Standard Grade Maths Watch

bob 911
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#101
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#101
I've got answers no working though!
I quickly went through the paper so its not error proof.

1. 41miles
2. 3x^3-14x^2+7x+4
3. 1.3 and -2.8
4. £4500
5. No it doesnt pass regulations as it is only 0.879m
6. No it will not be double as the area is only 7754.6cm^2
7. 108 degrees
b). 1.62cm
8. R=10.6cm
9.1200cm^2
b) 130cm
10.£11 an hour for 15 hours = £165
11. 4.55cm
12. (90,1)
b) X coordinate of Q = 48.6 R=131.4
13. 12seconds
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tilal6991
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#102
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#102
(Original post by bob 911)
I've got answers no working though!
I quickly went through the paper so its not error proof.

1. 41miles
2. 3x^3-14x^2+7x+4
3. 1.3 and -2.8
4. £4500
5. No it doesnt pass regulations as it is only 0.879m
6. No it will not be double as the area is only 7754.6cm^2
7. 108 degrees
b). 1.62cm
8. R=10.6cm
9.1200cm^2
b) 0.13cm
10.£11 an hour for 15 hours = £165
11. 4.55cm
12. (90,1)
b) X coordinate of Q = 48.6 R=131.4
13. 12seconds
All fine except 9 b) - I think it's 130 cm = 156000/1200
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JaminaKhan
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#103
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#103
Can anyone post the General Paper for 2011 please?

Thanks!
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bob 911
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#104
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#104
(Original post by tilal6991)
All fine except 9 b) - I think it's 130 cm = 156000/1200
Yeah thats what its should be left it as 156 instead of 156000.
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animelover123
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#105
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(Original post by cakes1995)
can only see answers to paper 1. but in saying that i am still in shock. sooooo difficult :confused:
I really hope they lower the pass mark for a 2. I think ive stuggled to get a 2 in both collumbs, it was really hard. I dont want to think more about it , beaucse it dreads me with fear. I thought Maths was my strong subjects so i expected a less stressful ride before the dreaded Rnglish, but after that shock , i dread what the examiners have done to this years papers
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tomctutor
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(Original post by bob 911)
Yeah thats what its should be left it as 156 instead of 156000.
Just to confirm P2 Q9
(a) area = mean width x height (for trapz)
= 0.5x(68+32) x 24 = 1200 sq. cm. (or cm^2)

NB. you could also have added the area of rect + 2x area of tri.
(b) 1lt = 1000 c.c. (or cm^3)

length = Vol / area = 156 x 1000 / 1200 = 130 cm.

Also Q2 wont get marks if you put (3x+1)(x-1)(x-4) factorised as they told you to expand!

PAPER 1 Soln---
Just been burned by the Boss for putting up paper!
The original scans and solutions have been removed so here's my original answers
to SGMaths Credit Paper 1,
Q1. 2.491
Q2. 2(m-3)(m+3)
Q3. -4
Q4. -7/5
Q5. 17 cakes
Q6. (a) median=4
. (b) P(5 or 6) =14/30
Q7. (a) £cost, 2a + 4c = 56.00
. (b) .. 1a + 3c = 36.00
. (c) cost of child ticket is £8.00, adult is £12.00
Q8. (a) y = -3/2 .x + 12
. (b) P(8/3, 8)
Q9 (a) 2/a^3
. (b) x = 2
Q10. use sine-rule, 10/sin(150) = 4/sin(B) so sin(B) = 1/5 as reqd.
Q11. (a) F = k s/d^2 .. const k,
. (b) F -> F/8
Q12. (a) S(10) = 1+2+3..+10 = 0.5(10x11) = 55
. (b) S(n) = 1+2+3..+ n = 0.5(n)(n+1) = 1/2 .(n^2 + n)
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Zahra0786
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#107
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#107
hi
i sat my credit maths yeaterday see the cake question in paper 1 see how you get 17.5 do you write 17
in paper 2 see the car question is that 4500 and is it okay if your workings are messy is it me or did you guys find it hard tooo.
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JaminaKhan
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#108
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#108
Has anyone got the worked solutions for paper 2?




(Original post by tomctutor)
i presume you mean p2 q8 of the credit paper-
the length of the chord is 18cm,

use pythagoras, r^2 - (r-5)^2 = (18/2)^2
gives... 10r - 25 =81 so r = 10.6cm

awaiting the unbelievers full worked solution to this!
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JaminaKhan
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#109
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#109
Has anyone got worked solutions for paper 2?
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seinfeld73
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#110
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#110
Paper 2 Question 1

1st week 28+10% = 30.8
2nd week 30.8+10% = 33.88
3rd week 33.88+10% = 37.268
4th week 37.268+10% = 40.994

OR

28 x (1.1)^4 = 40.994

ANSWER 41 miles
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tomctutor
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(Original post by Zahra0786)
hi
i sat my credit maths yeaterday see the cake question in paper 1 see how you get 17.5 do you write 17
in paper 2 see the car question is that 4500 and is it okay if your workings are messy is it me or did you guys find it hard tooo.
Would you pay the supermarket full price for half a cake!
A cake means a cake, a full cake and nothing but a cake so 17 is the correct answer.

The car (P2 Q4) is (1 - 0.16) x last yr price = £3780 gives £4500 exactly.
[aka % working backwords]

I think some of the questions were probably quite difficult esp P1- Q8, 9(b) also P2- Q8, Q12 so
if you struggled with these the markers might be a wee bit forgiving - we will only know when the final mi's are released!
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seinfeld73
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Paper 2 Question 2

(3x + 1)(x^2 - 5x +4)
3x(x^2 - 5x + 4)+1(x^2 -5x +4)
3x^3 - 15x^2 + 12x + x^2 - 5x +4
3x^3 - 14x^2 + 7x + 4
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mathsisfunny
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#113
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#113
(Original post by tomctutor)
Would you pay the supermarket full price for half a cake!
A cake means a cake, a full cake and nothing but a cake so 17 is the correct answer.

The car (P2 Q4) is (1 - 0.16) x last yr price = £3780 gives £4500 exactly.
[aka % working backwords]
In tesco you can buy 1/2 cakes...
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seinfeld73
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2x^2 + 3x - 7 = 0

Using formula from formula sheet

ax^2 + bx + c = 0

a= 2
b=3
c= -7

substitute into formula

x= (-b + squareroot(b^2 - 4ac))/2a

and

x= (-b - squareroot(b^2 - 4ac))/2a


you get x =1.3 and x= -2.8
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seinfeld73
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Paper 2 Question 4

since value is 16% less than last year

3780 = 84% (100-16)

3780/84 = 45 (this is what 1% would be)

45 x 100 = 4500 (100% = last years value)

ANSWER £4500
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seinfeld73
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Paper 2 Question 5

the ration of the angle of the staircase to the full circle is equivalent to the ration of the arc of the staircase to the full circle:

since full circle is 360 degrees
and the circumfrence of the circle is 2 x pi x r
length of the arc we need to find is BC

42/360 = BC/2 x pi x r

42/360 = BC/2 x pi x 1.2

multiply throughout by 2 x pi x 1.2 to get BC on its own

(42 x 2 x pi x 1.2)/360 = BC

BC = 0.879m

The safety regulations state the arc must be at least 0.9 m, therefore the staircase will not pass the safety regulations.
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seinfeld73
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Paper 2 Question 6

ratio (125/90)

ration of areas (125/90)^2

Area of B (125/90)^2 x 4020 = 7754.629 cm^2

This is not double the area of A therefore the salesman not justified
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seinfeld73
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Paper 2 Question 7

Angles on a straight line add up to 180

straight line EF has one angle 72 degrees the other angle (CDE) is therefore (180 - 72) = 108 degrees

All angles in a regular pentagon are equal therefore angle ABC is also 108 degrees. ANSWER to a)


For b) you need to use the Cosine Rule from formula sheet.

a^2 = b^2 + c^2 - 2bcCosA

a^2 = 1^2 + 1^2 - (2 x 1 x 1 x cos108)

a^2 = 2 - 2Cos108

a^2 = 2 + 0.6180339887

a = squareroot of 2.618033989

a = 1.62metres

The length of AC is 1.62 metres
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christinaaxox
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(Original post by seinfeld73)
Paper 2 Question 7

Angles on a straight line add up to 180

straight line EF has one angle 72 degrees the other angle (CDE) is therefore (180 - 72) = 108 degrees

All angles in a regular pentagon are equal therefore angle ABC is also 108 degrees. ANSWER to a)


For b) you need to use the Cosine Rule from formula sheet.

a^2 = b^2 + c^2 - 2bcCosA

a^2 = 1^2 + 1^2 - (2 x 1 x 1 x cos108)

a^2 = 2 - 2Cos108

a^2 = 2 + 0.6180339887

a = squareroot of 2.618033989

a = 1.62metres

The length of AC is 1.62 metres
do you think I will get marks in the b) if I carried a mistake from a)?
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seinfeld73
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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
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