Test Level 3: Time & Work - 1 - CAT MCQ

Since A and C together work twice as much as B
A + C = 2B ... (1)
And A and B together work thrice as much as C
A + B = 3C ... (2)
Equation (1) - equation (2) gives 
C - B = 2B - 3C
Or 3B = 4C
Or B = (4/3)C
Substituting this relation in equation (1), we get 
A = (5/3)C 
A + B + C = (5/3)C + (4/3)C + C = 4C
This means that 4C can complete the work in 12 days or C can complete in 12 x 4 = 48 days.

LCM of 15, 20 and 30 is 60.
Let the capacity of tank be 60 liters.
A and B fill the tank at a rate of 3 liters per minute and 2 liters per minute, respectively. C empties the tank at a rate of 4 liters per minute.
If opened successfully for 1 minute each, 1 liter will be filled in 3 minutes.
So, 55 liters will be filled in 55 × 3 = 165 minutes
After the 165th minute, tap A and B will fill the left capacity of 5 liters in 2 minutes.
So, the total time taken = 167 minutes

10 persons can clean 10 floors in 10 days means 1 person can clean one floor in 10 days
So 8 persons can clean one floor in 10/8 days
8 persons can clean 8 floors in

According to this question,

Subtracting equation (1) from equation (3), we get

And by putting this value in equation (2), we get

Putting the value of 1/B in equation (3), we get

 ⇒ Both A and C can do the work in

After 10 days, he still has to do the work of 20 days.
Since efficiency of the man has increased by 25%, he can do the work in 20 × 100/125 = 16 days.
So, he will complete the work 4 days before the scheduled time. 

Let initially, man can do 'x' amount of work in one day.
In 60 days work done = 60x
Man can do 'x' amount of work on the first day. On the second day, he will do 2x, on the third day 4x, and so on.
ATQ,
x + 2x + 4x + 8x + 16x + 32x > 60x
So, he can complete the work on the 6th day.

Let the capacity of the tank = LCM of 14 and 16 = 112 L
So, the rates of filling of the two pipes = 112/14 = 8 L/hr and 112/16 = 7 L/hr
So, effective filling rate = 8 + 7 = 15 L/hr
Ideally, time taken to fill the tank 

Actual time taken = 32 minutes + 7 hours 28 minutes = 8 hours
Thus, actual filling rate = 112/8 = 14 L/hr

Let the efficiencies of A and B be 5 and 4, respectively.
Let the efficiency of C be x.
Let the time taken by A to finish the work be 3 days.
So, time taken by B and C to finish the work = 2 days
So, total work = 3 × 5 = 15 units
So, (4 + x)2 = 15
Hence, x = 3.5

In 1st hr part filled = 1/20 = 2/40
In 2nd hr part filled

In 3rd hr part filled 

In 4th hr part filled = 2/40
In 5th hr part filled = 4/40
From 6th hr part filled = 3/40, and it continues.
So, at the end of 5th hr, 12/40 part of tank is filled.
Still 28/40 is left. It takes 28/3 hours.
So, the answer is  hours and 20 minutes.

Let the volume of the tank be 100v litres.
Let volumetric flow rate through tap A be A litres per hour.
So, 10A = 100v or A = 10v ...(i)
Let volumetric flow rate through tap C be C litres per hour.
So, 20C = 100v or C = 5v ...(ii)
Let volumetric flow rate through tap D be D litres per hour.
So, 25D = 100v or D = 4v ...(iii)
Now, fluid drained through taps C and D in 10 hours = 10(C + D) litres = 10(5v + 4v) litres
= 10(9v) litres = 90v litres
Volume of fluid left to be filled by tap A = 90v litres
Suppose tap A takes t hours to fill the tank.
tA = 90v ...(iv)
Plugging in the value of A in terms of 'v' from equation (i) into equation (iv), we get
t × 10v = 90v
Or, t = 9
Thus, tap A takes 9 hours to full the tank.
Thus, option (2) is correct.