Time and Work problems

C alone can finish the work in 40 days.
As given C does half as much work as A and B together
=> (A + B) can do it in 20 days
(A + B)s 1 days wok = 1/20.
A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)
A's 1 day?s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]
B's 1 days work = (1/20) x (2/3) = 1/30
(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40

All the three together will finish it in 40/3 = 13 and 1/3 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.

(A+B+C)'s 1 day's work = (1/24 + 1/6 + 1/12) = 7/24

so, A,B and C together will complete the work in 24/7 days.

Let efficiency of every man and every woman be 'm' unit/day and 'w' unit/day respectively

15×12×m = 10×16×w

? m/w = 8/9

Total work = 15 × 12 × 9 = 1620 units

In 2 days, total work done = 15 x 9 + 16 x 8 = 263 units

So, in 10 days work done will be = 263 × 5 = 1315 units

Remaining work will be done in = (1620-1315)/(15×8) = 5/2 days

Total days = 10 5/2 days.

A + B= 70%   

B + C =50%   

  ? ( A + B ) + ( B + C ) - ( A + B + C ) = B

=> B= 20%     A= 50%        and   C=30%

Hence A is most efficient

M x T / W = Constant
where, M= Men (no. of men)
T= Time taken
W= Work load
So, here we apply
M1 x T1/ W1 = M2 x T2 / W2
Given that, M1 = 4 men, T1 = 7 hours ; T2 = 2 hours, we have to find M2 =?
Note that here, W1 = W2 = 1 road, ie. equal work load.
Clearly, substituting in the above equation we get, M2 = 14 men.

Let A's 1 day's work=x and B's 1 day's work=y

Then x+y = 1/40 and 20x+60y=1 

Solving these two equations , we get : x= 1/80 and y= 1/80

Therefore B's  1 day work = 1/80

Hence,B alone shall finish the whole work in 80 days

From the given data,

=> (2 M + 3W) 8 = (3M + 2W)7

=> 16M + 24W = 21M + 14 W

=> 10W = 5M

=> 2W = M

=> 14W × ? = 7W × 8

? = 4 days

Given that

(10M + 15W) x 6 days = 1M x 100 days

=> 60M + 90W = 100M

=> 40M = 90W

=> 4M = 9W.

From the given data,

1M can do the work in 100 days

=> 4M can do the same work in 100/4= 25 days.

=> 9W can do the same work in 25 days.

=> 1W can do the same work in 25 x 9 = 225 days.

Hence, 1 woman can do the same work in 225 days.

A can do the work = 18 days
B can do the work = 18/2 = 9 days
(A + B)'s 1 day work = 1/18 + 1/9 = 1/6
=> In 3 days = 3x1/6 = 1/2 work is completed.