Test Level 2: Time & Work - 2 - CAT MCQ

Stuart Little can eat 12.5 sweets in 1 day and Mickey Mouse can eat 7 sweets in 1 day.
It implies that together they can eat 19.5 sweets in 1 day.
Hence, 507 sweets will be eaten in 26 days.

Let the remaining work be completed in x days. Then, we get
54 × 90 = 54 × 62 + (54 – 18) × x
x = 42 days

Assume, C can do x work in 1 day; B can do 2x work in 1 day; and A can do 4x work in 1 day.
Total work = 12(x + 2x + 4x) = 84x
A and B work for 8 days, 8(2x + 4x) = 48x will be completed.
B and C can do 3x work in 1 day.
So, the remaining 36x can be completed in 12 days.

In 1 hour, portion of tank filled 


The tank will be filled in 15/2 hours = 7 hours and 30 minutes.

A can finish the work in 20 days.
B can finish the work in 30 days
Ratio of their one-day work = (1/20) : (1/30) = 3 : 2
If B gets Rs. x, then A will get Rs. 3/2x.
∴ 3/2x + x + (x - 1500) = 9000
⇒ x = 3000
A, B and C will get Rs. 4500, Rs. 3000 and Rs. 1500, respectively.

Suppose work is 550 units.
Work done by P and Q in 1 day = 50 + 55 = 105 units
Work done by P and R in 1 day = 50 + 10 = 60 units
Total work done in 2 days = 165 units
Total work done in 6 days = 165 x 3 = 495 units
Remaining work after 6 days = 550 - 495 = 55 units; which is completed on 7th day
Thus, the work will be completed on 7th day.

Let x, y, a, and b denote the volumes of X, Y, A, and B, respectively.
a = 1/2 litres
⇒ b = 1/4 litres
40 operations must have transferred (a - b) × 40 litres from Y to X, i.e.,

Now, there are 10 litres in X and 10 litres in Y.
Total volume = 10 + 10 = 20 litres
Capacity of Y = 20 litres

Let the work done by 1 man in 1 hour be m.
Thus, work done by 20 men working 4 hours a day in 30 days = 20 × 4 × m × 30 = Total work = W ... (i)
Now, working 3 hours a day for 20 days, work done by 30 men = 3 × 20 × 30 × m
Working 10 hours a day, let the last man take d more days.
Work done by the last man = 10 × d × m
Thus, total work done = 3 × 20 × 30 × m + 10 × d × m = W ... (ii)
So, from (i) and (ii), we have 20 × 4 × m × 30 = 3 × 20 × 30 × m + 10 × d × m
Or d = 60
Thus, the last man takes 60 days.
Hence, answer option 4 is correct.

Let work done by 1 employee in 1 hour = w
Work done by 40 employees, working 6 hours a day, for 60 days = Total work = 40 × 6 × w × 60
Work actually done in first 20 days = 20 × 20 × 6 × w
Work left to be done = (40 × 6 × w × 60) - (20 × 20 × 6 × w) = 12,000w ...(i)
Let the 40 employees now work h hours per day for the rest of the 40 days.
So, work done by these 40 employees = 40 × 40 × h × w ...(ii)
On equating (i) and (ii), and solving for h, we get h = 7.5.
Thus, the employees now have to work 7.5 hours per day.
Thus, answer option 1 is correct.

Amounts of work done in one day by A, B, C and D are 1/4, 1/8, 1/16, and 1/32, respectively.
Using options, we note that the pair of B and C does 3/16 of work in one day.
The pair of A and D does  of the work in one day.
Hence, A and D take 32/9 days.

Two pipes can fill a cistern in 20 minutes and 10 minutes, respectively.
In one minute, pipes can fill 1/20 and 1/10 of the cistern, respectively.
Third pipe can empty it in 25 minutes.
In one minute, third pipe can empty 1/25 of the cistern.
1/5th of the cistern is already full, it means 4/5th remains empty.
In one minute, part of the cistern filled is:

The respective ratio of the efficiencies of A and B is 1.6 : 1 or 8 : 5.
Thus, times taken by A and B will be in the ratio of 1/8 : 1/5 or 5 : 8.
Let the times taken by A and B be 5x and 8x, respectively.
According to the problem, 5x = 8x - 9 or x = 3.
Thus, A and B can do the work in 15 days and 24 days, respectively.
If they work together, it will take them 

Let us assume that 'B' can destroy the wall in 'd' days.

⇒ d = 60 days
So, if both 'A' and 'B' work to construct the wall,
In 1 day, they can do 

So, it takes 20 days to construct the wall, if both 'A' and 'B' work towards constructing the wall.

Since five men can complete the work in 8 hours, (8:00 a.m. - 4:00 p.m.),
Number of man-hours required = 5 × 8 = 40
40 = 5 × 4 + 6 × H
⇒ H = 10/3 , or 3 hr 20 min
So, after 12:00 noon, it takes 3 hours and 20 minutes to complete the work.
Therefore, work will be completed at 3:20 p.m.

If B can do 'x' work per hour, A can do 1.2x.
Since A works only for 5 hours and B works for 6 hours, both can do the same amount of work in a day.
They together take 6 days to complete the work.
So, A alone takes 12 days working 5 hours a day.
So, he takes  working 3 hours a day, i.e. 20 days.

In the second case, if the number of men, their efficiency and their working hours are all increased by 25% (i.e. 5/4 times), then the number of days required will be 
i.e. 64/125 times.
If 'd' is the number of estimated days, then according to question,

⇒ d = 125

1 day-work of 15 men = 1/210
10 day-work of 15 men = 1/21
At the end of 10 days, 1/21 part will be finished; in the next 10 days, additional 2/21 part, and in the following 10 days, additional 3/21 part, and so on, will be finished.

From Arithmetic Progression formula, 

Or n = 6
Thus, the work will be finished in 10 × 6 = 60 days.

6 men earn 10% of $750 in 10 days.
Amount 1 man earns in 1 day

8 men earn 20% of $350 in x days. 
Amount 1 man earns in 1 day 

A, B and C complete the work in x, y and z days, respectively.
A, B and C complete respectively 1/x, 1/y and 1/z work in a day.
Work completed by A and B together 

Work completed by B and C together

Work completed by C and A together

From the above three equations, we get

A, B and C together would complete the work in 20 days.

Mohan can do half the work in 8 days.
∴ He can do the complete work in 16 days.
Sohan can do one-third of work in 6 days.
∴ He can do the complete work in 18 days.
Total time taken by both to finish the work together