Question
'A' can complete a task in 24 days, while 'B' can finish
the same task in 36 days. However, 'A' works at only 2/3 of his usual efficiency, and 'B' operates at 50% of his normal efficiency. Given these conditions, determine the number of days required for both of them, working together, to complete 75% of the task.Solution
Let the total work be 72 units (LCM of 24 and 36) .
Efficiency of 'A' = (72/24) = 3 units/day
Efficiency of 'B' = (72/36) = 2 units/day
New efficiency of 'A' = (2/3) X 3 = 2 units/day
New efficiency of 'B' = 0.5 X 2 = 1 unit/day
Therefore, required time taken = (0.75 X 72) ÷ (2 + 1) = (54/3) = 18 days
More Arithmetic Problem Questions
(√1296 – 12) × 5 = ? + 40
Simplify: 0.004 × 0.5
(21 X 5) + ? = (480 - 120) ÷ 3
8 × (25 % of 720) – 50 % of 135 % of 840 = ?
135÷ 15 x 19 + 14807 = ? + √3249 - √9604
5/13 × 104 + 1(2/9) × 198 = 133 + ?
What will come in the place of question mark (?) in the given expression?
√(? - 212) - 84 = 81 - 13 X 9
(64/25)? × (125/512)?-1 = 5/8
540 ÷ 6 + 25 % of 120 + ? * 8 = 72 * √9
- What will come in place of the question mark (?) in the following questions?
75% of 240 + 30 = ?
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