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04 Feb 2021, 04:30
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C
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Difficulty:
45%
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Question Stats:
73% (03:21) correct
27%
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wrong
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12 men can finish a project in 20 days. 18 women can finish the same project in 16 days and 24 children can finish it in 18 days, 8 women and 16 children worked for 9 days and then left. In how many days will 10 men complete the remaining project?
A. 11 1/2
B. 10 1/2
C. 10
D. 9
E. 8
A. 11 1/2
B. 10 1/2
C. 10
D. 9
E. 8
18 Feb 2021, 14:31
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Bunuel
12 men can finish a project in 20 days. 18 women can finish the same project in 16 days and 24 children can finish it in 18 days, 8 women and 16 children worked for 9 days and then left. In how many days will 10 men complete the remaining project?
A. 11 1/2
B. 10 1/2
C. 10
D. 9
E. 8
A. 11 1/2
B. 10 1/2
C. 10
D. 9
E. 8
Solution:
We see that the rate of the 12 men is 1/20 per day. Thus, the rate of each man is (1/20) / 12 = 1/240 per day. Similarly, the rate of the 18 women is 1/16 per day, and so the rate of each woman is (1/16) / 18 = 1/288 per day. The rate of 24 children is 1/18 per day, and the rate of each child is (1/18) / 24 = 1/432 per day. Therefore, the portion of work done by 8 women and 16 children in 9 days is (8 x 1/288 + 16 x 1/432) x 9 = (1/36 + 1/27) x 9 = 1/4 + 1/3 = 7/12. Therefore, the remaining 5/12 of the work can be done by 10 men in (5/12) / (10 x 1/240) = (5/12)/(1/24) = 5/12 x 24 = 10 days.
Answer: C
General Discussion
04 Feb 2021, 10:30
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\(12m*20=18w*16=24c*18\)
\(5m=6w=9c\)
We need to find the efficiency of \(m,w,c\). We have many ways to approach this like taking \(LCM(5,6,9)=90\). \(m=\frac{90}{5}=18\). Similarly, \(w=15\) and \(c=10\)
Another approach would be; \(5m=6w => \frac{m}{w}=\frac{6}{5}\) and \(6w=9c => \frac{w}{c}=\frac{3}{2}\)
\(\frac{m}{w}=\frac{6*3}{5*3}=\frac{18}{15}\) or \(\frac{w}{c}=\frac{3*5}{2*5}=\frac{15}{10}\)
Find the \(total \ work = 12*18*20 = 4320\)
Work done by \((8w+16c)*9 = (8*15+16*10)*9=(120+160)*9 = 2520\)
Remaining work \(= 4320 - 2520=1800\)
Completed by 10m, so efficiency of \(10m = 10*18 = 180\)
No of days required \(= \frac{1800}{180 }=10\) days
Ans C
\(5m=6w=9c\)
We need to find the efficiency of \(m,w,c\). We have many ways to approach this like taking \(LCM(5,6,9)=90\). \(m=\frac{90}{5}=18\). Similarly, \(w=15\) and \(c=10\)
Another approach would be; \(5m=6w => \frac{m}{w}=\frac{6}{5}\) and \(6w=9c => \frac{w}{c}=\frac{3}{2}\)
\(\frac{m}{w}=\frac{6*3}{5*3}=\frac{18}{15}\) or \(\frac{w}{c}=\frac{3*5}{2*5}=\frac{15}{10}\)
Find the \(total \ work = 12*18*20 = 4320\)
Work done by \((8w+16c)*9 = (8*15+16*10)*9=(120+160)*9 = 2520\)
Remaining work \(= 4320 - 2520=1800\)
Completed by 10m, so efficiency of \(10m = 10*18 = 180\)
No of days required \(= \frac{1800}{180 }=10\) days
Ans C
