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

You can practice all previous year OMET questions from the topic Arithmetic - Time & Work. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2020 QADI | Arithmetic - Time & Work**

X, Y, and Z are three software experts, who work on upgrading the software in a number of identical systems. X takes a day off after every 3 days of work, Y takes a day off after every 4 days of work and Z takes a day off after every 5 days of work.

Starting afresh after a common day off,

i) X and Y working together can complete one new upgrade job in 6 days

ii) Z and X working together can complete two new upgrade jobs in 8 days

iii) Y and Z working together can complete three new upgrade jobs in 12 days

If X, Y and Z together start afresh on a new upgrade job (after a common day off), exactly how many days will be required to complete this job?

- A.
2 days

- B.
3.5 days

- C.
2.5 days

- D.
4 days

- E.
5 days

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).

Workspace:

**IIFT 2019 QA | Arithmetic - Time & Work**

A group of women in a society decided to execute interior and exterior decoration of the society in a week’s time. Since 11 women dropped out every day from the second day, the entire decoration was completed on 12th day. How many women participated at the beginning? (Answer to the nearest integer)

- A.
137

- B.
141

- C.
145

- D.
148

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).

Workspace:

**IIFT 2019 QA | Arithmetic - Time & Work**

Rohit purchased a cistern which had a leakage. The cistern can be filled by two inlet pipes, which can individually fill the cistern in 12 min and 15 min respectively. Despite leakage, the two inlet pipes together can fill the cistern in 20 min. How long will it take to empty the completely full the cistern due to leakage?

- A.
10 mins

- B.
12 mins

- C.
15 mins

- D.
16 mins

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).

Workspace:

**XAT 2018 QADI | Arithmetic - Time & Work**

Abdul, Bimal, Charlie and Dilbar can finish a task in 10, 12, 15 and 18 days respectively. They can either choose to work or remain absent on a particular day. If 50 percent of the total work gets completed after 3 days, then, which of the following options is possible?

- A.
Each of them worked for exactly 2 days.

- B.
Bimal and Dilbar worked for 1 day each, Charlie worked for 2 days and Abdul worked for all 3 days.

- C.
Abdul and Charlie worked for 2 days each, Dilbar worked for 1 day and Bimal worked for all 3 days.

- D.
Abdul and Dilbar worked for 2 days each, Charlie worked for 1 day and Bimal worked for all 3 days.

- E.
Abdul and Charlie worked for 1 day each, Bimal worked for 2 days and Dilbar worked for all 3 days.

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).

Workspace:

**IIFT 2018 QA | Arithmetic - Time & Work**

Ram, Ravi and Ratan can alone finish an assignment in 9 days, 12 days and 15 days respectively. They decide to complete a work by working in turns. Ram works alone on Monday, Ravi does the work alone on Tuesday, followed by Ratan working alone on Wednesday & so on. What proportion of the complete work is done by Ravi?

- A.
2/9

- B.
12/47

- C.
1/3

- D.
4/9

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.

Workspace:

**IIFT 2018 QA | Arithmetic - Time & Work**

Nitin installed an overhead tank on the roof of his newly constructed house. Three taps are connected to the tank: 2 taps A and B to fill the tank and one tap C to empty it. Tap A alone can fill the tank in 12 hours, while tap B alone takes one and a half times more time than tap A to fill the tank completely. Tap C alone can empty a completely filled tank in 36 hours. Yesterday, to fill the tank, Nitin first opened tap A, and then after 2 hours opened tap B also. However after 6 hours he realised that tap C was open from the very beginning. He quickly closes tap C. What will be the total time required to fill the tank?

- A.
8 hours 48 minutes

- B.
8 hours 30 minutes

- C.
9 hours 12 minutes

- D.
9 hours 36 minutes

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.

Workspace:

**XAT 2017 QADI | Arithmetic - Time & Work**

Four two-way pipes A, B, C and D can either fill an empty tank or drain the full tank in 4, 10, 12 and 20 minutes respectively. All four pipes were opened simultaneously when the tank is empty. Under which of the following conditions the tank would be half filled after 30 minutes?

- A.
Pipe A filled and pipes B, C and D drained

- B.
Pipe A drained and pipes B, C and D filled

- C.
Pipes A and D drained and pipes B and C filled

- D.
Pipes A and D filled and pipes B and C drained

- E.
None of the above

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).

Workspace:

**IIFT 2017 QA | Arithmetic - Time & Work**

Somesh, Tarun and Nikhil can complete a work separately in 45, 60 and 75 days. They started the work together but Nikhil left 5 days after the start and Somesh left 2 days before the completion of the work. In how many days will the work be completed?

- A.
$25\frac{1}{7}$

- B.
$50\frac{1}{7}$

- C.
$35\frac{5}{7}$

- D.
$40\frac{5}{7}$

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.

Workspace:

**IIFT 2017 QA | Arithmetic - Time & Work**

An overhead tank, which supplies water to a settlement, is filled by three bore wells. The first two bore wells operating together fill the tank in the same time as taken by the third bore well to fill it. The second bore well fills the tank 10 hours faster than the first one and 8 hours slower than the third one. The time required by the third bore well to fill the tank alone is:

- A.
9 hours

- B.
12 hours

- C.
18 hours

- D.
20 hours

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)

∴ 2x^{2} − 46x + 180 = x^{2} − 10x

∴ x^{2} − 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.

Workspace:

**XAT 2016 QADI | Arithmetic - Time & Work**

A water tank has M inlet pipes and N outlet pipes. An inlet pipe can fill the tank in 8 hours while an outlet pipe can empty the full tank in 12 hours. If all pipes are left open simultaneously, it takes 6 hours to fill the empty tank. What is the relationship between M and N?

- A.
M : N = 1 : 1

- B.
M : N = 2 : 1

- C.
M : N = 2 : 3

- D.
M : N = 3 : 2

- E.
None of the above

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).

Workspace:

**IIFT 2016 QA | Arithmetic - Time & Work**

In the marketing management course of an MBA programme, you and your roommate can complete an assignment in 30 days. If you are twice as efficient as your roommate, the time required by each to complete the assignment individually is

- A.
45 days and 90 days

- B.
30 days and 60 days

- C.
40 days and 120 days

- D.
45 days and 135 days

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.

Workspace:

**IIFT 2015 QA | Arithmetic - Time & Work**

A tank is connected with both inlet pipes and outlet pipes. Individually, an inlet pipe can fill the tank in 7 hours and an outlet pipe can empty it in 5 hours. If all the pipes are kept open, it takes exactly 7 hours for a completely filled-in tank to empty. If the total number of pipes connected to the tank is 11, how many of these are inlet pipes?

- A.
2

- B.
4

- C.
5

- D.
6

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.

Workspace:

**IIFT 2015 QA | Arithmetic - Time & Work**

Three carpenters P, Q and R are entrusted with office furniture work. P can do a job in 42 days. If Q is 26% more efficient than P and R is 50% more efficient than Q, then Q and R together can finish the job in approximately:

- A.
11 days

- B.
13 days

- C.
15 days

- D.
17 days

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.

Workspace:

**XAT 2015 QA | Arithmetic - Time & Work**

Three pipes are connected to an inverted cone, with its base at the top. Two inlet pipes, A and B, are connected to the top of the cone and can fill the empty in 8 hours and 12 hours, respectively. The outlet pipe C, connected to the bottom, can empty a filled cone in 4 hours. When the cone is completely filled with water, all three pipes are opened. Two of the three pipes remain open for 20 hours continuously and the third pipe remains open for a lesser time. As a result, the height of the water inside the cone comes down to 50%. Which of the following options would be possible?

- A.
Pipe A was open for 19 hours.

- B.
Pipe A was open for 19 hours 30 minutes.

- C.
Pipe B was open for 19 hours 30 minutes.

- D.
Pipe C was open for 19 hours 50 minutes.

- E.
The situation is not possible.

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.

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

∴ π*r*^{2}h = 72x

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

Workspace:

**IIFT 2013 QA | Arithmetic - Time & Work**

A mother along with her two sons is entrusted with the task of cooking Biryani for a family get-together. It takes 30 minutes for all three of them cooking together to complete 50 percent of the task. The cooking can also be completed if the two sons start cooking together and the elder son leaves after 1 hour and the younger son cooks for further 3 hours. If the mother needs 1 hour less than the elder son to complete the cooking, how much cooking does the mother complete in an hour?

- A.
33.33%

- B.
50%

- C.
66.67%

- D.
None of the above

Answer: Option B

**Explanation** :

Let the number of days required to complete the work by mother, elder son and younger son be m, e and y hours respectively.

Therefore, work done by mother; elder son and younger son in one day is 1/m, 1/e and 1/n respectively.

Also m = e – 1 (∵ Mother take one hour less than the elder son)

Based on the conditions given,

$\frac{1}{m}$ + $\frac{1}{e}$ + $\frac{1}{y}$ = 1 ...(1)

$\frac{1}{e}$ + $\frac{1}{y}$ + $\frac{3}{y}$ = 1 ...(2)

$\frac{1}{y}$ = $\frac{e-1}{4e}$

Also m = e – 1

Substituting values of (1/m) and (1/y) in (1)

$\frac{1}{e-1}$ + $\frac{1}{e}$ + $\frac{e-1}{4e}$ = 1

Solving,

e = 3

Therefore, m = 2 and work done in 1 hour = 50%

Hence, option 2.

Workspace:

**IIFT 2013 QA | Arithmetic - Time & Work**

Capacity of tap Y is 60% more than that of X. If both the taps are opened simultaneously, they take 40 hours to fill the rank. The time taken by Y alone to fill the tank is

- A.
60 hours

- B.
65 hours

- C.
70 hours

- D.
75 hours

Answer: Option B

**Explanation** :

Tap X does a units of work in 1 hour.

∴ Tap Y does 1.6a units of work in 1 hour.

∴ In 1 hour Tap X and Y together do 2.6a units of work.

∴ Work done by Tap X and Y in 40 hours = 2.6a × 40

∴ Time taken by Tap Y alone to do this work

= $\frac{2.6a\times 40}{1.6a}$ = 65 hours

Hence, option 2.

Workspace:

**IIFT 2012 QA | Arithmetic - Time & Work**

12 men can complete a work in ten days. 20 women can complete the same work in twelve days. 8 men and 4 women started working and after nine days 10 more women joined them. How many days will they now take to complete the remaining work?

- A.
2 days

- B.
5 days

- C.
8 days

- D.
10 days

Answer: Option A

**Explanation** :

Let one man complete the piece of work in m days and let one woman complete the piece of work in n days.

∴ $\frac{12}{m}$ = $\frac{1}{10}$ and $\frac{20}{w}$ = $\frac{1}{12}$

∴ $\frac{1}{m}$ = $\frac{1}{120}$ and $\frac{1}{w}$ = $\frac{1}{240}$

Let n be the number of days after 9 days that the work takes to get over.

∴ 9$\left(\frac{8}{m}+\frac{4}{w}\right)$ + n$\left(\frac{8}{m}+\frac{4}{w}+\frac{10}{w}\right)$ = 1

∴ 9$\left(\frac{1}{15}+\frac{1}{60}\right)$ + n$\left(\frac{1}{15}+\frac{1}{60}+\frac{1}{24}\right)$ = 1

∴ $\frac{3}{4}$ + $\frac{15n}{120}$ = 1

∴ n = 2

Hence, option 1.

Workspace:

**IIFT 2011 QA | Arithmetic - Time & Work**

In Bilaspur village, 12 men and 18 boys completed construction of a primary health center in 60 days, by working for 7.5 hours a day. Subsequently the residents of the neighbouring Harigarh village also decided to construct a primary health center in their locality, which would be twice the size of the facility build in Bilaspur. If a man is able to perform the work equal to the same done by 2 boys, then how many boys will be required to help 21 men to complete the work in Harigarh in 50 days, working 9 hours a day?

- A.
45 boys

- B.
48 boys

- C.
40 boys

- D.
42 boys

Answer: Option D

**Explanation** :

A man is able to perform the work equal to the same done by 2 boys.

∴ 18 boys ≡ 9 men

Work done by 12 men and 18 boys, working 7.5 hours a day, in 60 days = (12 + 9) × 7.5 × 60

= 21 × 7.5 × 60 man hours

As the facility in Harigarh is twice that in Bilaspur, man hours required to build health facility in Harigarh

= 2 × 21 × 7.5 × 60

Let a boys be required to complete the work in 50 days, working 9 hours a day.

∴ we have,

(21 + a/2) × 9 × 50 = 2 × 21 × 7.5 × 60

∴ a = 42

Hence, option 4.

Workspace:

**IIFT 2011 QA | Arithmetic - Time & Work**

A contract is to be completed in 56 days and 104 men are set to work, each working 8 hours a day. After 30 days, 2/5th of the work is finished. How many additional men may be employed so that work may be completed on time, each man now working 9 hours per day?

- A.
56 men

- B.
65 men

- C.
46 men

- D.
None of the above

Answer: Option A

**Explanation** :

Work done by 104 men, working 8 hours a day, in 30 days = 104 × 8 × 30 man hours.

This is 2/5th of the total work.

= 104 × 8 × 30 × $\frac{5}{2}$ × $\frac{3}{5}$ = 104 × 4 × 90 man hours.

∴ Remaining work, which is 3/5th of the total work

Let a more men be employed to complete this work.

∴ (104 + a) × 9 × (56 – 30) = 104 × 4 × 90

∴ a = 56

Hence, option 1.

Workspace:

**IIFT 2010 QA | Arithmetic - Time & Work**

Three Professors Dr. Gupta, Dr. Sharma and Dr. Singh are evaluating answer scripts of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work.

- A.
7.2 days

- B.
9.5 days

- C.
11.5 days

- D.
None of these

Answer: Option A

**Explanation** :

Let Dr. Gupta take x days to complete the evaluation work.

∴ Dr. Sharma takes x + 10 days

As Dr. Gupta is 40% more efficient than Dr. Sharma, we have

$\frac{1}{x}$ = $\frac{1.4}{x+10}$

∴ x = 25

∴ x + 10 = 35

Also, Dr. Sharma is 20% more efficient than Dr. Singh.

∴ If Dr. Singh takes y days to complete the evaluation work,

$\frac{1}{35}$ = $\frac{1.2}{y}$

∴ y = 42

Now, let Dr. Singh complete the evaluation work in n days after Dr. Gupta has worked for 10 days and Dr. Sharma has worked for 15 days.

∴ $\frac{10}{25}$ + $\frac{15}{35}$ + $\frac{n}{42}$ = 1

∴ n = 7.2

Hence, option 1.

Workspace:

**IIFT 2010 QA | Arithmetic - Time & Work**

Three pipes A, B and C are connected to a tank. These pipes can fill the tank separately in 5 hours, 10 hours and 15 hours respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at 3/4th of their normal rates for the first hour after which they supplied water at the normal rate. Pipe C supplied water at 2/3rd of its normal rate for first 2 hours, after which it supplied at its normal rate. In how much time, tank would be filled.

- A.
1.05 Hours

- B.
2.05 Hours

- C.
3.05 Hours

- D.
None of these

Answer: Option C

**Explanation** :

Let the pipes work for n hours.

By the given conditions,

$\frac{1}{5\times {\displaystyle \frac{4}{3}}}$ + $\frac{1}{10\times {\displaystyle \frac{4}{3}}}$ + (n - 1) $\left[\frac{1}{5}+\frac{1}{10}\right]$ + $\frac{2}{15\times {\displaystyle \frac{3}{2}}}$ + $\frac{n-2}{15}$ = 1

∴ $\frac{3}{20}$ + $\frac{3}{40}$ + (n - 1) $\frac{3}{10}$ + $\frac{4}{45}$ + $\frac{n-2}{15}$ = 1

∴ $\frac{9}{40}$ + $\frac{3n-3}{10}$ + $\frac{4}{45}$ + $\frac{n-2}{15}$ = 1

∴ $\frac{9n-9}{30}$ + $\frac{2n-4}{30}$ = 1 - $\frac{9}{40}$ - $\frac{4}{45}$

∴ $\frac{11n-13}{30}$ = $\frac{360-81-32}{360}$

∴ 11n - 13 = $\frac{247\times 30}{360}$

∴ n ≈ 3.05

Hence, option 3.

Workspace:

**IIFT 2009 QA | Arithmetic - Time & Work**

Because of economic slowdown, a multinational company curtailed some of the allowances of its employees. Rashid, the marketing manager of the company whose monthly salary has been reduced to Rs.42000 is unable to cut down his expenditure. He finds that there is a deficit of Rs.2000 between his earnings and expenses in the first month. This deficit, because of inflationary pressure, will keep on increasing by Rs.500 every month. Rashid has a saving of Rs.60000 which will be used to fill this deficit. After his savings get exhausted, Rashid would start borrowing from his friends. How soon will he start borrowing?

- A.
10th month

- B.
11th month

- C.
12th month

- D.
13th month

Answer: Option D

**Explanation** :

If Rashid’s savings fill his deficit for n months,

60000 = $\frac{n}{2}$[4000 + (n - 1)500]

∴ n^{2} + 7n - 240 = 0

∴ n ≈ 12.38

∴ Rashid savings last for 12 months, after which he has to start borrowing from the 13^{th} month.

Hence, option 4.

Workspace:

**IIFT 2009 QA | Arithmetic - Time & Work**

Aditya, Vedus and Yuvraj alone can do a job in 6 weeks, 9 weeks and 12 weeks respectively. They work together for 2 weeks. Then Aditya leaves the job. Vedus leaves the job a week earlier to the completion of the work. The job would be completed in:

- A.
4 weeks

- B.
5 weeks

- C.
7 weeks

- D.
None of the above

Answer: Option A

**Explanation** :

Aditya, Vedus and Yuvraj do (1/6), (1/9) and (1/12) of the work respectively in a week. Let the total time taken to complete the work be n weeks.

Then,

2$\left(\frac{1}{6}\right)$ + (n - 1)$\left(\frac{1}{9}\right)$ + n$\left(\frac{1}{12}\right)$ = 1

∴ $\frac{12+4(n-1)+3n}{36}$ = 1

∴ n = 4

Hence, option 1.

Workspace:

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