Time & Work Sums

Some short tricks are given below to remember to solve the problems of Time & work with in a minute.
a) If A can do a work in X days and B can do the same work in Y days then Both can do the same work together in \frac{x*y}{x+y}.
b) If A and B together can do a work in x days and A alone can do the same work in y days then B alone can do the same work in \frac{x*y}{x-y}.
c)A can do a work in x days, B and C can do the same work in y and Z days respectively. Then A, B and C working together can do the work in \frac{xyz}{xy+yz+xz}.
d) A and B together can do a work in x days, B and C in y days and C and A in Z days then working together they can \frac{2xyz}{xy+yz+xz}.
e) A can do a work in x days B is y% more efficient then A. B can do the same work in \frac{100}{100+y}.*x
f) A can do a work in x days and B can do in y days. both start the work working together. A left the work after t days then time taken to complete the work in \frac{(x-t) * y}{x}.
g) A can do a work in x days, B in y days and C in Z days then ratio of their working efficiency is yz : xz: xy.
Type-1
1)A can complete a work in 6 days and B alone can complete the same work in 12 days. How many days can A and B together complete the same work.a)5,b)4,c)3,d)2 .
 

 
A and B together can do in \frac { A*B }{ A+B }
\frac { 6*12 }{ 6+12 }

= 4 days.
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2)A can do a work in 12 days, B can do the same work in 15 days and C can do the same work in 20 days. Working together in how many days they can do the same work. a)10,b)5,c)3,d)8.

In One Day they can do work=
\frac { 1 }{ 12 } +\frac { 1 }{ 15 } +\frac { 1 }{ 20 }
i.e \frac{1}{5} part of work 
so total work completed in 5 days.
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3)A and B working together can do a piece of work in 6 days. B alone can do it in 8 days, in how many days A alone could finish. a)12,b)10,c)24,d)16

A alone can do it in \frac { 8*6 }{ 8-6 } ;
= 24 days.
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4)A and B can finish a field work in 30 days, B and C in 40 days while C and A in 60 days. How long will they take to finish it together?.a)41/3,b)20/3,c)80/3,d)60/3.

2(A+B+C) in one day work = \frac { 1 }{ 30 } +\frac { 1 }{ 40 } +\frac { 1 }{ 60 }
=\frac { 3 }{ 40 } part.
(A+B+C) one day work =\frac{3}{80} part
So they can finish work in \frac{80}{3} days.
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5)A and B together can do a piece of work in 15 days. B alone can do it in 20 days. In how many days can A alone do it. a)40,b)50,c)60,d)65.

A alone can do it \frac{20*15}{20-15}
= 60 days
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6)Pipe P and Q can fill a tank in 10 and 12 hours respectively and R can empty it in 6 hours. If all the three are opened at 7 AM, At what time will one fourth of the leak tank be filled?.
a) 10AM, b)10PM,c)11AM,d)11PM.

Part of tank filled in 1 hour = \frac{1}{P} +\frac{1}{Q} -\frac{1}{R}
=\frac{1}{10} +\frac{1}{12} -\frac{1}{6}
= \frac{1}{60} part
SO 1/4 part of tank filled in 60 * \frac{1}{4}
15 hours
The time is 7AM + 15 hrs
i.e 10 PM.  
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7)A and B can do a piece of work in 36 days, B and C can do it in 60 days, A and C can do it in 45 days. C alone can do it in. a)90,b)180,c)120,d)150days

2(A+B+C) one day work=\frac{1}{36} +\frac{1}{60}+\frac{1}{45}
=\frac{12}{180}
=\frac{1}{15}
(A+B+C) one day work =\frac{1}{30}
So they can Complete work in 30 days.
Now C can complete work alone in \frac{36*30}{36-30}
= 180 days.
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8) A and B can do a piece of work in 10 days and 6 days respectively. After they have worked together for 3 days both of them left and C alone completed the rest work in 2 Days. C can alone complete the same work in.a)6,b)8,c)10,d)16.

Part of work by A & B together in 3 days = 3 * \frac{10+6}{10*6}
= \frac{4}{5} part
Remaining 1/5 part is completed by C in 2 days
C alone can complete in 2*5 =10 days.
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9) A can do a piece of work in 20 days and B in 30 days. They work together for 7 days and then leave the work. Then C alone finishes the remaining work in 10 days. In how many days will C finish the full work?a)25,b)30,c)24,d)20.

(A+B) in 7 days completed the part of work =\frac{30+20}{30∗20}*7
= \frac{7}{12}
Remaining \frac {5}{12} part is completed in 10 days by C alone.
C can complete the whole work in \frac{12}{5}*10
=24 days.
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10) A and B can together finish the work in 30 days. They worked together for 20 days and then left. After another 20 days, A finished the remaining work. The number of days in which B alone can finish the work is.a)54,b)60,c)50,d)48. 

In 20 Days the part of work is done (A+B) =\frac{2}{3} part
Remaining \frac{1}{3} is completed by A in 20 days
So A can complete the whole work in 20*3 = 60 days.
B can complete the whole work in \frac{60−30}{60∗30} days
= 60 days.

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Type-2
1)   A,B and C can do a piece of working in 20, 30 and 60 days respectively. In how many days A can do the work If he is assisted by B and C on every third day?,a)15,b)20,c)10,d)12.
 

 
2 days work done by A =  \frac {2}{20}
\frac {1}{10} part.
On every third day part of work done by (A+B+C) = \frac{1}{20}+\frac{1}{30}+\frac{1}{60}
\frac {1}{10} part.
therefore total part of work in 3 days =  \frac {1}{10}+ \frac {1}{10}
\frac {1}{5} part
total work completed in  = 5*3
=15 days
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2) Three pipes A,B,C can fill a tank in 12,15 &20 Hrs respectively. If A is open all the time and B & C are opened for 1 hr each alternately. The tank will be full in.a)3 and 2/3, b)4 and 2/5, c)5 hrs, d)7 hrs.

 
In first hour  part of tank fill = \frac{12+15}{12*15}
= \frac{3}{20}
in 2nd hour part of tank fill = \frac{12+20}{12*20}
=\frac{2}{15}
\frac{17}{60} part
In 3 round i.e 6 hours part of tank filled = 3* \frac{17}{60}
=  \frac{17}{20}
remaining part = 1- \frac{17}{20}
\frac{3}{20} part
which is filled by A and B in next 1 hour
so total time taken = 6+1
=7 hours.
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3) Two pipes A and B can fill a tank in 4 and 5 hrs respectively. If they are turned up alternatively for one hrs each, find the time taken to fill the tank.
 

Part of tank filled in 1st 2 hours = \frac {9}{20}
In two rounds i.e 4 hours tank is filled = 2*\frac {9}{20}
=\frac {9}{10}
Remaining \frac {1}{10} part is filled by A =  4*\frac {1}{10}
=\frac {2}{5}.
=24 min
so total time taken = 4 hours and 24 min.
=24 min
so total time taken = 4 hours and 24 min.
=24 min
so total time taken = 4 hours and 24 min.
 
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4) A tank has two pipes. The first can fill it in 40 minutes and the second can empty it in 1 Hour. If the pipes are opened alterable minutes, find the time taken to fill the tank.
 

In first 2 min part of tank filled = \frac{60-40}{60*40}</strong><br><strong>= [latex]\frac{1}{120}
so total time taken to fill = \frac{120}{2}
= 60 hours.
 
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5) A pipe can fill a tank in 12 minutes and another pipe in 15 minutes, but a third pipe can empty it in 6 minutes. The first two pipes are kept open for 5 minutes in the beginning and then the third pipe is also opened. In what time is the cistern emptied. a)40,b)45,c)35,d)30.

In 5 min part of tank filled = \frac{12+15}{12*15}* 5
=\frac{3}{4} part
All the 3 pipes open together the tank will be empty in
\frac { \frac { 20 }{ 3 } *6 }{ \frac { 20 }{ 3 } -6 }
= 60 min
so \frac{3}{4} part will be empty in
\frac{3}{4}*60
=45 mins
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6) Two taps can fill a cistern in 6 and 7 min, respectively, If  these taps are opened alternatively for a minute in what time will the cistern be filled.

In 2 min the part of cistern is filled = \frac{7+6}{7*6}
therefore in 3 rounds i.e  6 min part of cistern filled= 3*\frac{13}{42}
= \frac{13}{14}
remaining part = \frac{1}{14} was filled by 2nd pipe in \frac{1}{14}*6
\frac{3}{7}
total time taken = 6 &\frac{3}{7} min
 
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7) Alen and Border can do a piece of work individually in 21 and 42 days respectively. In how many days they can complete the work, working together alternatively. a)14,b)28,c)42,d)35


In 2 days Alen and Border can complete the part of work = \frac{42+21}{42*21}
= \frac{1}{14}
so alternatively time is taken =2*14
=28 days.
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Type – 3
1) A can do a piece of work in 6 days. B is 25% more efficient than A. How long would B alone take to finish this work? a)5and1/4,b)2and2/3,c)4 and 4/5 ,d)3and1/3
 

Let B will complete in 100 days then A will 125.
A:B = 125 : 100
6 : B  = 125 : 100
So B = 24/5
= 4 and 4/5.
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2)  A can do a certain job in 12 days. B is 60% more efficient than A. To do the same job B alone would take. a)8,b)10,c)7,d)7and1/2.

 

Let B will Complete in 100 days so A will in 160 days.
A : B  = 160 : 100
12 : B = 8 : 5
so B = 15/2
= 7 and 1/2
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3) A take twice as much as B and thrice as much as C to finish a piece of work.Working together, they can finish the work in 2days. The number of day required by B to do the work alone is:-a)4,b)6,c)8,d)12
 

Let C will work in = X days
So A will work= 3X days
and B will work In \frac{3X}{2} days.
All together they will work = 2 days
\frac{1}{3x}+\frac{2}{3x}+\frac{1}{x}=\frac{1}{2}
so X = 4
Time taken by B = 6 days.
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4) A is thrice as fast as B, and is therefore able to finish a work in 60 days less then B. Find the time in which they can do  working together. a)6,b)7,c)8,d)22 and 1/2.
 

Let B will complete in 90 days and So A will in 30 days.
Working together they will take = \frac{90*30}{120}
=22 \frac{1}{2}
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5) A does half as much work as B in three- fourth of the time. If together they take 18 days to complete a work. How much time shall B take to do it. A)30,b)45,c)35,d)40
  

Let B do whole work in X days
Therefore A will do half of the work in 3X/4 days.
So, A will do full work in 3X/2 days.
Together they take 18 days to complete
\frac{X*\frac{3X}{2}}{X+\frac{3X}{2}}
X= 30 Days.
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6) A takes twice as much as B and thrice as much time as C to finish a piece of work . Working together, They can finish the work in 2 days. The number of days required to B to do the work alone is. a) 4,b) 6,c) 8,d)12
  

Let C will work in = X days
So A will work= 3X days
and B will work In \frac{3X}{2} days.
All together they will work = 2 days
\frac{1}{3x}+\frac{2}{3x}+\frac{1}{x}=\frac{1}{2}
so X = 4
Time taken by B = 6 days.
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7)Babu and Asha can do a job together in 7 day. Asha is 1 and ¾ time as efficient as Babu. The same job can be done by Asha alone in. 
a) 49/4,b) 49/3,c)11,d)10 and1/2.

 

Let Asha Will do in X days then Babu will do in 7X/4 days.
Working together they will do in 7 days
\frac{X*\frac{7X}{4}}{X+\frac{7X}{4}} = 7
So X= 11 Days.
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8) Peter can do a piece of work in 12 days. Mohan is found to be 50% more efficient than Peter. If Mohan is given the piece of work how long will he take to do it.a)9,b)10,c)8,d)6.
 

let Mohan will do in 100 days
then peter will do 150 days
Ratio of both are
12 : X  = 150 : 100
so X = 8 days.

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10)Raju is twice as good as Ajay, Together they finish the work in 14 days. In how many days can Raju alone do the same work. a)16,b)21,c)32,d)22.
 


Let  Raju will do in X days
Then Ajay will do in  2X days
together they will work in 14 days
so \frac{X*2X}{X+2X} = 14
therefore X= 21 days.

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Type – 4
1)      If 4 men or 6 women can do a piece of work in 12 days working 7 hrs a day. How many days will it take to complete a work twice as large with 10 men and 3 women working together 8 hrs
a day. A)6,b)7,c)8,d)10.


6 Women = 4 Men
so 3 Women = 2 Men
4 Men can do a work in 12*7 
10 Men+ 3 women i.e 12 men in 8 hrs can do \frac{12*7*4}{12*8}
= 7/2 days
Twice work can do in 7 days.
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2) 1 man or 2 woman or 3 boys can complete a piece of work in 88 days, then 1 man, 1 woman and 1 boy together will complete it in. a)36,b)42,c)48,d)54.

1 man = 3 boy & 1 woman = 3/2 boy.
3 boy can complete in 88 days
so 1 man +1 woman+ 1 boy i.e 11/2 boy can complete in \frac{88*3*2}{11}
= 48 days.
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3) 3 men or 4 boys or 6 women take 10 days to do a job, how much time would they take to  complete the same job working together.a)10/3,b)4,c)11/3,d)none of these.


3 men = 4 boy =6 women
total 9  men
3 men can do  a job in 10 days
so 9 men can do in \frac{10*3}{9}
= 10/3 days
.
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4) 3 men or 4 women can plough a field in 43 days, how long will 7 men and 5 women take to plough it. a)10,b)11,c)9,d)12.

3 men = 4 women
so 7 men = 28/3 women
now, 4 women can plough in 43 days
(5+28/3) women can plough = \frac{43*4*3}{43}
= 12 days.
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5)A wall of 100 meters can be built by 7 men or 10 women in 10 days. How many days 14 men and 20 women take to build a wall of 600 meters.a)15,b)20,c)25,d)30.

7 men = 10 women 
so, 14 men = 20 women
20 women can built 100 meters wall in 10 days
so, (14 men + 20 women) ie 40 women can built 600 meters wall in \frac{10*10}{100}*\frac{600}{40}
= 15 days.

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6)1 man or 2 woman or 3 boys can do a work in 44 days. Then in how many days will 1 man, 1 woman and 1 boy do the work. a)42,b)20,c)24,d)34.

1 man = 3 boy & 1 women = 3/2 boy.
Now, 3 boys can do a work in 44 days
(1 man+ 1 Woman+ 1 boy) i.e 11/2 boy can do work in \frac{44*2*3}{11}
=24 days

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7) If 10 men or 20 boys can make 260 mats in 20 days, then how many mats will be made by 8 men and 4 boys in 20 days.a)260,b)240,c)280,d)520.

10 Men = 20 Boys
20 Boys can make 260 mats in 20 days.
(8 men + 4 boys) i.e 20 boys made in 20 days = 260 mats.

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8) if 3 men or 6 women can do a piece of work in 16 days, in how many days can 12 men and 8 women do the same work.
a) 4,b) 5,c)3,d)2.

3 men = 6 women
so 12 men = 24 women.
Now, 6 women can do a piece of work in 16 days
(12 men+ 8 women) i.e 32 women can do it in \frac{16*6}{32}
= 3 days.

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9) If 2 men or 3 women or 4 boys can do a piece of work in 52 days, then the same piece of work will be done in 1 man, 1 woman and 1 boy in a)48,b)36,c)45,d)None of these.


2 Men = 4 Boy so 1 men = 2 boy.
1 women = 4/3 boy.
Now, 4 boys can do a piece of work in 52 days
so( 1 man + 1 women + 1 Boy)i.e 13/3 boy can do in \frac{52*4*3}{13}
= 48 days

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Type -5
1) 50 persons can do a piece of work in 12 days, working 6 hr/day. 60 persons can do the same work in 8 days working X hr/day. Find the value of X a)15,b)5,c)7and1/2,d)5and2/3.

M1 * T1 * W2 = M2 * T2 * W1
50 * 12 * 6 = 60 * 8 *X
X = 15/2 days.

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2) A certain work can be done in a certain times by 25 men, with 5 men less, it could have done in 3 days more. In what time can it be done by 40 men. A)5and1/2,b)7and1/2,c)9and1/2,d)12and1/2.

Let time taken = X
25 * X = 20 * (X+3)
X = 12 days.
Let 40 men do it in Y days
25 * 12 = 40 * y
so, Y = 15/2 day
s
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3) A certain work can be done in a certain time by 36 men, But had there been 9 men more, it could have been done in 5 days less. In how many days can be 20 men do it. a)30,b)35,c)40,d)45.


Let time taken = X
Now, 36 * X = 45  (X-5)
X= 25 days.
Let 20 men do it in Y days.
So, 36 * 25 = 20 * Y
Y = 45 days.

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4) To complete a task in 45 days, A contractor employs  45 people for the same. Upon reviewing the work after 30 days, he notices that only half of the task is completed. In order to meet the deadline of 45 days how many extra people must be employ now. a)25,b)15,c)60,d)45.

M1 * T1 * W2 = M2 * T2 * W1
45 * 30 * 1/2 = X * 15 * 1/2
X = 90 men
Extra people = 45 .

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5)A can finish a work in 12 days and B can do it in 15 days. After A had worked for 3 days, B also joined to A to finish the remaining work. In how many days, the remaining work will be finished .a)3,b)4,c)5,d)6.
 

A and B together can do part of work in one day =  \frac{12+15}{12*15}
= \frac{3}{20}
Now part of work done by A in 3 days = 1/4
remaining part of work = 3/4
remaining work will be finished in = \frac{20/3}{3/4}
= 5 days

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6) If 8 women collect 200 kg of tea leaves in 10 hrs. How many more of tea leaves will 12 women collect in 8 hrs. a)24kg,b)40kg,c)50kg,d)100kg.
 

M1 * T1 * W2 = M2 * T2 * W1
8 * 10 * X = 12 * 8 * 200
So X = 240 kg0
40 kg more tea leaves.

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7) A contractor undertook to complete the work in 40 days and he deployed 20 men for his work. 8 days before the scheduled time he realized that 1/3rd of the work was still to be done. How many more men were required to complete the work in stipulated time.

A)16,b)15,c)20,d)25
M1 * T1 * W2 = M2 * T2 * W1
20 * 32 * 1/3  =  X * 8 * 2/3
so X = 20 men 
Required more men = 20 .

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8)A group of worker can complete a job in 120 days. If there were 4 more such workers then the work could be finished in 12 days less. What was the actual strength of workers. a)30,b)40,c)42,d)36.

Let X workers can complete in 120 days.
so (X+4) worker can complete in 108 days.
120 X = (X+4)*108
12 X = 4*108
X= 36 days.

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