A team of 30 men is supposed to do a work in 38 days. After 25 days, 5 more men were employed and the work was finished one day earlier than the scheduled time. How many days would it have been delayed if 5 more men were not employed? MCQ with Answer and Explanation
A team of 30 men is supposed to do a work in 38 days. After 25 days, 5 more men were employed and the work was finished one day earlier than the scheduled time. How many days would it have been delayed if 5 more men were not employed?
A. 2 days
B. 4 days
C. 1 day
D. 3 days
Answer: Option C
Solution (By JKSSB Mock Tests)
Scheduled time = 38 days. Work finished 1 day earlier, so it took 37 days. For the first 25 days, 30 men worked. For the remaining 37 - 25 = 12 days, 35 men worked. Remaining work = 35 * 12 = 420 man-days. If 5 extra men were not joined, 30 men would take 420 / 30 = 14 days to complete this remaining work. Total days taken without extra men = 25 + 14 = 39 days. Delay = 39 - 38 = 1 day.
Explanation:
Total work = LCM(20, 30) = 60 units. Efficiency of A = 3, B = 2. In 4 days, work completed = 4 * (3 + 2) = 20 units. Remaining work = 60 - 20 = 40 units. Time taken by B to finish remaining work = 40 / 2 = 20 days. Total days = 4 + 20 = 24 days.
A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
Explanation:
Total work = LCM(10, 50) = 50 units. Efficiency of (A + B) = 5, efficiency of C = 1. Total efficiency of (A + B + C) = 5 + 1 = 6. Given A = B + C. So, (B + C) + B + C = 6 => 2(B + C) = 6 => B + C = 3. Since C = 1, B = 2. Time taken by B alone = 50 / 2 = 25 days.
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