Standard Grade Maths
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Original post by seinfeld73
Paper 2 Question 8
you need to use pythagorus on the triangle that goes from the centre O down to the level of the water, along to B and back up the r marked on the diagram to O.
(quite hard to explain as can not work out how to draw a diagram on here but hope you are following me)
The length of the side from 0 down to the water level is equal to the radius of the circle minus the 5cm of water = (r-5)
the length of the side along the water level is 9cm (half of the length AB = 18cm)
therefore
r^2 = 9^2 + (r-5)^2
r^2 = 81 + (r-5)(r-5)
r^2 = 81 + r^2 - 5r - 5r +25
r^2 = 106 + r^2 - 10r
r^2 - r^2 = 106-10r
0 = 106 - 10r
10r = 106
r = 106/10
r = 10.6 cm
you need to use pythagorus on the triangle that goes from the centre O down to the level of the water, along to B and back up the r marked on the diagram to O.
(quite hard to explain as can not work out how to draw a diagram on here but hope you are following me)
The length of the side from 0 down to the water level is equal to the radius of the circle minus the 5cm of water = (r-5)
the length of the side along the water level is 9cm (half of the length AB = 18cm)
therefore
r^2 = 9^2 + (r-5)^2
r^2 = 81 + (r-5)(r-5)
r^2 = 81 + r^2 - 5r - 5r +25
r^2 = 106 + r^2 - 10r
r^2 - r^2 = 106-10r
0 = 106 - 10r
10r = 106
r = 106/10
r = 10.6 cm
Someone should put up a scanned pdf or something with P2 workings
for all to see!
Paper 2 Question 9
couple of ways to work this out
one way would be to work out the area of the rectangle using length 32 and breadth 24.
Area of rectangle would be l x b = 32 x 24 = 768
The area of one of the triangle sections would be 1/2 x base x height
base would be (68 cm - 32 cm)/ 2(as 2 triangles) = 18 cm
Area of triangle is 1/2 x 18 x 24 = 216 (we have 2 triangles so we will use this twice)
Total area of cross section = 768 + 216 + 216 = 1200 cm^2
b) Volume = area of cross section x l
156 litres is 156000 for our calculation
156000 = 1200 x l
l = 156000/1200
l = 130 cm
couple of ways to work this out
one way would be to work out the area of the rectangle using length 32 and breadth 24.
Area of rectangle would be l x b = 32 x 24 = 768
The area of one of the triangle sections would be 1/2 x base x height
base would be (68 cm - 32 cm)/ 2(as 2 triangles) = 18 cm
Area of triangle is 1/2 x 18 x 24 = 216 (we have 2 triangles so we will use this twice)
Total area of cross section = 768 + 216 + 216 = 1200 cm^2
b) Volume = area of cross section x l
156 litres is 156000 for our calculation
156000 = 1200 x l
l = 156000/1200
l = 130 cm
here is a picture of #8, not very good quality but my camera is broke ):
Original post by tomctutor
Someone should put up a scanned pdf or something with P2 workings
for all to see!
for all to see!
Yes good idea.....I've only got my laptop at moment so just trying to help until someone can scan in the answers
Paper 2 Question 10
using x = hourly rate
15x + 8x + 12(4/3)x = 429
15x + 8x + 16x = 429
39x = 429
x = 429/39
x = £11 per hour
Tom worked for 15 hours therefore was paid 15 x 11 = £165
using x = hourly rate
15x + 8x + 12(4/3)x = 429
15x + 8x + 16x = 429
39x = 429
x = 429/39
x = £11 per hour
Tom worked for 15 hours therefore was paid 15 x 11 = £165
Original post by seinfeld73
Paper 2 Question 10
using x = hourly rate
15x + 8x + 12(4/3)x = 429
15x + 8x + 16x = 429
39x = 429
x = 429/39
x = £11 per hour
Tom worked for 15 hours therefore was paid 15 x 11 = £165
using x = hourly rate
15x + 8x + 12(4/3)x = 429
15x + 8x + 16x = 429
39x = 429
x = 429/39
x = £11 per hour
Tom worked for 15 hours therefore was paid 15 x 11 = £165
I spent atleast 20 minutes on this, doing it by trial and error. What even.
Paper 2 Question 11
The area of the amount of paper it takes to go once around the cylinder would be
3000/3 = 1000 cm^2
If this paper is layed flat it would be a rectangle shape with one side equal to 70cm and the other equal to the circumfrence of the circle (pi x D)
Area of rectangle = l x b
1000 = pi x D x 70
D = 1000/(70 x PI)
D = 4.55 cm
The area of the amount of paper it takes to go once around the cylinder would be
3000/3 = 1000 cm^2
If this paper is layed flat it would be a rectangle shape with one side equal to 70cm and the other equal to the circumfrence of the circle (pi x D)
Area of rectangle = l x b
1000 = pi x D x 70
D = 1000/(70 x PI)
D = 4.55 cm
Paper 2 Question 13
h = 48 + 8t - t^2
height 0 when flare reaches the ground
0 = 48 + 8t - t^2
rearrange to get
t^2 - 8t - 48 = 0
(t + 4)(t - 12) = 0
t = -4 (time can not be negative so we can not use this answer)
t = 12 seconds
It takes 12 seconds to enter sea.
h = 48 + 8t - t^2
height 0 when flare reaches the ground
0 = 48 + 8t - t^2
rearrange to get
t^2 - 8t - 48 = 0
(t + 4)(t - 12) = 0
t = -4 (time can not be negative so we can not use this answer)
t = 12 seconds
It takes 12 seconds to enter sea.
oops sorry missed q 12
Paper 2 Question 12
a) Point P is where y = 4sinx - 3 is at its maximum
maximum point of sin x = 1 when x = 90 degrees
y = 4 x 1 - 3
y = 4-3
y = 1
P is at the point (90, 1)
b) y = 0 at the points Q and R
0 = 4sinx - 3
3= 4sin x
sin x = 3/4 (since this is positive we are looking for angles in
quadrant 1 and quadrant 2)
x = inverse sin (3/4)
x = 48.59 (quadrant 1)
x = 180 - 48.59 = 131.41 (quadrant 2)
Q (48.6 , 0)
R (131.4, 0)
Paper 2 Question 12
a) Point P is where y = 4sinx - 3 is at its maximum
maximum point of sin x = 1 when x = 90 degrees
y = 4 x 1 - 3
y = 4-3
y = 1
P is at the point (90, 1)
b) y = 0 at the points Q and R
0 = 4sinx - 3
3= 4sin x
sin x = 3/4 (since this is positive we are looking for angles in
quadrant 1 and quadrant 2)
x = inverse sin (3/4)
x = 48.59 (quadrant 1)
x = 180 - 48.59 = 131.41 (quadrant 2)
Q (48.6 , 0)
R (131.4, 0)
Does anyone know where I could find the general paper to have a look at? :-)
OMG why did they have to make foundation so hard 
after i seen that pipe question it just made me feel like taking the pipe and shuvin it up my bum 
foundation paper - Q 1, i still cant get it can someone help me out? it was 1+1
Original post by Dado Prso
the pipe question was no complete the square question for sure.
sg maths 08 was the daddy
sg maths 08 was the daddy
Have to agree with you there. 2008 paper is alot more difficult than this years.
If anyone can find me a copy of this year's Standard Grade Maths exam, I as an A1 Higher candidate would gladly look over it for anyone :P
Credit 2011 - Paper 2 - Question 4a)
4a) "A car is valued at £3780.
This is 16% less than last year's value.
What was the value of the car last year?"
ANSWER = £4500
Working :-
Since this years is 16% less then this years value is 84%
84% = £3780
1% = £3780/84 = £45
100% = £45 x 100
100% = £4500
4a) "A car is valued at £3780.
This is 16% less than last year's value.
What was the value of the car last year?"
ANSWER = £4500
Working :-
Since this years is 16% less then this years value is 84%
84% = £3780
1% = £3780/84 = £45
100% = £45 x 100
100% = £4500
Original post by mckinnon94
If anyone can find me a copy of this year's Standard Grade Maths exam, I as an A1 Higher candidate would gladly look over it for anyone :P
All the answers have been done. Paper 1 answers are in an attatchment earlier on and all the answers to paper 2 are also earlier.
Original post by eastkilbride#1
Credit 2011 - Paper 2 - Question 4a)
4a) "A car is valued at £3780.
This is 16% less than last year's value.
What was the value of the car last year?"
ANSWER = £4500
Working :-
Since this years is 16% less then this years value is 84%
84% = £3780
1% = £3780/84 = £45
100% = £45 x 100
100% = £4500
4a) "A car is valued at £3780.
This is 16% less than last year's value.
What was the value of the car last year?"
ANSWER = £4500
Working :-
Since this years is 16% less then this years value is 84%
84% = £3780
1% = £3780/84 = £45
100% = £45 x 100
100% = £4500
everyone got that question right? including me and i was in foundation!!!!!!!!!!!!
Is anybody taking maths higher next year ?!?!?!?!
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