Jump to content

Maths HL


Barbra

Recommended Posts

Papers 1 and 2 each worth 30% of your grade; IA and Paper 3 worth 20% each. As long as your overall weighted percentage of the course is over 43% you should get a 4, 56% to get a 5 . 43% is the highest 3/4 boundary I have found.

Link to post
Share on other sites

2 hours ago, kw0573 said:

Papers 1 and 2 each worth 30% of your grade; IA and Paper 3 worth 20% each. As long as your overall weighted percentage of the course is over 43% you should get a 4, 56% to get a 5 . 43% is the highest 3/4 boundary I have found.

So would that mean, if i get about 30 points each in both papers and 17 on my IA, I would need to get about 20 points to get a 4 and 30 points to get a 5 in paper 3?? Or is that totally wrong? I'm really sorry about this, I am just quite confused!

Link to post
Share on other sites

Let me try this again.
Let's say the exam is split into IA and non-IA. The weight is 20% IA and 80% non-IA. You need 43% weighted average to get a 4. Let's say you keep your 17/20 for IA

So 17/20 * 20% + w * 80% = 43%.
17/20 * 0.2 + 0.8 w = 0.43, x = 0.325. That mean on average (or on weighted average), you need about to score on weighted average w = 0.325 = 32.5% on each exam to get a 4. That is about 40/120 for each of P1,P2 and 20/60 for P3. 

If you want you can repeat the calculation to split non-IA components into the 3 papers,
So x / 120 * 30% + y / 120 * 30% + z/60 * 20% = 0.325 * 80%. Plug in the raw marks you think you get in paper 1 and 2 for x and y and you'll know how much you need on paper 3 to get 32.5% in the papers. In practice you should aim higher (because you should aim higher) to allow for lowering marks in IA due to moderation. 

Link to post
Share on other sites

10 minutes ago, kw0573 said:

Let me try this again.
Let's say the exam is split into IA and non-IA. The weight is 20% IA and 80% non-IA. You need 43% weighted average to get a 4. Let's say you keep your 17/20 for IA

So 17/20 * 20% + w * 80% = 43%.
17/20 * 0.2 + 0.8 w = 0.43, x = 0.325. That mean on average (or on weighted average), you need about to score on weighted average w = 0.325 = 32.5% on each exam to get a 4. That is about 40/120 for each of P1,P2 and 20/60 for P3. 

If you want you can repeat the calculation to split non-IA components into the 3 papers,
So x / 120 * 30% + y / 120 * 30% + z/60 * 20% = 0.325 * 80%. Plug in the raw marks you think you get in paper 1 and 2 for x and y and you'll know how much you need on paper 3 to get 32.5% in the papers. In practice you should aim higher (because you should aim higher) to allow for lowering marks in IA due to moderation. 

oh okay gotcha. thanks, and yeah i know about the IAs. does it get moderated down quite often?

Link to post
Share on other sites

It generally would get moderated down 1-2 marks for high grades and stay same or even increase for low grades. Also I forgot about number of total marks in HL were changed last year so my last equation it should be x  / 100 * 30% + y /100 *30% + z / 50*20% instead. 

Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...