(A + B)’s 1 day’s work = \(\displaystyle \frac{1}{m}+\frac{1}{n}=\frac{{n+m}}{{mn}}=\frac{{m+n}}{{mn}}\)
Required time = \(\displaystyle \frac{{mn}}{{m+n}}\)
A takes three times as long as B and C together to do a job. B takes four times as long as A and C together to do the work. If all the three, working together can complete the job in 24 days, then the number of days, A alone will take to finish the job is
A, B and C can do a work separately in 16, 32 and 48 days respectively. They started the work together but B left off 8 days and C six days before the completion of the work. In what time is the work finished?
(a) 10 days
(b) 9 days
(c) 12 days
(d) 14 days
(e) 18 days
Answer for this MCQ Time and Work is (c)
Let the work be completed in x days.
According to the question, \(\displaystyle \frac{x}{{16}}+\frac{{x-8}}{{32}}+\frac{{x-6}}{{48}}=1\)
\(\displaystyle \frac{{6x+3x-24+2x+12}}{{96}}=1\)
11x – 36 = 96
11x = 96 + 36 = 132
\(\displaystyle x=\frac{{132}}{{11}}=12days\)
X can do a piece of work in ‘p’ days and Y can do the same work in ‘q’ days. Then the number of days in which X and Y can together do that work is
(a) \(\displaystyle \frac{{p+q}}{2}\)
(b) \(\displaystyle \frac{1}{p}+\frac{1}{q}\)
(c) \(\displaystyle \frac{{pq}}{{p+q}}\)
(d) pq
(e) qp
Answer for this MCQ Time and Work is (c)
X’s 1 day’s work = \(\displaystyle \frac{1}{p}\)
Y’s 1 day’s work = \(\displaystyle \frac{1}{q}\)
(X + Y)’s 1 day’s work = \(\displaystyle \frac{1}{p}+\frac{1}{q}=\frac{{q+p}}{{pq}}\)
Required time = \(\displaystyle \frac{{pq}}{{p+q}}\)
