Tank & Pipes Question
A tank is supplied by two pipes A and B and emptied by a third pipe C. If the tank is empty and all pipes are opened, the tank can be filled in 20 hours. If the tank is full and A and C are opened, the tank can be emptied in 4 hours. If the tank is full and B and C are opened, the tank can be emptied in 2 hours. Pipe A supplies 50 liters per minute more than B.
*Find the rate of pipe A in liters per minute.
*Find the rate of pipe C in liters per minute.
*find the capacity of the tank in liters.
*Find the rate of pipe A in liters per minute.
*Find the rate of pipe C in liters per minute.
*find the capacity of the tank in liters.
Replies
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ISHAN TOPRERe: ?????
The volume of tank=[C-(a+b)]*20*60=[(b+50+b)-c]*20*60=[2b-c+50]20*60---------1
If A and C are opened volume of tank= (c-a)*4*60--------2
If B and C are opened volume of tank=(c-b)*2*60------------3
Equating 2 and 3 with 1 we get two equations as
21b-11c+500=0 and 11b-6c+300=0
a=110lit/min, b=60 lit/min c=100 lit/min
vol of tank= 1200liters.
Am I right? -
maria florRe: ?????
I think your correct....but something wrong on your answer in pipe B.
Find the rate of pipe A in liters per minute.
a. 120 b. 130 c.110 d. 140
Find the rate of pipe C in liters per minute.
a. 170 b. 160 c. 150 d. 140
find the capacity of the tank in liters.
a. 1200 b. 12500 c. 11500 d. 13000 -
ISHAN TOPREYup got it Maria. ๐
a=110lit/min, c=160lit/min volume of tank= 1200 liters. It was just a calculation mistake. ๐ -
maria florFind the rate of pipe A in liters per minute. Answer: 110
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maria florHere's the next question:
A cylindrical tank of radius 1.5m. and height of 5m. is filled with water to depth of of 2m.
*If one spherical ball having a diameter of 2m. is dropped into tank slowly, by how much is the level of water raised?
*If two spherical balls having diameter of 2m. each is dropped into the tank slowly, by how much is the level of the water raised. -
ISHAN TOPREYup Maria.
Volume of water initially in tank= Pi*1.5*1.5*2=14.137m3
Volume of ball=4*Pi*8/3=33.510m3
When the ball is dropped in tank Vol=vol of tank initially+vol of ball.
Dividing this total volume by Pi*1.5*1.5 we will get the vertical height.
Vertical height-2 will give total rise in water level.
The rise in tank level will be 4.7407m and 9.481m respectively.
Am I right? -
maria flor
It's wrong Ishu...ishutopreYup Maria.
Volume of water initially in tank= Pi*1.5*1.5*2=14.137m3
Volume of ball=4*Pi*8/3=33.510m3
When the ball is dropped in tank Vol=vol of tank initially+vol of ball.
Dividing this total volume by Pi*1.5*1.5 we will get the vertical height.
Vertical height-2 will give total rise in water level.
The rise in tank level will be 4.7407m and 9.481m respectively.
Am I right? -
maria florHere's another question but not related to pipe or pipe. This is just easy.
Two circles are tangent to a third circle internally and are tangent to each other externally. The distance between their centers are 10 inches, 7 inches and 5 inches. Find their radii. -
ISHAN TOPREOh sorry I calculated the answers yesterday but forgot to post them. ๐
If you will draw the diagram you will get the three equations.
x+y=10
y+z=5
x+z=7
Solving them we get the radii of those circle as x=6, y=4, z=1.
Am I right? -
maria florHere's next:
A frog is trapped in a 24m well. Every minute it jumps 3m and slips back 2m. After how many jumps will it reach the top of the well???? -
ISHAN TOPREThe frog jumps 1m on each jump. So he will jump 22 times. Because after the 21st jump, the frog will get out on 22nd jump.
How's that? ๐ -
maria florOK. Pls. show your solution
Next:
The perimeter of a triangle with integral sides is 24 units. If it is a right triangle, what must be its are? Show your solution.๐ -
ISHAN TOPREAs you say, if the triangle is an isosceles right angled triangle.
Now hypotenuse=(underroot2)*length of side.
hence let length of side be x.
x+x+(root 2)*x=24
x=7.0294
Now area of triangle=0.5*base*height.
Area=0.5*7.0294*7.0294
=24.706 square unit -
maria florNext question:
A mixture compound from equal parts of two liquids, water and oil was placed in a hemispherical bowl. The total depth of the two liquids is 6cm. After standing for a short time in the mixture separated. Thethickness of the segment of oil is 2cm.Find the base radius of the segment containing water. -
maria flor
Just to remind you, your answer is wrong.ishutopreYup Maria.
Volume of water initially in tank= Pi*1.5*1.5*2=14.137m3
Volume of ball=4*Pi*8/3=33.510m3
When the ball is dropped in tank Vol=vol of tank initially+vol of ball.
Dividing this total volume by Pi*1.5*1.5 we will get the vertical height.
Vertical height-2 will give total rise in water level.
The rise in tank level will be 4.7407m and 9.481m respectively.
Am I right? -
ISHAN TOPREOK Maria. Assuming the hemispherical container is filled full. The original depth is full. So the radius of hemispherical container is 8 cm. Agreed?
Now the upper 2 cm is the oil layer. So if you draw the diagram and chop off above 2 cm from hemisphere, then you get the radius of immediate base radius of segment.as 7.746 cm. ๐
Also the diameter is 15.492 cm. ๐
Am I right? -
maria flor
your answer is wrong..ishutopreOK Maria. Assuming the hemispherical container is filled full. The original depth is full. So the radius of hemispherical container is 8 cm. Agreed?
Now the upper 2 cm is the oil layer. So if you draw the diagram and chop off above 2 cm from hemisphere, then you get the radius of immediate base radius of segment.as 7.746 cm. ๐
Also the diameter is 15.492 cm. ๐
Am I right?
You are reading an archived discussion.
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