CAT Time and Work Questions

Let us assign the amount of work done by Renu in one hour is R
And the amount of work done by Seema in one hour is S

We are told that a certain task with different time durations for Seema and Renu
Renu=15 days with 4 hours each day, that is total work done by Renu is $$15\times\ 4\times\ R=60R$$
Similarly for Seema, 8 days and 5 hours each day, total work done by Seema is $$40S$$
We know that $$60R=40S$$ or $$S=1.5R$$

For this task, let us assume that the amount of days that Seema decides to work is X and the hours per day that Renu decides to work is Y
We are then told that, Seema works for 2Y hours per day and Renu works for 2X days, 
Work done by them will be $$2XYR$$ and $$2XYS$$

We are told that Y=2, Making this $$4XR$$ and $$4XS$$
$$S=1.5R$$
Total work will be, $$4XR$$+$$6XR$$=$$60R$$
We get the value of $$X=6$$

X is the number of days Seema will work, which is 6. 

We are given that Sam completes a piece of work in 20 days. We are also given that Mohit is twice as fast, so he should take only 10 days. Mohit is thrice as fast as Ayna, so he would take 30 days. 

Let's take the total work to be 60 units; this would give the work done per day for Mohit, Sam, and Ayna to be 6, 3 and 2, respectively. 

On the first day Sam and Mohit work: doing 9 units
On the second day Sam and Ayan work: doing 5 units
On the third day, Mohit and Ayan work: doing 8 units

Essentially doing 22 units in a 3 days cycle. 
After two such cycles, there will be 60-44 = 16 units of work left

On day 7, Sam and Mohit would work 9 units, leaving 7 units 
On day 8, Sam and Ayan would work 5 units, leaving 2 units
And on day 9, Ayan and Mohit would complete the remaining work. 

So Sam worked for a total of 2+2+2= 6 days and on each day he did 3 units of work, completing 18 units of work.

The ratio of work done by Sam would be $$\frac{18}{60}=\frac{3}{10}$$

Therefore, Option B is the correct answer. 

We can take the work done by Amal, Vimal and Sunil to be A, V and S, respectively. 

Let's take the total work they did to be T.

We are given the equations:
150A + 150V = T      ...(1)
100V + 100S = T      ...(2)

75A + 135V + 45S = T    ...(3)

Adding (1) and (2), we get: 150A + 250V + 100S = 2T   ...(4)

And multiplying (3) with 2 we get: 150A + 270V + 100S = 2T  ...(5)

Subtracting (5) from (4), we get 10S = 20V or simply S = 2V

Using this in (2), we get the total work T to be 300V, and using that result in (1), we get A=V

Therefore, the work done by A, V and S equals V, V, and 2V units per day. 

Now, in the question, we are given work cycles. To simplify the calculations, we should consider a time duration that is the LCM of the period taken by the three agents, which in this case would be 6

In 6 days, Amal will work for 6 days, doing 6V units of work.
Vimal will work for 3 days, doing 3V units of work. 
Sunil will work for 2 days, doing 4V units of work. 

So, in one 6-day cycle, 13V units of work will be done. 
Dividing the total work (300V) by 13V we can see that in 23 cycles 299V units of work will be done. 

These 23 cycles will be $$23\times\ 6\ =\ 138$$ days
The remaining 1V units of work will be done the next day. 

Therefore, a total of 139 days will be required. 

Let the time taken by A to fill the tank alone be x hours, which implies the time taken by B to empty the tank alone is (x-1) hours (B is the drainage pipe), and the time taken by C to fill the tank is y hours.

It is given that when pipes A, B, and C are turned on together, the empty tank is filled in two hours. 

Hence, $$\frac{1}{x}-\frac{1}{x-1}+\frac{1}{y}=\frac{1}{2}$$ .... Eq(1)

It is given that if pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank.

Hence, B worked for 1 hour, and C worked for 2 hours 15 minutes, which is equal to $$\frac{9}{4}$$ hours.

In 1 hour, B worked $$-\frac{1}{x-1}$$ units, and in $$\frac{9}{4}$$ hours, C worked $$\frac{9}{4y}$$ units.

Hence, $$\frac{9}{4y}-\frac{1}{x-1}=1$$ .... Eq(2)

Solving both equations, we get $$y=\frac{3}{2}$$, and $$x=3$$

Hence, the time taken by C is $$\frac{3}{2}$$ hours, which is equal to $$90$$ minutes.

The correct option is A

Let the work done by Rahul, Rakshita, and Gurmeet be a, b, and c units per day, respectively, and the total units of work are W.

Hence, we can say that 7(a+b+c) < W ( Rahul, Rakshita, and Gurmeet, working together, would have taken more than 7 days to finish a job).

Similarly, we can say that 15(a+c) > W ( Rahul and Gurmeet, working together would have taken less than 15 days to finish the job)

Now, comparing these two inequalities, we get: 7(a+b+c) < W < 15(a+c)

It is also known that they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. Therefore, the total units of work done is: W = 6(a+b+c)+3b

Hence, we can say that 7(a+b+c) < 6(a+b+c)+3b < 15(a+c)

Therefore, (a+b+c) < 3b => a+c < 2b, and 9b < 9(a+c) => b < a+c

=> a+b+c < 3b => 7(a+b+c) < 21b , and 15b < 15(a+c)

Hence, The number of days required for b must be in between 15 and 21 (both exclusive).

Hence, the correct option is D

Let 'g' and 's' be the efficiencies of Gautam and Suhani. Let W is the total amount of work.

=> g + s = W/20 (1 day work) ----(1)

Also Gautam doing only 60% => 3g/5 and Suhani doing 150% => 3s/2

=> 3g/5 + 3s/2 = W/20 (1 day work)

=> $$g+s=\dfrac{3g}{5}+\dfrac{3s}{2}$$

=> $$\dfrac{s}{g}=\dfrac{4}{5}$$ => Gautam is the more efficient person.

Now, from the 1st equation

=> $$g+\dfrac{4g}{5}=\dfrac{W}{20}$$

=> $$\dfrac{9}{5}g=\dfrac{W}{20}$$

=> $$g=\dfrac{W}{36}$$

=> Gautam takes 36 days to finish the complete work.

Let us assume the efficiencies of Amal, Sunil, and Kamal are a, s, and k, respectively.

Given that they are in H.P.

=> $$\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{1}{k}$$ ---(1)

Also, given that Kamal takes twice as much time as Amal to do the same amount of job

=> a = 2k

Given that when Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job.

=> If W is the total work => 4a + 9s + 16k = W.

from (1)$$\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{2}{a}$$ => $$a=\dfrac{3}{2}s$$ and $$k=\dfrac{3}{4}s$$

=> $$4\left(\dfrac{3s}{2}\right)+9s+16\left(\dfrac{3s}{4}\right)=W$$

=> $$6s+9s+12s=W$$

=> $$27s=W\ =>\ s\ =\ \dfrac{W}{27}$$

=> Sunil will take 27 days to finish the work when working alone.

Let the unit of work done by 1 man in 1 hour and 1 day be 1 MDH unit (Man Day Hour).

Thus, in 7 hours per day for 10 days, the work done by N people =$$N\times\ 7\times\ 10$$ MDH units.

Since this is equal to 35% of the total work,

35% of the total work = $$N\times\ 7\times\ 10$$ MDH units.

Total work = $$\frac{\left(N\times\ 100\times\ 7\times\ 10\right)}{35}=200\times N$$ MDH units.

The work left = $$200N-70N=130N$$ MDH units.

Now, 10 people left the job. So, the number of people left = (N-10)

Since (N-10) people completed the rest of work in 14 days by working 10 hours a day,

$$(N-10)\times\ 14\times\ 10=130N$$

$$10N=1400$$

N = 140

Thus, the correct option is D.

Let the time taken by Anu, Tanu and Manu be 5x, 8x and 10x hours.

Total work = LCM(5x, 8x, 10x) = 40x

Anu can complete 8 units in one hour

Tanu can complete 5 units in one hour

Manu can complete 4 units in one hour

It is given, three of them together can complete in 32 hours.

32(8 + 5 + 4) = 40x

x = $$\frac{68}{5}$$

It is given,

Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day, i.e. 36 + 4 = 40 hours

40(8 + 5) + y(4) = 40x

4y = 24

y = 6

Manu alone will complete the remaining work in 6 hours.

Let the efficiency of Bob be 3 units/day. So, Alex's efficiency will be 6 units/day, and Cole's will be 2 units/day.

Since Bob can finish the job in 40 days, the total work will be 40*3 = 120 units.

Since Alex and Bob work on the first day, the total work done = 3 + 6 = 9 units.

Similarly, for days 2 and 3, it will be 5 and 8 units, respectively.

Thus, in the first 3 days, the total work done = 9 + 5 + 8 = 22 units.

The work done in the first 15 days = 22*5 = 110 units.

Thus, the work will be finished on the 17th day(since 9 + 5 = 14 units are greater than the remaining work).

Since Alex works on two days of every 3 days, he will work for 10 days out of the first 15 days.

Then he will also work on the 16th day.

The total number of days = 11.

Let Rahul work at a units/hr and Gautam at b units/hour
Now as per the condition :
8a+6b =7.5a+7.5b
so we get 0.5a=1.5b
or a=3b
Therefore total work = 8a +6b = 8a +2a =10a
Now Rahul alone takes 10a/10 = 10 hours.

Let the total amount of work be 60 units.

Then Anu, Vinu, and Manu do 4, 5, and 3 units of work per day respectively.

On the 1st day, Anu and Vinu work. Work done on the 1st day = 9 units

On the 2nd day, Manu and Vinu work. Work done on the 2nd day = 8 units

This cycle goes on. And in 6 days, the work completed is 9+8+9+8+9+8 = 51 units.

On the 7th day, again Anu and Vinu work and complete the remaining 9 units of work. Thus, the number of days taken is 7 days.

Let Entire work be W 
Now Anil worked for 24 days 
Bimal worked for 14 days and Charu worked for 14 days .
Now Anil Completes W in 60 days 
so in 24 days he completed 0.4W 
Bimal completes W in 84 Days
So in 14 Days Bimal completes = $$\frac{W}{6}$$
Therefore work done by charu = $$W-\frac{W}{6}-\frac{4W}{10}$$= $$\frac{26W}{10}$$=$$\frac{13W}{30}$$
Therefore proportion of Charu = $$\frac{13}{30}\times\ 21000$$=9100

Let the total work be 48 units. Let Amar do 'm' work, Akbar do 'k' work, and Anthony do 'n' units of work in a month.

Amar and Akbar complete the project in 12 months. Hence, in a month they do $$\frac{48}{12}$$=4 units of work.

m+k = 4.

Similarly, k+n = 3, and m+n = 2.

Solving the three equations, we get $$m=\frac{3}{2},\ k=\frac{5}{2},\ n=\frac{1}{2}$$.

Here, Amar works neither the fastest not the slowest, and he does 1.5 units of work in a month. Hence, to complete the work, he would take $$\frac{48}{1.5}=32$$months.

Now Anil Paints in 12 Days 
Barun paints in 16 Days 
Now together Arun , Barun and Chandu painted in 6 Days 
Now let total work be W 
Now each worked for 6 days 
So Anil's work = 0.5W 
Barun's work = $$\frac{6W}{16}=\frac{3W}{8}$$
Therefore Charu's work = $$\frac{W}{2}-\frac{3W}{8}=\ \frac{W}{8}$$
Therefore proportion of charu =$$\frac{24000}{8}=\ 3,000$$

Let A fill the tank at x liters/hour and B drain it at y liters/hour 
Now as per Condition 1 : 
We get Volume filled till 10pm = 8x-7y       (1) .
Here A operates for 8 hours and B operates for 7 hours .
As per condition 2
We get Volume filled till 6pm = 4x-2y        (2)
Here A operates for 4 hours and B operates for 2 hours .
Now equating (1) and (2)
we get 8x-7y =4x-2y
so we get 4x =5y 
y =4x/5
So volume of tank = $$8x-7\times\ \frac{4x}{5}=\frac{12x}{5}$$
So time taken by A alone to fill the tank = $$\frac{\frac{12x}{5}}{x}=\frac{12}{5}hrs\ $$
= 144 minutes

Let Jack take "t" days to complete the work, then John will take "2t" days to complete the work. So work done by Jack in one day is (1/t) and John is (1/2t) . 

Now let Jim take "m" days to complete the work. According to question, $$\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}\ or\ \frac{1}{m}=\frac{1}{2t\ }or\ m=2t$$ Hence Jim takes "2t" time to complete the work.

Now let the three of them complete the work in "p" days. Hence John takes "p+3" days to complete the work.

$$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$$

$$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$$

or m=1. Hence JIm will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work

Let the desired efficiency of each worker '6x' per day.

140*6x*200= 6 km ...(i)

In 60 days 60/200*6=1.8 km of work is to be done but actually 1.5km is only done.

Actual efficiency 'y'= 1.5/1.8 *6x =5x.

Now, left over work = 4.5km which is to be done in 140 days with 'n' workers whose efficiency is 'y'.

=> n*5x*140=4.5 ...(ii)

(i)/(ii) gives,

$$\frac{\left(140\cdot6x\cdot200\right)}{\left(n\cdot5x\cdot140\right)}=\frac{6}{4.5}$$

=> n=180.

.'. Extra 180-140 =40 workers are needed. 

Assuming A completes a units of work in a day and B completes B units of work in a day and the total work = 1 unit

Hence, 12(a+b)=1.........(1)

Also, 9($$\ \frac{\ a}{2}$$+3b)=1  .........(2)

Using both equations, we get, 12(a+b)= 9($$\ \frac{\ a}{2}$$+3b)

=> 4a+4b=$$\ \frac{\ 3a}{2}$$+9b

=> $$\ \frac{\ 5a}{2}$$=5b

=> a=2b

Substituting the value of b in equation (1),

12($$\ \frac{\ 3a}{2}$$)=1

=> a=$$\ \frac{\ 1}{18}$$

Hence, the number of days required = 1/($$\ \frac{\ 1}{18}$$)=18

Consider the work done by a man in a day = a and that by a machine = b

Since, 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, hence the efficiency will be double.

=> 3a+8b = 2(3b+8a)

=> 13a=2b

Hence work done by 13 men in a day = work done by 2 machines in a day.

=> If two machines can finish the job in 13 days, then same work will be done 13 men in 13 days. 

Hence the required number of men = 13

Let the total work be LCM of 20, 40 = 40 units 

Efficiency of Anil and Sunil is 2 units and 1 unit per day respectively.

Anil works alone for 3 days, so Anil must have completed 6 units.

Bimal completes 10% of the work while working along with Anil and Sunil.

Bimal must have completed 4 units.

The remaining 30 units of work is done by Anil and Sunil

Number of days taken by them 30/3=10

The total work is completed in 3+10=13 days

Let the efficiency of type A pipe be 'a' and the efficiency of type B be 'b'.
In the first case, 10 type A and 45 type B pipes fill the tank in 30 mins.
So, the capacity of the tank  = $$\dfrac{1}{2}$$(10a + 45b)........(i)
In the second case, 8 type A  and 18 type B pipes fill the tank in 1 hour.
So, the capacity of the tank = (8a + 18b)..........(ii)
Equating (i) and (ii), we get
10a + 45b = 16a + 36b
=>6a = 9b
From (ii), capacity of the tank = (8a + 18b) = (8a + 12a) = 20a
In the third case, 7 type A and 27 type B pipes fill the tank.
Net efficiency = (7a + 27b) = (7a + 18a) = 25a
Time taken = $$\dfrac{20\text{a}}{25\text{a}}$$ hour = 48 minutes.
Hence, 48 is the correct answer.

Let the efficiency of humans be 'h' and the efficiency of robots be 'r'.
In the first case, 
Total work = (15h + 5r) * 30......(i)
In the second case,
Total work = (5h + 15r) * 60......(ii)
On equating (i) and (ii), we get
(15h + 5r) * 30 = (5h + 15r) * 60
Or, 15h + 5r = 10h + 30r
Or, 5h = 25r
Or, h = 5r
Total work = (15h + 5r) * 30 = (15h + h) * 30 = 480h
Time taken by 15 humans = $$\dfrac{\text{480h}}{\text{15h}}$$ days= 32 days.
Hence, option C is the correct answer.

Let us assume that A can complete 'a' units of work in a day and B can complete 'b' units of work in a day. 
A works alone till half the work is completed. 
A and B work together for 4 days and B works alone to complete the last 5% of the work. 
=> A and B in 4 days can complete 45% of the work.

Let us assume the total amount of work to be done to be 100 units.
4a + 4b = 45 ---------(1)
B needs 25% more time than A to finish a job. 
=> 1.25*b = a ----------(2)

Substituting (2) in (1), we get, 
5b+4b = 45
9b = 45
b = 5 units/day

B alone can finish the job in 100/5 = 20 days. 
Therefore, option A is the right answer.

Let 't' pm be the time when the tank is emptied everyday. Let 'a' and 'b' be the liters/hr filled by pump A and pump B respectively.

On Monday, A alone completed filling the tank at 8 pm. Therefore, we can say that pump A worked for (8 - t) hours. Hence, the volume of the tank = a*(8 - t) liters. 

Similarly, on Tuesday, B alone completed filling the tank at 6 pm. Therefore, we can say that pump B worked for (6 - t) hours. Hence, the volume of the tank = b*(6 - t) liters. 

On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. Therefore, we can say that pump A worked for (5 - t) hours  and pump B worked for 2 hours. Hence, the volume of the tank = a*(5 - t)+2b liters. 

We can say that  a*(8 - t) =  b*(6 - t) =  a*(5 - t) + 2b

a*(8 - t) = a*(5 - t) + 2b

$$\Rightarrow$$ 3a = 2b   ... (1)

a*(8 - t) =  b*(6 - t)

Using equation (1), we can say that 

$$a*(8-t)=\dfrac{3a}{2}\times(6-t)$$

$$t = 2$$

Therefore, we can say that the tank gets emptied at 2 pm daily. We can see that A takes 6 hours and pump B takes 4 hours alone. 

Hence, working together both can fill the tank in = \dfrac{6*4}{6+4} = 2.4 hours or 2 hours and 24 minutes. 

The pumps started filling the tank at 2:00 pm. Hence, the tank will be filled by 4:24 pm.  

Let the efficiency of filling pipes be 'x' and the efficiency of draining pipes be '-y'.
In the first case, 
Capacity of tank = (6x - 5y) * 6..........(i)
In the second case,
Capacity of tank = (5x - 6y) * 60.....(ii)
On equating (i) and (ii), we get
 (6x - 5y) * 6 =  (5x - 6y) * 60
or, 6x - 5y = 50x - 60y
or, 44x = 55y
or, 4x = 5y
or, x = 1.25y

Capacity of the tank = (6x - 5y) * 6 = (7.5y - 5y) * 6 = 15y
Net efficiency of 2 filling and 1 draining pipes = (2x - y) = (2.5y - y) = 1.5y
Time required = $$\dfrac{\text{15y}}{\text{1.5y}}$$hours = 10 hours.

Hence, 10 is the correct answer.

Let 'R' and 'G' be the amount of work that Ramesh and Ganesh can complete in a day. 

It is given that they can together complete a work in 16 days. Hence, total amount of work = 16(R+G) ... (1)

For first 7 days both of them worked together. From 8th day, Ramesh worked at 70% of his original efficiency whereas Ganesh worked at his original efficiency. It took them 17 days to finish the same work. i.e. Ramesh worked at 70% of his original efficiency for 10 days.

$$\Rightarrow$$ 16(R+G) = 7(R+G)+10(0.7R+G)

$$\Rightarrow$$ 16(R+G) = 14R+17G

$$\Rightarrow$$ R = 0.5G ... (2)

Total amount of work left when Ramesh got sick = 16(R+G) - 7(R+G) = 9(R+G) = 9(0.5+G) = 13.5G

Therefore, time taken by Ganesh to complete the remaining work = $$\dfrac{13.5G}{G}$$ = 13.5 days.

Let the rate of work of a person be x units/day. Hence, the total work = 120x.

It is given that one first day, one person works, on the second day two people work and so on.

Hence, the work done on day 1, day 2,... will be x, 2x, 3x, ... respectively.

The sum should be equal to 120x.

$$120x = x* \frac{n(n+1)}{2}$$

$$n^2 + n - 240 = 0$$

n = 15 is the only positive solution.

Hence, it takes 15 days to complete the work.

Let the time taken by the outlet pipe to empty = x hours
Then, $$\frac{1}{8} - \frac{1}{x} = \frac{1}{10}$$
=> $$x = 40$$
Hence time taken by the outlet pipe to make the tank half-full = 40/2 = 20 hour

Let the time take by kamal to complete the task be x days.
Hence we have $$\frac{1}{10} + \frac{1}{8} + \frac{1}{x} = \frac{1}{4}$$
=> x = 40 days.
Ratio of the work done by them = $$\frac{1}{10} : \frac{1}{8} : \frac{1}{40}$$ = 4 : 5 : 1
Hence the wage earned by Kamal = 1/10 * 1000 = 100

After 1 min the cans will contain following amount of chemicals : A - 1060 B - 1030 C - 970 D - 950

So, we can see that the can D loses 50 ltrs in 1 min which is highest. So the can D will lose 1000 ltrs in 20*1 = 20 mins.

Let the work done by each technician in one hour be 1 unit.
Therefore, total work to be done = 60 units.
From 11 AM to 5 PM, work done = 6*6 = 36 units.
Work remaining = 60 - 36 = 24 units.
Work done in the next 3 hours = 7 units + 8 units + 9 units = 24 units.
Therefore, the work gets done by 8 PM.

Let the work done by the big pump in one hour be 3 units.
Therefore, work done by each of the small pumps in one hour = 2 units.
Let the total work to be done in filling the tank be 9 units.
Therefore, time taken by the big pump if it operates alone = 9/3 = 3 hours.
If all the pumps operate together, the work done in one hour = 3 + 2*3 = 9 units.
Together, all of them can fill the tank in 1 hour.
Required ratio = 1/3

A takes 4 days to complete the work.
So, B takes 8 days to complete the same work.
C takes 16 days to complete the work.
D takes 32 days to complete the same work.

In order to measure, let the total work be of 64 units. Hence, the speed of working of each of the four persons is given below.

A - 16 units/hr
B - 8 units/hr
C - 4 units/hr
D - 2 units/hr

From the given options, we need to find two pairs in such a way that their speeds are in the ratio 3:2. Note that A+D=18 while B+C=12 and the ratio is 3:2

Hence, the first pair is A and D and the second pair is B and C

Let the time taken working together be t.
Time taken by Arnold = t+1
Time taken by Asit = t+6
Time taken by Afzal = 2t
Work done by each person in one day = $$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}$$
Total portion of workdone in one day $$=\frac{1}{t}$$
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$$
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}=\frac{2-1}{2t}$$
$$2t+7=\frac{(t+1)\cdot(t+6)}{2t}$$
$$3t^2-7t+6=0 \longrightarrow\ t=\frac{2}{3} $$or $$t=-3$$
Therefore total time = $$\frac{2}{3}$$hours = 40mins

Alternatively,
$$\frac{1}{(t+1)}+\frac{1}{(t+6)}+\frac{1}{2t}=\frac{1}{t}$$
From the options, if time $$= 40$$ min, that is, $$t = \frac{2}{3}$$
LHS = $$\frac{3}{5} + \frac{3}{20} + \frac{3}{4} = \frac{(12+3+15)}{20} = \frac{30}{20} = \frac{3}{2}$$
RHS = $$\frac{1}{t}=\frac{3}{2}$$
The equation is satisfied only in case of option C
Hence, C is correct

If they work together, then total work done in single day = $$\frac{1}{3} + \frac{1}{4} + \frac{1}{6} = \frac{9}{12}$$
So $$\frac{9}{12}$$ work is done in 1 day
Hence unit work will be done in $$\frac{12}{9}$$ or $$\frac{4}{3}$$ days.
 

Raja takes 7 days to complete the job.

Feb 25, 1996 is a Sunday.

If T and J start working on Sunday, they can complete the work by wednesday because T would have worked for 3 days and J would worked for 4 days, thereby matching the number of days worked by Raja.

Hence, they can complete the job in 4 days.

The number of days taken by Raja to complete the work is 5+31+2 = 38
So, cumulative number of days needed by T and S to complete the work is 38.

Both of them take two days off in a week as S takes of on Saturday and Sunday and T takes off on Tuesday and Thursday.

So, total number of man-working days per week by the duo is 10.
Hence, after three weeks, they finish 30 man working days.
i.e by end of 17th March 1996 (Sunday), 30 man working days are finished.

Both of them work on Monday,
S works on Tuesday
Both of them work on Wednesday
S works on Thursday
Both of them work on Friday and the remaining 8 man working days are also over.

Hence, the required date is 17+5 = 22 March 1996 (Friday)

Suppose he appointed x persons and y of them didn't come. Hence work done by each of them increases by 32.

So $$\frac{480}{x-y} - \frac{480}{x} = 32$$

Now we can check options by putting in the above eq. 

x=10 and y=4 will be our answer

As machine B's efficiency is twice as of A's, Hence, it will complete its work in 30 hours.
And C's efficiency is putting A and B together i.e. = 20 hours $$( (\frac{1}{60} + \frac{1}{30})^{-1})$$
Now if all three work together, then it will be completed in x (say) days.

$$\frac{1}{x} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60}$$

or x = 10 hours

Money will be distributed in the ratio of work done in an hour.
i.e. $$\frac{1}{6} : \frac{1}{8} : \frac{1}{15}$$
or 20:15:8
Hence part of A will be = $$\frac{20}{43} \times 94.6 = 44$$
part of B will be = $$\frac{15}{43} \times 94.6 = 33$$
part of C will be = $$\frac{8}{43} \times 94.6 = 17.60$$