Barbra Posted May 3, 2018 Report Share Posted May 3, 2018 How many points do you need to get in paper 1, 2 and 3 if you want to get a four or a five in maths HL with 17 points predicted in your IA? Reply Link to post Share on other sites More sharing options...
kw0573 Posted May 3, 2018 Report Share Posted May 3, 2018 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. Reply Link to post Share on other sites More sharing options...
Barbra Posted May 3, 2018 Author Report Share Posted May 3, 2018 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! Reply Link to post Share on other sites More sharing options...
kw0573 Posted May 3, 2018 Report Share Posted May 3, 2018 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. Reply Link to post Share on other sites More sharing options...
Barbra Posted May 3, 2018 Author Report Share Posted May 3, 2018 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? Reply Link to post Share on other sites More sharing options...
kw0573 Posted May 3, 2018 Report Share Posted May 3, 2018 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. Reply Link to post Share on other sites More sharing options...
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