A and B can do a piece of work in 34/3 days working alternatively with B working first. A can complete the work alone in 48 days in how many days B does 4 times of the work?
Answer & explanation
Correct answer: option 1
let total work = 1
A and B are working alternately with B start first
total work time=34/3=11 +1/3 days
for 5 -5 days both will work .on 11th day B will work and A will work for next \(\frac{1}{3}\)rd day .
Thus B works for =6 days and A works for 5+1/3 days
A can complete the work in 48 days
A's 1 day work = \(\frac{1}{48}\)
A's \(\frac{16}{3}\) day work =\(\frac{16}{48 × 3}\)
= \(\frac{1}{9}\)
So B’s work in 6 days= \(\frac{8}{9}\)
B’s per day work=\(\frac{8}{54}\) = \(\frac{4}{27}\)
So for B the days required to complete 4 time work
=\(\frac{27 × 4}{4}\) = 27 days