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10 years ago

Does anyone know how to answer the following questions, I have attempted both several times but can't seem to get an answer.

Q1). Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers. - Taken from March 2013 Edexcel GCSE linear paper question 21.

Q2) There are three different types of sandwiches on a shelf.

There are :

4 egg sandwiches

5 cheese sandwiches

and 2 ham sandwiches

Erin takes at random 2 of these sandwiches

Work out the probability that she takes 2 different types of sandwiches. Question 24 of the same paper...

I really want an A* for maths so if anyone could list me a few more A-A* topics and help me answer these questions it would be highly appreciated, thanks

Q1). Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers. - Taken from March 2013 Edexcel GCSE linear paper question 21.

Q2) There are three different types of sandwiches on a shelf.

There are :

4 egg sandwiches

5 cheese sandwiches

and 2 ham sandwiches

Erin takes at random 2 of these sandwiches

Work out the probability that she takes 2 different types of sandwiches. Question 24 of the same paper...

I really want an A* for maths so if anyone could list me a few more A-A* topics and help me answer these questions it would be highly appreciated, thanks

Reply 1

10 years ago

Am not quite sure if its correct but I'm guessing 3 over 11 (3/11). The different combination of sandwiches are 1 Ham + 1 cheese. 1 Egg + 1 Ham. 1 Egg + 1 cheese. 3 different combination out of 11 options.

Revision topics

Changing the subject of a formula -

Equations with fractions -

Simultaneous equations -

inequality -

Parallel and perpendicular lines -

Cubic graphs and graphs of circles -

Graphs and proportion -

Factorising quadratic equations -

equations -

quadratic equations -

Posted from TSR Mobile

Revision topics

Changing the subject of a formula -

Equations with fractions -

Simultaneous equations -

inequality -

Parallel and perpendicular lines -

Cubic graphs and graphs of circles -

Graphs and proportion -

Factorising quadratic equations -

equations -

quadratic equations -

Posted from TSR Mobile

Original post by Chocolatecake123

Does anyone know how to answer the following questions, I have attempted both several times but can't seem to get an answer.

Q1). Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers. - Taken from March 2013 Edexcel GCSE linear paper question 21.

Q2) There are three different types of sandwiches on a shelf.

There are :

4 egg sandwiches

5 cheese sandwiches

and 2 ham sandwiches

Erin takes at random 2 of these sandwiches

Work out the probability that she takes 2 different types of sandwiches. Question 24 of the same paper...

I really want an A* for maths so if anyone could list me a few more A-A* topics and help me answer these questions it would be highly appreciated, thanks

Q1). Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers. - Taken from March 2013 Edexcel GCSE linear paper question 21.

Q2) There are three different types of sandwiches on a shelf.

There are :

4 egg sandwiches

5 cheese sandwiches

and 2 ham sandwiches

Erin takes at random 2 of these sandwiches

Work out the probability that she takes 2 different types of sandwiches. Question 24 of the same paper...

I really want an A* for maths so if anyone could list me a few more A-A* topics and help me answer these questions it would be highly appreciated, thanks

Q1

For these proof questions you need to represent the possible integers in algebra. In this case n and n+1 will do. Now square them and subtract one from the other and show it equals their sum.

If you had to have even numbers you couldn't just use n and would use 2n and 2n+2 (since 2n divides by 2 it must be even). For an odd number you would use 2n+1, 2n+3.

Q2

You should work through this as a probability tree problem 'without replacement'.

The initial probability of choosing an egg is 4/11, the probablity that she then chooses a second egg is 3/10 since one of them has already gone. The probability of Egg&Egg is therefore 4/11 x 3/10

work out the probabilities of both cheese and both ham. Use these results to calculate the probability both the sandwiches are the same and from there the probability that both are different.

(edited 9 years ago)

Reply 3

10 years ago

simple enough for ques 1, let the two integers be

x= n+1, y = n

now y^2 = n^2, x^2 = (n+1)^2

as we know,

x^2 - y^2 = (x+y)*(x-y)

this implies x^2 - y^2 = (n+1+n)(n+1-n)

which equals (x+y)

hence x^2+y^2 = x+y which is the required proof

x= n+1, y = n

now y^2 = n^2, x^2 = (n+1)^2

as we know,

x^2 - y^2 = (x+y)*(x-y)

this implies x^2 - y^2 = (n+1+n)(n+1-n)

which equals (x+y)

hence x^2+y^2 = x+y which is the required proof

(edited 10 years ago)

1) We take an integer to be n ( can be any letter)

Therefore the next integer is n+1

Now it tells you the difference between the squares = the sum of the integers.

Just sub in what you know:

To find the difference between it is always easier to take the smallest away from the biggest.

Therefore (n+1)^2 - n^2 = n + (n+1)

So (n^2 + 2n + 1) - n^2 = 2n + 1

2n+1= 2n+1

2n+1 - (2n+1) = 0

If you take equal numbers away from eachother it always equals 0, and 2n+1 - (2n+1) = 0 so that is sufficient proof right?

Therefore the next integer is n+1

Now it tells you the difference between the squares = the sum of the integers.

Just sub in what you know:

To find the difference between it is always easier to take the smallest away from the biggest.

Therefore (n+1)^2 - n^2 = n + (n+1)

So (n^2 + 2n + 1) - n^2 = 2n + 1

2n+1= 2n+1

2n+1 - (2n+1) = 0

If you take equal numbers away from eachother it always equals 0, and 2n+1 - (2n+1) = 0 so that is sufficient proof right?

2) 11 sandwiches

4 egg = 4/11

5 Cheese = 5/11

2 Ham = 2/11

Use the and rule, to get two sandwiches they will take one AND then another. Now remember, if you take one there will only be 10 left 1 won't magically appear. Therefore the 2nd fraction should be x/10. Unless the question explicitly states that they put it back or replace it remember this. Also its two DIFFERENT sandwhiches, therefore what that other guy said about Egg & another Egg is incorrect

Possible combinations:

egg 1st cheese 2nd, = 4/11 x 5/10 = 20/110

egg 1st ham 2nd, = 4/11 x 2/10 = 8/110

cheese 1st egg 2nd, = 5/11 x 4/10 = 20/110

cheese 1st ham 2nd, = 5/11 x 2/10 = 10/110

ham 1st egg 2nd, = 2/11 x 4/10 = 8/110

ham 1st cheese 2nd 2/11 x 5/10 = 10/110

Add these all together you get 76/110.

4 egg = 4/11

5 Cheese = 5/11

2 Ham = 2/11

Use the and rule, to get two sandwiches they will take one AND then another. Now remember, if you take one there will only be 10 left 1 won't magically appear. Therefore the 2nd fraction should be x/10. Unless the question explicitly states that they put it back or replace it remember this. Also its two DIFFERENT sandwhiches, therefore what that other guy said about Egg & another Egg is incorrect

Possible combinations:

egg 1st cheese 2nd, = 4/11 x 5/10 = 20/110

egg 1st ham 2nd, = 4/11 x 2/10 = 8/110

cheese 1st egg 2nd, = 5/11 x 4/10 = 20/110

cheese 1st ham 2nd, = 5/11 x 2/10 = 10/110

ham 1st egg 2nd, = 2/11 x 4/10 = 8/110

ham 1st cheese 2nd 2/11 x 5/10 = 10/110

Add these all together you get 76/110.

(edited 10 years ago)

Original post by sneakbo2

2) 11 sandwiches

4 egg = 4/11

5 Cheese = 5/11

2 Ham = 2/11

Use the and rule, to get two sandwiches they will take one AND then another. Now remember, if you take one there will only be 10 left 1 won't magically appear. Therefore the 2nd fraction should be x/10. Unless the question explicitly states that they put it back or replace it remember this. Also its two DIFFERENT sandwhiches, therefore what that other guy said about Egg & another Egg is incorrect

Possible combinations:

egg 1st cheese 2nd, = 4/11 x 5/10 = 20/110

egg 1st ham 2nd, = 4/11 x 2/10 = 8/110

cheese 1st egg 2nd, = 5/11 x 4/10 = 20/110

cheese 1st ham 2nd, = 5/11 x 2/10 = 10/110

ham 1st egg 2nd, = 2/11 x 4/10 = 8/110

ham 1st cheese 2nd 2/11 x 5/10 = 10/110

Add these all together you get 76/110.

4 egg = 4/11

5 Cheese = 5/11

2 Ham = 2/11

Use the and rule, to get two sandwiches they will take one AND then another. Now remember, if you take one there will only be 10 left 1 won't magically appear. Therefore the 2nd fraction should be x/10. Unless the question explicitly states that they put it back or replace it remember this. Also its two DIFFERENT sandwhiches, therefore what that other guy said about Egg & another Egg is incorrect

Possible combinations:

egg 1st cheese 2nd, = 4/11 x 5/10 = 20/110

egg 1st ham 2nd, = 4/11 x 2/10 = 8/110

cheese 1st egg 2nd, = 5/11 x 4/10 = 20/110

cheese 1st ham 2nd, = 5/11 x 2/10 = 10/110

ham 1st egg 2nd, = 2/11 x 4/10 = 8/110

ham 1st cheese 2nd 2/11 x 5/10 = 10/110

Add these all together you get 76/110.

What the other guy said about the Egg&Egg is not incorrect - it is a different way to reach the same result and is more efficient.

You have fewer calculations to do and therefore to make mistakes with if you work out Egg&Egg + Cheese&Cheese + Ham&Ham and than take that result away from 1

Another efficient alternative is to work out:

Egg 1st x NotEgg 2nd + Cheese x NotCheese + Ham x NotHam

All 3 methods are acceptable and earn full marks

(edited 10 years ago)

Reply 7

9 years ago

Its just simple probability! ;D You need to revise PROBABILTY TREE DIAGRAMS

Solution: (Pretend the lines are tree diagram (You will understand if you revise the topic)

. ..............<----------- 2/9 (EGG) ((4-2) and only 9 in total (11 become 9))

4/11 (EGG) < -------------- 5/9 (CHEESE) (Total doesn't change (Sum Total does))

. ..............<---------- 2/9 (HAM) (Same goes for ham)

...................<------------- 4/9 (EGG) (Total doesn't change (Sum Total does))

5/11 (CHEESE) < ---------------- 3/9 (CHEESE) (5 - 2 = 3 left over sandwhiches)

....................<------------ 2/9 (HAM) (Total doesn't change (Sum Total does))

................<----------------- 4/9 (EGG)

2/11 (HAM) < ------------- 5/9 (CHEESE) ....etc. (Self explanitary)

................<--------------- 0/9 (HAM)

So... Probability (Egg AND Cheese) = 4/11 x 5/9 = 20/99 >--- Revise Multiplying fractions AND & OR Probability (Topics)

Probability (Egg AND Ham) = 4/11 x 2/9 = 8/99 The Question asks for all the 2 different sandwiches

Probability (Cheese AND Ham) = 5/11 x 2/9 = 10/99 each time and these are all the solution.

Not finished yet... Then... Since it wants all the combinations you do: 20/99 + 8/99 + 10/99 = 38/99 (Topic) Adding Fractions

...... 38/99 is your answer!

Hope this helped Don't think to complicatively remeber your basics (Probability tree) btw i'm 14. Not all Questions are as hard as they seem break it down!

THIS WAS FOR Q2) ONLY BY THE WAY!

Solution: (Pretend the lines are tree diagram (You will understand if you revise the topic)

. ..............<----------- 2/9 (EGG) ((4-2) and only 9 in total (11 become 9))

4/11 (EGG) < -------------- 5/9 (CHEESE) (Total doesn't change (Sum Total does))

. ..............<---------- 2/9 (HAM) (Same goes for ham)

...................<------------- 4/9 (EGG) (Total doesn't change (Sum Total does))

5/11 (CHEESE) < ---------------- 3/9 (CHEESE) (5 - 2 = 3 left over sandwhiches)

....................<------------ 2/9 (HAM) (Total doesn't change (Sum Total does))

................<----------------- 4/9 (EGG)

2/11 (HAM) < ------------- 5/9 (CHEESE) ....etc. (Self explanitary)

................<--------------- 0/9 (HAM)

So... Probability (Egg AND Cheese) = 4/11 x 5/9 = 20/99 >--- Revise Multiplying fractions AND & OR Probability (Topics)

Probability (Egg AND Ham) = 4/11 x 2/9 = 8/99 The Question asks for all the 2 different sandwiches

Probability (Cheese AND Ham) = 5/11 x 2/9 = 10/99 each time and these are all the solution.

Not finished yet... Then... Since it wants all the combinations you do: 20/99 + 8/99 + 10/99 = 38/99 (Topic) Adding Fractions

...... 38/99 is your answer!

Hope this helped Don't think to complicatively remeber your basics (Probability tree) btw i'm 14. Not all Questions are as hard as they seem break it down!

THIS WAS FOR Q2) ONLY BY THE WAY!

(edited 9 years ago)

Reply 8

9 years ago

Original post by Chocolatecake123

Does anyone know how to answer the following questions, I have attempted both several times but can't seem to get an answer.

Q1). Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers. - Taken from March 2013 Edexcel GCSE linear paper question 21.

Q2) There are three different types of sandwiches on a shelf.

There are :

4 egg sandwiches

5 cheese sandwiches

and 2 ham sandwiches

Erin takes at random 2 of these sandwiches

Work out the probability that she takes 2 different types of sandwiches. Question 24 of the same paper...

I really want an A* for maths so if anyone could list me a few more A-A* topics and help me answer these questions it would be highly appreciated, thanks

Q1). Prove algebraically that the difference between the squares of any two consecutive integers is equal to the sum of these two integers. - Taken from March 2013 Edexcel GCSE linear paper question 21.

Q2) There are three different types of sandwiches on a shelf.

There are :

4 egg sandwiches

5 cheese sandwiches

and 2 ham sandwiches

Erin takes at random 2 of these sandwiches

Work out the probability that she takes 2 different types of sandwiches. Question 24 of the same paper...

I really want an A* for maths so if anyone could list me a few more A-A* topics and help me answer these questions it would be highly appreciated, thanks

Q1:

This may help:

https://www.youtube.com/watch?v=HAxR4iP-WVU

Reply 9

9 years ago

Original post by MunaFuzzyG

Its just simple probability! ;D You need to revise PROBABILTY TREE DIAGRAMS

Solution: (Pretend the lines are tree diagram (You will understand if you revise the topic)

. ..............<----------- 2/9 (EGG) ((4-2) and only 9 in total (11 become 9))

4/11 (EGG) < -------------- 5/9 (CHEESE) (Total doesn't change (Sum Total does))

. ..............<---------- 2/9 (HAM) (Same goes for ham)

...................<------------- 4/9 (EGG) (Total doesn't change (Sum Total does))

5/11 (CHEESE) < ---------------- 3/9 (CHEESE) (5 - 2 = 3 left over sandwhiches)

....................<------------ 2/9 (HAM) (Total doesn't change (Sum Total does))

................<----------------- 4/9 (EGG)

2/11 (HAM) < ------------- 5/9 (CHEESE) ....etc. (Self explanitary)

................<--------------- 0/9 (HAM)

So... Probability (Egg AND Cheese) = 4/11 x 5/9 = 20/99 >--- Revise Multiplying fractions AND & OR Probability (Topics)

Probability (Egg AND Ham) = 4/11 x 2/9 = 8/99 The Question asks for all the 2 different sandwiches

Probability (Cheese AND Ham) = 5/11 x 2/9 = 10/99 each time and these are all the solution.

Not finished yet... Then... Since it wants all the combinations you do: 20/99 + 8/99 + 10/99 = 38/99 (Topic) Adding Fractions

...... 38/99 is your answer!

Hope this helped Don't think to complicatively remeber your basics (Probability tree) btw i'm 14. Not all Questions are as hard as they seem break it down!

THIS WAS FOR Q2) ONLY BY THE WAY!

Solution: (Pretend the lines are tree diagram (You will understand if you revise the topic)

. ..............<----------- 2/9 (EGG) ((4-2) and only 9 in total (11 become 9))

4/11 (EGG) < -------------- 5/9 (CHEESE) (Total doesn't change (Sum Total does))

. ..............<---------- 2/9 (HAM) (Same goes for ham)

...................<------------- 4/9 (EGG) (Total doesn't change (Sum Total does))

5/11 (CHEESE) < ---------------- 3/9 (CHEESE) (5 - 2 = 3 left over sandwhiches)

....................<------------ 2/9 (HAM) (Total doesn't change (Sum Total does))

................<----------------- 4/9 (EGG)

2/11 (HAM) < ------------- 5/9 (CHEESE) ....etc. (Self explanitary)

................<--------------- 0/9 (HAM)

So... Probability (Egg AND Cheese) = 4/11 x 5/9 = 20/99 >--- Revise Multiplying fractions AND & OR Probability (Topics)

Probability (Egg AND Ham) = 4/11 x 2/9 = 8/99 The Question asks for all the 2 different sandwiches

Probability (Cheese AND Ham) = 5/11 x 2/9 = 10/99 each time and these are all the solution.

Not finished yet... Then... Since it wants all the combinations you do: 20/99 + 8/99 + 10/99 = 38/99 (Topic) Adding Fractions

...... 38/99 is your answer!

Hope this helped Don't think to complicatively remeber your basics (Probability tree) btw i'm 14. Not all Questions are as hard as they seem break it down!

THIS WAS FOR Q2) ONLY BY THE WAY!

Hi

The idea of the forum is to guide people rather than giving them the answer.

Don't think to complicatively

In text the answer given looks complicated.

Original post by MunaFuzzyG

Its just simple probability! ;D

Original post by m4ths/maths247

Hi

The idea of the forum is to guide people rather than giving them the answer.

.

The idea of the forum is to guide people rather than giving them the answer.

.

You do realize that this thread was from May 2013

Reply 11

9 years ago

Original post by davros

You do realize that this thread was from May 2013

What are we in now? is it 2014 already?

Original post by sneakbo2

2) 11 sandwiches

4 egg = 4/11

5 Cheese = 5/11

2 Ham = 2/11

Use the and rule, to get two sandwiches they will take one AND then another. Now remember, if you take one there will only be 10 left 1 won't magically appear. Therefore the 2nd fraction should be x/10. Unless the question explicitly states that they put it back or replace it remember this. Also its two DIFFERENT sandwhiches, therefore what that other guy said about Egg & another Egg is incorrect

Possible combinations:

egg 1st cheese 2nd, = 4/11 x 5/10 = 20/110

egg 1st ham 2nd, = 4/11 x 2/10 = 8/110

cheese 1st egg 2nd, = 5/11 x 4/10 = 20/110

cheese 1st ham 2nd, = 5/11 x 2/10 = 10/110

ham 1st egg 2nd, = 2/11 x 4/10 = 8/110

ham 1st cheese 2nd 2/11 x 5/10 = 10/110

Add these all together you get 76/110.

4 egg = 4/11

5 Cheese = 5/11

2 Ham = 2/11

Use the and rule, to get two sandwiches they will take one AND then another. Now remember, if you take one there will only be 10 left 1 won't magically appear. Therefore the 2nd fraction should be x/10. Unless the question explicitly states that they put it back or replace it remember this. Also its two DIFFERENT sandwhiches, therefore what that other guy said about Egg & another Egg is incorrect

Possible combinations:

egg 1st cheese 2nd, = 4/11 x 5/10 = 20/110

egg 1st ham 2nd, = 4/11 x 2/10 = 8/110

cheese 1st egg 2nd, = 5/11 x 4/10 = 20/110

cheese 1st ham 2nd, = 5/11 x 2/10 = 10/110

ham 1st egg 2nd, = 2/11 x 4/10 = 8/110

ham 1st cheese 2nd 2/11 x 5/10 = 10/110

Add these all together you get 76/110.

it says probs of getting 2 different sandwhiches it doesnt matter about order thats wrong

Original post by Life1278

it says probs of getting 2 different sandwhiches it doesnt matter about order thats wrong

Please don't resurrect 8 year old threads

Original post by davros

Please don't resurrect 8 year old threads

lol just saw that

Q2 I just did a paper with this q ! you need to use probability trees to answer this question, each sandwich being a branch, and then the second row being the second sandwich. you’d then find the probability that erin took the same sandwich twice (EE, CC, HH) and to do that you would take , for example the egg sandwich branches, and multiply them, 4/11 x 3/10 = 12 / 110 Do this for ham and cheese and then add all of these probabilities together. Now, do 1-(probability that it is the same) and then you’ll have the answer. HOPE THIS HELPS

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