Question
A alone can complete 40% of a work in 16 days while B
takes 12 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 = 16/0.4 = 40 days Time taken by B alone to complete the entire work = 40 + 12 = 52 days Let the total work = L.C.M of 40, 52 and 20 = 520 units Then, efficiency of B = (520/52) = 10 units/day Combined efficiency of B and C = (520/20) = 26 units/day So, efficiency of C alone = 26 – 10 = 16 units/day So, time taken by C alone to complete the entire work = (520/16) = 32.5 days
What will come in the place of question mark (?) in the given expression?
? = (40% of 80% of 6400) ÷ 64
420 ÷ 7 + 140 % of 20 + ? × 13 = 18 × 15
Determine the simplified value of the expression: 12 × 15 - 20 + 15 + 12 - 18 + 3 × 4 + 18.
15% of 1800 + 22 = ?Â
Solve the following:
523 + 523 x 523 ÷ 523
24% of 150% of 500 + 140 = ? × 8Â
[∛(3375/19683  )- ∛(125/1728  )  ] ÷ ∛(64/729)  = ? - 3/8Â
7, 8, 12, 21, 37, ?
What value should come in the place of (?) in the following questions.
336 ÷ 6 ÷ √16 * ? = 1400 ÷ 4
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