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?
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
2222 ÷ 22 + 992 ÷ 16 + 650 ÷ 25 = ?
30% of 8/5 × 5/7 × 2870 =?
(-251 × 21 × -12) ÷ ? = 158.13
464 + 181 +? = (154 × 25) - (15) 2
2852 + 7848 + 2962 + 4268 = ? – 1460
168 163 153 138 118 ?
...√256 × 25 – 15 × 14 =?
7/11 × 1034 + 1(4/7) × 2401 = 1230 +?
60% of 60 + 40% of ? = 75% of 160 - 30% of 120