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# [Official Thread] OCR MEI AS Mathematics 2016 (C1, C2 & S1)

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1. (Original post by Studious_Student)
It shouldn't be 4. Talk me through your method.
Okay a is 5 right . 5/3 is 1.66666 so remainder 0.66666 0.6666x5 is 3.33, (3.3333/5) x5 is something else anyway it doesn't come to 25, no way , 4 point something
2. (Original post by The person)
Everyone's saying that it was a hard paper. Everyone's complaining about the f(h+5) question. But if you actually revised properly and actually put in the effort, you could have answered these "impossible questions". You'll get the grade you deserve.
I found that question pretty simple because I revised that. I think it was tricky in the way that some of the questions were asked (like question 2 which I ****ed up. Also can someone confirm if you wrote the equation of the tangent in the form Y-y1=m(X-x1) would that still be right? I think I had something like y-15=8(x-5)
3. (Original post by ineedA)
Okay a is 5 right . 5/3 is 1.66666 so remainder 0.66666 0.6666x5 is 3.33, (3.3333/5) x5 is something else anyway it doesn't come to 25, no way , 4 point something
That's not the remainder, that's the multiple that you need to multiply by 3 to get 5, think of it in the way of mixed fractions.

5/3 = 1 and 2/3, The two is the remainder here because it takes 1 lot of 3 and a remaining 2 to make the 5. Thus the remainder is 2. For the second part, it would be zero, because 6 can be completely divided by 3 to make 2/1 with no remainder needed.
4. (Original post by studious_student)
that's not the remainder, that's the multiple that you need to multiply by 3 to get 5, think of it in the way of mixed fractions.

5/3 = 1 and 2/3, the two is the remainder here because it takes 1 lot of 3 and a remaining 2 to make the 5. Thus the remainder is 2. For the second part, it would be zero, because 6 can be completely divided by 3 to make 2/1 with no remainder needed.
shiiiiiiiiiiiiiiii, reckon ill get anny marks for my method, i know its unlikely -_-
5. (Original post by ineedA)
shiiiiiiiiiiiiiiii
Don't worry about it ahaha, I think a lot of people made stupid mistakes on this paper, I think I have lost around 8-10 altogether from them.
6. (Original post by Studious_Student)
Don't worry about it ahaha, I think a lot of people made stupid mistakes on this paper, I think I have lost around 8-10 altogether from them.
do you have any idea whether or not the angle in the triangle was specified to 3 sig fig?
7. (Original post by ineedA)
do you have any idea whether or not the angle in the triangle was specified to 3 sig fig?
By standard convention it is 3 significant figures. But I'm pretty sure they accept any reasonable form of rounding up (e.g. 2 Decimal places)
8. What do you guys expect the grade boundaries to be?
9. (Original post by reoshinwho)
What do you guys expect the grade boundaries to be?
I reckon low 50s for an A and then each grade below that the same distance apart as they usually do (5 marks between the grades?)
10. Guys where can I find the unofficial mark scheme??
11. (Original post by ghfijrwosbsv)
Guys where can I find the unofficial mark scheme??
12. (Original post by GO97)
guys for that question with "f(h+5) - f(5)/h" could you assume that h represents the the change in x and so when when x=5 the change in x would be 0 and so u sub that into the simplified "h+8" to geit "0+8" which gives you 8, the answer???? or am i way off
can someone help confirm if i would get any marks for this^
13. It was roughly 260.7 so you were right to round up to 261. Also, you can't have a decimal of a flu case so it had to be a whole integer.
14. (Original post by ineedA)
Okay a is 5 right . 5/3 is 1.66666 so remainder 0.66666 0.6666x5 is 3.33, (3.3333/5) x5 is something else anyway it doesn't come to 25, no way , 4 point something
Incorrect, 5/3 is 1 remainder 2 therefore it would be 2x5 rather then the .6 recurring x5
15. (Original post by Crozzer24)
The log graph was awful, felt impossible to draw a best line of fit, ended up with y intercept of 0.6 and gradient 0.2 which seems way off
nahh dw I got the same
16. (Original post by Studious_Student)
So then wouldn't the answer be a = 12.7? Because the unofficial markscheme says a = 8.96

r^2/2 x Theta = a^2/2 x sin(theta)

Sub values in:

144/2 x 0.8 = a^2/2 x Sin (0.8)

(288/5)/sin (0.8) = a^2/2 ---> 80.2948504 = a^2/2

160.5897008 = a^2

a = 12.67 = 12.7 (3 sig figs) ?
That's what I got but I think it's wrong😞
17. (Original post by e910b)
It was roughly 260.7 so you were right to round up to 261. Also, you can't have a decimal of a flu case so it had to be a whole integer.
But say in a question it's asking for the number of people with flu, you can't have 0.7 of a person with flu so surely you'd take the number of whole people (so 260)? You don't have 261 cases so can't use 261 as the number surely?
18. According to the unofficial mark scheme I think I got around 53 (give or take marks here or there depending on if i get method marks or if i missed some method marks), would that be a B?
19. (Original post by NickC98)
But say in a question it's asking for the number of people with flu, you can't have 0.7 of a person with flu so surely you'd take the number of whole people (so 260)? You don't have 261 cases so can't use 261 as the number surely?
All this rounding stuff is a pain haha. In s1, you're not allowed to round at all for an expectation, even when you can't have 0.7 of a person.
It's so annoying you lose marks for these things
20. (Original post by pandasandpens)
All this rounding stuff is a pain haha. In s1, you're not allowed to round at all for an expectation, even when you can't have 0.7 of a person.
It's so annoying you lose marks for these things
Well I sure am glad I'm not doing S1...

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