Question
A, B and C alone can complete a work in 20, 30 and 25
days respectively. All of them started working together but after 4 days from start A left the job and after 5 more days B also left the job. So for how many days did C work?Solution
Let the total work = 300 units (LCM of 20, 30 and 25) Amount of work done by A alone in one day = 300/20 = 15 units Amount of work done by B alone in one day = 300/30 = 10 units Amount of work done by C alone in one day = 300/25 = 12 units Amount of work done by A, B and C together in 4 days = 4 × (15 + 10 + 12) = 148 units Amount of work done by B and C together in 5 days = 5 × (10 + 12) = 110 units Remaining work = 300 – 148 – 110 = 42 units So, the time taken by C alone to complete 42 units work = 42/12 = 3.5 days So, C worked for 3.5 + 4 + 5 = 12.5 days
`sqrt(1297)` + 189.99 =?
90.004% of 9500 + 362 = ?
(74.76 ÷ 12.11 X ?)% of 239.89 = 600.19
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
1784.04 - 483.98 + 464.98 ÷ 15.06 = ?3
If tan θ + cot θ = 16, then find the value of tan2θ + cot2θ.
480 ÷ 10 + 18 % of 160 + ? * 9 = 60 * √36
(95.89% of 625.15 + 36.36% of 499.89) ÷ 6.02 = ? – 269.72
11.89 × 2.10 × 4.98 × 4.03 ÷ 7.98 of 15.03 = ?