GCSE Mathematics MathsMadeEasy - Maths, English and ... · 4 0 0 0 x 3 . 6 shift % = Interest for year 1 = 4000 x 3.6 ÷ 100 = £144 Add the interest to the £4000: £4000 + £144
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
ANSWERS Marks shown in brackets for each question (2)
A* A B C D E
88 75 60 45 25 15
3 Legend used in answers
Green Box - Working out Red Box and - Answer
Authors Note Every possible effort has been made to ensure that everything in this paper is accurate and the author cannot accept responsibility for any errors. Apart from any fair dealing for the purposes of research or private study as permitted under the Copyright, Designs and Patents Act 1988, this paper may only be reproduced, stored or transmitted in any form or by any means with the prior permission in writing of the author, or in the case of reprographic reproduction in accordance with the terms and licence by the CLA. Enquiries concerning reproduction outside these terms should be sent to the author. The right of David Weeks to be identified as the author of this work has been asserted by him in accordance with the Copyright, Designs and Patents Act 1988.
(2) b) Using your prime factor tree or otherwise work out the Lowest Common Multiple for 168 and 40
..............................
(2)
Start with your number at the top. See if it can be halved This halves into 2 and 84
A prime factor tree breaks a number down into its prime number factors.
84 can be halved again =2 × 42
42 can be halved again =2 × 21
2 is the first prime number
21 =3 × 7 both prime nimbers
168 = 2 × 2 × 2 × 3 × 7 Theses are all prime numbers
40 = 2 × 2 × 2 × 5
The lowest common multiple is the value which both 40 and 168 will divide into Look at the prime numbers that we got for 168 and 40 and mark the those that do not appear in both. Cross multiply these with the top of the opposite prime factor tree 168 = 2 × 2 × 2 × 3 × 7 21 × 40 = 840 40 = 2 × 2 × 2 × 5 5 × 168 = 840
2. A cuboid lines on the co-ordinate axes. Distances in cm Not drawn accurately
The point B has co-ordinates (7, 3, 4) a) What are the co-ordinates of the point A
............................. (1)
b) What are the co-ordinates of the point C
............................. (1)
c) What is the distance between A and C correct to 1 decimal place
.............................cm (3)
When you have co-ordinates of (x, y ) they are on a graph. If you add a z co-ordinate it makes it into a 3-D space. Each corner of the cuboid is given by the co-ordinate (x, y, z)
B is 5 along the x-axis, 3 long the y-axis and 2 along the z-axis
A is the same distance as B along the x-axis and z-axis but has no distance along the y-axis
7, 0, 4
A is zero along the x-axis 3 along the and y-axis and zero along the z-axis 0, 3, 0
To work out AC which is diagonally across the cuboid we need to draw a triangle as shown dotted. The base length of this triangle = √58 = 7.616cm Use Pythagoras with AC2 = base length2 + 42
5. There are 21 coloured buttons in a bag 6 buttons are blue 10 buttons are green 4 buttons are red. 1 black button If you take two buttons at random from the bag without replacement What is the probability that you pick a) Two red buttons
............................. (2)
b) A blue button and a green button
............................. (2)
Probability (1st red) = number of red buttons = 4 total number buttons 21
1 35
Total number buttons is 6 + 10 + 4 + 1 = 21
Probability (2nd red) = 3 20
REMEMBER: we have already taken one red out so have 3 left out of a total of 20
Probability (two reds) = 41 x 31 = 1 217 205 35
Probability ( blue - green) = 6 x 10 = 6 = 1 21 202 42 7
Probability (both events) = 1 + 1 = 2 7 7 7
Probability ( green - blue) = 10 x 6 = 1 21 20 7
2 7
We can select this in two ways: either a blue button followed by a green button OR a green button followed by a blue button. We work out the probability for each event and then add them
8. Diagram NOT drawn accurately In the diagram, above shows an isosceles triangle XYZ with measurements in centimetres. XY = 4a XZ = 4a YZ = 14 a) Find an expression in terms of a, for the Perimeter of the triangle in its simplest form
………………….…..…….(2) b) If the perimeter of the triangle is a2, calculate a, correct to 1 decimal point.
x = – b ± acb 42 − 2a
2
95.108
2
1208
2
56648
2
1414648 ±+=±+=+±+
=−××−±+
=x
2
95.2
2
95.18 −= orx
a = ……………… (3)
Perimeter is the distance around the outside of the triangle. From Y to Z to X and back to Y as shown
9. Laura paid £78 for an mp3 player in a sale with 35% off the original price. How much was the mp3 player originally?
£……………. (3)
Sale 35% off original price
It’s NOT 35% of £78 !!!! because £78 was NOT the original price
120
The original price is reduced by 35% so, £78 is 65% or 0.65 of the original price. If we say the original price is O and the sale price is S we can make an equation: O × 0.65 = S O = S ÷ 0.65 and O = 78 ÷ 0.65 = 120
Multiply by 3 so 4y becomes 12y then we can add the equations to eliminate it
–1½
Often we can just add or subtract equations to eliminate x or y. But this time It’s harder- first we have to multiply one equation so x or y is the same as the other equation
× 3
So x = 75÷ 30 = 2 ½
30x = 75
2 ½
Now add the two equations to eliminate y
It becomes
SUBSTITUTE x = 2 ½ into one of the equations
6x – 4y = 21 6 x 2 ½ – 4y = 21 15 – 4y = 21 (– 15) – 4y = 6 y = –1 ½
This is a SURDs question. We need to make the denominator rational. To do this multiply both top and bottom by 2 + √3. (sign change from original) 5 × 3 – √8 = 15 – 5√8 = 15 – 5√8 = 15 – 5√8 = 15 – 10√2 3 + √8 3 – √8 9 + 3√8 –3√8 –√8√8 9 – 8 Notice that these parts are opposite so cancel out.
15. A man starts at home H and runs for 5 km on a bearing of 1250 to I He then runs for 8 km on a bearing of 2300 for 8km to J He then runs home in a total time of 1 hour and nine minutes.
a) Work out how far away (D) from home he is, correct to one decimal place
........................0 (3)
b) Calculate his average speed in km per hour to one decimal place.
.........................km/hr (2)
Angle HIJ = 360 – 65 – 230 = 650
In triangle HIJ we have 2 sides and an angle and need to find side D – in other words, 3 sides and an angle so use the Cosine Rule: a2 = b2 + c2 –2bc cos A
Assume side D is ‘a’ and so angle A = 650 b = 8 and c = 5 Substituting: a2 = 82 + 52 – 2 × 8 × 5 × cos 65 a2 = 64 + 25 – 80 × cos 65 a2 = 89 – 33.809 a = √55.19 = 7.4 km
Angle NIH = 650 (supplementary with 1250(180=125+ 65)
Speed = total distance ÷ time Time taken = 1 hour 9 minutes. To convert minutes to hours divide by 60 9 ÷ 60 = 0.15 hours = ( 5 + 8 + 7.42) km ÷ 1.15 hours = 20.42 ÷ 1.15 hours = 17.756 or 17.8 km/hr 17.8
c) Calculate the area of the triangle HIJ Give your answer correct to three significant figures.
.........................km2 (2)
16. Cyril walks 50 metres in a time of 35.6 seconds The distance of 50 metres was measured to the nearest metre. The time of 35.6 seconds was measured to the nearest tenth of a second. a) What is the upper bound for the distance of 50 metres
…………………m (1)
b) What is the lower bound for the time of 35.6 seconds
..........................seconds (1)
8.45
We can’t use the normal formula for the area of a ∆ since we don’t have a right angled ∆ Use: area of ∆ = ½ a b sin C
The two sides a and b are either side of the angle.
C = 65, a = 5 and b = 8
Substitute into formula: Area = ½ × 5 × 8 × sin 65 = 20 × 0.4226 = 8.452
5 5 0 x x 8 5 6 x Sin = .
8.452 to 3 significant figures is 8.45
50.5
Upper and Lower Bounds are related to accuracy. Once you know the degree of accuracy to which a measurement has been rounded, you can then find the Upper and Lower bounds. The Upper Bound is the biggest possible value the measurement can have before it is rounded down. The Lower Bound is the smallest possible value the measurement can have before it was rounded up
We want 50m to the nearest metre. So Accuracy = 1m and Bounds will be ± 0.5m Upper bound = 50 + 0.5 = 50.5m.
Note: Measured value = V Take the level of accuracy A Divide by two: A/2 The Bounds = ± A/2 Lower Bound = V + A/2 Upper Bound = V – A/2
We want 35.6 seconds to the nearest second So Accuracy = 1/10s = 0.1 and Bounds will be ± 0.05s Lower bound = 35.6 – 0.05 = 35.55 s 35.55
18. In the diagram O is the centre of the circle and DCB = 1040
The shape ABOD is symmetrical
a) Work out the value of angle b
…………………0
(2)
b) Explain how you got your answer ………………………………………………………………………………………………... ………………………………………………………………………………………………...
(1) c) Work out the value of angle a
…………………0 (2)
152
ABCD is a cyclic quadrilateral – a four sided shape where each corner touches the circumference of a circle Opposite angles add up to 1800
DAB = 180 – 104 = 760
DOB is an angle at the centre of the circle DAB is an angle at the circumference. They both start at the same sector. Angle at centre = 2 x angle at the circumference.
DOB = 2 x DAB DOB (b) = 2 x 76 = 1520
Since ABOD is symmetrical we can draw a line through the shape Reflex angle DOB = 360 – b = 360 – 152 = 204 The line AO cuts reflex angle DOB in half AOB = 102 The line AO also cuts angle DAB in half OAB = 38 ABO(a) = 180 – 102 – 38 = 40
DAB = 76 0 - cyclic quadrilateral, sum opposite angles=180 B = 2 x DAB – angle at centre = 2x angle at circumf erence
A cone with base diameter of 6cm and height 5cm is dropped into a cylinder of diameter 8 cm part filled with water. a) Calculate the increase in the height of the water in the cylinder. Leave your answer as a fraction in its simplest form.
…………………cm (3)
Volume water pushed out (cylinder) = Volume cone
π r2h = 3
1 π r2h
π × 42 × h = 3
1 π × 3 2 × 5
π × 16 × h = π × 3 × 5 Rearrange h 15 π = 15 cm 16 π 16
b) The density of water is 1 gram per cm3. Calculate the mass of the water displaced by the cone. Leave your answer in terms of π
…………………gm (2)
23. David recorded how many maths questions students downloaded from his website.
He recorded downloads for each quarter from January 2007 to June 2008.
Year Months Number of downloads
2007 Jan – Mar 172
Apr – Jun 446
Jul – Sept 58
Oct – Dec 124
2008 Jan – Mar 195
Apr – Jun 479
a) What was the means number of downloads in 2007?
…………. (1)
b) Calculate the three-point moving averages for this information.
………….. ………….. ……………. …………. (2)
The units given for density give you a clue for it’s formula : Density of water is in grams or mass cm3 volume Rearrange mass = density × volume 1 × 15π gms