CAT Previous Year Questions: Time & Work - CUET

Let total work be 40 units
Anil's day work = 2 unit
Sunil's 1 day work = 1 unit
For the first 3 days work done = 6  unit
Work done by Bimal = 4 unit
Work done by A and S during this time = 30 unit
Days = 30 / 3 = 10 days
Total days = 10 + 3 = 13 days

Let the rate of each filling pipe be 'f ' litres/hr and the rate of each draining pipes be 'd' litres/hr
As per the first condition given in the question, Capacity of tank will be = (6f - 5d ) x 6 …(i)
And as per the second condition given in the question, Capacity of tank will be = (5f - 6d ) x 60 …(ii)
Since the capacity of tank is same in both the cases, on equating (i) and (ii), we have
    ► (6f - 5d) x 6 = (5f - 6d) x 60
    ► or, 6f - 5d = 50f - 60d
    ► or, 44f = 55d
    ► or, 4f = 5d
Therefore, f = 1.25d , putting this value in eq. (i),
Capacity of the tank = (6f - 5d ) x 6 = (7.5d - 5d) x 6 = 15d
When 2 filling and 1 draining pipes work simultaneously, the effective rate of filling the tank = (2f - d ),
Now putting f =1.25d in this eq. we have
Effective rate = (2.5d - d ) = 1.5d
Therefore time required to fill the empty tank = 15d / 1.5d = 10 hours.

Let the efficiency of humans be 'E' and the efficiency of robots be ' R'.
In the first case
Total work done = (15E + 5R) x 30 …(i)
In the second case,
Total work done = (5E + 15R) x 60 …(ii)
Since same work is finished in both the cases, so by equating (i) and (ii), we have :
    ► (15E + 5R) x 30 = (5E + 15R) x 60
    ► (15E + 5R) = (5E + 15R) x 2
    ► or, 15E + 5R = 10E + 30R
    ► or, 5E = 25R
    ► or, E = 5R
Putting this value of E in (I), we have
Total work = (15E + 5R) x 30 = (15E + R) x 30 = 480E
Time taken by 15 humans = 480E / 15E days = 32 days.

Let the work completed by A in a day be 'a' units and that by B be 'b' units.
As per question : A alone works till half the work is completed. Thus, 50% of work is done and rest 50% needs to be completed.
Afterwards, A and B work together for 4 days and B works alone to complete the remaining 5% of the work. This means A and B in 4 days can complete 45% of the work.
Now suppose that the total amount of work to be done is 100 units.
   ► So, 4 a + 4b = 45 …(i)
Because B  needs 25% more time than A to finish a job.
   ► So, 1.25 x b = a …(ii)
Substituting value of a from (ii) in (i), we get
   ► 5b + 4b = 45
   ► or, 9b = 45
   ► or, b = 5 units/day
Hence, B alone can finish the job in 100 / 5 = 20 days.

Let the tank be emptied everyday at 'T' pm. Let 'a' litres/hr' and 'b' litres/hr be the amount of water filled by pump A and pump B, respectively.
As per question, on Monday, A alone completed filling the tank at 8 pm. Therefore, the pump A worked for (8 - T ) hours.
So, the volume of the tank = a x (8 - T ) litres.
Similarly, on Tuesday, B alone completed filling the tank at 6 pm.
Therefore, the pump B worked for (6 - T ) hours.
Hence, the volume of the tank = b x (6 - T ) litres.
On Wednesday, A alone worked till 5 pm. and then B worked alone from 5 pm to 7 pm, to fill the tank.
This means pump A worked for (5 - T ) hours and pump B worked for rest of the 2 hours. Hence, the volume of the tank = a x (5 - t) + (2 x b) litres.
We can say that a x (8 - T ) = b x (6 - T ) = a x (5 - T ) + 2b …(i)
or, a x (8 - T ) = a x (5 - T ) + 2b
or, 3a = 2b…(ii)
Again from (i) a x (8 - T ) = b x (6 - T)
From (ii) we know that 3a = 2b
So, a x (8 -T ) = (3a / 2) x (6 - T)
or, T = 2
Thus, the tank gets emptied at 2 pm daily.
A alone fills the tank at 8pm, thus, A takes 6 hours and pump B alone fills the tank at 6pm thus B takes 4 hours.
Hence, working together, both can fill the tank in (6 x 4 ) / (6 + 4 ) = 2.4 hours or 2 hours and 24 minutes.

Let the rate at which each inlet of type A brings water be 'a' litres per minute and from B 'b' litres per minute.
Then, as per question 30 x (10a + 45b)
= 60 x (8a + 18b)
or, 300a + 450b = 480a + 1080b
or, 180a = 630b
or, 6a = 9b, Thus, a : b = 3 : 2 or b = (2 / 3)a …(i)
Total capacity of tank = 30 x (10a + 45b) = 30[10a + 45 x (2 / 3)a ] = 1200a
When 7 inlets of type A and 27 inlets of type B are open, the total efficiency
= 7a + 27b
= 7a + 27 x (2 / 3) a = 25a
So, time taken = capacity of tank/total efficiency = 1200a / 25a = 48 minutes
Hence, 48 min is the correct answer.

Let the efficiency of Ramesh be R units per day and that of Ganesh be G units per day.
As per question 16 ( R + G) = 17G + 7R + ( R - 0.3R) x 10
16R + 16G = 17G + 7R + 7R
So, G = 2R, So total efficiency when both work together = R + G = R + 2R = 3R
Hence, total work = 16 x 3R = 48R
In second case, where Ramesh got sick after 7 days
Work left after 7 days = 48R - 7 (3R) = 27R
Time Ganesh would take to complete this work alone = 27R / 2R = 13.5 days

Let the rate of work of a person by Y units/day.
Hence, the total work = 120Y .
Given : On first day, only one person works, on the second day two people work and so on.
Hence, the work done on day 1, day 2,… will be Y ,2Y ,3Y …respectively.
Let the total number of days required to complete the work is n.
Thus sum should be equal to 120Y.
So, Y + 2Y + 3Y +…+ nY = 120Y
or, 120Y = Y x n (n - 1) / 2
or, n² + n - 240 = 0 or (n - 15)(n + 16) = 0
So, n = 15 or n = -16
But, n = 15 is the only positive value.
Hence, it will take 15 days to complete the work.

Let the time taken by the outlet pipe to empty the tank fully when only outlet pipe is open = T hours.
Then, as per question [(1 / 8) - (1 / T )] = 1 /10
or, 1 / T = 1 / 8 - 1 /10 = 1 / 40,
therefore the outlet pipe can empty the tank fully in 40 hours.
Hence, time taken by the outlet pipe to make the tank half-full = 40 / 2 = 20 hours.

Amal, Bimal and Kamla can together complete the job in 4 days.
Amal's contribution in 1 day =1 / 10 and Bimal's contribution = 1 / 8.
So Kamal's contribution = 1 / 4 - (1 / 8 + 1 / 10) = 1 / 40

Ratio of their work
= 1 / 10 : 1 / 8 : 1 / 40 = 4 : 5 : 1
So, amount of share for Kamal = 1000 x 1 /10 = ₹ 100

Let the capacity of the tank
   ► = LCM (4 ,10,12,20) = 60
So, work done by A, B, C =15, 6, 5, 3 respectively.
Option 1 : Work Done in a minute
   ► = 15 - 6 - 5 - 3 = 1
So, after 30 minutes, the tank will be half filled.

As per question, M inlet pipes can fill the M / 8^th part of the tank in 1 hour and N outlet pipes can empty N / 12^th part of the tank in 1 hour. If all pipes are left open, in 1 hour, the tank will be filled = M / 8 - N /12 = 1 / 6 (as the tank will take 6 hours to fill). Then, 6 M - 4 N = 8, or, M = (8 + 4 N ) / 6. There will be infinite number of solutions ( N = 1,4 ,7,10,13,16 ...) but none of the above ratios are correct. For example, above equation is solvable when M = 2,N = 1 or M = 4, N = 4. But the ratios 2 : 1 (i.e., M = 2,N = 1 ) or 1 : 1 ( M = 4, N = 4 ) are not correct because the solution of the above equation will not be an integer for values like M = 4 ,N = 2 or M = 6, N = 6.

Let your efficiency of doing some work be t days, then the efficiency of your roommate is 2t days.
So, we have, 

t = 45 days thus, 2t = 90 days

The two inlet A and B and one outlet pipe C, fill or empty the tank as per rates tabulated below :

Given : Two of the pipes are kept open for 20 hours and the third pipe C for a lesser period.
The tank is initially full and finally (volume = (1 / 3) x π x r²h) the water level drops to 1 / 2 and so the volume will become (1 / 3) x π x (r / 2)² x  h / 2 = 1 / 8 {(1 / 3) x π x r²h )}, i.e., 1 / 8^th of the original volume.
This means 7 / 8^th volume of water must have drained out.
From the table above, we can see that net rate of emptying the tank = 3 + 2 - 6 = 1 liter per hr.
So, to drain the full tank out, when all the pipes are open, it will take 24 hrs.
So, to drain 7 / 8^th tank = 7 x 24 / 8 = 21 hrs.
Since, the 7 / 8^th water drained out in 20 hrs only, so one of the two inlet pipes must have opened for lesser time. Assuming that if all the three pipes were open for all 20 hrs, we can see that 1 extra unit was required to be drained or the water was not filled by closing one of the inlet pipes a little early.
From the table, we can see that pipe B if opened for 1 / 2 hrs less, will fill the water by 1 litre less.
So, B works for 19.5 hrs only.

Let there be i and o inlet and outlet pipes respectively.
∴ i + o = 11…(i)
Assume that the capacity of the tank is 35 units.
So, inlet pipe fills 5 units and the outlet pipes empties 7 units of the tank in one hour.
The completely filled tank empties in 7 hours.
∴ 7o - 5i = 5…(ii)
So, now we solve the equations, and get i and o and i = 6.

Let P do x units of work in one day.
∴ so, total work = 42x
So, Q and R do 1.26x and (1.26 x 1.5 =)1.89x units of work in one day, respectively.
∴ Total work done by Q and R in one day = 1.26x + 1.89x = 3.15x
So, Q and R can finish the work in (42x / 3.15x ≈) 13 days.