XAT Time and Work Questions

Let us assume the amount of work to be finished = LCM of {10, 12, 15, 18} = 180 units.

The amount of work which Abdul can complete in a day = $$\dfrac{180}{10}$$ = 18 units.

The amount of work which Bimal can complete in a day = $$\dfrac{180}{12}$$ = 15 units.

The amount of work which Charlie  can complete in a day = $$\dfrac{180}{15}$$ = 12 units.

The amount of work which Dilbar can complete in a day = $$\dfrac{180}{18}$$ = 10 units.

It is given that 50 percent of the total work gets completed after 3 days. Therefore, we can say that 90 units of work was completed in 3 days. 

Let us check options. 
Option A: Each of them worked for exactly 2 days.
In this case amount of work completed = 2*(10+15+12+18) = 110 units.

Option B: Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.
In this case amount of work completed = 1*(10+15)+2*(12)+3*(18) = 103 units.

Option C: Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.
In this case amount of work completed = 1*(10)+3*(15)+2*(18+12) = 115 units.

Option D: Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.
In this case amount of work completed = 1*(12)+3*(15)+2*(18+10) = 113 units.

Option E: Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.
In this case amount of work completed = 1*(18+12)+2*(15)+3*(10) = 90 units.

Therefore, we can say that option E is the correct answer. 

Let us assume the volume of the tank to be 60 litres. 
A can fill or empty 60/4 = 15 litres in a minute.
B can fill or empty 60/10 = 6 litres in a minute.
C can fill or empty 60/12 = 5 litres in a minute.
D can fill or empty 60/20 = 3 litres in a minute. 

We have to find the combination for which the tank will be half-full in 30 minutes (i.e., completely filled in 1 hour).
Therefore, the combination must result in a net input of 60/60 = 1 litre per minute.

Let us evaluate the options.
Option B:
Pipe A drained and pipes B, C and D filled

The net result will be -15 + 6 + 5 + 3 = -1 litre/minute. We can eliminate option B. 

Option C:
Pipes A and D drained and pipes B and C filled

The net result will be -15-3+6+5 = -7 litres/minute. We can eliminate option C as well. 

Option D:
Pipes A and D filled and pipes B and C drained
The net result will be 15+3-6-5 = 7 litres/minute. We can eliminate option D as well.

Option A:
Pipe A filled and pipes B, C and D drained

The net result will be 15-6-5-3 = 1 litre/minute. 
Therefore, option A is the right answer. 

M inlet pipe can fill $$(\frac{M}{8})^{th}$$ part of the tank in 1 hour.

Similarly, N outlet pipes can empty $$(\frac{N}{12})^{th}$$ part of the tank in 1 hour.

=> $$\frac{M}{8} - \frac{N}{12} = \frac{1}{6}$$

=> $$6 M - 4 N = 8$$

=> $$M = \frac{4 + 2 N}{3}$$

If, N = 1, => M = 2 => $$M : N = 2 : 1$$

If, N = 4, => M = 4 => $$M : N = 1 : 1$$

Thus, Ans - (E)

Height of cone comes down to 50%, => it becomes $$\frac{1}{2}$$

=> Volume would become $$\frac{1}{8}$$ as radius will also become half by similar triangles.

Let the capacity of cone = 24 litres

Volume of water run-off = $$24 - \frac{1}{8} \times 24 = 21$$ litres

Volume of water left in the cone = $$ \frac{1}{8} \times 24 = 3$$ litres

Pipe A's efficiency = $$\frac{24}{8} = 3$$ litres/hr

Pipe B's efficiency = $$\frac{24}{12} = 2$$ litres/hr

Pipe C's efficiency = $$\frac{24}{-4} = -6$$ litres/hr

All will run 19 hours simultaneously (going by the options)

=> Net effect = $$(3 + 2 - 6) \times 19 = -19$$ litres

This means that after 19 hours, 19 litres of water has been removed, we need to remove 2 more litres as per the requirement. Thus,  C will definitely run for another hour.

If we run A and C together for the 20th hour, net effect = $$(3 - 6) \times 1 = -3$$ litres

Run B for 30 minutes => $$2 \times \frac{1}{2} = 1$$ litres

$$\therefore$$ Volume of water removed = $$-19 -3 + 1 = -21$$ litres

Thus, Pipe B was open for 19 hours 30 minutes.

Let the total time required be 12T (LCM of 2, 3 and 4 for the ease of calculation)
No. of pages read by Albela in 12T minutes at 2 minute per page = 6T
No. of pages read by Bob in 12T minutes at 3 minute per page = 4T
No. of pages read by Chulbul in 12T minutes at 4 minute per page = 3T
Total no. of pages read by all three of them = (6T + 4T + 3T) = 13T
According to the question, total pages = 78
So, 13T = 78
Or, T = 6
Therefore, no. of pages read by Bob = 4T = 24
Hence, option  A is the correct answer.