AQA Maths Linear June 2011 Calculator paper (Specification A)

yes i got tht exactly!

anybody remember what they got when there was a page of algebra and you had to solve them... the very last one was one using division... i got 5w=20, then w = -4 but i dont think that was right. what did everyone else get?

Reply 62

Member737,514

10

Original post by Syeda786

anybody remember what they got when there was a page of algebra and you had to solve them... the very last one was one using division... i got 5w=20, then w = -4 but i dont think that was right. what did everyone else get?

This might be the one where I got 6.25... my memory fails me :s-smilie:

Reply 63

Bruno

Original post by glurblug

Everyone else seems to have gotten 12 so it's probably me who misread it... Can anyone else confirm whether it was 24pi cm^2 or 24cm^2?

It was 24pi cm^2

Reply 64

Bruno

Original post by Member737,514

But there are 36 different number combinations in total? So there are 2 possible combinations per time as there is only one of each card: 7+9 and 8+9 therefore wouldn't the answer be 2/36 or 1/18

You could get 7+9 and 9+7 so I think there are 4 different possibilities. Why are there 36 combinations?

Reply 65

glurblug

Original post by Bruno

It was 24pi cm^2

Damn... I completely missed that. How many marks was the question? Will I still get marks if I worked it out correctly for 24cm^2?

Reply 66

bluekipper

Don't worry - You'll only lose one mark.

Correct solution:
Area of sector = 24pi = pi r^2 x 60/360
pi cancels out to give: r^2 = 6 x 24 = 144
So r = 12

(edited 12 years ago)

Reply 67

Member737,514

10

Original post by Bruno

You could get 7+9 and 9+7 so I think there are 4 different possibilities. Why are there 36 combinations?

There are only 36 different number combinations. In terms of number as well as order in which they are picked, there are then 81. But frankly whether a 3 or a 5 is picked first is irrelevant as it's the total of the two numbers you're after. So if you counted 1 and 2, then 2 and 1, as two combinations, there would be 81 different card combinations but the each of the two card combinations give the same total. There are only 2 number combinations that give a total of 16+. You're picking them both out in one go, so 7+9 and 9+7 are effectively the same number combination.
I'm not 100% if I'm correct though; this was just my interpretation of the question. :smile:

Reply 68

zzz.

Original post by glurblug

Everyone else seems to have gotten 12 so it's probably me who misread it... Can anyone else confirm whether it was 24pi cm^2 or 24cm^2?

IT was def. 24pi cm^2.

Reply 69

sauce

13

Original post by Member737,514

I got this! 60degrees/360degrees=1/6. 24pi x 6 = total circle area= 144pi. area=pi x r^2 so 144pi/pi = 144. root144 = 12 :smile:

I got that too!

Reply 70

Lol_Fish

OP

12

Original post by glurblug

Actually, I don't think it specified explicitly whether it was with replacement or without. Maybe they'll give marks for both? Does anyone remember the exact wording of the question?

It said two cards that are "ATLEAST" 16, so that means equal to or greater than 16 so 9 + 7 and 9 + 8

Reply 71

Elmyra

wait so the sine and cosine questions had to be to 3 significant figures!

Reply 72

Chucky

Original post by Member737,514

There are many answers for this... I got 30= 1/2 x 10 x (2+4)
so a= 2, b=4 and h=10

Then I rearranged it for the following question to: b= 2A/h - a. Did anyone else get this?

yeah got the same and its right because i checked by substituting in my values after.

Reply 73

Chucky

pretty sure i got atleast 80 on both papers so the a star is still in sight. Personally i found this paper easier but on this paper hoping that the only ones i got wrong were the probability (4 marks) the final vector (4 marks) the 2nd to last equation (2marks) and the use the line to find the values (3 marks) i then may have possibly got the road area one wrong which was 4 marks but hopefully i got atleast 83 and then hopefully 80 on the other paper. Does anyone think 163 would be an a star?

Reply 74

Member737,514

10

Original post by Chucky

pretty sure i got atleast 80 on both papers so the a star is still in sight. Personally i found this paper easier but on this paper hoping that the only ones i got wrong were the probability (4 marks) the final vector (4 marks) the 2nd to last equation (2marks) and the use the line to find the values (3 marks) i then may have possibly got the road area one wrong which was 4 marks but hopefully i got atleast 83 and then hopefully 80 on the other paper. Does anyone think 163 would be an a star?

Yes probably, the A* boundary is usually around 80%

Reply 75

zzz.

Original post by Elmyra

wait so the sine and cosine questions had to be to 3 significant figures!

Reallyyy??

Reply 76

glurblug

Original post by Lol_Fish

It said two cards that are "ATLEAST" 16, so that means equal to or greater than 16 so 9 + 7 and 9 + 8

That isn't relevant to what I was talking about. If it was with replacement, you could get two 9s, two 8s etc. But I don't think it was with replacement. (Or at least I hope it wasn't cause that's how I answered the question).

Reply 77

Since95

Original post by Bruno

I got 11 years for the compound interest and rounded to 30 for the angle using the sine rule.

Me and a lot of mates, got 12 years for compound interest as it was 3000
how many years for it to double with an interest of 6.5%
as a multiplier 1.065 followed by the indices

with an indicie of 11 you would have been given = 5997.45 which is too low
meaning that 12 giving over 6000 gave an appropriate amount for double

3000x1.065(12) =6387.3

Reply 78

Since95

Original post by Member737,514

Yes probably, the A* boundary is usually around 80%

from my mocks they were closer to 70%

(just giving my fellow student room peers relief) :cool:

Reply 79

Since95

Original post by zzz.

Crap...

so it was 4/81?

the question was using the cards 1-9 , 9 cards, pick two cards to make atleast 16 what is the probability (not entirely sure whether it said it was replaced, but i'm assuming it did... as it definitely didn't say otherwise)

anyway first for the question you had to work out the pairs that mad 16+

7+9=16

8+9=17

7=1/9
8=1/9
9=1/9

as a fraction for probability

so first you'll find the probability of getting one of the sums - the and rule.

1/9 x 1/9 = 1/81 - probability of 7+9

and the same for 8+9

1/9 x 1/9 = 1/81

which gave the same probability.

then you would use the or rule of getting either pairs (adding the fractions)

1/81 + 1/81 = 2/81

the final probability should have been 2/81.

please let me know if you spot any error, but i feel it is correct (challenge me :confused:

)

AQA Maths Linear June 2011 Calculator paper (Specification A)

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