A can do a piece of work in 14 days and B in 21 days. They begin together but 3 days before the completion of the work, A leaves. The total number of days taken to complete the work is: MCQ with Answer and Explanation
A can do a piece of work in 14 days and B in 21 days. They begin together but 3 days before the completion of the work, A leaves. The total number of days taken to complete the work is:
A. 10(1/5) days
B. 8(3/5) days
C. 9(1/5) days
D. 7(2/5) days
Answer: Option A
Solution (By JKSSB Mock Tests)
Total work = LCM(14, 21) = 42 units. Efficiency of A = 3, B = 2. Let total days be x. A works for (x - 3) days and B works for x days. 3*(x - 3) + 2*x = 42 => 3x - 9 + 2x = 42 => 5x = 51 => x = 51/5 = 10(1/5) days.
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?
Explanation:
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.
A and B can complete a piece of work in 20 days and 30 days respectively. They work together for some days and then A leaves. If B completes the remaining work in 10 days, for how many days did they work together?
Explanation:
Total work = LCM(20, 30) = 60 units. Efficiency of A = 3, B = 2. B works alone for 10 days: work completed = 10 * 2 = 20 units. Remaining work = 60 - 20 = 40 units. This was done by A and B together. Combined efficiency = 3 + 2 = 5 units/day. Days worked together = 40 / 5 = 8 days.
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