Question
A can do (2/5) of a work in 6 days and B can do 30% of
the same work in 3 days. They worked together for 4 days. The remaining work was completed by C alone in 6 days. B and C together can complete 70% of the original work in:Solution
A can complete the whole work in 6 × 5/2 = 15 days B can complete the whole work in 3 × 100/30 = 10 days Let the total unit of work be 30 unit (LCM of 15 and 10) Efficiency of A = 30/15 = 2 unit/day Efficiency of B = 30/10 = 3 unit/day Work done by A and B in 4 days = (2 + 3) x 4 = 20 unit Remaining work = 30 - 20 = 10 units Let the remaining work completed by C in x days 6 x x = 10 x = 10/6 B and C together can complete 70% of the original work in 30/(3 + 10/6) x 70/100 = 4 (1/2) days
- A series is given with one term missing. Choose the correct alternatives from the given ones that will complete the series.
57, 59, 56, 61, 54, ___ - Which letter and number cluster will replace the question mark (?) to complete the given series?
LT6, KU12, IW24, FZ48, ____ - Which letter-cluster will replace the question mark (?) in the following series?
NPQR, OORQ, PNSP, ____, RLUN - Select the number from among the given options that can replace the question mark (?) in the following series.
17, 18, 22, 31, 47, ___ - Which letter-cluster will replace the question mark (?) in the following series?
RGV, UME, ?, AYW, DEF