Hi i have decided to make this post to go through most of the answers to AQA Maths Paper 1 NonCalculator Higher Tier. ( i have been through the answers with my Maths teacher and she gave me a copy of both papers and i understand all of the answers so if you don't understand a solution i can talk with you more through MSN or something.
Question 1
Amy, Beth and Chloe share a £5000 lottery win in the ratio 11 : 8 : 6
How much does each of them receive?
11 + 8 + 6 = 25 parts.
5000 / 25 = £200 per part.
Amy = 11 x 200 = £2200
Beth = 8 x 200 = £1600
Chloe = 6 x 200 = £1200
Question 2
a) Solve the inequality 7y < 3y + 6
7y  3y < 6
4y < 6
y < 1.5
b) Make r the subject of the formula p = 3 + 2r
p  3 = 2r
(p  3) / 2 = r
r = (p  3) / 2
c) Solve the equation 1/2 x  5 = 1/4 x + 3
1/2 x  1/4 x = 3 + 5
1/4 x = 8
x = 32
Question 3
a) Complete this table of values for y = (2 + x)(3  x)
When x = 2
y = (0)(5)
y = 0
When x = 4
y = (6)(1)
y = 6
b) (Plotted values of 2,0 1,4 0,6 1,6 2,4 3,0 4,6
Question 4
A box contains a number of counters.
Each counter is coloured red (R) or white (W).
Each counter is also numbered 1 or 2.
The table shows the probabilities of picking the different colours and numbers when a counter is picked at random from the box.
Number
12
CR 1/5 1/10
o
l
o W 1/4 9/20
u
r
a) Sam says that there are 50 counters in the box.
Explain why Sam must be wrong.
Sam says there is a probability of 1/4 of picking a W1. 1/4 x 50 = 12.5 and you can't have 12 and half counters.
b) Show that the probability of picking a red counter (R) at random from the box is 3/10.
R1 = 1/5 R2= 1/10 1/5 + 1/10 = 2/10 + 1/10 = 3/10
Question 5
The diagram shows an L shape.
Draw the locus of all points 2cm from the L shape.
At top of L, bottom left corner of L, and bottom right of L should be 2cm halfcircles ( 1/4 circle for bottom left corner ) and rest straight lines 2cm from L.
Question 6
1 million = 10^6
1 billion = 10^9
1 trillion = 10^12
a) How many millions are there in one trillion?
one trillion / one million
= 10^12 / 10^6
= 10^6
= 1 million.
b) Write 8 billion in standard form.
8 x 10^9
c) Work out 8 billion multiplied by 3 trillion.
= ( 8 x 10^9) x (3 x 10^12)
= 24 x 10^21
= 2.4 x 10^22.
Question 7
On monday Joe drinks 2+ 1/3 pints of milk.
On tuesday he drinks 1+ 3/4 pints of milk.
Work out the total amount of milk Joe drinks on Monday and Tuesday.
2+ 1/3 = 7/3
1+ 3/4 = 7/4
7/3 + 7/4 = 28/12 + 21/12
= 49/12
= 4+1/12 pints.
Question 8
In a test there are 30 questions.
All candidates start with 25 marks.
Every correct answer scores 4 marks, but for every wrong answer 1 mark is deducted.
a) Jack attempts every question, getting x of them correct.
Write down an expression, in x, for the number of questions he got wrong.
30 questions. x right, (30  x) are wrong.
b) Write down an expression for Jack's total mark.
(Remember that he starts with 25 marks).
Starts with 25. 25
x correct. 4 marks for every correct. 4x. 25 + 4x
mark deducted for every wrong answer. (30  x)= 30 + x.
25 + 4x 30 + x
= 5x 5.
c) Jack scores 80 marks in the test.
Use your answer to part (b) to calculate the value of x, the number of questions he got correct.
80 marks.
5x  5 = 80
5x = 85
x = 17
Question 9
A sprinter runs 200 metres in 20.42 seconds.
Estimate his average speed in kilometres per hour.
You must show your working.
20.42 ~ 20 seconds.
1 minute = 1/60 of hour.
1 second = 1/60 x 1/60 of hour.
= 1/ 3600 of hour
20 seconds = 20/3600 of hour = 1 / 180 of hour.
Speed = Distance / Time
= 0.2 / (1/180)
= 36 km/h .
Question 10
An inspector visits a large company to check their vehicles.
The company has 4 largeload vehicles, 136 light vans and 21 cars.
The inspector decides to sample 10% of the vehicles.
Each type of vehicle is to be represented in the sample.
a) What is this kind of sampling procedure called?
Stratified.
b) How many of each type of vehicle should be inspected?
10% x 4 = 0.4 ~ 0.
10% x 136 = 13.6 ~ 14.
10% x 21 = 2.1 ~ 2.
Question 11
The diagram shows a cylinder.
The diameter of the cylinder is 10cm.
The height of the cylinder is 10cm.
a) Work out the volume of the cylinder. Give your answer in terms of pi.
Volume = pi x r^2 x h
= pi x 5^2 x 10
= 25pi x 10
=250pi cm3.
b) Twenty of the cylinders are packed in a box of height 10cm.
The diagram shows how the cylinders are arranged inside the box.
The shaded area is the space between the cylinders.
Work out the volume inside the box that is not filled by the cylinders. Give your answer in terms of pi.
Shaded area (space not filled) = volume of box  volumes of 20 cylinders.
= (50 x 40 x 10)  (20 x 250pi)
= 20000  5000pi cm3.
Question 12
A, B and C are points on the circumference of a circle with centre O.
BD and CD are tangents.
Angle BDC = 40 degrees.
(a) (i) Work out the value of p.
p = 360 (sum of angles in quadrilateral)  90 (tangent)  90 (tangent)  40 (BDC) = 140 degrees.
ii) Hence write down the value of q.
q = p/2
= 140/2
= 70 degrees.
b) The tangent DB is extended to T.
The line AO is added to the diagram.
Angle TBA = 62 degrees.
i) Work out the value of x.
TBA = 62 degrees.
ABO = 90 (tangent)  62
= 28 degrees
BAO = 28 degrees (isoceles)
x = 180  28  28
= 124 degrees.
ii) Work out the value of y.
BOC = p = 140 degrees.
AOC = 360  140  x
= 360  140  124
= 96 degrees
2y (isoceles) = (180  96) / 2
= 84
y = 42 degrees.
Question 13
THE HARDEST QUESTION ON THE PAPER I THINK
The diagram shows the graph of the equation y = x² + px + q
The graph crosses the xaxis at A and B(2,0).
C (3,5) also lies on the graph.
a) Find the values of p and q.
Well basically, there were people on the topic for this paper cheering because they thought they had the right answer (6,0) for coordinate A. I'm sorry but that is wrong.
y = x² + px + q
Using the coordinate of C (3,5) we can make a simultaneous equation.
5 = 9 + 3p + q
5 = 9  3p + q
14 = 3p + q
q  3p = 14 (1)
Using the coordinate of B we can make the second simultaneous equation.
y = x² + px + q
0 = 4 + 2p + q
q + 2p = 24 (2)
q  3p = 14
q + 2p = 4
5p = 10
p = 2
q  3p = 14
q  6 = 14
q = 8
y = x² + px + q
y = x² + 2x  8
So factorising or using the quadratic forumla:
y= (x  2)(x + 4)
so the two solutions of x where the graph crosses the xaxis are
x  2 = 0
x = 2 (Co ordinate B)
and x + 4 = 0
x = 4
b) Coordinate of A = (4,0)
Question 14
Express the recurring decimal 0.4272727........ as a fraction.
Give your answer in its simplest form.
x = 0.42727......
1000x = 427.2727.....
10x = 4.2727....
990x = 427.2727....  4.2727....
= 423.
423 / 990
4 + 2 + 3 = 9 (9 is multiple of 3) so 423 can be divided by 3.
= 141/ 330
= 47/ 110.
Question 15
a) Simplify (2x^4y)³
2 x 2 x 2 = 8
x^4 x x^4 x x^4 = x^12
y x y x y = y^3
= 8x^12y³
b) Factorise fully 2x²  50y²
= 2(x²  25y²)
= 2 (x 5y)(x + 5y).
Question 16
a) i) Show that √20 = 2√5
√20 = √2 x √2 x √5
= 2√5
ii) Expand and simplify (√2 + √10)²
= 2 + √20 + √20 + 10
= 12 + 2√5 + 2√5
= 12 + 4√5
b) Is this triangle rightangled?
Does (√3)² + (2 + √5)² = (√2 + √10)² ?
√3 x √3 = 3.
(2 + √5)² = 4 + 2√5 + 2√5 + 5
= 9 + 4√5.
(√2 + √10)² = 12 + 4√5
3 + 9 + 4√5 = 12 + 4√5
Yes, the triangle is right angled.
Question 17
In triangle ABC, M lies on BC such that BM = 3/4 BC.
AB = s and AC = t
Find A > M in terms of s and t.
BM = 3/4 BC.
BC= s + t
= ts
BM = 3/4 (ts)
= 3/4t  3/4s
AM = s + 3/4t  3/4s
AM = 1/4s + 3/4t.
Question 18
sin 70 = 0.9397
a) Write down another solution of the equation sin x = 0.9397
90  70 = 20
90 + 20 = 110 degrees.
b) Solve the equation sin x = 0.9397
70 +180 = 250 degrees
110 + 180 = 290 degrees
Question 19
A straight line has the equation y = 2x  3
A curve has the equation y² = 8x  16
(2x  3) ² = 8x  16
4x² 6x 6x + 9 = 8x  16
4x²  12x + 9 = 8x  16
4x² 20x + 25 = 0
By use of the quadratic formula or factorisation:
(2x  5)(2x  5) = 0
2x  5 = 0
2x = 5
x = 2.5 .
y = 2x  3
y = 5  3
y = 2.
b) Sketch 2 is the right sketch because it shows 1 intersection above the xaxis (2.5, 2)
Sketch 1 is wrong because there is 2 points of intersection on the graph.
Sketch 3 is wrong because there is 0 points of intersection on the graph.
Question 20
At the end of a training programme students have to pass and exam to gain a certificate.
The probability of passing the exam at the first attempt is 0.75
Those who fail are allowed to resit.
The probability of passing a resit is 0.6
No further attempts are allowed.
Probability students fails = 0.25 x 0.4 = 0.1
Probability all 3 gain certificate = 0.9 x 0.9 x 0.9 = 0.729
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