CAT Previous Year Questions: Time & Work (July 1) - CAT MCQ

Let the efficiency of Bob be 3 units/day. So, Alex's efficiency will be 6 units/day, and Cole's will be 2 units/day.

Since Bob can finish the job in 40 days, the total work will be 40*3 = 120 units.

Since Alex and Bob work on the first day, the total work done = 3 + 6 = 9 units.

Similarly, for days 2 and 3, it will be 5 and 8 units, respectively.

Thus, in the first 3 days, the total work done = 9 + 5 + 8 = 22 units.

The work done in the first 15 days = 22*5 = 110 units.

Thus, the work will be finished on the 17th day(since 9 + 5 = 14 units are greater than the remaining work).

Since Alex works on two days of every 3 days, he will work for 10 days out of the first 15 days.

Then he will also work on the 16th day.

The total number of days = 11.

Let the time taken by Anu, Tanu and Manu be 5x, 8x and 10x hours.

Total work = LCM(5x, 8x, 10x) = 40x

Anu can complete 8 units in one hour

Tanu can complete 5 units in one hour

Manu can complete 4 units in one hour

It is given, three of them together can complete in 32 hours.

32(8 + 5 + 4) = 40x

X = 68/5

It is given,

Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day, i.e. 36 + 4 = 40 hours

40(8 + 5) + y(4) = 40x

4y = 24

y = 6

Manu alone will complete the remaining work in 6 hours.

Let the work done by Rahul, Rakshita, and Gurmeet be a, b, and c units per day, respectively, and the total units of work are W.

Hence, we can say that 7(a + b + c) < W ( Rahul, Rakshita, and Gurmeet, working together, would have taken more than 7 days to finish a job).

Similarly, we can say that 15(a + c) > W ( Rahul and Gurmeet, working together would have taken less than 15 days to finish the job)

Now, comparing these two inequalities, we get: 7(a + b + c) < W < 15(a + c)

It is also known that they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job.
Therefore, the total units of work done is: W = 6(a + b + c) + 3b

Hence, we can say that 7(a + b + c) < 6(a + b + c) + 3b < 15(a + c)

Therefore, (a + b + c) < 3b ⇒ a + c < 2b, and 9b < 9(a + c) => b < a + c

⇒ a + b + c < 3b => 7(a + b + c) < 21b , and 15b < 15(a + c)

Hence, The number of days required for b must be in between 15 and 21 (both exclusive).

Hence, the correct option is D

“The amount of job that Amal, Sunil and Kamal can individually do in a day, are in harmonic progression.” This implies that the amount of time taken individually by Amal, Sunil and Kamal to finish a job are in A.P.
“Kamal takes twice as much time as Amal to do the same amount of job.” Since the amount of time taken individually by Amal, Sunil and Kamal to finish a job are in A.P, this simply means that Sunil Should take 1.5 times the time as Amal to do the same amount of job.
So, to do the same amount of job individually, the times taken by Amal, Sunil and Kamal will be in the ratio, 1 : 1.5 : 2 or 2 : 3 : 4
Amal, Sunil and Kamal worked for 4, 9 and 16 days respectively to finish the job.
Sunil does in 3 days what Amal does in 2 days.
Therefore, Sunil does in 6 days what Amal does in 4 days.
Sunil does in 3 days what Kamal does in 4 days.
Therefore, Sunil does in 12 days what Kamal does in 16 days.
So, to finish the entire job on his own, Sunil would require, 6 + 9 + 12 = 27 days.

Now Anil Paints in 12 Days
Barun paints in 16 Days
Now together Arun , Barun and Chandu painted in 6 Days
Now let total work be W
Now each worked for 6 days
So Anil's work = 0.5W
Barun's work = 6W / 16 = 3W/8
Therefore Charu's work =W/2 - 3W/8 = W/8
Therefore proportion of charu = 24000/8 = 3,000

Let Rahul work at a units/hr and Gautam at b units/hour
Now as per the condition :
8a + 6b =7.5a + 7.5b
so we get 0.5a = 1.5b
or a = 3b
Therefore total work = 8a + 6b = 8a + 2a = 10a
Now Rahul alone takes 10a/10 = 10 hours.

Let Entire work be W
Now Anil worked for 24 days ( 10 days alone and 14 days with bimal and charu both )
Bimal worked for 14 days and Charu worked for 14 days .
Now Anil Completes W in 60 days
so in 24 days he completed 0.4W
Bimal completes W in 84 Days
So in 14 Days Bimal completes = W/6
Therefore work done by charu = 
Therefore proportion of Charu = 13/30 x 21000 = 9100

Let the total amount of work be 60 units.
Then Anu, Vinu, and Manu do 4, 5, and 3 units of work per day respectively.
On the 1^st day, Anu and Vinu work. Work done on the 1^st day = 9 units
On the 2^nd day, Manu and Vinu work. Work done on the 2^nd day = 8 units
This cycle goes on. And in 6 days, the work completed is 9 + 8 + 9 + 8 + 9 + 8 = 51 units.
On the 7^th day, again Anu and Vinu work and complete the remaining 9 units of work. Thus, the number of days taken is 7 days.

Let the total work be 48 units. Let Amar do 'm' work, Akbar do 'k' work, and Anthony do 'n' units of work in a month.
Amar and Akbar complete the project in 12 months. Hence, in a month they do 48/12 = 4 units of work.
m + k = 4.
Similarly, k+n = 3, and m+n = 2.
Solving the three equations, we get m = 3/2, k = 5/2, n = 1/2
Here, Amar works neither the fastest not the slowest, and he does 1.5 units of work in a month. Hence, to complete the work, he would take 48/1.5 = 32months.

Let Total work be 108 days
At normal efficiency work is done in 12 days
Let A' s day work = A & B' s day work = B
      ► A + B = 9 & 
      ► A + B = 9…(1)
      ► A + 6B = 24…(2)
Solving (1) & (2), A = 6 B = 3
At usual efficiency A will do the work in 108 / 6 = 18 days.