Rohit crosses a 700 metres wide road in 35 seconds | Time and Work Questions & Answers

Q:

A and B can do a work in 4 hours and 12 hours respectively. A starts the work at 6 AM and they work alternately for one hour each. When will the work be completed?

A) 4 days	B) 5 days
C) 6 days	D) 7 days

Answer & Explanation Answer: C) 6 days

Explanation:

Work donee by A and B in the first two hours, working alternatively = First hour A + Second hour B = (1/4) + (1/12) = 1/3.

Thus, the total time required to complete the work = 2 (3) = 6 days

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Q:

Two pipes A and B can fill a tank in 24 hours and $17 \frac{1}{7}$ hours respectively. Harihar opened the pipes A and B to fill an empty tank and some times later he closed the taps A and B , when the tank was supposed to be full. After that it was found that the tank was emptied in 2.5 hours because an outlet pipe "C" connected to the tank was open from the beginning. If Harihar closed the pipe C instead of closing pipes A and B the remaining tank would have been filled in :

A) 2 hours	B) 8 hours
C) 6 hours	D) 4 hours

Answer & Explanation Answer: B) 8 hours

Explanation:

Efficiency of Inlet pipe A = 4.16% $(\frac{100}{24})$

Efficiency of Inlet pipe B = 5.83% $(\frac{100}{17 \frac{1}{7}})$

Therefore, Efficiency of A and B together = 100 %

Now, if the efficiency of outlet pipe be x% then in 10 hours the capacity of tank which will be filled = 10 * (10 - x)

Now, since this amount of water is being emptied by 'C' at x% per hour, then

$\frac{10 x (10 - x)}{x} = 2.5 h r s$

=> x = 8

Therefore, in 10 hours 20% tank is filled only. Hence, the remaining 80% of the capacity will be filled by pipes A and B in 80/10 = 8 hours

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Q:

There are three boats B1, B2 and B3 working together they carry 60 people in each trip. One day an early morning B1 carried 50 people in few trips alone. When it stopped carrying the passengers B2 and B3 started carrying the people together. It took a total of 10 trips to carry 300 people by B1, B2 and B3. It is known that each day on an average 300 people cross the river using only one of the 3 boats B1, B2 and B3. How many trips it would take to B1, to carry 150 passengers alone?

A) 15	B) 30
C) 25	D) 10

Answer & Explanation Answer: A) 15

Explanation:

Combined efficiency of all the three boats = 60 passenger/trip

Now, consider option(a)

15 trips and 150 passengers means efficiency of B1 = 10 passenger/trip

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

Therefore the efficiency of B2 and B3 = 250/5 = 50 passenger/trip

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

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Q:

A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of A, B and C respectively (in Rs):

A) 200, 250, 300	B) 300, 200, 250
C) 200, 300, 400	D) None of these

Answer & Explanation Answer: B) 300, 200, 250

Explanation:

A's 5 days work = 50%

B's 5 days work = 33.33%

C's 2 days work = 16.66% [100- (50+33.33)]

Ratio of contribution of work of A, B and C = $50 : 33 \frac{1}{3} : 16 \frac{2}{3}$ = 3 : 2 : 1

A's total share = Rs. 1500

B's total share = Rs. 1000

C's total share = Rs. 500

A's one day's earning = Rs.300

B's one day's earning = Rs.200

C's one day's earning = Rs.250

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Q:

Amit can do a piece of work in 45 days, but Bharath can do the same work in 5 days less, than Amit, when working alone. Amit and Bharath both started the work together but Bharath left after some days and Amit finished the remaining work in 56 days with half of his efficiency but he did the work with Bharath with his complete efficiency. For how many days they had worked together?

A) 6	B) 8
C) 9	D) 12

Answer & Explanation Answer: B) 8

Explanation:

$\begin{matrix} A m i t & B h a r a t \\ N o . o f D a y s & 45 & 40 \\ E f f i c i e n c y & 2.22 P e r c e n t (= \frac{1}{45}) & 2.5 P e r c e n t (= \frac{1}{40}) \end{matrix}$

Amit did the work in 56 days = $56 \times \frac{1}{45 \times 2} = \frac{28}{45}$

Therefore, Rest work 17/45 was done by Amit and Bharath = $\frac{\frac{17}{45}}{\frac{17}{360}}$ = 8 days

( since Amit and Bharath do the work in one day = $\frac{1}{45} + \frac{1}{40} = \frac{17}{360}$ )

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Q:

Two pipes A and B can fill a cistern in 4 minutes and 6 minutes respectively . If these pipes are turned on alternately for 1 minute each how long will it take to the cistern to fill?

Answer

As the pipes are operating alternatively, thus their 2 minutes job is = $\frac{1}{4} + \frac{1}{6} = \frac{5}{12}$

In the next 2 minutes the pipes can fill another $\frac{5}{12}$ part of cistern.

Therefore, In 4 minutes the two pipes which are operating alternatively will fill $\frac{5}{12} + \frac{5}{12} = \frac{5}{6}$ Remaining part = $1 - \frac{5}{6} = \frac{1}{6}$

Pipe A can fill $\frac{1}{4}$ of the cistern in 1 minute

Pipe A can fill $\frac{1}{6}$ of the cistern in = $\frac{2}{3}$ min

Therefore, Total time taken to fill the Cistern

4 + $\frac{2}{3}$ minutes.

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28 13880

Q:

A contractor undertakes to complete a work in 130 days. He employs 150 men for 25 days and they complete 1/4 of the work . He then reduces the number of men to 100, who work for 60 days, after which there are 10 days holidays.How many men must be employed for the remaining period to finish the work?

Answer

150 men in 25 days do = $\frac{1}{4}$ work

Let 1 man in 1 day does = x work

Total work done by 150 men in 25 days = 150x * 25 = $\frac{1}{4}$ work => x = $\frac{1}{15000}$

Therefore, 100 men in 60 days do = 100 * 60x = 6000x work = 6/15 = 2/5

Total work done = $\frac{1}{4} + \frac{2}{5} = \frac{13}{20}$

Therefore, Remaining work = 1 - $\frac{13}{20}$ = $\frac{7}{20}$

Remaining time = 130 - (25+60+10) = 35 days

Therefore, work is done in 25 days by 150 men.

Therefore, Work is done in 35 days by 150 men.

Hence, he should employ 50 more men.

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Q:

A and B can do a piece of work in 40 and 50 days. If they work at it an alternate days with A beginning in how many days, the work will be finished ?

Answer

(A+B)'s two days work = $\frac{1}{40} + \frac{1}{50} = \frac{9}{200}$

Evidently, the work done by A and B duing 22 pairs of days

i.e in 44 days = $22 \times \frac{9}{200} = \frac{198}{200}$

Remaining work = $1 - \frac{198}{200}$ = 1/100

Now on 45th day A will have the turn to do 1/100 of the work and this work A will do in $40 \times \frac{1}{100} = \frac{2}{5}$

Therefore, Total time taken = 44 $\frac{2}{5}$ daya

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A) 54	B) 76
C) 72	D) 84

Rohit crosses a 700 metres wide road in 35 seconds. What is his speed (in km/hr)?

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