Time and Work
Exam Duration: 45 Mins Total Questions : 30
A can type a report in 16 h and B can do the same job in 12 h. With the help of C they complete the report in 6 h. Then, C alone can type the report in how many hours?
- (a)
48 h
- (b)
42 h
- (c)
45 h
- (d)
49 h
A's 1 h work \(=\frac { 1 }{ 16 } \)
B's 1 h work \(=\frac { 1 }{ 12 } \)
(A+B+C)'s 1 h work \(=\frac { 1 }{ 6 } \)
C's 1 h work \(=\frac { 1 }{ 6 } -\frac { 1 }{ 16 } -\frac { 1 }{ 12 } =\frac { 1 }{ 48 } \)
Working together P and Q can do a job is 6 days, Q and R can do the same job in 10 days while while P and R can do it in 5 days. How long will it take if all of them work together to complete the job?
- (a)
\(4\frac { 2 }{ 7 } \)
- (b)
\(4\frac { 3 }{ 7 } \)
- (c)
\(4\frac { 4 }{ 7 } \)
- (d)
\(4\frac { 5 }{ 7 } \)
(P+Q)'s 1 day's work \(=\frac { 1 }{ 6 } \)
(Q+R)'s 1 day's work \(=\frac { 1 }{ 10 } \)
(P+R)'s 1 day's work \(=\frac { 1 }{ 5 } \)
Adding 2(P+Q+R)'s 1 day's work \(=\frac { 1 }{ 6 } +\frac { 1 }{ 10 } +\frac { 1 }{ 5 } =\frac { 7 }{ 15 } \)
(P+Q+R)'s 1 day's work \(=\frac { 7 }{ 30 } \)
(P+Q+R) can do the entire work in \(=\frac { 30 }{ 7 } \)
\(=4\frac { 2 }{ 7 } \) days
Sudha can do a work in 10 days and Rani can do the same work in 40 days respectively. If they work at it on alternate days Sudha beginning the work, then in how many days the work, then in how many days the work will be completed?
- (a)
16 days
- (b)
8 days
- (c)
12 days
- (d)
14 days
Sudh's 1 day's work \(=\frac { 1 }{ 10 } \)
Rani's 1 day's work \(=\frac { 1 }{ 40 } \)
(Sudha + Rani)'s 2 day's work \(=\frac { 1 }{ 10 } +\frac { 1 }{ 40 } =\frac { 1 }{ 8 } \)
Time taken to complete the work = 8x2=16 days
A is twice as good a workmen as B and together they finish a piece of work in 24 days. In how many days will B alone finish the work?
- (a)
63 days
- (b)
75 days
- (c)
60 days
- (d)
72 days
(A's 1 day work) : (B's 1 days work) = 2:1
(A+B)'s 1 day work \(=\frac { 1 }{ 24 } \)
B's 1 days work \(=\frac { 1 }{ 3 } \times \frac { 1 }{ 24 } =\frac { 1 }{ 72 } \)
A can do a piece of work in 2 days and B in 16 days. After working for 4 days with A, B leaves the work. In how many more days will A finish the remaining work?
- (a)
5 days
- (b)
12 days
- (c)
8 days
- (d)
9 days
(A+B)'s 4 days work \(=4\left( \frac { 1 }{ 12 } +\frac { 1 }{ 16 } \right) =\frac { 7 }{ 12 } \)
Remaining work which is done by A alone
\(=1-\frac { 7 }{ 12 } =\frac { 5 }{ 12 } \)
\(\frac { 5 }{ 12 } \) of the work will be done by A in \(=\frac { 5 }{ 12 } \times 12\)
= 5 days
Anita can do a work in 30 days. Sunita is 25% more efficient, then Anita. The number of days taken by Sunita to do the same work is
- (a)
24 days
- (b)
30 days
- (c)
27 days
- (d)
33 days
Let, the efficency of Anita = 100
The efficency of Sunita = 125
Ratio of efficency of Anita:Sunita =100:125=4:5
Ratio of time of Anita:Sunita = 5:4
So, if Anita takes 5 days, Sunita takes 4 days to do the same work
Now, if Anita takes 30 days, Sunita will take 24 days
Two friends A and B workng together completed a work in 26 days. Thei skills of doing the work are in the ratio 8:5. How many days will B take, if engaged alone?
- (a)
10 days
- (b)
16 days
- (c)
12 days
- (d)
18 days
Skill ratio A and B = 8:5
Work done by B is \(\frac { 5 }{ 13 } \) of the whole which he does in = 26 days
number of days taken by B to do the whole work \(=\frac { 26\times 5 }{ 13 } =10days\)
B working alone can do the work in 10 days.
A can do a piece of work in 120 days and B can do it in 150 days. They work together for 20 days. Then, B leaves and A continues the work in completed in 48 more days. In how days can C do it, if he works alone?
- (a)
240 days
- (b)
225 days
- (c)
250 days
- (d)
260 days
(A+B)'s 20 day's work \(=20\left( \frac { 1 }{ 120 } +\frac { 1 }{ 150 } \right) =\frac { 3 }{ 10 } \)
A's 12 day's work \(=\frac { 12 }{ 120 } =\frac { 1 }{ 10 } \)
Work completed \(=\frac { 3 }{ 10 } +\frac { 1 }{ 10 } =\frac { 2 }{ 5 } \)
Work remaining \(=1-\frac { 2 }{ 5 } =\frac { 3 }{ 5 } \)
Let C do the work in x days.
(A+C)'s 48 days's work = Remaining work
\(48\left( \frac { 1 }{ 120 } +\frac { 1 }{ C } \right) =\frac { 3 }{ 5 } \Rightarrow \frac { 1 }{ C } =\frac { 1 }{ 240 } \)
or C can do the whole work in 240 days.
X can do a work in 4 days whle Y can do the same work in 9 days. Both of them finish the worj together and get Rs. 2600. Find the sahres of X and Y respectively.
- (a)
Rs 1800, Rs 800
- (b)
Rs 2000, Rs 600
- (c)
Rs 1500, Rs 1100
- (d)
Rs 1900, Rs 700
The ratio which they share the money is the ratio of work done per day i.e., the ratio of their efficiencies = 9:4
X's share \(=\frac { 9 }{ 13 } \times 2600=Rs.1800\)
Y's share \(=\frac { 4 }{ 13 } \times 2600=Rs.800\)
A and B workng together can finish a job in T days. If A works alone and completes the job he will take (T+5) days. If B works alone and completes the same job, he will take (T+45) days. Find T.
- (a)
15 days
- (b)
20 days
- (c)
16 days
- (d)
18 days
A's 1 day's work \(=\frac { 1 }{ T+5 } days\)
B's 1 day's work \(=\frac { 1 }{ T+45 } days\)
(A+B)'s 1 day's work \(=\frac { 1 }{ T+5 } +\frac { 1 }{ T+5 } =\frac { 1 }{ T } \)
\(=\frac { 1 }{ (T+5) } +\frac { 1 }{ (T+45) } =\frac { 1 }{ T } \)
On solving, T=15 days
A completes 60% of a task in 15 days and then, takes the help of B and C. B is 50% efficient as A and C is 50% as efficient as B is. In how many more days will they complete the work?
- (a)
\(5\frac { 5 }{ 7 } day\)
- (b)
\(4\frac { 2 }{ 5 } days\)
- (c)
\(6\frac { 2 }{ 7 } days\)
- (d)
\(3\frac { 2 }{ 7 } days\)
Let the efficency of A be 4x, B be 2x and C be x. Ratio of efficency of A, B and C=4x:2x:x=4:2:1
Ratio of time of A, B and C \(=\frac { 1 }{ 4 } :\frac { 1 }{ 2 } :1\) =1:2:4
A completes \(\frac { 60 }{ 100 } \) of work in 15 days
A can do the whole work in \(=\frac { 15\times 100 }{ 60 } =25days\)
If A takes 25 days, B takes 50 days and C takes 100 days to complete the work.
Work completed by A =60% = \(\frac { 60 }{ 100 } =\frac { 3 }{ 5 } \)
Remaining work \(=\frac { 2 }{ 5 } \)
(A+B+C)'s 1 day's work \(=\frac { 1 }{ 25 } +\frac { 1 }{ 50 } +\frac { 1 }{ 100 } =\frac { 7 }{ 100 } \)
(A+B+C) will do \(\frac { 2 }{ 5 } th\) of the remaining work in \(\frac { 2 }{ 5 } \times \frac { 100 }{ 7 } =\frac { 40 }{ 7 } =5\frac { 5 }{ 7 } days\)
4 men and 6 boys can do a piece of work in 8 days and 6 men and 4 boys can do it 7 days. If the daily wages of a voy be Rs 20. What will be the weekly wages of a man?
- (a)
Rs.260
- (b)
Rs.280
- (c)
Rs.240
- (d)
Rs.290
(4m+6b) can do a iece of work in 8 days.
(32m+48b) can do the work in 1 day.
Also,(*6m+4b) can do a piece of work in 7 days.
(42m,+28b) can do the work in 1 day.
32m+48b=42m+28b
10m=20b, 1m=2b
Wages of a man = 2 x 20 =Rs.40
or Rs 40x7 i.e., Rs.280 per week.
A man, a woman and boy can together complete a piece of work in 5 days. If a man alone can do it in 12 days and a boy alone in 15 days, how long will a woman take to complete the work?
- (a)
20 days
- (b)
18 days
- (c)
21 days
- (d)
15 days
(1 man+ 1 woman + 1 boy)'s 1 days work \(=\frac { 1 }{ 5 } \)
Let a woman alone do it in x days.
Then, \(\quad \quad \quad \frac { 1 }{ 12 } +\frac { 1 }{ 15 } +\frac { 1 }{ x } =\frac { 1 }{ 5 } \\ \Rightarrow \quad \quad \frac { 1 }{ x } =\frac { 1 }{ 5 } -\left( \frac { 1 }{ 12 } +\frac { 1 }{ 15 } \right) =\frac { 1 }{ 20 } \)
So, 1 woman can complete the work in 20 days.
A tap can fill a cistern in 12 h. After half the tank is filled, 2 more similar taps are opened. What is the total time taken to fill the tank complete bys?
- (a)
9 h
- (b)
6 h
- (c)
8 h
- (d)
4 h
\(\frac { 1 }{ 2 } \) of the tank can be filled by the tap in 5 h.
Part filled by one tap in 1 \(h=\frac { 1 }{ 2 } \)
Part filled by three taps in \(h=\frac { 1 }{ 4 } \)
\(\frac { 1 }{ 2 } \) tank can be filled by the three pipes \(\frac { 1 }{ 4 } \times 2=2h.\)
Total time taken to fill the tank completely =6+2=8h
An electric pump can fill a reservior in 36 h. Bacause of a leak in the reservior it took 45 h to fill the reservior. When the reservior is full, how much time will the leak take to empty it?
- (a)
180 h
- (b)
175 h
- (c)
185 h
- (d)
170 h
Work done by the leak in \(1h=\left( \frac { 1 }{ 36 } -\frac { 1 }{ 45 } \right) =\frac { 1 }{ 180 } \)
Leak will empty the reservior in 180 h.
A cistern has a leak which would empty it in 10 h. A tap is turened on which admits 4 l a minute into the cistern and it is now emptied in 15 h. How many litres does the cistern hold?
- (a)
7400 L
- (b)
7200 L
- (c)
7100 L
- (d)
7500 L
When both pipes are working the part of the cistern fill in 1th \(=\frac { 1 }{ 15 } \)
The leak can empty \(\frac { 1 }{ 10 } \) of the cistern in 1h.
Work done by a filling pipe in \(1h=\frac { 1 }{ 10 } -\frac { 1 }{ 15 } =\frac { 1 }{ 30 } \)
So, the cistern is full by the filling pipe in 30 h.
Capacity of the cistern = water admitted by the pipe for 30 h at the rate of 4 L/min
=30x4x60L=7200 L
A water tank of 3000 L capacity is connected to a tap which can fill it at the rate of 35L/min and water is let out at the same time at the rate of 5 L/min. After 2 h. the outlet is shut off. Find how long will it take now for the tank to become full?
- (a)
25 min
- (b)
26 min
- (c)
22 min
- (d)
24 min
Water filled in the tank in 120 min = 25 x 120 = 3000L
Water let out from the tankl in 120 min = 5 x 120 = 600 L
Remaining water left in the tank = (3000-600) L = 2400L
Ramaining space in the tank = (3000-2400) L = 600 L
Time taken to fill up 600 L of water = 600+25
=24 min
A alone can do a piece of work in 12 days. B alone can do it in 15 days. If the total wages for the entire job are Rs. 9081. If they both finish the work together what is the share of A?
- (a)
5045
- (b)
5046
- (c)
5047
- (d)
5048
Ratio of work done by A and B \(=\frac { 1 }{ 12 } :\frac { 1 }{ 15 } =5:4\)
A's share \(=\frac { 5 }{ 9 } \times Rs.9081=Rs.5045\)
Working together A and B can complete a piece of work in 5 days. If A had worked twice as fast the work would have been completed in 4 days. In how many days can A alone complete the work?
- (a)
20 days
- (b)
\(6\frac { 2 }{ 3 } days\)
- (c)
18 days
- (d)
\(6\frac { 1 }{ 4 } days\)
Let A do the work in a days
A's 1 day's work \(=\frac { 1 }{ a } \)
Let B do the work in b days.
\(\frac { 1 }{ a } +\frac { 1 }{ b } =\frac { 1 }{ 5 } \quad \quad \quad ...(i)\\ \frac { 2 }{ a } +\frac { 1 }{ b } =\frac { 1 }{ 4 } \quad \quad \quad ...(ii)\)
Solving Eqs.(i) and (ii), we get
a = 20 days
Construction of a raod was entrusted to a civil engineer. He was to finish the work in 124 days for which he employees 120 workmen. Two-thirds of the work was completed in 4 days. In how many days can A alone complete the work?
- (a)
80
- (b)
64
- (c)
56
- (d)
24
Total work = 124 x 120 = 14880 men-days
Work completed in 64 days \(=\frac { 2 }{ 3 } \times 14880\)
\(=9920\)
Remaining work for 60 days \(=\frac { 1 }{ 3 } \times 14880\)
=4960 men-days
By formula,\(\frac { { M }_{ 1 }{ D }_{ 1 } }{ { W }_{ 1 } } =\frac { { M }_{ 2 }{ D }_{ 2 } }{ { W }_{ 2 } } ,\frac { 120\times 64 }{ 9920 } =\frac { { M }_{ 2 }\times 60 }{ 4960 } \)
M2 = 64
Men to be reduced =120-64 = 56
Working 7 h daily 24 men can complete a piece of work in 27 days. In how many days would 14 men complete the same piece of work working 9h daily?
- (a)
29 days
- (b)
39 days
- (c)
36 days
- (d)
30 days
Days required \(=\frac { 24\times 27\times 7 }{ 14\times 9 } =36 days\)
In what time would a cistern be filled by three pipes whose diameters are 2 cm, 3 cm, 4 cm, running together, when the largest alone can fill it is 58 min? The amount of water flowing in each pipe is propotional to the square iun each pipe is propotional to the aquare of its diameter.
- (a)
36 min
- (b)
32 min
- (c)
23 min
- (d)
28 min
Pipe whose diameter is 4 cm fills the cistern in 58 min.
Let the capacity of cistern bexL.
Then, flow rate of 4 cm diameter pipe \(=\frac { x }{ 58 } L/min\)
\(\frac { x }{ 58 } \alpha ({ 4) }^{ 2 },\frac { x }{ 58 } =K\times 16,k=\frac { x }{ 928 } \)
Flow rate of 3cm diameter pipe \(=\frac { x }{ 928 } \times { (3) }^{ 2 }L/min\)
Time required by 3cm diameter pipe to fill the cistern
\(=\frac { x }{ 4x } \times 928=232min\)
Time required by all the three pipes working together
\(=\frac { 58\times 103\frac { 1 }{ 9 } \times 232 }{ 58\times 103\frac { 1 }{ 9 } +58\times 232+103\frac { 1 }{ 9 } \times 232 } \\ =\frac { 58\times 928\times 232 }{ 58\times 103+58\times 232\times 9+928\times 232 } =32min\)
There is a leak in the bottom of a cistern. Before the leak, it could be filled in \(4\frac { 1 }{ 2 } h\) . It now takes \(\frac { 1 }{ 2 } h\) longer. If the cistern is full, in how much time would the leakage empty the full cistern?
- (a)
45 h
- (b)
40 h
- (c)
36 h
- (d)
56 h
Time taken to fill the cistern before the leak \(=\frac { 9 }{ 2 } h\)
Work done by the cistern when there is not leak \(=\frac { 2 }{ 9 } \)
Time taken to fill the cistern after the leak = 5 h
Work done by the cistern when there is a leak \(=\frac { 1 }{ 5 } \)
Net work done \(=\frac { 2 }{ 9 } -\frac { 1 }{ 5 } =\frac { 1 }{ 45 } \)
Time taken to empty the full cistern = 45-5 =40 h.
If 15 men or 24 women or 36 boys can do a piece of work in 12 days, working 8 h a day, how many men must be associated with 12 women and 6 boys to do another piece of work \(2\frac { 1 }{ 4 } \) times as great in 30 days working 6 h a day?
- (a)
4
- (b)
8
- (c)
6
- (d)
10
15M = 24W = 36B
\(\Rightarrow \quad 1W=\frac { 5 }{ 8 } M\quad and\quad 1B=\frac { 5 }{ 12 } M\\ \quad \quad \quad \quad 12W+6B=\frac { 5 }{ 8 } \times 12+\frac { 5 }{ 12 } \times 6=10M\)
Now, \({ m }_{ 1 }\times { d }_{ 1 }\times { t }_{ 1 }\times { w }_{ 2 }={ m }_{ 2 }\times { d }_{ 2 }\times t_{ 2 }\times w_{ 1 }\)
Let the number of additional men required be x.
\(\quad 15\times 12\times 8\times \frac { 9 }{ 4 } =(10+x)\times 30\times 6\\ \Rightarrow \quad (10+x)=\frac { 15\times 3\times 8\times 9 }{ 36\times 6 } \Rightarrow x=18-10=8\)
A bath tube can be filled by a cold water pipe in 20 min and by a hot water pipe in 30 min. A person leaves the bathroom after turning on both pipes simultaneously and returns at the moment when the bath tub should be full. Finding however, that the waster pipe has ben open, he now closes it. IN 3 min more the bath tub is full. In what time would the waste pipe empty it?
- (a)
38 min
- (b)
45 min
- (c)
43 min
- (d)
48 min
The usual time ewquired to fill the tank when both the pipes are opened \(=\left( \frac { 1 }{ 20 } +\frac { 1 }{ 30 } \right) =\frac { 1 }{ 12 } \)
It means work done by all the three pipes for 12 min+work done by both the pipes for 3 min = 1 Let the waste pipe takes x min to empty the tank.
Then,
\(=\left( \frac { 1 }{ 20 } +\frac { 1 }{ 30 } \right) =\frac { 1 }{ 12 } \\ \quad \quad 12\left[ \frac { 1 }{ 20 } +\frac { 1 }{ 30 } -\frac { 1 }{ x } \right] +3\left[ \frac { 1 }{ 20 } +\frac { 1 }{ 30 } \right] =1\\ \Rightarrow \quad \quad 12\left[ \frac { 1 }{ 12 } -\frac { 1 }{ x } \right] +3\left[ \frac { 1 }{ 12 } \right] =1\\ \quad \quad \quad 12\left[ \frac { x-12 }{ 12x } \right] =\frac { 3 }{ 4 } \Rightarrow 16x-192=12x\\ or\quad \quad 4x=192\Rightarrow x=\frac { 192 }{ 4 } =48\\ \quad \quad \quad \quad \quad x=48\quad min\)
Two coal loading machines each working 12 h/day for 8 days handle 900 tonnes of coal with an efficency of 90; while 3 other coal loading machines at an efficiency of 80% are set to handle 12000 tonnes of coal in 6 days. Find how many hours per day each should work?
- (a)
20 h/day
- (b)
18 h/day
- (c)
16 h/day
- (d)
14 h/day
\({ m }_{ 1 }\times { d }_{ 1 }\times { t }_{ 1 }\times { w }_{ 2 }\times { e }_{ 1 }={ m }_{ 2 }\times { d }_{ 2 }\times t_{ 2 }\times w_{ 1 }\times { e }_{ 2 }\)
\(\Rightarrow \quad 2\times 12\times 8\times \times 12000\times 0.9=3\times 6\times x\times 9000\times 0.8\\ \Rightarrow \quad \quad x=\frac { 2\times 12\times 8\times 12000\times 0.9 }{ 3\times 6\times 9000\times 0.8 } =16h/days\)
If 36 men can dig a trench 200m long, 3 m wide and 2 m deep in 6 days working 10 h a day, in how many days working 8 h/day will 10 men dig a trench 100 m ling, 4 m wide and 3 m deep?
- (a)
15 days
- (b)
27 days
- (c)
20 days
- (d)
54 days
\(\quad \quad { m }_{ 1 }\times { d }_{ 1 }\times { t }_{ 1 }\times { w }_{ 2 }={ m }_{ 2 }\times { d }_{ 2 }\times { t }_{ 2 }\times { w }_{ 1 }\\ \quad 36\times 6\times 10\times 1200=10\times { d }_{ 2 }\times 8\times 1200\\ \quad \quad \quad \quad { d }_{ 2 }=\frac { 36\times 6\times 10\times 1200 }{ 10\times 8\times 1200 } =27days\)
In a factory men, women and children were employed in the ratio of 8:5:1 to finish a job and their individua wages were in the ratio of 5:2:3. When 20 women were employed, the total daily wages of all amounted to Rs.318. Find the total daily wages in rupees paid to each category.
- (a)
210, 70 and 38
- (b)
240, 60, 18
- (c)
200, 90, 28
- (d)
Cannot be determined
When there are 20 women.
Then, number of men=32
And number of children = 4
Then, wages of each category will be in the following ration= 160:40:12
Then, wages of men \(=\frac { 160 }{ 160+40+12 } \times 318\) =Rs.240
Wages of women \(=\frac { 40 }{ 160+40+12 } \times 318\) =Rs.60
Wages of children \(=\frac { 12 }{ 160+40+12 } \times 318\) =Rs.18
Sita takes tice as much time as Gita to complete a work and Rita does it in the same time as Sita and Gita together. If all three working together can finish the work in 6 days, then the time taken by Gita, Sita and Rita to finifh the work, is
- (a)
18, 36 and 12 days
- (b)
20, 38 and 14 days
- (c)
24, 42 and 18 days
- (d)
None of the above
Let Geeta takes x days to complete the work. Sita takes 2x days to complete the work.
Work done by Geeta in 1 day \(=\frac { 1 }{ x } \)
Work done by Sita in 1 day \(=\frac { 1 }{ 2x } \)
Total work done by them together in one day
\(=\frac { 1 }{ x } +\frac { 1 }{ 2x } =\frac { 3 }{ 2x } \)
Total work will be done in \(\frac { 2x }{ 3 } \) days.
Rita takes \(\frac { 2x }{ 3 } \) days to complete the work.
All three working together takes 6 days to complete the work.
\(\Rightarrow \quad \quad \frac { 1 }{ x } +\frac { 1 }{ 2x } +\frac { 3 }{ 2x } =\frac { 1 }{ 6 } ,\frac { (2+1+3) }{ 2x } =\frac { 1 }{ 6 } =\frac { 6 }{ 2x } =\frac { 1 }{ 6 } \)
x=18 days
2x=36 days,\(\frac { 2x }{ 3 } \) 12days
A cistern is provided by two taps A and B. A can fill it in 20 min and B in 25 min. Both the taps are kept open for 5 min an then the tap B is turned off. The cistern will be completely filled in another
- (a)
11 min
- (b)
10 min
- (c)
15 min
- (d)
12 min
A takes 20 min to fill it
In 1 min, it fills \(=\frac { 1 }{ 20 } th\)
B takes 25 min to fill it
In 1 min, it fills \(=\frac { 1 }{ 25 } th\)
Both together can fill it \(\left( \frac { 1 }{ 20 } +\frac { 1 }{ 25 } \right) \) in one min.
In 5 min, they will fill,
\(\left( \frac { 1 }{ 20 } +\frac { 1 }{ 25 } \right) 5=\frac { (5+4)5 }{ 100 } =\frac { 9 }{ 20 } \)
Work left to be done \(=1-\frac { 9 }{ 20 } =\frac { 11 }{ 20 } \)
A can completely fill it in
\(\frac { 1 }{ 20 } \times x=\frac { 11 }{ 20 } \\ \quad \quad \quad x=\frac { 11 }{ 20 } \times 20\\ \quad \quad \quad \quad =11\quad min\)