Time and Work Questions

From the given data,

12 children 16 days work,
One child’s one day work = 1/192.

8 adults 12 days work,
One adult’s one day’s work = 1/96.

Work done in 3 days = ((1/96) x 16 x 3) = 1/2

Remaining work = 1 – 1/2 = 1/2

(6 adults+ 4 children)’s 1 day’s work = 6/96 + 4/192 = 1/12

1/12 work is done by them in 1 day.

1/2 work is done by them in 12 x (1/2) = 6 days.

Given the ratio of efficiencies of P, Q & R are 3 : 8 : 5

Let the efficiencies of P, Q & R be 3x, 8x and 5x respectively

They can do work for 12 days.

=> Total work = 12 x 16x = 192x

Now, the required time taken by Q to complete the job alone = 162x8x = 24days.

Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total. 

Work done by P and Q, working together in one day = 1/6 + 1/18  = 4/18 = 2/9 

They would take 9/2 days = 4 (1/2) days to complete the work working together.

From the above formula i.e (m1 x t1/w1) = (m2 x t2/w2)

so, [(34 x 8 x 9)/(2/5)] = [(M x 6 x 9)/(3/5)]

so, M = 136 men

Number of men to be added to finish the work = 136-34 = 102 men.

One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)

One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)

Divide both sides by 2 in (I) and then multiply both sides by 5

Now we get, 15b + 20g = 1/4.

Therefore, 15 boys and 20 girls can do the same work in 4 days.

Combined efficiency of all the three boats = 60 passenger/trip 

Now, consider option(a)

15 trips and 150 passengers means efficiency  of B1 =  10 passenger/trip 

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

Therefore the efficiency of B2 and B3 = 250/5 = 50 passenger/trip  

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

Now, Total work = LCM(16, 8) = 48

A's one day work = + 48/16 = + 3

B's one day work = - 48/8 = -6

Given A worked for 5 days to build the wall => 5 days work = 5 x 3 = + 15

2days B joined with A in working = 2(3 - 6) = - 6

Remaining Work of building wall = 48 - (15 - 6) = 39

Now this remaining work will be done by A in = 39/3 = 13 days.

We know that Time is inversely proportional to Efficiency

Here given time ratio of P & P+Q as

P/P+Q = 150/100 = 3:2

Efficiency ratio of P & Q as

P:Q = 2:1 ....(1)

Given Efficiency ratio of Q & R as

Q:R = 3:1 ....(2)

From (1) & (2), we get

P:Q:R = 6:3:1

P alone can finish the work in

22.5(3+1)/6 = 15 days.