A can complete a piece of work in 14 days and B in 21 days. They begin together but A leaves 3 days before the completion of the work. Find the total number of days taken to complete the work. MCQ with Answer and Explanation
A can complete a piece of work in 14 days and B in 21 days. They begin together but A leaves 3 days before the completion of the work. Find the total number of days taken to complete the work.
A. 10(1/5) days
B. 9(1/5) days
C. 11 days
D. 8(3/5) 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, B works for x days. 3(x - 3) + 2x = 42 => 3x - 9 + 2x = 42 => 5x = 51 => x = 51/5 = 10(1/5) days.
A, B and C can do a piece of work in 10, 12 and 15 days respectively. They began the work together but A left after 2 days and B left 3 days before the completion of the work. How long did the work last?
Explanation:
Total work = LCM(10, 12, 15) = 60 units. Efficiency of A = 6, B = 5, C = 4. Let the total time be x days. A worked for 2 days. B worked for (x - 3) days. C worked for x days. Total work = 2*6 + (x - 3)*5 + x*4 = 60 => 12 + 5x - 15 + 4x = 60 => 9x - 3 = 60 => 9x = 63 => x = 7 days.
A and B can complete a work in 10 days and 15 days respectively. They started together, but A left after some days and B finished the remaining work in 5 days. After how many days did A leave?
Explanation:
Total work = LCM(10, 15) = 30 units. Efficiency of A = 3, B = 2. B worked alone for 5 days: work done = 5 * 2 = 10 units. Remaining work = 30 - 10 = 20 units. This was completed by A and B together. Combined efficiency = 3 + 2 = 5 units/day. Time they worked together = 20 / 5 = 4 days. Thus, A left after 4 days.
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