Subject: General Aptitude · Chapter: Time and Work · Exam: · Marks: · Difficulty:
A can do in one day three times the work done by B in one day. They together finish $\frac{2}{5}$ of the work in 9 days. The number of days by which B can do the work alone is [SSC-CHSL (10+2) Exam , 2015]
(a)90 days
(b)120 days
(c)100 days
(d)30 days
Answer
Answer (as printed): A
Explanation
Let time taken by A alone in doing the work be x days. $\therefore$ Time taken by B alone = 3x days. A's 1 day's work = $\frac{1}{x}$. B's 1 day's work = $\frac{1}{3x}$. A and B together finish $\frac{2}{5}$ of the work in 9 days. $\therefore$ Time taken by A and B in doing the whole work = $\frac{9 \times 5}{2} = \frac{45}{2}$ days. According to the given information: $\frac{1}{x} + \frac{1}{3x} = \frac{2}{45} \Rightarrow \frac{3+1}{3x} = \frac{2}{45} \Rightarrow \frac{4}{3x} = \frac{2}{45}$. By cross-multiplying: $2 \times 3x = 4 \times 45 \Rightarrow x = \frac{4 \times 45}{2 \times 3} = 30$ days. Time taken by A = 30 days. $\therefore$ Time taken by B = $3 \times 30 = 90$ days.
Explanation as extracted from the printed page; notation may be imperfect.