55889.Two outlet pipes A and B are connected to a full tank. Pipe A alone can empty the tank in 10 minutes and pipe B alone can empty the tank in 30 minutes. If both are opened together, how much time will it take to empty the tank completely?
7 minutes
7 minutes 30 seconds
6 minutes
6 minutes 3 seconds
Explanation:
Let the capacity of the tank be LCM(10, 30) = 30 units.
=> Efficiency of pipe A = 30 / 10 = 3 units / minute
=> Efficiency of pipe A = 30 / 30 = 1 units / minute
=> Combined efficiency of pipe A and pipe B = 4 units / minute
Therefore, time required to empty the tank if both pipes work = 30 / 4 = 7 minutes 30 seconds
Let the capacity of the tank be LCM(10, 30) = 30 units.
=> Efficiency of pipe A = 30 / 10 = 3 units / minute
=> Efficiency of pipe A = 30 / 30 = 1 units / minute
=> Combined efficiency of pipe A and pipe B = 4 units / minute
Therefore, time required to empty the tank if both pipes work = 30 / 4 = 7 minutes 30 seconds
55890.Three pipes A, B and C were opened to fill a cistern. Working alone, A, B and C require 12, 15 and 20 minutes respectively.After 4 minutes of working together, A got blocked and after another 1 minute, B also got blocked. C continued to work till the end and the cistern got completely filled. What is the total time taken to fill the cistern ?
6 minutes
6 minutes 15 seconds
6 minutes 40 seconds
6 minutes 50 seconds
Explanation:
Let the capacity of the cistern be LCM(12, 15, 20) = 60 units.
=> Efficiency of pipe A = 60 / 12 = 5 units / minute
=> Efficiency of pipe B = 60 / 15 = 4 units / minute
=> Efficiency of pipe C = 60 / 20 = 3 units / minute
=> Combined efficiency of pipe A, pipe B and pipe C = 12 units / minute
Now, the cistern is filled with the efficiency of 12 units / minute for 4 minutes.
=> Pool filled in 4 minutes = 48 units
=> Pool still empty = 60 – 48 = 12 units Now, A stops working.
=> Combined efficiency of pipe B and pipe C = 7 units / minute
Now, the cistern is filled with the efficiency of 7 units / minute for 1 minute.
=> Pool filled in 1 minute = 7 units
=> Pool still empty = 12 – 7 = 5 units
Now, B also stops working.
These remaining 5 units are filled by C alone.
=> Time required to fill these 5 units = 5 / 3 = 1 minute 40 seconds
Therefore, total time required to fill the pool = 4 minutes + 1 minutes + 1 minute 40 seconds = 6 minutes 40 seconds
Let the capacity of the cistern be LCM(12, 15, 20) = 60 units.
=> Efficiency of pipe A = 60 / 12 = 5 units / minute
=> Efficiency of pipe B = 60 / 15 = 4 units / minute
=> Efficiency of pipe C = 60 / 20 = 3 units / minute
=> Combined efficiency of pipe A, pipe B and pipe C = 12 units / minute
Now, the cistern is filled with the efficiency of 12 units / minute for 4 minutes.
=> Pool filled in 4 minutes = 48 units
=> Pool still empty = 60 – 48 = 12 units Now, A stops working.
=> Combined efficiency of pipe B and pipe C = 7 units / minute
Now, the cistern is filled with the efficiency of 7 units / minute for 1 minute.
=> Pool filled in 1 minute = 7 units
=> Pool still empty = 12 – 7 = 5 units
Now, B also stops working.
These remaining 5 units are filled by C alone.
=> Time required to fill these 5 units = 5 / 3 = 1 minute 40 seconds
Therefore, total time required to fill the pool = 4 minutes + 1 minutes + 1 minute 40 seconds = 6 minutes 40 seconds
55891.Working alone, two pipes A and B require 9 hours and 6.25 hours more respectively to fill a pool than if they were working together. Find the total time taken to fill the pool if both were working together.
6
6.5
7
7.5
Explanation:
Let the time taken if both were working together be n hours.
=> Time taken by A = n + 9
=> Time taken by B = n + 6.25
In such kind of problems, we apply the formula : n2 = a x b,
where a and b are the extra time taken if both work
individually than if both work together.
Therefore, n2 = 9 x 6.25 => n = 3 x 2.5 = 7.5
Thus, working together, pipes A and B require 7.5 hours.
Let the time taken if both were working together be n hours.
=> Time taken by A = n + 9
=> Time taken by B = n + 6.25
In such kind of problems, we apply the formula : n2 = a x b,
where a and b are the extra time taken if both work
individually than if both work together.
Therefore, n2 = 9 x 6.25 => n = 3 x 2.5 = 7.5
Thus, working together, pipes A and B require 7.5 hours.
55892.Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
5/11
7/11
6/11
8/11
Explanation:
LCM(30,20,10)=60
Suppose capacity of the tank is 60 litre. Then,
quantity filled by pipe A in 1 minute =60/30=2 litre.
quantity filled by pipe B in 1 minute =60/20=3 litre.
quantity filled by pipe C in 1 minute =60/10=6 litre.
Quantity filled by pipe C in 3 minutes
=3×6=18 litre.
Quantity filled by pipe A,B,C together in 3 minutes
=3(2+3+6)=33 litre.
Required proportion =18/33=6/11
LCM(30,20,10)=60
Suppose capacity of the tank is 60 litre. Then,
quantity filled by pipe A in 1 minute =60/30=2 litre.
quantity filled by pipe B in 1 minute =60/20=3 litre.
quantity filled by pipe C in 1 minute =60/10=6 litre.
Quantity filled by pipe C in 3 minutes
=3×6=18 litre.
Quantity filled by pipe A,B,C together in 3 minutes
=3(2+3+6)=33 litre.
Required proportion =18/33=6/11
55893.Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
4 hours
2 hours
6 hours
3 hours
Explanation:
Suppose pipe A alone can fill the cistern in x hours.
Then pipe B alone can fill the cistern in (x+6) hours.
Part filled by pipe A in 1 hr =1/x
Part filled by pipe B in 1 hr =1/x+6
Part filled by pipe A and pipe B in 1 hr =1/x+1/x+6
It is given that pipes A and B together can fill the cistern in 4 hours.
i.e., Part filled by pipes A and B in 1 hr =1/4
⇒1/x+1/x+6=1/4
From here, it is better to find the value of x from the choices which will be easier. Or we can solve it as follows.
4(x+6)+4x=x(x+6)
4x+24+4x=x^2+6x
x^2−2x−24=0
(x−6)(x+4)=0
x=6 or −4
Since x cannot be negative, x=6
i.e.,pipe A alone can fill the cistern in 6 hours
Suppose pipe A alone can fill the cistern in x hours.
Then pipe B alone can fill the cistern in (x+6) hours.
Part filled by pipe A in 1 hr =1/x
Part filled by pipe B in 1 hr =1/x+6
Part filled by pipe A and pipe B in 1 hr =1/x+1/x+6
It is given that pipes A and B together can fill the cistern in 4 hours.
i.e., Part filled by pipes A and B in 1 hr =1/4
⇒1/x+1/x+6=1/4
From here, it is better to find the value of x from the choices which will be easier. Or we can solve it as follows.
4(x+6)+4x=x(x+6)
4x+24+4x=x^2+6x
x^2−2x−24=0
(x−6)(x+4)=0
x=6 or −4
Since x cannot be negative, x=6
i.e.,pipe A alone can fill the cistern in 6 hours
55894.Two pipes A and B attached to a swimming pool can fill the pool in 20 minutes and 30 minutes respectively working alone. Both were opened together but due to malfunctioning of motor of pipe A, it had to be shut down after two minutes but B continued to work till the swimming pool was filled completely. Find the total time taken to fill the pool.
20
22
25
27
Explanation:
Let the capacity of the pool be LCM(20, 30) = 60 units.
=> Efficiency of pipe A = 60 / 20 = 3 units / minute
=> Efficiency of pipe B = 60 / 30 = 2 units / minute
=> Combined efficiency of pipe A and pipe B = 5 units / minute
Now, the pool is filled with the efficiency of 5 units / minute for two minutes.
=> Pool filled in two minutes = 10 units
=> Pool still empty = 60 - 10 = 50 units
This 50 units is filled by B alone.
=> Time required to fill these 50 units = 50 / 2 = 25 minutes
Therefore, total time required to fill the pool = 2 + 25 = 27 minutes
Let the capacity of the pool be LCM(20, 30) = 60 units.
=> Efficiency of pipe A = 60 / 20 = 3 units / minute
=> Efficiency of pipe B = 60 / 30 = 2 units / minute
=> Combined efficiency of pipe A and pipe B = 5 units / minute
Now, the pool is filled with the efficiency of 5 units / minute for two minutes.
=> Pool filled in two minutes = 10 units
=> Pool still empty = 60 - 10 = 50 units
This 50 units is filled by B alone.
=> Time required to fill these 50 units = 50 / 2 = 25 minutes
Therefore, total time required to fill the pool = 2 + 25 = 27 minutes
55895.Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
6 2/3hours
7 hours
6 hours
7 1/2hours
Explanation:
LCM(12,15,20)=60
Take capacity of the tank as 60 litre. Then,
Quantity filled by pipe A in 1 hour =60/12=5 litre.
Quantity filled by pipe B in 1 hour =60/15=4 litre.
Quantity filled by pipe C in 1 hour =60/20=3 litre.
In first hour, A and B are open and fill 5+4=9 litre. In second hour, A and C are open and fill 5+3=8 litre.
Therefore, in 2 hours, 9+8=17 litre is filled.
This pattern goes on.
In 2×3=6 hours, 17×3=51 litre is filled.
In 7th hour, A and B are open and fill 9 more litre. Thus, tank is totally filled.
(∵51+9=60)
Therefore, required time =7 hour.
LCM(12,15,20)=60
Take capacity of the tank as 60 litre. Then,
Quantity filled by pipe A in 1 hour =60/12=5 litre.
Quantity filled by pipe B in 1 hour =60/15=4 litre.
Quantity filled by pipe C in 1 hour =60/20=3 litre.
In first hour, A and B are open and fill 5+4=9 litre. In second hour, A and C are open and fill 5+3=8 litre.
Therefore, in 2 hours, 9+8=17 litre is filled.
This pattern goes on.
In 2×3=6 hours, 17×3=51 litre is filled.
In 7th hour, A and B are open and fill 9 more litre. Thus, tank is totally filled.
(∵51+9=60)
Therefore, required time =7 hour.
55896.Three pipes A, B and C were opened to fill a tank. Working alone, A, B and C require 10, 15 and 20 hours respectively. A was opened at 7 AM, B at 8 AM and C at 9 AM. At what time the tank would be completely filled, given that pipe C can only work for 3 hours at a stretch, and needs 1 hour standing time to work again.
12 : 00 PM
12 : 30 PM
1 : 30 PM
1 : 00 PM
Explanation:
Let the capacity of the tank be LCM (10, 15, 20) = 60
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 15 = 4 units / hour
=> Efficiency of pipe C = 60 / 20 = 3 units / hour
=> Combined efficiency of all three pipes = 13 units / hour
Till 9 AM, A works for 2 hours and B work for 1 hour.
=> Tank filled in 2 hours by A = 12 units
=> Tank filled in 1 hour by B = 4 units
=> Tank filled till 9 AM = 16 units
=> Tank still empty = 60 - 16 = 44 units
Now, all three pipes work for 3 hours with the efficiency of 13 units / hour.
=> Tank filled in 3 more hours = 39 units
=> Tank filled till 12 PM = 16 + 39 units = 55 units
=> Tank empty = 60 - 55 = 5 units
Now, C is closed for 1 hour and these remaining 5 units would be filled by A and B working together with the efficiency 10 units / hour.
=> Time taken to fill these remaining 5 units = 5 / 10 = 0.5 hours
Therefore, time at which the tank will be completely filled = 12 PM + 0.5 hours = 12 : 30 PM
Let the capacity of the tank be LCM (10, 15, 20) = 60
=> Efficiency of pipe A = 60 / 10 = 6 units / hour
=> Efficiency of pipe B = 60 / 15 = 4 units / hour
=> Efficiency of pipe C = 60 / 20 = 3 units / hour
=> Combined efficiency of all three pipes = 13 units / hour
Till 9 AM, A works for 2 hours and B work for 1 hour.
=> Tank filled in 2 hours by A = 12 units
=> Tank filled in 1 hour by B = 4 units
=> Tank filled till 9 AM = 16 units
=> Tank still empty = 60 - 16 = 44 units
Now, all three pipes work for 3 hours with the efficiency of 13 units / hour.
=> Tank filled in 3 more hours = 39 units
=> Tank filled till 12 PM = 16 + 39 units = 55 units
=> Tank empty = 60 - 55 = 5 units
Now, C is closed for 1 hour and these remaining 5 units would be filled by A and B working together with the efficiency 10 units / hour.
=> Time taken to fill these remaining 5 units = 5 / 10 = 0.5 hours
Therefore, time at which the tank will be completely filled = 12 PM + 0.5 hours = 12 : 30 PM
55897.A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Time required by the first pipe to fill the tank is
6 hours
15 hours
10 hours
30 hours
Explanation:
Suppose the first pipe alone can fill the tank in x hours. Then,
second pipe alone can fill the tank in (x−5) hours,
third pipe alone can fill the tank in (x−5)−4=(x−9) hours.
Part filled by first pipe and second pipe together in 1 hr
= Part filled by third pipe in 1 hr
⇒1/x+1/x−5=1/x−9
From here, better to find the value of x from the given choices which will be easier. Or we can solve it as follows.
(x−5)(x−9)+x(x−9) = x(x−5)
x^2−14x+45+x^2−9x=x^2−5x−14x+45+x^2−9x=−5x
x2−18x+45=0
(x−15)(x−3)=0
x=15 or 3
We can not take the value x=3 because, (x−9) becomes negative which is not possible, because the third pipe can fill the tank in (x−9) hours.
Hence, x=15
Suppose the first pipe alone can fill the tank in x hours. Then,
second pipe alone can fill the tank in (x−5) hours,
third pipe alone can fill the tank in (x−5)−4=(x−9) hours.
Part filled by first pipe and second pipe together in 1 hr
= Part filled by third pipe in 1 hr
⇒1/x+1/x−5=1/x−9
From here, better to find the value of x from the given choices which will be easier. Or we can solve it as follows.
(x−5)(x−9)+x(x−9) = x(x−5)
x^2−14x+45+x^2−9x=x^2−5x−14x+45+x^2−9x=−5x
x2−18x+45=0
(x−15)(x−3)=0
x=15 or 3
We can not take the value x=3 because, (x−9) becomes negative which is not possible, because the third pipe can fill the tank in (x−9) hours.
Hence, x=15
55898.A tap can fill a tank in 4 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
1 hr 30 min
2 hr 30 min
2 hr
3 hr
Explanation:
A tap can fill a tank in 4 hours.
Therefore the tap can fill half the tank in 2 hours.
Remaining part =1/2
After half the tank is filled, three more similar taps are opened.
Hence, total number of taps becomes 4.
Part filled by one tap in 1 hour =1/4
Part filled by four taps in 1 hour =4×1/4=1
i.e., 4 taps can fill remaining half in 30 minutes.
Total time taken
= 2 hour + 30 minute = 2 hour 30 minutes
A tap can fill a tank in 4 hours.
Therefore the tap can fill half the tank in 2 hours.
Remaining part =1/2
After half the tank is filled, three more similar taps are opened.
Hence, total number of taps becomes 4.
Part filled by one tap in 1 hour =1/4
Part filled by four taps in 1 hour =4×1/4=1
i.e., 4 taps can fill remaining half in 30 minutes.
Total time taken
= 2 hour + 30 minute = 2 hour 30 minutes