TITA Based Questions: Time and Work | Quantitative Aptitude (Quant) - CAT PDF Download

Ans: 27
Let Amir = A , Sunil = S and Kashish = K and 
let the time taken by A and K to complete a work is x and 2x days respectively.
Work done in a day is the efficiency of a person. 
Hence, if efficiencies of A, S and K are in Harmonic Progression, time taken by them to finish a work will be in Arithmetic Progression.
∴ Time taken by S alone is arithmetic mean of time taken by A and K alone.
⇒ Time taken by S alone = (x + 2x)/2 = 1.5x
Using unitary method
⇒ x = 4 + 6 + 8
⇒ x = 18
∴ S will take 1.5 × 18 = 27 days to finish the job alone.

Ans: 36 
Let work done per day (efficiency) of Garimaand Suhani are 'g' and 's' units.
A shortfall of 40% for Garima is compensated by 50% extra work done by Suhani.
⇒ 0.4 × g = 0.5 × s
⇒ g/s = 5/4
⇒ Let g = 5x and s = 4x
Together they can complete the work in 20 days.
⇒ Total work to be done = 20 × (5x + 4x) = 180x units
The faster among the two is Garima whose efficiency is 5x.
∴ Time required by Garima alone = 180x / 5x = 36 days.

Ans: 6
Ratio of time taken by Anjali, Tanya and Maya is 5 : 8 : 10.
⇒ Ratio of efficiencies of Anjali, Tanya and Maya

Let their efficiencies be 8x, 5x and 4x respectively per hour.
Total work done by them in 4 days = (8x + 5x + 4x) × 4 × 8 = 17x × 32 = 544x
Now, Anjali and Tanya worked by 6 days working 6 hours 40 minute i.e.
⇒ Time taken by Maya to complete the remaining work =

Ans: 11
Bob takes 40 days to finish the work. 
Alex takes is twice as fast as Bob and hence will take half the time taken by Bob i.e., 20 days. 
Alex is also thrice as fast as Claire, hence Claire will take thrice the time taken by Alex, i.e., 60 days.
Time take by
Alex = 20 days
Bob = 40 days
Claire = 60 days
Let the total work to be done = 120 units.
∴ Efficiency of 
Alex = 6 units/day
Bob = 3 units/day
Claire = 2 units/day

Work done/cycle = 22 units.
∴ Work done in 5 cycles = 5 × 22 = 110 units
Work left after 5 cycles (15 days) = 120 – 110 = 10 units.
On 16th day Alex and Bob will together complete 9 units of work while remaining 1 unit of work will be completed by Bob and Claire on 17th day.
∴ Alex worked for 10 days in 5 complete cycles + on 16th day i.e., total 11 days.

Ans: 32
Let the total work to be done = LCM (12, 16, 24) = 48 units.
Let the efficiency of Amit, Adil and Aditya be ‘x’, ‘y’ and ‘z’ units/month⇒ x + y = 48/12 = 4   …(2)
⇒ y + z = 48/16 = 3   …(1)
⇒ z + x = 48/24 = 2   …(3)
Adding all these equations
⇒ x + y + z = 9/2 = 4.5   …(4)
Solving these four equations we get,
x = 1.5, y = 2.5 and z = 0.5
∴ The person who is neither slowest not fastest is Amit with efficiency of 1.5 units/month.
∴ Time required by Amit to complete the task alone
= 48/1.5 = 32 months.

Ans: 3000
Let the area to be painted = LCM(12, 16) = 48 units.
⇒ Efficiency of Anil = 48/12 = 4 units/day

and Efficiency of Bhanu = 48/16 = 3 units/day
Work done by Anil in 6 days = 6 × 4 = 24 units
Work done by Anil in 6 days = 6 × 3 = 18 units
∴ Remaining work done by Chinmay = 48 – 24 – 18 = 6 units.
⇒ Payment received by Chinmay = 24000/48 × 6 = Rs. 3,000
Alternately, 
Fraction of work done by Anil in 6 days = 6/12 = ½
Fraction of work done by Bhanu in 6 days = 6/16 = 3/8
⇒ Fraction of work done by Chinmay in 6 days = 1 - ½ - 3/8 = 1/8
Since, Chinmay completes 1/8th of the work, he will receive 1/8th of the payment.
∴ Chinmay’s payment = 1/8 × 24,000 = Rs. 3,000

Ans: 4
Let John’s efficiency be 1 unit/day.
∴ Jack’s efficiency is 2 units/day.
Jack and Jim’s efficiency is thrice John’s efficienc.
∴ 2 + e_Jim = 3 × 1 
⇒ e_Jim = 1
John takes three days more than that taken by three of them working together
Let the time taken by all three together is d.
∴ Total work to be done = 4 × d = 1 × (d + 3)
⇒ d = 1
∴ John finished the work in d + 3 = 4 days.
Since, John and Jim have same efficiency, Jim will also finish the work in 4 days.

Ans: 40
In 60 days 1.5 kms out of 6 kms is built.
∴ In 60 days 1.5/6 = 1/4th work is done.
To complete the whole work, it would take 4 × 60 = 240 days, i.e., 180 more days.
Days remaining now is 200 – 60 = 140 days.
∴ To complete the remaining work in 140 days, contractor will have to hire more people.
 Assuming he hires x more men.
The work which 140 men would take 180 more days, now need to be completed by 140 + x men in 140 days.
∴ 180 × 140 = 140 × (140 + x)
⇒ x = 40
∴ Contractor hires 40 more people.

Ans: 13
Let the work done by one man and one machine per day be x and y respectively.
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job.
Since efficiency is inversely proportional to the time taken, so the efficiency of 3 men and 8 machines is twice that of 8 men and 3 machines.
∴ (3x + 8y) = 2(8x + 3y)
∴ 13x = 2y.
So, work done by 13 men in a day = work done by 2 machines in a day.
∴ If two machines can finish the job in 13 days, same work will be done by 13 men in 13 days.

Ans: 12
If John works the same number of regular and over-time hours say 'p'
The income would be 57p and 114p
Let's say that he works 'x' hours regular and 'y' hours overtime...
So, the income would be 57x and 114y
we are told that 114y is 15% of 57x
114y = 0.15 × 57x
y = 0.075x
we also know that x + y = 172
therefore, x + 0.075x = 1.075x = 172
x = 160
y = 172 - 160 = 12
Therefore, the number of hours he worked overtime is 12 hours.

Ans: 10
Let a filling pipe fills the tank at ‘a’ liters per hour and a draining pipe drains at ‘b’ liters per hour.
Work done by 6 filling and 5 emptying pipes in 6 hours = 6(6a - 5b)
Work done by 5 filling and 6 emptying pipes in 60 hours = 60(5a - 6b)
6(6a – 5b) = 60(5a – 6b)
∴ 6a – 5b = 50a – 60b
∴ 44a = 55b
∴ a = 5b/4
Let the tank gets filled completely in ‘m’ hours when one draining pipe and two filling pipes are on.
Work done by 2 filling and 1 emptying pipes in m hours = m(2a - b)
∴ m(2a – b) = 6(6a – 5b)Solving this, we get m = 10

Ans: 6
Step 1: Find total work in hours (assuming the task is measured in work hours).

Total work by Ritika = 15 days × 4 hours/day = 60 hours.

Total work by Shruti = 8 days × 5 hours/day = 40 hours.

Since both complete the same task, total work is the same. So, we consider the total work as 60 hours (the higher number).
Step 2: Calculate work rates per hour for Ritika and Shruti.

Ritika's rate = Total work / Total time = 60 hours / (15 days × 4 hours) = 60 / 60 = 1 unit/hour.

Shruti's rate = 60 hours / (8 days × 5 hours) = 60 / 40 = 1.5 units/hour.

Step 3: Define variables for new working hours and days:

Let Ritika work x hours per day.

Ritika works for D days.

Then Shruti works double the number of hours as Ritika = 2x hours/day.

Shruti works half the number of days as Ritika = D/2 days (since Ritika works double the days as Shruti).

Step 4: Given Ritika works 2 hours per day, so x = 2.

Ritika's total work done = (rate) × (hours per day) × (days) = 1 unit/hour × 2 hours/day × D days = 2D units.

Shruti's total work done = 1.5 units/hour × 2x hours/day × (D/2) days
Since x = 2,
Shruti's work = 1.5 × 4 × (D/2) = 3D units.

Step 5: Total work done by both = total work (60 units)
2D + 3D = 60
5D = 60
D = 12 days

Step 6: Find days Shruti works:
Shruti works D/2 = 12/2 = 6 days.

Ans: 5 
Let k be the part of work labourers can do in one day and x -10, x, y be the number of days for which they remained present. Then
(x − 10) k + xk = 2/3 and yk = 1/3 
(2x − 10) 1/3y = 2/3 ⇒ 2x − 10 = 2y ⇒ x − 5 = y 
The required number of days= y − (x − 10) = 5

Ans: 2 : 1
12 men + 16 boys can do the work in 5 days.
5 x (12 men + 16 boys) can do the work in 1 day.
Similarly 4 x (13 men + 24 boys) can do the same work in 1 day.
⇒ 52 men + 96 boys = 60 men + 80 boys.
⇒ 8 men = 16 boys or 1 man = 2 boys, i.e. 2 : 1

Ans: 2
5 men + 3 boys can reap 23 hectares in 4 days.
3 men + 2 boys can reap 7 hectares in 2 days.
14 (5 men + 3 boys) can reap 23 x 14 hectares in 4 days.
23 (3 men + 2 boys) can reap 7 x 2 x 23 hectares in 4 days.
14 (5 men + 3 boys) = 23 (3 men + 2 boys).
1 man = 4 boys. Now 5 men + 3 boys = 23 boys.
23 boys can reap 23 hectares in 4 days.
30 boys can reap 45 hectares in 6 days.
But 30 boys = 28 boys + 2 boys = 7 men + 2 boys.
Hence 2 boys must assist 7 men.