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Time and Work Questions

1. Vijay can do a piece of work in 24 days. Rakesh can do the same work in 30 days and Vinod in 40 days. Vijay and Vinod worked for 4 days and handed it to Rakesh. Rakesh worked for some days and handed it again to Vijay and Vinod 6 days before completing the work. For how many days did Rakesh work?




SHOW ANSWER
Correct Ans:10 days
Explanation:
Combined work done by Vijay and Vinod in 4 + 6 days (4 initial days and last 6 days)

= 10/24 + 10/40 = 2/3

Remaining work = 1 – 2/3 = 1/3

Rakesh works for ( 1/3 * 30 ) = 10 days

Hence, option B is correct.
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2. Arnab is twice as fast as Bhanu, and Bhanu is one-third as fast as Chandu. If together they can complete work in 30 days, in how many days can Arnab, Bhanu and Chandu do the work respectively?




SHOW ANSWER
Correct Ans:90, 180, 60
Explanation:
Let's Chandu can do the work in 2x days.

Bhanu can do it in 6x days and Arnab can do it in 3x days.

1/2x + 1/ 6x +1/ 3x = 1/30 ⇒ 3 + 1 + 2/6x = 1/30 ⇒ x = 30

So, the time taken to complete the work by Arnab, Bhanu and Chandu is 90, 180, 60 days respectively.

Hence, option C is correct.
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3. A contractor hire Amitabh Arora to complete the work and Amitabh can do the work in 25 days. Amitabh worked for 5 days and after that Bindu singh completed it in 20 days. In how many days will Amitabh and Bindu together finish the work?




SHOW ANSWER
Correct Ans:(25/2) days
Explanation:
Amitabh can do a piece of work in 25 days.

In 5 days he can do 1/5th work.

Rest work = 1/5 – 1/5 = 4

Bindu can complete 4/5th of the work in 20 days so he can do the complete work in 25 days.
Amitabh's 1 day work = 1/25

Bindu's 1 day work = 1/25

(Amitabh and Bindu)'s 1 day work = 1/25 + 1/25 = 2/25

Amitabh and Bindu together can complete the work in = 25/2 days

Hence, option A is correct.

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4. It takes 6 workers a total of 10 hours to assemble a computer, with each working at the same rate. If six workers start at 9.00 am, and one worker per hour is added beginning at 3.00 pm, at what time will the computer assembled?




SHOW ANSWER
Correct Ans:6.00 PM
Explanation:
--> 6 workers complete some work in 10 hours.

--> 1 worker completes the same work in 60 hours.

--> Let the total work be equivalent to 60 man-hours.

--> From 9.00 am to 3.00 pm, all 6 workers work together for 6 hours.

--> Amount of work done by 6 workers in 6 hours is 6 * 6 = 36 man-hours

--> From 3.00 pm to 4.00 pm, 7 man-hours of work will be done.

--> From 4.00 pm to 5.00 pm, 8 man-hours of work will be done.

--> From 5.00 pm to 6.00 pm, 9 man-hours of work will be done.

--> Thus, total amount of work done up to 6.00 pm is 36 + 7 + 8 + 9 = 60 man-hours

--> Thus, the computer will be assembled at 6.00 pm.

--> Hence, option C is correct.
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5. A and B undertake a project worth Rs. 54000 . A alone can do the work in 10 days. They work together for 3 days. After 3 days, B works alone for 3 days and A completes the remaining work in 3 more days.What is the share of B in the earnings?




SHOW ANSWER
Correct Ans:Rs. 21600
Explanation:
----> A can do 10% work in a day. A has worked 6 days in total. And so has B

----> A completed 60% work in 6 days and B did 40% in 6 days.

----> Efficiency of A and B = 6 : 4

----> B's share = 4 /10 * 54000 = 21600
----> Hence, option A is correct.
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6. Radhe does 70% of some work in 15 days. Later, with Shyam’s help, she completes the remaining work in 4 days. In how many days can Shyam alone complete the entire work?         




SHOW ANSWER
Correct Ans:35.3 days
Explanation:
----> Let total work = 150 units

----> Since Radhe does 70% of the work (i.e. 105 units) in 15 days,

----> Radhe = 105/15 = 7 units per day
----> Work left = 150 − 105 = 45 units

----> Let Shyam do x units of work per day. Shyam and Radhe finish the pending work in 4 days.

----> 4(x + 7) = 45

----> 4x = 45 − 28 = 17 i.e. x = 4.25 units

----> Time taken by Shyam to complete the work = 150/4.25 = 35.29 ≈ 35.3 days

----> Note : that 0.3 is equivalent to 5/17 (among the options).

----> Hence, option C is correct.
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7. The hourly wages of a mason have increased by 25%. Since the increase, the number of hours he works daily has reduced by 16%. If he was earning Rs. 120 per day before the increase, how much (in Rs.) is he earning now?




SHOW ANSWER
Correct Ans:126
Explanation:
-----> Daily wages = hourly wages * work hours.

-----> Let the original hourly wages and work hours be Rs. x and y hours respectively.

-----> Since he used to earn Rs. 120 earlier, xy = 120

-----> New hourly wages = Rs. (1.25x) and new working hours = 0.84y

-----> New daily wages = (1.25x)(0.84y) = 1.05xy = 1.05 * 120 = Rs. 126

Hence, option C is correct.
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8. Four examiners can examine a certain number of papers in 10 days by working for 5 hours a day. For how many hours in a day can 2 examiners must work in order to examine twice the number of papers in 20 days?




SHOW ANSWER
Correct Ans:10 hours
Explanation:
---> Formula used M1D1H1W2 = M2D2H2 W1
---> 4*10*5*2P = 2*20*H*P
---> H = 10 hours
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9. If a father or his two daughters or his three kids can finish a work in 88 days, then how many days will a man, a daughter and a kid together take to finish the same work?




SHOW ANSWER
Correct Ans:48 days
Explanation:
According to the question,
2 daughters = 1 father
1 daughter = ½ father and
3 kids = 1 father
1 kid = 1/3 father
1 father + 1 daughter + 1 kid = 1+(1/2)+(1/3) = (11/6) of father
1 father can finish a work in 88 days
(11/6) of father will finish a work in ((88 *6)/11) = 48 days
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10. Pipe A is kept open throughout the time while Pipe B is open for first 8 minutes. After two minutes of closing of pipe B, pipe C is opened and kept open till the tank is full. Pipe D is opened for last 15 minutes. Each pipe fill an equal share of the tank. Pipe C and Pipe D can fill the tank together is 24 minutes. In how much time pipe A alone can fill the tank of same capacity.




SHOW ANSWER
Correct Ans:80 minutes
Explanation:
Let the pipe A, B, C and D alone can fill the tank in 'a' minutes, 'b' minutes, 'c' minutes and 'd' minutes respectively.
Let the pipe A kept opened for 't' minutes.
Time that pipe B kept opened = 8 minutes
Time that pipe C kept opened = t - (8 + 2) = (t - 10) minutes
Time that pipe D kept opened = 15 minutes
As per the question,
(t/a) + (8/b) + ((t - 10)/c) + (15/d) = 1

Also given that each pipe fill an equal share of the tank,
(t/a) = (8/b) = ((t - 10)/c) = (15/d) = 1/4
8/b = 1/4
=> b = 32 minutes
15/d = 1/4
=> d = 60 minutes

Pipe C and Pipe D can fill the tank together is 24 minutes.
(24/c) + (24/d) = 1
(24/c) + (24/60) = 1
24/c = 1 - (24/60)
24/c = 36/60
c = (24 x 60)/36
c = 40 minutes

((t - 10)/c) = 1/4
((t - 10)/40) = 1/4
t - 10 = 10
t = 20 minutes.
So, t/a = 1/4
20/a = 1/4
a = 80 minutes.
Hence, pipe A alone can fill the tank in 80 minutes.
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11. A and B together can complete a work in 14(2/5) days while B and C together can complete the same work in 10(2/7) days. A alone starts work and after 8 days B replaced him. B did the work for next 12 days and the remaining work is completed by C in next 5 days, then find time taken by A, B & C together to complete that work, if C work with 50% of his usual efficiency?




SHOW ANSWER
Correct Ans:10(2/7) days
Explanation:
A and B together can work for = 72/5 days
One day work of (A + B) = 5/72
B and C together can work for = 72/7 days
One day work of (B + C) = 7/72
As per question,
(A + B)8days + (B + C)4days + (C)1 days = Total work
(5*8)/72 + (7*4)/72 + 1/C = 1
(5/9) + (7/18) + 1/C = 1
(10 + 7)/18 + 1/C = 1
1/C = 1 - (17/18)
1/C = 1/18
C's alone can work = 18 days

B's alone can work = 7/72 - 1/18
= (7 - 4)/72
= 3/72
= 1/24
B's alone can work = 24 days.

A's alone can work = 5/72 - 1/24
= (5 - 3)/72
= 2/72
= 1/36
A's alone can work = 36 days.

Total work unit = LCM of (18, 24, 36) = 72 units
Efficiency of A = 72/36 = 2 units/day
Efficiency of B = 72/24 = 3 units/day
Efficiency of C = 72/18 = 4 units/day

Given that if C work with 50% of his usual efficiency.
New efficiency of C = 4*(50/100) = 2 units/day
So, time taken by A, B & C together to complete = 72/(2 + 3 + 2)
= 72/7
= 10(2/7) days.
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12. A road is built by P, Q & R, who are able to build the road in 8 days, 12 days and 10 days respectively when working alone. Every day along with P, Q works for half day, while R works for 2/3 rd of the day. If after four days P & R stopped working, then find the time taken by Q to build the remaining road while working full day?




SHOW ANSWER
Correct Ans:4/5 days
Explanation:
Let the total work be 1 unit.
One day work of P = 1/8
One day work of Q = 1/(12*2) = 1/24
One day work of R = 2/(3*10) = 2/30 = 1/15
One day work done of all together = (1/8) + (1/24) + (1/15)
= (15 + 5 + 8)/120
= 28/120
= 7/30 units

Total part of road built by P, Q and R in 4 days = 4*(7/30)
= 28/30 unit
Remaining part of road = 1 - (28/30)
= (30 - 28)/30
= 2/30
= 1/15 units
Let, Q will take 'X' days to build the remaining part of the road.
(1/12)*X = 1/15
X = 12/15 = 4/5 days.
Hence, Q will take 4/5 days to build the remaining part of the road.
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13. A can complete a work in 10 days, B in 12 days & C in 15 days. All of them began working together. But A left after 2 days of work and B 3 days before completion of the work. How long did the work last?




SHOW ANSWER
Correct Ans:7
Explanation:
Let the total work be 60. (LCM of 10, 12 & 15)
Therefore efficiency of A working alone = 60/10 = 6 work per day
Similarly efficiency of B working alone = 60/12 = 5 work per day
& efficiency of C working alone = 60/15 = 4 work per day
For the first 2 days all three worked.
Therefore work completed by them in 2 days = 2 *(6+5+4) = 30 work
For last 3 days C worked alone = 12 work
Remaining work = 60 - (30+12) = 18 work
This work was done by B+C = 18/(5+4) = 18/9 = 2 days
Hence total time taken = 2 + 2+ 3 = 7 days
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14. Either 6 men or 17 women can paint a wall in 33 days. The number of days required to paint three such walls by 12 men and 32 women working at same rate?




SHOW ANSWER
Correct Ans:25.5
Explanation:
6 men = 17 women
1 m = 17/6 women
Total work = 17*33
One day efficiency of 12 men and 32 women = 12 m + 32 w
= (12*17)/6 w + 32 w = 34 w + 32 w = 66 w
Days required for 12 men and 32 women = (17*33)/66 = 8.5 days
The number of days required to paint three such walls = 8.5*3 = 25.5 days
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15. "˜B"™ is 50% less efficient than "˜A"™ and "˜C"™ is 25% more efficient than "˜B"™. "˜B"™ alone can complete the work in 60 days. If "˜A"™, "˜B"™, "˜C"™ together started the work, "˜A"™ left 1 day before and "˜B"™ left 6 days before completion of work, then find number of days for which "˜A"™ worked?




SHOW ANSWER
Correct Ans:15 days
Explanation:
Let efficiency of A be 200x
Efficiency of B = 200x * (50/100) = 100x
Efficiency of C = 100x * (125/100) = 125x

Ratio of their efficiency =
A : B : C = 200x : 100x : 125x = 8 : 4 : 5
Work = 1/Efficiency
Ratio of their work = A : B : C = 1/8 : 1/4 : 1/5

Given that, B alone complete the work in 60 days
A alone can complete work in = 60 * (4/8) = 30 days
C alone can complete work in = 60 * (4/5) = 48 days

Total unit = LCM (60, 30, 48) = 240 units
On last day, C alone completes work = 5 unit
Work done by A and C together in 5 days = (8 + 5) * 5 = 65 unit
Remaining work was completed by A, B and C
= (240 - 65 - 5) / (8 + 4 + 5)
= 170/17 = 10 days
Number of days for which A worked = 10 + 5 = 15 days
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16. A tank has four inlet pipes. Through the first three inlets pipes opened together, tank can be filled in 12 min and through the last three inlets, tank can be filled in 15 minutes and through the first and last inlets, tank can be filled in 24 minutes. Then find the time taken by last pipe to fill half of the tank?




SHOW ANSWER
Correct Ans:40 min
Explanation:
Let the four inlet pipes be A, B, C and D respectively.
Given, A + B + C together can fill the tank in = 12 min
B + C + D together can fill the tank in = 15 min
A + D together can fill the tank in = 24 min
Thus, the Total capacity of the tank = 120 units [---> L.C.M of time taken by all 4 inlet pipes be A, B, C and D]

Now, Efficiency of inlet pipes A + B + C = Total capacity of the tank/ Time taken by A + B + C together to fill the tank
---> Efficiency of inlet pipes A + B + C = 120/12 = 10 units/minute
Efficiency of inlet pipes B + C + D = 120/15 = 8 units/minute
Efficiency of inlet pipes A + D = 120/24 = 5 units/minute

On Adding all these efficiencies, we get,
2 (A + B + C + D) = 23 units/minute
---> Efficiency of (A + B + C + D) = (23/2) units/minute
---> Efficiency of D = (23/2) - Efficiency of (A + B + C)
----> Efficiency of D = (23/2) - 10
----> Efficiency of D = 3/2 units/minute

Now, Time taken by last pipe (i.e., Pipe D) to fill half of the tank = (1/2) * [Total capacity of the tank/Efficiency of pipe D]
= (1/2) * [120/(3/2)]
= (1/2) * [(120 * 2)/3]
= 40 min
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17. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:




SHOW ANSWER
Correct Ans:15 hours
Explanation:
Let the first pipe alone takes ‘x’ hours to fill the tank.
Given, second pipe fills the tank 5 hours faster than the first pipe
---> second pipe alone takes ‘(x - 5)’ hours to fill the tank.

Given, second pipe fills the tank 4 hours slower than the third pipe.
---> Third pipe fills the tank 4 hours faster than the second pipe i.e, (x - 5 - 4)
---> Third pipe alone takes ‘(x - 9)’ hours to fill the tank.

Given, Time taken by first and second pipe to fill the tank = Time taken by third pipe alone
----> [1/x] + [1/(x - 5)] = [1/(x - 9)]
----> [(x - 5) + x] / [x (x - 5)] = [1/(x - 9)]
---> [2x - 5] / [x (x - 5)] = [1/(x - 9)]
---> [2x - 5] * (x - 9)= [x (x - 5)]
----> 2x2 - 18x - 5x + 45 = x2 - 5x
---> x2 - 18x + 45 = 0
---> x2 - 15x - 3x + 45 = 0
---> x(x - 15) - 3(x - 15) = 0
---> (x - 15) (x - 3) = 0
[neglecting x = 3]
----> x = 15
Thus, The time required by the first pipe to fill the tank = 15 hours.
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18. A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m3. The emptying capacity of the pump is 10 m3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?




SHOW ANSWER
Correct Ans:50 m3 per minute
Explanation:
Let the filling capacity of the pump = x m3 per minute
Then the emptying capacity of the pump = (x + 10) m3 per minute

Given, capacity of the tank is 2400 m3

Time required for filling the tank = Total capacity of the tank / filling capacity of the pump = (2400/x) minutes
Time required for emptying the tank = Total capacity of the tank / emptying capacity of the pump = [2400/ (x + 10)] minutes

Given, Pump needs 8 minutes lesser to empty the tank than it needs to fill it
---> Time required for emptying the tank = Time required for filling the tank - 8 minutes
---> [2400/ (x + 10)] = (2400/x) - 8
----> (2400/x) - [2400/ (x + 10)] = 8
----> (300/x) - [300/ (x + 10)] = 1
---> [300(x + 10) - 300x] / x(x + 10) = 1
---> [300(x + 10) - 300x] = x(x + 10)
---> 300x + 3000 - 300x = x2 + 10x
---> 3000 = x2 + 10x
---> x2 + 10x - 3000 = 0
---> x2 + 60x - 50x - 3000 = 0
---> x(x + 60) - 50(x + 60) = 0
---> (x + 60) (x - 50) = 0
---> x = -60, 50
Since, the value of x (ie., filling capacity of the pump) can not be negative, x = 50
Thus, the filling capacity of the pump = 50 m3 per minute.
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19. A can do a work in 21 days and B can do the same work in 14 days, working together in how many days they would complete the work?




SHOW ANSWER
Correct Ans:(42/5) days
Explanation:
Let the total work = 42 units ( L.C.M of 21, 14)
----> Efficiency of A = (42/21) = 2 units/day
----> Efficiency of B = (42/14) = 3 units/day
----> Total efficiency of (2+3) = 5 units
----> No. of days taken by them to complete the work = ( 42/5) days

Hence the answer is : ( 42/5) days
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20. 18 men can earn Rs. 360 in 5 days. How much money will 15 men earn in 9 days?




SHOW ANSWER
Correct Ans:540
Explanation:
Use this formula
--->((W1/M1 D1)) = ((W2/M2 D2))
----> (360/ (18 * 5)) = ( W2/ ( 15 *9))
----> W2 = 540

Hence 15 men earn Rs.540 in 9 days
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