IIFT questions on Arithmetic - Time & Work

From the above information, we can say that Rahul, Sohan and Ram take 12, 18 and 10 days respectively to complete the work alone.
LCM(12, 18, 10) = 180
Let the total work be 180 units
So Work done by Rahul in 1 day = 15 units
Work done by Sohan in 1 day = 10 units
Work done by Ram in 1 day = 18 units
Let the time taken by Sohan to complete the remaining work be x days.
Therefore, 180 = 3*(15 + 10 + 18) + 10x
or, 180 = 129 - 10x
or, x = 5.1 days

Time taken by A alone to finish the work = 18*9 hrs
Time taken by B alone to finish the work = 22*9 hrs
Let the total work be LCM(18*9, 22*9) = 198*9 units
Work done by A in 1 hour = 11 units
Work done by B in 1 hour = 9 units

Let the number of hours B worked be 'x'
work done by A + work done by B = Total work done
(6)(11) + (13)(9)(11) + (3)(11) + x(9) = 198*9
x = 44 hours
Number of days B worked = 4 days 8 hours
Therefore, B stopped working on 30th Nov at 11 am.
Answer is option C.

Let the number of hours needed by Surojit working alone be S hours and the number of hours required by Debjani when she works alone be D.
The time required to finish the work when they work together = $\frac{S\bullet D}{S+D}$
Hence, we can frame the given 2 conditions as:

S - $\frac{S\bullet D}{S+D}$ = $\frac{21}{5}$          ...(1)

D - $\frac{S\bullet D}{S+D}$ = $\frac{141}{5}$        ..(2)

Solving the two equations simultaneously, we get S = 15 hours approx

Let the tank capacity be 64 liters.
Hence, capacity of pipes A and B resp are 4 lit/min and 2 lit/min
Also, the capacity of the draining pipe = -1 lit/min
Now, when all pipes are open for 09 mins, tank filled = (4 + 2)*9 - (9*1) = 45 liters
Now, the tank travels for (6/36)*60 min = 10 minutes
During this time, liquid lost = 10*1 lit/min = 10 liters
Liquid left = 35 liters
hence, percentage filled = 35/64 = 55% approx

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).

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).

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 (c).

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 (c).

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 (a).

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 (b). 

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 (a).

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 (d).

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 (b).

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 (b).

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 (b).

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 (a).

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 (d).

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 (a).

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 (a).

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 (c).

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

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

∴ n² + 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 (d).

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 (a).