Efficiency of Inlet pipe A = 4.16%
Efficiency of Inlet pipe B = 5.83%
Therefore, Efficiency of A and B together = 100 %
Now, if the efficiency of outlet pipe be x% then in 10 hours the capacity of tank which will be filled = 10 * (10 - x)
Now, since this amount of water is being emptied by 'C' at x% per hour, then
=> x = 8
Therefore, in 10 hours 20% tank is filled only. Hence, the remaining 80% of the capacity will be filled by pipes A and B in 80/10 = 8 hours
Ratio of times taken by A and B = 100 : 130 = 10 : 13.
Suppose B takes x days to do the work.
Then, 10 : 13 :: 23 : x => x = ( 23 x 13/10 ) => x = 299 /10.
A's 1 day's work = 1/23 ;
B's 1 day's work = 10/299 .
(A + B)'s 1 day's work = ( 1/23 + 10/299 ) = 23/299 = 113 .
Therefore, A and B together can complete the work in 13 days.
Work donee by A and B in the first two hours, working alternatively = First hour A + Second hour B = (1/4) + (1/12) = 1/3.
Thus, the total time required to complete the work = 2 (3) = 6 days
A : B : C
Ratio of efficiency 3 : 1 : 2
Ratio of No.of days 1/3 : 1/1 : 1/2
or 2 : 6 : 3
Hence A is correct.
A + B = C + D
| | | |
Ratio of efficiency 10x + 5x 9x + 6x
|________| |_________|
15x 15x
Therefore , ratio of efficiency of A:C =10:9
Therefore, ratio of days taken by A:C = 9:10
Therefore, number of days taken by A = 18 days
Let 1 woman's 1 day work = x.
Then, 1 man's 1 day work = x/2 and 1 child's 1 day work x/4.
So, (3x/2 + 4x + + 6x/4) = 1/7
28x/4 = 1/7 => x = 1/49
1 woman alone can complete the work in 49 days.
So, to complete the work in 7 days, number of women required = 49/7 = 7.
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%
=> B= 20% A= 50% and C=30%
Hence A is most efficient
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