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IGCSE MATHS P4 DOUBTS

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Step one: consider a triangle as your polygon, so "n" will be 3, so the number of diagonals will, obviously, be 0
step two: now substitute in the equation, so you'll have 1/p x 3(3-q)=0
step 3: simplify it as follow 1/p x 9-3q=0
9-3q/p=0
step 4: cross multiply ( meaning, solve it), p x 0=9-3q
0=9-3q
3q=9
q=9/3=3
step 5: substitute the "q" in the equation, but this time, use 4 for "n", so number of diagonals will be 2
so, 1/p x 4(4-3)=2
1/p x 4(1)=2
1/p x 4 =2
4/p = 2
p=2



Then use the equation for the rest of the questions
 
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@Razo513

for c (i) first substitute the new values, so you'll have something that looks like this: 1/3 x π x (3x)^2 x 3h
first equation was: 1/3 x π x x^2 x h= W
if we want to find how much the new W is compared with the first one, we have to divide the second equation by the first

so, 1/3 x π x 9x^2 x 3h divide by 1/3 x π x x^2 x h= W

now all you have to do is cancel the corresponding term from both equations, at the end you'll find yourself with only 9 x 3 which gives you 27

PS: working for this question is not needed to be shown under the question

for c(ii), its pretty much the same thing, you substitute in the equation, then divide by the original one, and cancel the terms, again, this is a one mark question, they DO NOT need the working.

Hope I was of Help
 
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