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
The OA is B.
Source: Economist GMAT
Amit and Ian paint a wall in alternating shifts. First Amit
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Amit's 1st shift: He paints 1/2 of the room, so ½ of the room is left unpainted.swerve wrote: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
Ian's 1st shift: He paints 1/3(1/2) = 1/6 of the room, so 1/2 - 1/6 = 1/3 of the room is left unpainted.
Amit's 2nd shift: He paints 1/2(1/3) = 1/6 of the room, so 1/3 - 1/6 = 1/6 of the room is left unpainted.
Ian's 2nd shift: He paints 1/3(1/6) = 1/18 of the room, so 1/6 - 1/18 = 1/9 of the room is left unpainted.
Amit's 3rd shift: He paints 1/2(1/9) = 1/18 of the room, so 1/9 - 1/18 = 1/18 of the room is left unpainted.
Ian's 3rd shift: He paints 1/3(1/18) = 1/54 of the room, so 1/18 - 1/54 = 1/27 of the room is left unpainted.
Amit's 4th shift: He paints 1/2(1/27) = 1/54 of the room, so 1/27 - 1/54 = 1/54 of the room is left unpainted.
Alternate Solution:
Let's say the room is 6^3 = 216 ft^2. In his first shift, Amit will paint 216/2 = 108 ft^2 and there will be 216 - 108 = 108 ft^2 left to be painted.
In his first shift, Ian will paint 108/3 = 36 ft^2 and there will be 108 - 36 = 72 ft^2 left to be painted.
In his second shift, Amit will paint 72/2 = 36 ft^2 and there will be 72 - 36 = 36 ft^2 left to be painted.
In his second shift, Ian will paint 36/3 = 12 ft^2 and there will be 36 - 12 = 24 ft^2 left to be painted.
In his third shift, Amit will paint 24/2 = 12 ft^2 and there will be 24 - 12 = 12 ft^2 left to be painted.
In his third shift, Ian will paint 12/3 = 4 ft^2 and there will be 12 - 4 = 8 ft^2 left to be painted.
Finally, in his fourth shift, Amit will paint 8/2 = 4 ft^2 and there will be 8 - 4 = 4 ft^2 left to be painted.
Thus, after Amit's fourth shift, 4/216 = 1/54 of the room will be left to be painted.
Answer: B
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Say the P is the area of the total area to be painted and R is the total area remained unpainted after Amit has completed his 4th shiftswerve wrote: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
The OA is B.
Source: Economist GMAT
Area remained unpainted after Amit's 1st shift = P(1 - 1/2); we know that Amit does 1/2 of the remaining
Area remained unpainted after Amit's 1st and Ian's 1st shift = P(1 - 1/2)(1 - 1/3); we know that Ian does 1/3 of the remaining
Thus,
R = P*[(1 - 1/2)^4]*[(1 - 1/3)^3]; we are given that Amit does 4 shifts, thus, Ian does only 3 shifts
R = P*(1/2)^4*(2/3)^3
R = P*(2^3/2^4)*(1/3^3)
R = P*(1/2)*(1/27)
R = P/54
R/P = 1/54 --> fraction of area left unpainted.
The correct answer: B
Hope this helps!
-Jay
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During each of Amit's 4 shifts, he paints 1/2 of the room, leaving 1/2 the room unpainted.swerve wrote: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
Since 1/2 of the room is left unpainted FOUR TIMES, the total fraction left unpainted as a result of Amit's 4 shifts = (1/2)�.
Since Ian works after each of Amit's first 3 shifts, Ian paints a total of 3 times.
During each of Ian's 3 shifts, he paints 1/3 of the room, leaving 2/3 of the room unpainted.
Since 2/3 of the room is left unpainted THREE TIMES, the total fraction left unpainted as a result of Ian's 3 shifts = (2/3)³.
Multiplying the two values in blue, we get:
Total fraction left unpainted = (1/2)�(2/3)³ = 1/54.
The correct answer is B.
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Fraction of unpainted area after the first shift of Amit =
$$\frac{1}{1}-\ \frac{1}{2}=\ \frac{\left(2-1\right)}{2}=\ \frac{1}{2}$$
Fraction of unpainted area after the front shift of Ian =
$$\frac{1}{2}-\ \frac{1}{3}\cdot\ \frac{1}{2}$$
$$Work\ remaining\ after\ Amit\ completes\ shift\ =\ \frac{1}{2}$$
$$Work\ remaining\ after\ Ian\ completes\ shift\ =\ \frac{2}{3}of\ work$$
$$Remaining\ after\ Ian\ completes\ shift\ =\ \frac{2}{3}of\ \ \frac{1}{2}$$
Fraction of unpainted area after completes fourth shift =
$$\frac{\left(\left(\frac{1}{2}\cdot\ \frac{2}{3}\right)\right)}{shift\ 1}\cdot\ \frac{\left(\left(\frac{1}{2}\cdot\frac{2}{3}\right)\right)}{shift\ 2}\ \cdot\ \frac{\left(\left(\frac{1}{2}\cdot\frac{2}{3}\right)\right)}{shift\ 3}\cdot\frac{\left(\left(\frac{1}{2}\right)\right)}{shift\ 4}$$
$$\frac{2}{6\ }\cdot\ \frac{2}{6}\cdot\ \frac{2}{6}\ \cdot\frac{1}{2}=\ \frac{4}{432}=\frac{1}{54}$$
Option B is CORRECT.
$$\frac{1}{1}-\ \frac{1}{2}=\ \frac{\left(2-1\right)}{2}=\ \frac{1}{2}$$
Fraction of unpainted area after the front shift of Ian =
$$\frac{1}{2}-\ \frac{1}{3}\cdot\ \frac{1}{2}$$
$$Work\ remaining\ after\ Amit\ completes\ shift\ =\ \frac{1}{2}$$
$$Work\ remaining\ after\ Ian\ completes\ shift\ =\ \frac{2}{3}of\ work$$
$$Remaining\ after\ Ian\ completes\ shift\ =\ \frac{2}{3}of\ \ \frac{1}{2}$$
Fraction of unpainted area after completes fourth shift =
$$\frac{\left(\left(\frac{1}{2}\cdot\ \frac{2}{3}\right)\right)}{shift\ 1}\cdot\ \frac{\left(\left(\frac{1}{2}\cdot\frac{2}{3}\right)\right)}{shift\ 2}\ \cdot\ \frac{\left(\left(\frac{1}{2}\cdot\frac{2}{3}\right)\right)}{shift\ 3}\cdot\frac{\left(\left(\frac{1}{2}\right)\right)}{shift\ 4}$$
$$\frac{2}{6\ }\cdot\ \frac{2}{6}\cdot\ \frac{2}{6}\ \cdot\frac{1}{2}=\ \frac{4}{432}=\frac{1}{54}$$
Option B is CORRECT.