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Hard maths questions for higher maths GCSE

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Original post by notnek
Forget Hannah's sweets, here is the worst answered question in the last 10 years for all the KS4 maths exams:



This is a treat for all the GCSE students revising hard :smile:


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Original post by Kvothe the arcane

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That's what I got also and would be accepted by the mark scheme.

It's a tricky one - I can understand why so few got all the marks.
Original post by Kvothe the arcane

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God that question is complete and utter bs waste of time.
Original post by HapaxOromenon
Correct. It was from the 2015 Senior Mathematical Challenge.


Are you allowed any way of answering (i.e. f(5*100))?
Original post by Zacken
This 6 mark question came up in my GCSE exam and I quite enjoyed it.

Integers aa and bb are such that (a+35)2+a−b5=51(a + 3\sqrt{5})^2 + a - b\sqrt{5} = 51. Find the possible values of aa and the corresponding values of bb.


hi i've given it a go and for a i got
a=√36-6a√5-b√5
2
i know its probably wrong but any hints would be appreciated
also which exam board is this from?
thanks
Original post by OHH_MY_DAYZ:b
hi i've given it a go and for a i got
a=√36-6a√5-b√5
2
i know its probably wrong but any hints would be appreciated
also which exam board is this from?
thanks


Nopes, expand them out and compare the rational and irrational parts.

CIE IGCSE.
Original post by Zacken
Nopes, expand them out and compare the rational and irrational parts.

CIE IGCSE.


okay so i expanded the brackets and got
a2 +3a√5 +3a√5 +a -b√5=51
and then the 3a√5 +3a√5 become 6a√5
and then i'm lost....
Original post by OHH_MY_DAYZ:b
okay so i expanded the brackets and got
a2 +3a√5 +3a√5 +a -b√5=51
and then the 3a√5 +3a√5 become 6a√5
and then i'm lost....


So you get a^2 + a = 51 and 6a - b = 0
Without:

a) using a calculator or
b) using long division or
c) multiplying the factors provided

show that:

9,999,999,999,999,999=100000001×10001×101×11×99,999,999,999,999,999 = 100000001 \times 10001 \times 101 \times 11 \times 9

Hint:

Spoiler

Original post by atsruser
Without:

a) using a calculator or
b) using long division or
c) multiplying the factors provided

show that:

9,999,999,999,999,999=100000001×10001×101×11×99,999,999,999,999,999 = 100000001 \times 10001 \times 101 \times 11 \times 9

Hint:

Spoiler



Spoiler

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