Question: 16 -
A is twice as good a workman as B and is therefore able to finish a piece of work in 30 days less than B. In how many days they can complee the whole work; working together?
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12
-
10
-
22
-
20
Answer:
20
Solution:
Ratio of times taken by A and B = 1 : 2.
The time difference is (2 - 1) 1 day while B take 2 days and A takes 1 day.
If difference of time is 1 day, B takes 2 days.
If difference of time is 30 days, B takes 2 x 30 = 60 days.
So, A takes 30 days to do the work.
A's 1 day's work = 1/30
B's 1 day's work = 1/60
(A + B)'s 1 day's work = 1/30 + 1/60 = 1/20
A and B together can do the work in 20 days
Ratio of times taken by A and B = 1 : 2.
The time difference is (2 - 1) 1 day while B take 2 days and A takes 1 day.
If difference of time is 1 day, B takes 2 days.
If difference of time is 30 days, B takes 2 x 30 = 60 days.
So, A takes 30 days to do the work.
A's 1 day's work = 1/30
B's 1 day's work = 1/60
(A + B)'s 1 day's work = 1/30 + 1/60 = 1/20
A and B together can do the work in 20 days
Question: 17 -
There are three boats B1, B2 and B3 working together they carry 60 people in each trip. One day an early morning B1 carried 50 people in few trips alone. When it stopped carrying the passengers B2 and B3 started carrying the people together. It took a total of 10 trips to carry 300 people by B1, B2 and B3. It is known that each day on an average 300 people cross the river using only one of the 3 boats B1, B2 and B3. How many trips it would take to B1, to carry 150 passengers alone?
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10
-
15
-
25
-
20
Answer:
15
Solution:
Combined efficiency of all the three boats = 60 passenger/trip
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 = 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.
Combined efficiency of all the three boats = 60 passenger/trip
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 = 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.
Question: 18 -
An air conditioner can cool the hall in 40 miutes while another takes 45 minutes to cool under similar conditions. If both air conditioners are switched on at same instance then how long will it take to cool the room approximately ?
-
19 minutes
-
22 minutes
-
18 minutes
-
24 minutes
Answer:
22 minutes
Solution:
Let the two conditioners be A and B
'A' cools at 40min
'B' at 45min
Together = (a x b)/(a + b)
= (45 x 40)/(45 + 40)
= 45 x 40/85
= 21.1764
= 22 min (approx).
Let the two conditioners be A and B
'A' cools at 40min
'B' at 45min
Together = (a x b)/(a + b)
= (45 x 40)/(45 + 40)
= 45 x 40/85
= 21.1764
= 22 min (approx).
Question: 19 -
A does half as much work as Band C does half as much work as A and B together. If C alone can finish the work in 40 days, then together ,all will finish the work in?
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17 + 4/7 days
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15 + 3/2 days
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13 + 1/3 days
-
16 days
Answer:
13 + 1/3 days
Solution:
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.
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.
Question: 20 -
A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days ?
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13 days
-
9 days
-
15 days
-
11 days
Answer:
13 days
Solution:
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.
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.