\)
=> Remaining part after Ian's first work: \(\frac{1}{2} - \frac{1}{ 6} = \frac{2}{3}\)
=> \(\frac{2}{3}\) of work will be left 3 times. The total fraction that remains unpainted after 4 shifts of Amit is \((\frac{2}{3})^3\).
Total fraction left: \((\frac{1}{2})^4 * (\frac{2}{3})^3 = \frac{1}{16} * \frac{8}{27} = \frac{1}{54}\)
Answer B
=> \(\frac{1}{2}\) of work will be left 4 times. The total fraction that remains unpainted after 4 shifts of Amit is (\(\frac{1}{2})^4\).
Ian works after Amit so he will paint 3 times. He paints \(\frac{1}{3}\) of unpainted part(\(\frac{1}{2}) = \frac{1}{3} * \frac{1}{2} = \frac{1}{6}\)
=> Remaining part after Ian's first work: \(\frac{1}{2} - \frac{1}{ 6} = \frac{2}{3}\)
=> \(\frac{2}{3}\) of work will be left 3 times. The total fraction that remains unpainted after 4 shifts of Amit is \((\frac{2}{3})^3\).
Total fraction left: \((\frac{1}{2})^4 * (\frac{2}{3})^3 = \frac{1}{16} * \frac{8}{27} = \frac{1}{54}\)
Answer B
