Question
A alone can complete 70% of a work in 21 days while B
takes 15 days more than A to complete it. If B and C together can complete the work in 20 days, then find the time taken by C alone to complete the same work.Solution
Time taken by A alone to complete the entire work = 21/0.7 = 30 days Time taken by B alone to complete the entire work = 30 + 15 = 45 days Let the total work = L.C.M of 30, 45 and 20 = 180 units Then, efficiency of B = (180/45) = 4 units/day Combined efficiency of B and C = (180/20) = 9 units/day So, efficiency of C alone = 9 – 4 = 5 units/day So, time taken by C alone to complete the entire work = (180/5) = 36 days
(√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 = ?
