CMAT Time and Work Questions

Let slower pipe fill the tank at the rate x units/ minute.

Faster pipe will fill tank at the rate of 3x units/ minute

Number of units filled by both the pipes in a minute = 4x

Total time taken to fill tank = 36 minutes

capacity of tank = 36$$\times\ $$4x = 144x

Time taken by slower pipe to fill the tank = $$\frac{144x}{x}$$ = 144 minutes

Answer is option D.

Anuj takes 1 day to complete 
Bharat takes 2 days 
Chetan takes 4 days and Dhiraj takes 8 days to complete
option A chetan and Dhiraj together will take $$\frac{4\times\ 8}{4+8}=\frac{32}{12}=\frac{8}{3\ }days$$so A is correct.
Now calculating for all we get AB in 2/3 days AC in 4/5 days AD in 8/9 days BC in 8/6 days BD in 16/10 days
So we can say BD is 2nd slowest pair
Hence A and C are correct

Let the volume be 20 units 
So A fills at 4 units/hr 
B fills at 1 unit/hr
Now together they fill 4+1 = 5 units/ hour
Now time taken to fill completely : 20/5 = 4 hrs

Let L does a units of work/day 
M does b units of work/ day 
N does c units of work/ day 
Now as per given conditions :
Let work be W
So we get
W = 72(a+b) =120(b+c) =90(a+c)
we get
6a+6b = 10b +10c
we get 6a-4b-10c=0   (1)
also
4a+4b =5a+5c
we get a-4b+5c =0   (2)
From (1) and (2)
we get a=3 , b=2 and c=1
Now therefore W = 360
Now time taken by L alone to complete the work = 360/3 = 120 days

Let the efficiency of a man, woman and child be $$3x,2x,x$$ espectively.

Thus, 1 day wage = $$(3x\times20)+(2x\times30)+(x\times36)=78$$

=> $$156x=78$$

=> $$x=\frac{1}{2}$$

$$\therefore$$ Wages of 15 men, 21 women and 30 children for 18 weeks = $$[(15\times3x)+(21\times2x)+(30\times x)]\times18\times7$$

= $$117\times\frac{1}{2}\times126$$

= $$117\times63=Rs.$$ $$7,371$$

=> Ans - (A)

Water capacity for 50 days is

33000 x 50= 1650000.-----------(1)

water capacity for 35 days is

37000 x 35= 1295000------------(2)

Let the minimum water to be used is K litres

so, K x15=355000

K=$$\ \frac{355000\ }{15}$$

K= 23666.66litres.

Therefore, estimation of how much minimum water should be used daily so that it could last 50 days is 23666.66liters.

Given, Rajesh can check the quality of 1000 items in 5 hours

$$\Rightarrow$$ Efficiency of Rajesh = 200 items/hr.

Also given, Rakesh can complete the 75% of same job in 3 hours

$$\Rightarrow$$ time taken by Rakesh to complete 100% of the job = $$\frac{100}{75}\times 3$$ = 4 hours.

$$\Rightarrow$$ Efficiency of Rakesh = 250 items/hr.

Given in the process of checking 1300 items, Rakesh spends only 2 hours. Let say time spent by Rajesh be 't' hours, which is also the total time taken.

Total Work = Efficiency $$\times$$ time taken 

$$\Rightarrow 1300 = 250\times 2 + 200\times t$$

$$\Rightarrow 200t = 800$$

$$\Rightarrow t = 4$$

Hence the time taken in the process of checking 1300 items is 4 hours.

Let the capacity of the cistern be 600 units.

From the given data, the efficiencies of pipes A and B are 5 units/ min and 4 units/min respectively.

Let the efficiency of outlet pipe C be 'k' units/min.

Given the time taken to fill the cistern when all the three pipes are open = 100 minutes

$$\Rightarrow Efficiency  of  pipes\times time  taken = Capacity  of  cistern$$.

$$\Rightarrow (5+4-k)\times 100 = 600$$

$$\Rightarrow 9-k = 6$$

$$\Rightarrow k = 3$$

Therefore the time taken(t)  by pipe C to empty the cistern = $$Capacity  of  the  cistern\div efficiency  of  pipe  C$$

$$\Rightarrow t = 600\div 3 = 200 minutes = 3 hrs 20 min$$.

As per the given conditions, weeks 1 and 3 will have similar pay structure and weeks 2 and 4 will have similar pay structure.

For weeks 1 and 3

wage/hr for working = 60

wage/hr during rest = 30

Total payment per day = (60*6 )+ (30*3) = 450

Total payment for weeks 1 and 3 = 450 * 12 = 5400

For weeks 2 and 4

wage/hr for working = 20

wage/hr during rest = 10

Total payment per day = (20*3 )+ (10*6) = 120

Total payment for weeks 2 and 4 = 120 * 12 = 1440

$$\therefore $$ Total monthly salary = 5400 + 1440 = 6840

Hence, option A.

Let amount eaten by each horse and each bullock in one day be 'h' units and 'b' units respectively.

So Total fodder = Total effciency$$\times$$Total number of days 

$$\Rightarrow$$ Total work =  26$$\times$$h$$\times$$170 = 20$$\times$$b$$\times$$170

$$\Rightarrow$$ b = 1.3h............(1)

The amount of fodder eaten by 10 horses and 8 bullocks in one day = 10h + 8b = 10h + 8(1.3h) = 20.4h

Time taken by them to eat the same amount of fodder = total fodder/amount eaten by them in one day

= $$\frac{ 26\times h\times 170}{20.4h}$$

= 216.67 days

Given Anil is twice as good a student as Bharat

$$\Rightarrow$$ Efficiency  of  Anil : Efficiency  of  Bharat = 2:1

and also, Efficiency is inversely proportional to Time taken,

$$\Rightarrow$$ Time  taken  by  Anil : Time  taken  by  Bharat = 1:2...............................(1)

Given that Time taken by Anil is 30 min less than Bharat's time.

Let say, Time taken by Bharat be 't' minutes.

Then the time taken by Anil = t-30 minutes

Substituting these in equation (1), we get

$$\frac{t-30}{t} = \frac{1}{2}$$

$$\Rightarrow$$ t = 60  minutes.

Therefore the time taken by Anil and Bharat are 30 minutes and 60 minutes respectively.

Let Efficiency of Bharat be 'x', then Efficiency of Anil will be '2x'

$$\Rightarrow$$ Total Work = Efficiency\times Time taken = $$(2x)\times 30 (or) (x)\times 60 = 60x$$  units.

Efficiency when Anil and Bharat are working together = x+2x = 3x

Total Work = 60x units

Time taken(T) by Anil and Bharat together to complete the work = Total work/ Efficiency when Anil and Bharat work together

$$\Rightarrow T = \frac{60x}{3x}$$

$$\Rightarrow$$ T = 20  minutes.

Let the efficiencies of Rakesh, Prakash, and Ashok be 'r' 'p' and 'a' respectively.

GIven that Rakesh can do a job an hour quicker than Prakash.

So let time taken by Prakash be 't' hours, then time taken by Rakesh will be 't-1' hours.

Total work(W) = Efficiency$$\times$$ Time taken = p$$\times$$ t = r$$\times$$ (t-1)

$$\Rightarrow$$ t = $$\frac{r}{r-p}$$....................(1)

Given that, Working together, Rakesh, Prakash and Ashok can finish the same job in an hour.

$$\Rightarrow$$ Total work(W) = (r+p+a) (1) units.............(2)

Also given that,  if Prakash works for an hour, and then Ashok works for four hours, the job will be completed.

$$\Rightarrow$$ Total work(W) = p(1) + a(4) units...........(3)

Equating (2) and (3), we get

(r+p+a) (1) = p(1) + a(4)

$$\Rightarrow$$ r = 3a..............(4)

Substituting this value in equation (1), we get

t = $$\frac{3a}{3a-p}$$.............(5)

As the Total work is always constant, p$$\times$$ t = p(1) + a(4)

$$\Rightarrow$$ t = 1 + 4$$\frac{a}{p}$$.......(6)

Equating (5) and (6), we get

 $$\frac{3a}{3a-p} = 1 + 4\frac{a}{p}$$

Let $$\frac{a}{p}$$ = 'k'

$$\Rightarrow \frac{3k}{3k-1} = 1 + 4k$$

$$\Rightarrow 3k = 12k^2 + 3k - 4k -1$$

$$\Rightarrow 12k^2 - 4k - 1 = 0$$

Solving for k, we get k = $$\frac{1}{2} or -\frac{1}{6}$$[which is not possible]

Hence k = $$\frac{1}{2}$$

$$\Rightarrow$$ p =2a.............(7) 

Substituting (4) and (7) in equation (2) we get,

Total work(W) = 6a units.

Time taken by Ashok alone to do the job = Total work/ Efficiency of Ashok

= 6a/a

=6 hours.