Working alone, A takes \(2\frac{2}{7}\) days more and B takes \(\frac{7}{25}\) day more to do a piece of work than A and B take working together. In how many days will A and B together finish the work?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{4}{5}\) day
Let A and B together take \(T\) days. Writing each person's time as \(T\) plus the extra time and adding their one-day work gives an equation in \(T\).
A alone takes \(T+a\) days and B alone \(T+b\) days, with \(a=\frac{16}{7}\) and \(b=\frac{7}{25}\)
\(\displaystyle \frac{1}{T+a}+\frac{1}{T+b}=\frac{1}{T}\)
Cross-multiplying: \(T(T+b)+T(T+a)=(T+a)(T+b)\)
\(2T^{2}+(a+b)T=T^{2}+(a+b)T+ab\), so \(T^{2}=ab\)
\(\displaystyle T^{2}=\frac{16}{7}\times\frac{7}{25}=\frac{16}{25}\), so \(\displaystyle T=\frac{4}{5}\) day (time cannot be negative)
Check: A takes \(\displaystyle \frac{4}{5}+\frac{16}{7}=\frac{108}{35}\) days, B takes \(\displaystyle \frac{4}{5}+\frac{7}{25}=\frac{27}{25}\) days, and \(\displaystyle \frac{35}{108}+\frac{25}{27}=\frac{135}{108}=\frac{5}{4}\) of the work per day
So together they finish the work in \(\frac{4}{5}\) day.