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Completely stuck on this Maths question!! NEED HELP!!

photo.jpgI've literally attempted this question a million times and still to no avail. Anyone who can help me solve it is a life saver!
Reply 1
Well, brute forcing it is surely possible.

1. Find y=f(x)
2. Find y=g(h)
3. f(x)=g(h) then get h by itself.
Original post by SpacVelociraptor
photo.jpgI've literally attempted this question a million times and still to no avail. Anyone who can help me solve it is a life saver!


Write out the volumes. It should be easy to see how you can eliminate y.
It's better to show your attempts and allow people to point out errors, rather than have someone come along and give you the answer. But I'm sure you'll get there :smile:
Reply 4
Original post by SpacVelociraptor
photo.jpgI've literally attempted this question a million times and still to no avail. Anyone who can help me solve it is a life saver!


HINT: the volume of a cone is 1/3 the volume of a cylinder with the same height and radius.
Reply 5
Write out the volumes of each shape (maybe called the spheres 1 and 2 as they are separate).
Eg. cylinder is pi x r^2 x h

Then plug in the relevant substitutions. Eg for the cylinder, the radius (r) is 2y. And so on.

You'll come to 4 volumes for each shape each in terms of y.

Put the correct volumes equal to one another. (Cylinder volume equal to sphere 1 etc).

Get y= for both.

Solve and you should get h=Ax, with A being a number people can confirm once you've given it a go. :smile:
Reply 6
I remember doing this question for my GCSE maths exam!
This is as far as I get before I get completely stuck! photo (1).jpg
Original post by SpacVelociraptor
This is as far as I get before I get completely stuck! photo (1).jpg


r=3yr2=(3y)23y2r= 3y \Rightarrow r^2 = (3y)^2 \neq 3y^2

same for 2y.

So you have

π(2y)2x=43π(2y)3\pi (2y)^2 x = \frac{4}{3}\pi(2y)^3

and

π(3y)2h=43π(3y)3\pi (3y)^2 h = \frac{4}{3}\pi(3y)^3
(edited 11 years ago)
Hint:

Divide the top equation through by

π(2y)2\pi(2y)^2

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