A can do a certain work in 15 days. B is 10% less efficient than A, and C is 20% more efficient than A. All the three work together for 3 days. The remaining work is completed by D alone in  days. D alone can complete the original work in:

Options:
a) 15 days
b) 18 days
c) 25 days
d) 20 days

Solution

As per question
Let A efficient = 100
B is 10% less efficient = 90
C is 20% more efficient than A = 120
Total efficient = 100 + 90 + 120 = 310
Total work = A efficient x days
Total work = 100 x 15 = 1500

3 days work of A , B , C = 3( A + B + C ) = 3(100+90+120)
3 days work of A , B , C = 3 x 310 = 930
Remain work = 1500 – 930 = 570
570 unit done by D in days
1500 unit will be done in = = 25 days

Ans: c) 25 days