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http://examsolutions.net/a-level-mat...uary/paper.php
See q2b and the corresponding mark scheme for this question part.
I did this
(1/2)(2)(8) + 8T + (1/2)[75-(T+5)](8)=500
20 + 8T + 4(70-T)=500
...
T=50s
The mark scheme used the trapezium method to get area, but I used a different method and still got the right answer. So do I get full marks still? Aren't the A2 methods there for the method which is right or what
?
Also, for the same paper, see q6d. Would 3.97m (3SF) be okay?
Thanks
See q2b and the corresponding mark scheme for this question part.
I did this
(1/2)(2)(8) + 8T + (1/2)[75-(T+5)](8)=500
20 + 8T + 4(70-T)=500
...
T=50s
The mark scheme used the trapezium method to get area, but I used a different method and still got the right answer. So do I get full marks still? Aren't the A2 methods there for the method which is right or what
?Also, for the same paper, see q6d. Would 3.97m (3SF) be okay?
Thanks
0
reply
Also, in general for M1 past papers, I'm guessing no marks are given for your diagrams, right?
Also, I tend to see, for example, this
20cos30° = T cos60° M1 A2, in MS
but I'd write down, 20cos30 - Tcos60 = 0 instead, I'm guessing this is still okay, right?
Thanks, better to be safe than sorry.
Also, I tend to see, for example, this
20cos30° = T cos60° M1 A2, in MS
but I'd write down, 20cos30 - Tcos60 = 0 instead, I'm guessing this is still okay, right?
Thanks, better to be safe than sorry.
0
reply
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#5
(Original post by krisshP)
http://examsolutions.net/a-level-mat...uary/paper.php
See q2b and the corresponding mark scheme for this question part.
I did this
(1/2)(2)(8) + 8T + (1/2)[75-(T+5)](8)=500
20 + 8T + 4(70-T)=500
...
T=50s
The mark scheme used the trapezium method to get area, but I used a different method and still got the right answer. So do I get full marks still? Aren't the A2 methods there for the method which is right or what?
Also, for the same paper, see q6d. Would 3.97m (3SF) be okay?
Thanks
http://examsolutions.net/a-level-mat...uary/paper.php
See q2b and the corresponding mark scheme for this question part.
I did this
(1/2)(2)(8) + 8T + (1/2)[75-(T+5)](8)=500
20 + 8T + 4(70-T)=500
...
T=50s
The mark scheme used the trapezium method to get area, but I used a different method and still got the right answer. So do I get full marks still? Aren't the A2 methods there for the method which is right or what
?Also, for the same paper, see q6d. Would 3.97m (3SF) be okay?
Thanks
I'm not sure whether they would accept 3.97. I've always thought that the number of S.F must be less than or equal to the least accurate figure given.
0
reply
Report
#6
(Original post by krisshP)
Also, in general for M1 past papers, I'm guessing no marks are given for your diagrams, right?
Also, I tend to see, for example, this
20cos30° = T cos60° M1 A2, in MS
but I'd write down, 20cos30 - Tcos60 = 0 instead, I'm guessing this is still okay, right?
Thanks, better to be safe than sorry.
Also, in general for M1 past papers, I'm guessing no marks are given for your diagrams, right?
Also, I tend to see, for example, this
20cos30° = T cos60° M1 A2, in MS
but I'd write down, 20cos30 - Tcos60 = 0 instead, I'm guessing this is still okay, right?
Thanks, better to be safe than sorry.
0
reply
(Original post by LilacLily)
You would still get full marks. You have just divided the trapezium into a 2 triangles and a square. If it makes sense and if you clearly show the working then you would still get the method marks.
I'm not sure whether they would accept 3.97. I've always thought that the number of S.F must be less than or equal to the least accurate figure given.
You would still get full marks. You have just divided the trapezium into a 2 triangles and a square. If it makes sense and if you clearly show the working then you would still get the method marks.
I'm not sure whether they would accept 3.97. I've always thought that the number of S.F must be less than or equal to the least accurate figure given.
(Original post by LilacLily)
This is perfectly fine. It's the same thing.
This is perfectly fine. It's the same thing.
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