N people finish 35% of the project by working 7 hours a day for 10 days.
N people finish 35% of the project by working 70 hours.
N people finish 5% of the project by working 10 hours.
N people finish 65% of the project by working 130 hours.
65% of the project is done in N * 130 man hours.
The remaining 65% was actually done by (N-10) people in 14 days by working 10 hours a day.
(N-10) people finish 65% of the project by working 140 hours.
65% of the project is done in (N – 10) * 140 man hours.
N * 130 = (N – 10) * 140
13N = 14N – 140
N = 140

 Time taken by Anu, Tanu and Manu to complete a job is in the ratio 5 : 8 : 10

This means that their efficiencies are in the ratio 1/5:1/8:1/10

Or, their efficiencies are in the ratios 8 : 5 : 4

This means, for instance, if Anu can wash 8 plates in an hour, Tanu and Manu can wash 5 & 4 plates respectively in one hour.

So, let us remodel the entire question over this imaginary scenario of washing plates…

All three of them finish the job in 4 days working 8 hours per day.

Anu, Tanu and Manu can wash 8, 5 and 4 plates respectively in an hour.

So the total number of plates = (8+5+4)×4×8=17×4×8

Anu and Tanu work together for 6 days, working 6 hours 40 minutes per day.

The total number of plates washed by them = (8+5)×6×6*2/3=13×40

So the remaining plates to be washed =17×4×8−13×40

=4(17×8−130)

=4(80+56−130)

=4×6

So, the question is, in how man hours can Manu wash 4×6

 plates?

We know that Manu can wash 4 plates in one hour.

Therefore, he requires 6 hours to wash 4×6

 plates.

The answer is ‘6’

 Let R be the fraction of work done by Rahul in 1 hour.

and G be the fraction of work done by Gautam in 1 hour.

Initially Rahul works from 9AM to 5PM (8 hours) and Gautam works for 2 hours less.

8R + 6G = 1 whole unit of work

If they start together they finish 30 minutes earlier or if they start at 9 AM, they finish at 4:30PM (7.5 hours)

7.5R + 7.5G = 1 whole unit of work

8R + 6G = 7.5R + 7.5G

0.5R = 1.5G

R = 3G

This means Rahul is thrice as efficient as Gautam.

8R + 6G = 1 whole unit of work

8R + 2R = 1 whole unit of work

10R = 1 whole unit of work

R is the fraction of work done by Rahul in 1 hour.

Since 10R = 1

R alone takes 10 hours to finish the job.

 Let the filling rate of pipe A be ‘a’ and the rate of emptying the tank of pipe B be ‘b’.

Then from the given information we can say that, when pipe A is kept open from 2PM to 10PM, i.e. for 8 hours and pipe B drains from 3PM till 10PM (7 hours), the tank gets filled up.

So,

8a – 7b = 1 [1]

Also, from the second statement,

A is kept open from 2 PM to 6 PM (4 hours), and B is kept open for 2 hours (4 PM to 6PM), the tank gets filled up.

4a – 2b = 1 [2]

Multiplying equation [2] and subtracting [1] from it we get

8a – 4b – (8a – 7b) = 2 – 1

7b – 4b = 1

b = 1/3

From here we can compute the value of a

Putting the value of b in eq [2]

4a – 2(1/3) = 1

4a – 2/3 = 1

4a = 1 + 2/3

4a = 5/3

a = 5/12

Hence the rate of filling is 5/12.

If only pipe a is kept open, the tank will get filled in, say ‘n’ hours

n a = 1

n (5/12) = 1

So, n = 12/5

= 2.4 hours

= 144 minutes

 Let’s consider the efficiencies of Amar, Akbar, and Anthony to be x, y, z respectively.

x + y = 1/12

y + z = 1/16

z + x = 1/24

If we add the above equations, we will get 2(x + y + z) = 3/16

x + y + z = 3/32

From the above equations, we can get x = 1/32

, y = 5/96

, z = 1/96

Hence, x is neither the fastest nor the slowest.

Amar can complete the project in 32 months.

 The contractor finishes 1.5 km in 60 days with 140 people.

That means, he finishes, 0.5 km in 20 days with 140 people.

He has 200 – 60 = 140 days left to finish the job.

Let’s say 20 days is 1 unit of time.

So the contractor finishes 0.5 km in 1 unit of time with 140 people, He has 4.5 kms to finish in 7 units of time(7 × 20 days) with n people.

The working capacity of the peopl should still be the same…

Therefore, 0.5/140 * 1 = 4.5/n * 17

n = 180.

Therefore, the contractor needs, 180 – 140 = 40 people more.

  John = x units/day

Jack = 2x units/day

Jim = x units/day (1/3rd of John does = 3x units, out of which jack does 2x)

Total = 4x units/day

John’s time taken = x(n+3) = 4x × n

Then n = 1 day for 4x units

For x units, Jim will take 4 days

 Anil in one day can do th of the work

Sunil in one day can do th of the work

Anil starts the job and Sunil joins him after three days.

So, Anil would have done th of the work by the time Sunil joins

After Sunil joins, they both would be doing th of work everyday

Now, it is known that Bimal joins them after some days and finishes 10 % of the work.

Now, Anil alone had done th of the work and Bimal completes th of the work

So, in total they would have done = th of the work (is the work done by Bimal)

Remaining work = , which would be done by Anil and Sunil in 10 days.

Combining everything,

Total number of days = 3 + 10 = 13 days

 If John works the same number of regular and over-time hours say ‘p’

The the income would be 57p and 114p

Let’s say that he works ‘x’ hours regular and ‘y’ hours overtime…

So, the income would be 57x and 114y

we are told that 114y is 15% of 57x

114y = 0.15 * 57x

y = 0.075x

we also know that x+y = 172

therefore, x + 0.075x = 1.075x = 172

x = 160

y = 172 – 160 = 12

Therefore, the number of hours he worked Overtime is 12.

Let Machines be referred as R and men be referred as M

It is given that, three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job.

From the given data,

3M + 8R = 2 x (3R + 8M)

2R = 13M

R = M

Therefore, if two machines can finish a job in 13 days, 2 Robos can finish the job in 13 days. 

  Given that a tank is emptied everyday at a fixed time point and after that either pump A or B or both start working to fill the tank.

On Monday, A alone completed filling the tank at 8 pm.

On Tuesday, B alone completed filling the tank at 6 pm.

On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank.

We have to find at what time will the tank filled on thursday if both were used simultaneously all along

A is doing 3 hours less work on Wednesday which B completes in 2 hours therefore B = 1.5 A

⟹ nn−2

 = 32

⟹ 2n = 3n – 6

⟹ n = 6

A takes 6 hours and B takes 4 hours or the tank is closed at 2 pm

When they are together open

⟹ 16

 + 14

 = 2+312

⟹ Rate is 512

 or they can fill the entire tank in 125

 hours which is 2 hours 24 minutes

Starting from 2 pm, they fill the entire tank by 4:24 pm

Here the break through is figuring out the ratio or efficiency of the rate at which they fill is B : A = 3 : 2

The conventional way is

⟹ n−3n

 + 2n−2

 = 1

n-3 hours at the rate of 1n

 per hour

2 hours at the rate 1n−2

 per hour

Solving this we get n = 6 and we can find solution using this n.

The question is “A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?”

Hence, the answer is 4 : 24 PM

 Given that Ramesh and Ganesh can together complete a work in 16 days

After seven days of working together, Ramesh got sick and his efficiency fell by 30%.

To the original schedule of 16 days there are 9 more days and in this 9 days Ramesh fall short by 9 × 0.3x = 2.7x where x is the work done by Ramesh in a day

So the change = 2.7x (i.e. the gap because of drop in efficiency)

This 2.7x is compensated by one day of Ganesh + another 0.7 of Ramesh

2.7x = G + 0.7 x (Because the last day also he works with reduced efficiency)

2x = G

Hence Ganesh works twice as that of Ramesh.

If Ganesh had worked alone after Ramesh got sick means in the 7 days Ramesh would have completed 7/48

 of the task

Ganesh on each day can do 1/24

 of the task

Task to be completed by Ganesh is 41/48

 completely

One day he can complete 1/24

 so 41/48

 × 24/1

 = 20.5 days

If Ganesh had worked alone after Ramesh got sick, the no.of days needed for completing the remaining work is 20.5 – 7 = 13.5 days

 Given that a person can complete a job in 120 days.

He works alone on Day 1.

On Day 2, he is joined by another person who also can complete the job in exactly 120 days.

On Day 3, they are joined by another person of equal efficiency.

Everyday a new person with the same efficiency joins the work

So by first day 1 person

Similarly by second day 2

Third day 3 and it goes on…….until it makes a total of 120

We have to find how many days are required to complete the job

So 15×16/2

 = 120 [by n(n+1)/2]

So 1 + 2 + 3 + 4 …… till 15 days are required to complete this Job.

This will get completed in 15 days.

Hence 15 days are required to complete the job

 Given that Amal can complete a job in 10 days ⟹ Rate = 1/10

Bimal can complete it in 8 days ⟹ Rate = 1/8

Amal, Bimal and Kamal together complete the job in 4 days ⟹ Rate = 1/4

Kamal’s rate of doing work is,

Amal + Bimal + Kamal = 1/4

⟹ 1/10 + 1/8+ Kamal = 1/4

⟹ Kamal = 1/4

 – 1/8

 − 1/10

⟹ Kamal = 10/40 – 5/40− 4/40

Kamal’s rate of doing work = 1/40

Kamal would have taken 40 days to complete the entire task.

Amal : Bimal : Kamal = 1/10 : 1/8: 1/40

 = 4 : 5 : 1

Total amount of Rs.1000 is paid as remuneration

Amal : Bimal : Kamal = 4/10 : 5/10 : 1/10

Kamal’s share = 110

 × 1000 = Rs.100