I might be silly but you have already answered your own question on the first post.
D=3X or a ratio 1 to 3. I think you have labled the hight and diameter of the cone up the wrong way against the cylinder.

IGCSE, looking for the original question now that I apparently have two maths buffs at my disposal

Hi 24digital
The idea of the question is to show you know how to calculate volume and cancel out like terms to arrive at a common ratio it would be toooo easy to label diameter D and height H etc

Thought you might be, we did AQA gcse for maths everything else has to be IGCSE

as your on to quads try this site for making worksheets

http://www.math.com/students/worksheet/algebra_sp.htm

Thanks hen, scanner went on the blink so I had to take a photo of it



As he is an external candidate, they have to be IGCSE with no course work. Bit of a nightmare as we spent two years on the CIE spec for Religious Studies and then discovered three weeks ago that they do not offer a June exam!! We have had to switch to Edexcel at the last minute and the spec is very different. Last year he did the three sciences and history, this year is the last lot but includes Maths and Latin - ouch!

My son is external as well we did AQA 4306 linear which is now 4365 linear and comes in foundation and higher tier, further maths comes as a linear exam also

We are doing higher tier but he drew the line at further and pure maths - thanks goodness.

I thought I had seen this type of question before taken from 2010 GCSE

I'll give him that one to try tomorrow
A palinka on me for anyone who solves the original question

Son began,
1/3(pidy) = piyx
pidy = 3piyx
dy = 3yx
d = 3x
x = d/3

how do we get to a ratio of 1:3 ??

third line down d is equal to y so cancel out which leaves y = 3x


I also have the answer for tomorrows question

So long ago that I can't be certain but I think I said that and son said that we would be assuming that d = y and I couldn't see how to prove that the diameters were equal

for a ratio there must be one variable the question wants y and x which must be between objects or there would be no point to the question

so if we say diameter is 10 and height of cone 20 we get 209.33 volume

then divide 209.33 by pi x 10 we get 6.66

so Y =20 and 3X = 20

From dy=3xy, the y variables must be equal and therefore cancel each other but I remembered that the original equation, as Fib pointed out, had been written wrongly and I put d instead of r²

Looking at it now

pi y (d/2)² = x pi (y/2)²
3

= y (d/2)² = x (y/2)²
3

= y (d/2)² = 3x(y/2)²

= square root then multiply by two? yd = 3xy

This shoud then be simplified to y=3x but it doesn't as you must clear the y variables leaving d=3x. Even if you were to accept the y=3x that would be an x:y ratio of 3:1 and not 1:3

Now I know why I gave up on this

appologies confusing my piD with pir2

it still works the same

if you substitute 5 for radius and 20 for height of cone it will come out at height of cylinder at 6.66, so Y height of cone is equal to 3 times X height of cylinder

why the question is confusing is they have used Y two mean two different planes if they had called both diameters D, cone height Y and cylinder height X

but the ration of x to y is 3:1 not 1:3

did you do the question from the exam above?

Not had time yet - bit of a mad weekend....

I think the original question had labeled d and y wrongly and asked for the ratio of x:y but gave the answer of y:x which is a pain as you end up spending longer on the question than is necessary. It has been a problem throughout the book and we will be glad to get the exam over and done with. Did you homeschool your son?