Arithmetic - Time & Work - Previous Year CAT/MBA Questions

Answer: Option C

Explanation :

Let efficiency per day of X, Y and Z be x, y and z respectively.

Let the amount of work required for an upgrade = T units.

(i) X and Y working together can complete one new upgrade job in 6 days. In 6 days X will work for 5 days and Y will work for 5 days.

∴ 5x + 5y = T   …(1)

(ii) Z and X working together can complete two new upgrade jobs in 8 days. X will work for 6 days and Z will work for 7 days.

∴ 6x + 7z = 2T   …(2)

(iii) Y and Z working together can complete three new upgrade jobs in 12 days. Y will work for 10 days and Z will work for 10 days.

∴ 10y + 10z = 3T   …(3)

Solving (1) , (2) and (3) we get

x = T/10, y = T/10 and z = T/5

If X, Y and Z work together their combined efficiency = T/10 + T/10 + T/5 = 2T/5

∴ Time required by three of them together to complete 1 upgrade = T/(2T/5) = 2.5 days.

Hence, option (c).

Answer: Option C

Explanation :

Let, the number of women who started the work is N,

∴ Total work to be done = N × 7 woman days.

Day 1: Number of women worked = N

Day 2:  Number of women worked = N - 11

Day 3: Number of women worked = N - 22 and so on till

Day 12: Number of women worked = N – 11 × 11 = 121

∴ Total work done = N + (N – 11) + (N – 22) + … + (N – 121) = 12N - 726 woman days.

⇒ 12N – 726 = 7N

⇒ N = 726/5 = 145.2 = 145 (nearest integer)

Hence, option (c).

Answer: Option A

Explanation :

Let, the total capacity of cistern be LCM (12,15,20) = 60 units

∴ Rate of the two inlet pipes are 60/12 = 5 units per minute and 60/15 = 4 units per minute.

Also, rate of filling of cistern along with the leakage = 60/20 = 3 units per minute

∴ Rate of leakage (L) = 3 – (5 + 4) = -6 units per minute. (-ve indicates that leakage is emptying the tank)

∴ Time taken by leakage to empty the cistern = 60/6 = 10 mins

Hence, option (a).

Answer: Option E

Explanation :

Let the total units of work = LCM of (10, 12, 15 and 18) = 180

Therefore, efficiencies of Abdul, Bimal, Charlie and Dilbar individually in one day will be 18 units, 15 units, 12 units and 10 units per day respectively. 

Let us check options now:

Option (a): Total work done = 2 × (18 + 15 + 12 + 10) 
= 110 units

Option (b): Total work done = 1 × (15 + 10) + 2 × 12 + 3 × 18 
= 25 + 24 + 54 = 103 units

Option (c): Total work done = 2 × (18 + 2) + 1 × 10 + 3 × 15 
= 40 + 10 + 45 = 95 units

Option (d): Total work done = 2 × (18 + 10) + 1 × 12 + 3 × 15 
= 56 + 12 + 45 = 113 units

Option (e): Total work done = 1 × (18 + 12) + 2 × 15 + 3 × 10 
= 30 + 30 + 30 = 90 units

Hence, option (e).

Answer: Option C

Explanation :

Let the total work be 180 units (LCM of 9, 12 and 15) and let Ram, Ravi and Ratan do a, b and c units of work per day.

∴ a = 180/9 = 20; b = 180/12 = 15 and c

= 180/15 = 12

Ram works alone on Monday, Ravi works alone on Tuesday and Ratan works alone on Wednesday.

∴ Work done in three days = a + b + c

= 20 + 15 + 12 = 47

∴ Work done in nine days = 3(47)

= 141 units

On the 10th day, Ram does 20 units, thereby taking the completed total work to 141 + 20 = 161 units.

On the 11th day, Ravi does 15 units, hence, taking the completed total work to 176 units.

The work gets completed on the 12th day.

Hence, Ravi has done work equivalent to four days when the work is completed.

∴ Ravi’s total work = 4(15) = 60 units

∴ Required proportion = 60/180 = 1/3

Hence, option 3.

Answer: Option C

Explanation :

Tab B takes 1.5 times more than tap A to fill the tank.

A takes 12 hours to fill the tank alone.

Hence, time taken by tap B alone = 1.5 × 12 = 18 hours

Let the total capacity of the tank be 180 litres.

Let taps A and B fill a and b litres per hour and let tap C empty c litres per hour.

∴ a = 180/12 = 15; b = 180/18 = 10 and c = 180/36 = 5

For the first two hours, only A and C are working; for the next four hours, all three of A, B and C are working and for the next n hours, only A and B are working.

∴ 2(a – c) + 4(a + b – c) + n(a + b) = 180

∴ 2(15 – 5) + 4(15 + 10 – 5) + n(15 + 10) = 180

∴ 2(10) + 4(20) + 25n = 180

∴ 25n = 180 – 20 – 80 = 80

∴ n = 80/25 = 3.2 hours

∴ Total time = 6 + 3.2 = 9.2 hours

i.e. 9 hours and 12 minutes

Hence, option 3.

Answer: Option A

Explanation :

Let the capacity of the tank = LCM(4, 10, 12, 20) = 60 units
∴ Efficiency of A, B, C and D = 15, 6, 5, 3 units/minute respectively.

We need to check options now,
Option 1: Work Done in a minute = 15 – 6 – 5 – 3 = 1 unit
∴ After 30 minutes, the tank will be half filled.

We don’t have to check other options.
Hence, option (a).

Answer: Option A

Explanation :

Let the total work be equivalent to 900 units and let Somesh, Tarun and Nikhil respectively complete a, b and c units of work per day. 

∴ a = 900/45 = 20; b = 15 and c = 12

Let the work be completed in n days.

For the first five days, all three people were working. For the last two days, only Tarun was working.   

Hence, for the remaining n − 5 − 2 i.e. n − 7 days in between, both Somesh and Tarun were working. 

Hence, considering amount of work done in each day: 

5(20 + 15 + 12) + (n − 7)(20 + 15) + 2(15) = 900

∴ 235 + 35n − 245 + 30 = 900

∴ 35n = 880 i.e. n = 25.14 days 

Hence, option 1.

Answer: Option B

Explanation :

Let the first bore well fill the tank in x hours. 

Hence, the second fills it in (x − 10) hours and the third fills it in (x − 10) − 8 = (x − 18) hours. 

Let the total capacity of the tank to be filled be (x)(x − 10)(x − 18) and let the individual quantity filled by the three respective bore wells be a, b and c. 

∴ a = (x − 10)(x − 18); b = (x)(x − 18); c = (x)(x − 10) 
 
Since the first two bore wells fill the tank in the same time as the third bore well alone, amount of work done by first two bore wells per hour is the same as amount of work done by the third bore well per hour. 

i.e. a + b = c

∴ (x − 10)(x − 18) + (x)(x − 18) = (x)(x − 10)

∴ (x − 18)(2x − 10) = (x)(x − 10)

∴ 2x² − 46x + 180 = x² − 10x

∴ x² − 36x + 180 = 0 

∴ x = 30 or x = 6
 
Since (x − 18) is number of hours, this value has to be positive. Hence, x > 18 i.e. x = 30

∴ Time taken by third borewell = (x − 18) = 12 hours

Hence, option 2. 

Answer: Option E

Explanation :

Let the total capacity of the tank be 24 units.

∴ The work done by an inlet pipe = 24/8 = 3 units/hour and 
Work done by an outlet pipe = 24/12 = 2 units/hour

Now, work done by M inlet and N outlet pipes in an hour = 3M – 2N

It taken 6 hours for these pipes to fill the tank completely.

∴ 6(3M – 2N) = 24

∴ 3M – 2N = 4

Now when N = 1, M = 2.

Also when N = 4, M = 4

Similarly we get infinite values of N for which we get different values of M and the ratio M : N cannot be determined uniquely.

Hence, option (e).

Answer: Option A

Explanation :

If “you” are twice as efficient as your roommate, the roommate will take double the time taken by you i.e. the number of days for each will be in the ratio 1 : 2.
Hence, options 3 and 4 can be eliminated.

Let the total work correspond to 90 units.

If your roommate does x units of work per day, you can do 2x units of work.

∴ Total work done per day when both of you work together = x + 2x = 3x units

Since both of you can together complete the work in 30 days, work done per day = 90/30 = 3 units

∴ 3x = 3 i.e. x = 1

Hence, you and your roommate can individually do 2 units and 1 unit per day.

Hence, time taken by you = 90/2

= 45 days and time taken by roommate

= 90/1 = 90 days

Hence, option 1.

Answer: Option D

Explanation :

Let there be x and y inlet and outlet pipes respectively.

∴ x + y = 11                   … (i)

Assume that the capacity of the tank is 35 units.

So, inlet pipe fills 5 units and the outlet pipes empties 7 units of the tank in one hour.

The completely filled tank empties in 7 hours.

∴ 7y – 5x = 5                  … (ii)

Solving the two equations, we get y = 5 and x = 6

Hence, option 4.

Answer: Option B

Explanation :

Assume that P does p units of work in one day.

∴ Total work = 42p

So, Q and R do 1.26p and (1.26 × 1.5 =)1.89p units of work in one day respectively.

∴ Total work done by Q and R in one day = 1.26p + 1.89p = 3.15p

So, Q and R can finish the work in (42p/3.15p ≈) 13 days.

Hence, option 2.

Answer: Option C

Explanation :

Let the capacity of the tank be 24x litres.

Pipes A and B fill 3x and 2x litres per hour while pipe C empties 6x litres in an hour.

Let radius of the cone be r and height be h.

$\frac{1}{3}\pi r^{2}h=24x$

∴ πr²h = 72x

For first 19 hours, water inside the cone = 24x + 57x + 38x – 114x = 5x litres

​​​​​​​

∆ABE ∼ ∆ACD

If AC = 2AB, CD = 2BE

∴ BE = r/2 and AB = h/2

After 50% reduction in the height of the water, volume

$=\frac{1}{3}\pi {(r/2)}^2(h/2)=\frac{\pi r^{2}h}{24}=\frac{72x}{24}=3x$

Option 1: Pipe A was open for 19 hours.

i.e., B and C were open for 1 more hour.

∴ 2x – 6x = –4x

The cone will have 5x – 4x = x litres of water.

∴ Option 1 is eliminated.

Option 2: Pipe A was open for 19 hours 30 minutes.

i.e., B and C were open for 1 more hour and A for 30 more minutes.

∴ 2x – 6x + 1.5x = –2.5x

The cone will have 5x – 2.5x = 2.5x litres of water

∴ Option 2 is eliminated.

Option 3: Pipe B was open for 19 hours 30 minutes.

i.e., A and C were open for 1 more hour and B for 30 more minutes.

∴ 3x – 6x + 1x = –2x

The cone will have 5x – 2x = 3x litres of water.

∴ Option 3 would be the possible option.

Hence, option 3.