A tap can empty a tank in 30 minutes. A second tap can empty it in 45 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?
(a) 18 minutes
(b) 14 minutes
(c) 15 minutes
(d) 30 minutes
(e) None of these
Answer for this Time and Work question is (a)
Work done by \(\displaystyle {{1}^{{st}}}\) tap in one minute = \(\displaystyle \frac{1}{{30}}\)
Work done by \(\displaystyle {{2}^{{nd}}}\) tap in one minute = \(\displaystyle \frac{1}{{45}}\)
Both tap one minute work = \(\displaystyle \frac{1}{{30}}+\frac{1}{{45}}=\frac{{45+30}}{{1350}}=\frac{{75}}{{1350}}=\frac{1}{{18}}\)
Both tap will empty the tank in 18 minutes.
A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days, 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day ?
(a) 80
(b) 81
(c) 82
(d) 83
(e) None of these
Answer for this Time and Work question is (b)
Let x additional men employed.
117 men were supposed to finish the whole work in 46 × 8 = 368 hours.
But 117 men completed \(\displaystyle \frac{4}{7}\) of the work in 33 × 8 = 264 hours
Therefore, 117 men could complete the work in 462 hours.
Now (117 + x) men are supposed to do \(\displaystyle \frac{3}{7}\) of the work, working 9 hours a day, in 13 × 9 = 117 hours, so as to finish the work in time.
i.e. (117 + x) men are supposed to complete the whole work in \(\displaystyle 117\times \frac{1}{3}=273hours\)
\(\displaystyle \Rightarrow \) x + 117 = 3 × 66 = 198 Þ x = 81
Hence, required number of additional men to finish the work in time = 81.
Two pipes A and B can fill a cistern in 30 minutes and 40 minutes respectively. Both the pipes are opened. Find when the second pipe B must be turned off so the cistern may just be full in 10 minutes
(a) \(\displaystyle 26\frac{2}{3}\ \) min
(b) 25 min
(c) \(\displaystyle 40\frac{2}{3}\ \) min
(d) \(\displaystyle 42\frac{2}{3}\ \) min
(e) None of these
Answer for this Time and Work question is (a)
Let A and B together work for x minutes than amount of water filled in the period = \(\displaystyle x(\frac{1}{{30}}+\frac{1}{{40}})=\frac{{7x}}{{120}}\)
Remaining part = \(\displaystyle 1-\frac{{7x}}{{120}}=(\frac{{120-7x}}{{120}})\)
Work done by A in (10 – x) minutes = \(\displaystyle \frac{{120-7x}}{{120}}=1-\frac{{7x}}{{120}}\)
\(\displaystyle \frac{{7x}}{{120}}+\frac{{10-x}}{{30}}=1\) or 7x + 40 – 4x = 120
3x = 120 – 40 = 80
x = \(\displaystyle 26\frac{2}{3}\min \)
Two workers A and B working together completed a job in 5 days. If A worked twice as efficiently as he actually did and B worked \(\displaystyle \frac{1}{3}\) as efficiently as he actually did, the work would have completed in 3 days. Find the time for A to complete the job alone.
(a) \(\displaystyle 6\frac{1}{4}\) days
(b) \(\displaystyle 5\frac{3}{4}\) days
(c) 5 days
(d) 3 days
(e) None of these
Answer for this Time and Work question is (a)
(A + B)’s one day’s work = \(\displaystyle \frac{1}{5}thwork\)
Let A can do job in x days. Then,
A’s one day’s work = \(\displaystyle \frac{1}{x}thwork\)
and B’s one day’s work = \(\displaystyle \frac{1}{5}-\frac{1}{x}=\frac{{x-5}}{{5x}}thwork\)
Now , (2)A’s work + \(\displaystyle (\frac{1}{3})\) B’s work = \(\displaystyle \frac{1}{3}rdwork\)
Pipe A can fill a tank in 30 minutes while pipe B can fill it in 45 minutes. Another pipe C can empty a full tank in 60 minutes. If all three pipes are opened simultaneously, the empty tank will be filled in
(a) \(\displaystyle \frac{2}{7}\) hour
(b) \(\displaystyle \frac{3}{7}\) hour
(c) \(\displaystyle \frac{4}{7}\) hour
(d) \(\displaystyle \frac{5}{7}\) hour
(e) None of these
Answer for this Time and Work question is (b)
Part of tank filled by all three pipes in 1 minute
A can build up a wall in 8 days while B can break it in 3 days. A has worked for 4 days and then B joined to work with A for another 2 days only. In how many days will A alone build up the remaining part of wall?
(a) \(\displaystyle 13\frac{1}{3}\) days
(b) \(\displaystyle 7\frac{1}{3}\) days
(c) \(\displaystyle 6\frac{1}{3}\) days
(d) 7 days
(e) None of these
Answer for this Time and Work question is (b)
B’s one day’s work \(\displaystyle \frac{1}{3}rd\) work
Therefore, A’s 4 day’s work = \(\displaystyle 4\times \frac{1}{8}=\frac{1}{2}nd\) work
Hence, In next two days, total wall = \(\displaystyle \frac{1}{2}+2(\frac{1}{8})-2(\frac{1}{3})=\frac{1}{{12}}th\) work
Now, \(\displaystyle \frac{1}{8}th\) wall is built up by A in one day
\(\displaystyle \frac{{11}}{{12}}th\) wall is build up by A in \(\displaystyle 8\times \frac{{11}}{{12}}=7\frac{1}{3}days\)
12 men can finish a project in 20 days. 18 women can finish the same project in 16 days and 24 children can finish it in 18 days. 8 women and 16 children worked for 9 days and then left. In how many will 10 men complete the remaining project ?
(a) \(\displaystyle 8\times \frac{{11}}{{12}}=7\frac{1}{3}\) days
