CAT — Time & Work
9 questions, free to view. Click any question to see the answer and explanation.
Teams A, B, and C consist of five, eight, and ten members, respectively, such that every member within a team is equally productive. Working separately, teams A, B, and C can complete a certain job in 40 hours, 50 hours, and 4 hours, respectively. Two members from team A, three members from team B, and one member from team C together start the job, and the member from team C leaves after 23 hours. The number of additional member(s) from team B, that would be required to replace the member from team C, to finish the job in the next one hour, is
4
2
1
3
2
Step 1: Find the work rate of each member
Assume the total work is 200 units (LCM of 40, 50 and 4).
Team A completes 200 units in 40 hours → rate = 5 units/hour → each member = 1 unit/hour.
Team B completes 200 units in 50 hours → rate = 4 units/hour → each member = 1/2 unit/hour.
Team C completes 200 units in 4 hours → rate = 50 units/hour → each member = 5 units/hour.
Step 2: Find the work done in the first 23 hours
2 members from Team A: 2 × 1 = 2 units/hour
3 members from Team B: 3 × (1/2) = 3/2 units/hour
1 member from Team C: 5 units/hour
Total rate = 2 + 3/2 + 5 = 17/2 units/hour
Work done in 23 hours = 23 × 17/2 = 391/2 units
Step 3: Find the remaining work
Remaining work = 200 − 391/2 = 9/2 units
This work has to be completed in the next 1 hour.
Step 4: Find the required additional members from Team B
After the Team C member leaves, existing rate = 2 + 3/2 = 7/2 units/hour
Let x = additional members from Team B (each contributing 1/2 unit/hour).
7/2 + x/2 = 9/2 → 7 + x = 9 → x = 2
Sam can complete a job in 20 days when working alone. Mohit is twice as fast as Sam and thrice as fast as Ayna in the same job. They undertake a job with an arrangement where Sam and Mohit work together on the first day, Sam and Ayna on the second day, Mohit and Ayna on the third day, and this three-day pattern is repeated till the work gets completed. Then, the fraction of total work done by Sam is
1/20
3/10
1/5
3/20
3/10
Here is the step-by-step breakdown to solve this problem without using LaTeX formatting:
1. Determine Efficiencies (Work Rates)
Let's find the ratio of work rates (efficiencies) for Sam, Mohit, and Ayna.
● Mohit is twice as fast as Sam, which means Sam's efficiency = Mohit's efficiency / 2.
● Mohit is thrice as fast as Ayna, which means Ayna's efficiency = Mohit's efficiency / 3.
To keep the calculations clean and avoid fractions early on, let's assume Mohit's efficiency is a common multiple of 2 and 3.
● Mohit's efficiency = 6 units/day
Using this assumption:
● Sam's efficiency = 6 / 2 = 3 units/day
● Ayna's efficiency = 6 / 3 = 2 units/day
2. Find Total Work
Sam can complete the entire job alone in 20 days.
● Total Work = Sam's efficiency * 20 days
● Total Work = 3 * 20 = 60 units
3. Analyze the 3-Day Work Cycle
The team works in a repeating 3-day pattern:
● Day 1 (Sam + Mohit): 3 + 6 = 9 units
● Day 2 (Sam + Ayna): 3 + 2 = 5 units
● Day 3 (Mohit + Ayna): 6 + 2 = 8 units
● Total work done in 1 full cycle (3 days): 9 + 5 + 8 = 22 units
4. Track Progress to Completion
Now, let's see how many full cycles fit into the total work of 60 units.
● After 2 full cycles (6 days): Work completed = 22 * 2 = 44 units
● Remaining work = 60 - 44 = 16 units
Now, we evaluate the subsequent days step-by-step:
● Day 7 (Sam + Mohit turn): They can complete 9 units.
● Remaining work = 16 - 9 = 7 units
● Day 8 (Sam + Ayna turn): They can complete 5 units.
● Remaining work = 7 - 5 = 2 units
● Day 9 (Mohit + Ayna turn): Only 2 units are left. Since their combined capacity is 8 units/day, they will finish this remaining work in 2/8 (or 1/4) of a day. Sam does not work on this day.
5. Calculate Sam's Contribution
Let's count the total number of days Sam actually worked:
● In the first 2 cycles (6 days): Sam works on Day 1 and Day 2 of each cycle.
● Days worked = 2 cycles * 2 days/cycle = 4 days
● Day 7: Sam works the full day. (+1 day)
● Day 8: Sam works the full day. (+1 day)
● Day 9: Sam does not work.
● Total days Sam worked: 4 + 1 + 1 = 6 days
Since Sam's efficiency is 3 units/day:
● Work done by Sam = 6 days * 3 units/day = 18 units
6. Find the Fraction of Total Work
● Fraction = Work done by Sam / Total Work
● Fraction = 18 / 60 = 3/10
Correct Answer:
B. 3/10
− Renu would take 15 days working 4 hours per day to complete a certain task whereas Seema would take 8 days working 5 hours per day to complete the same task. They decide to work together to complete this task. Seema agrees to work for double the number of hours per day as Renu, while Renu agrees to work for double the number of days as Seema. If Renu works 2 hours per day, then the number of days Seema will work, is
Step 1: Find their hourly efficiencies
Renu takes 15 × 4 = 60 hours to complete the work → efficiency = 1/60 work per hour.
Seema takes 8 × 5 = 40 hours to complete the work → efficiency = 1/40 work per hour.
Step 2: Let Seema work for x days
Then Renu works for 2x days.
Renu works 2 hours/day → Total hours by Renu = 2 × 2x = 4x hours.
Seema works double Renu's hours = 4 hours/day → Total hours by Seema = 4x hours.
Step 3: Form the work equation
Work by Renu = 4x × (1/60) = x/15
Work by Seema = 4x × (1/40) = x/10
x/15 + x/10 = 1 → 2x + 3x = 30 → 5x = 30 → x = 6
Pipes A and C are fill pipes while Pipe B is a drain pipe of a tank. Pipe B empties the full tank in one hour less than the time taken by Pipe A to fill the empty tank. When pipes A, B and C are turned on together, the empty tank is filled in two hours. If pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank. If Pipe A can fill the empty tank in less than five hours, then the time taken, in minutes, by Pipe C to fill the empty tank is
90
120
75
60
90
Let Pipe A fill the tank in a hours. Pipe B empties in (a−1) hours. Pipe C fills in c hours.
Step 1: Form the first equation
When A, B, C are opened together: 1/a − 1/(a−1) + 1/c = 1/2
Since 1/a − 1/(a−1) = −1/[a(a−1)]:
1/c = 1/2 + 1/[a(a−1)] ...(1)
Step 2: Form the second equation
B and C are opened together for 1 hour, then C fills remaining for 5/4 hours.
−1/(a−1) + 1/c + 5/(4c) = 1
−1/(a−1) + 9/(4c) = 1 ...(2)
Step 3: Substitute from equation (1) into (2)
9/4 × (1/2 + 1/[a(a−1)]) − 1/(a−1) = 1
Multiply by 4: 9/2 + 9/[a(a−1)] − 4/(a−1) = 4
1/2 = (4a − 9)/[a(a−1)]
a(a−1) = 2(4a−9)
a² − 9a + 18 = 0 → (a−3)(a−6) = 0
Since Pipe A fills in less than 5 hours: a = 3
Step 4: Find Pipe C's time
1/c = 1/2 + 1/(3×2) = 1/2 + 1/6 = 2/3
c = 3/2 hours = 90 minutes
The rate of water flow through three pipes A, B and C are in the ratio 4 : 9 : 36. An empty tank can be filled up completely by pipe A in 15 hours. If all the three pipes are used simultaneously to fill up this empty tank, the time, in minutes, required to fill up the entire tank completely is nearest to
73
78
76
71
73
Step 1: Pipe A fills tank in 15 hours → rate = 1/15 tank/hour.
A : B : C = 4 : 9 : 36 → 1 part = (1/15) ÷ 4 = 1/60 tank/hour.
Step 2: Rates
Pipe B = 9 × (1/60) = 3/20 tank/hour
Pipe C = 36 × (1/60) = 3/5 tank/hour
Step 3: Combined rate
= 1/15 + 3/20 + 3/5 = 4/60 + 9/60 + 36/60 = 49/60 tank/hour
Step 4: Time = 1 ÷ (49/60) = 60/49 hours
= (60/49) × 60 = 3600/49 ≈ 73.47 minutes ≈ 73 minutes
− Amal and Vimal together can complete a task in 150 days, while Vimal and Sunil together can complete the same task in 100 days. Amal starts working on the task and works for 75 days, then Vimal takes over and works for 135 days. Finally, Sunil takes over and completes the remaining task in 45 days. If Amal had started the task alone and worked on all days, Vimal had worked on every second day, and Sunil had worked on every third day, then the number of days required to complete the task would have been
Given:
● Amal and Vimal together complete the work in 150 days.
● Vimal and Sunil together complete the work in 100 days.
Amal works for 75 days.
Then Vimal works for 135 days.
Finally, Sunil works for 45 days and completes the task.
Find the number of days required if:
● Amal works every day.
● Vimal works on every second day.
● Sunil works on every third day.
Step 1: Let the daily work rates be
Amal = A
Vimal = V
Sunil = S
Then,
A + V = 1/150
V + S = 1/100
Step 2: Use the second work schedule
The total work done is
75A + 135V + 45S = 1
Using
S = 1/100 − V,
75A + 135V + 45(1/100 − V) = 1
75A + 90V = 11/20
Divide throughout by 15,
5A + 6V = 11/300
Step 3: Solve for the individual rates
From
A + V = 1/150,
A = 1/150 − V
Substitute into
5A + 6V = 11/300,
5(1/150 − V) + 6V = 11/300
1/30 + V = 11/300
V = 1/300
Therefore,
A = 1/150 − 1/300
= 1/300
Also,
S = 1/100 − 1/300
= 1/150
Step 4: Determine the work pattern
Every 6 days,
● Amal works all 6 days.
● Vimal works on days 2, 4 and 6 (3 days).
● Sunil works on days 3 and 6 (2 days).
Work done in 6 days
= 6A + 3V + 2S
= 6 × 1/300 + 3 × 1/300 + 2 × 1/150
= 6/300 + 3/300 + 4/300
= 13/300
Step 5: Find the number of complete cycles
After 23 cycles (138 days),
Work done
= 23 × 13/300
= 299/300
Remaining work
= 1/300
Step 6: Find the last day
On the 139th day,
Amal works.
Since Amal's daily work is
1/300,
the remaining work is completed.
Hence, the total number of days required is
139
Answer: 139
Arun, Varun and Tarun, if working alone, can complete a task in 24, 21, and 15 days, respectively. They charge Rs 2160, Rs 2400, and Rs 2160 per day, respectively, even if they are employed for a partial day. On any given day, any of the workers may or may not be employed to work. If the task needs to be completed in 10 days or less, then the minimum possible amount, in rupees, required to be paid for the entire task is
38400
38880
34400
47040
38400
Cost to complete task if each works alone: Arun = 24×2160 = ₹51840, Varun = 21×2400 = ₹50400, Tarun = 15×2160 = ₹32400. Tarun is cheapest per unit of work.
Tarun works all 10 days: completes 10/15 = 2/3 of work. Remaining = 1/3. Varun needs (1/3)÷(1/21) = 7 days.
Total cost = 10×2160 + 7×2400 = 21600 + 16800 = ₹38400.
The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression. Kamal takes twice as much time as Amal to do the same amount of job. If Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job. Then the number of days Sunil will take to finish the job working alone, is
Let the amount of work done per day by Amal, Sunil and Kamal be A, S and K respectively.
Since the work rates are in Harmonic Progression (HP), their reciprocals are in Arithmetic Progression (AP).
Kamal takes twice as much time as Amal.
Therefore, K = A/2
Taking reciprocals, 1/K = 2/A
Since 1/A, 1/S, 1/K are in AP,
2/S = (1/A + 1/K) = (1/A + 2/
A) = 3/A
Hence, S = 2A/3
Now use the work done.
Amal works for 4 days. Work done = 4A
Sunil works for 9 days. Work done = 9 × (2A/3) = 6A
Kamal works for 16 days. Work done = 16 × (A/2) = 8A
Together they complete the entire job.
So, 4A + 6A + 8A = 1 → 18A = 1 → A = 1/18
Therefore, S = 2A/3 = 2/3 × 1/18 = 1/27
Thus, Sunil alone completes the work in 1 ÷ (1/27) = 27 days
Answer: 27
Ankita is twice as efficient as Bipin, while Bipin is twice as efficient as Chandan. All three of them start together on a job, and Bipin leaves the job after 20 days. If the job got completed in 60 days, the number of days needed by Chandan to complete the job alone, is
Step 1: Assume Chandan's work rate. Let Chandan's work rate be 1 unit/day. Then, Bipin's work rate = 2 units/day and Ankita's work rate = 4 units/day. Together they work at 4 + 2 + 1 = 7 units/day.
Step 2: Work completed in the first 20 days. All three work together for 20 days. Work done = 20 × 7 = 140 units.
Step 3: Work completed in the remaining 40 days. After 20 days, Bipin leaves. Only Ankita and Chandan continue. Combined work rate = 4 + 1 = 5 units/day. They work for another 40 days. Work done = 40 × 5 = 200 units.
Step 4: Find the total work. Total work = 140 + 200 = 340 units. Since Chandan alone does 1 unit/day, he will complete 340 units in 340 days.
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