Edexcel June 2018 GCSE Maths Higher Tier Paper 1

Edexcel june 2018 Gcse maths higher tier paper 1

 

Hi and thanks for checking out this playlist from our youtube channel Garry Teaches Maths. If you visit our channel you can find many more maths related videos from GCSE and A level maths. In this playlist we cover all the questions from the Edexcel paper from start to finish. If you want to watch all the videos just press play, and watch from the beginning or if you are interested in a particular video the click on the contents on the top right hand side of the video, and you will see a list of contents from which you can select your chosen video. For even more help, if you are interested in maths tuition, we can help you find a tutor from our excellent list of highly qualified maths tutors.

Contents of our playlist

Question 1 Mixed Numbers and Fractions

This question is in two parts, the first part asks us to add two mixed numbers together. To do this we add the whole number parts and the fractions separately. The whole number part is easy, because we just add the numbers together. When we add the fractions together, we have to recognise that they have different denominators, and that we have to make them the same before we can add them. Once we have converted the fracions so that their denominators are the same, it is simply a case of adding the numerators of the fractions and keeping the denominators the same. We then put the whole number part and the fraction part back together, and the question is complete.

Question 2 Ratios

Here we are given two ratios, one of which is the number of house to the number of flats, and a second one which is the number of flats to bungalows. The number of bungalows is given, and we are asked to find the number of houses. Since we know the ratio of flats to bungalows, we can use it to find the number of flats. We then use the ratio of houses to flats, and the fact that we now know the number of flats to work out the number of houses. For more questions on ratios for practice, check out this list of questions

Question 3 Percentages

This is a question where we are asked to work out the percentage profit that someone makes when they are selling some sweets. We are given the initial cost to the seller when she buys the sweets in bulk, in this case the cost of 5kg being £10. We are then told that she splits up the sweets into smaller bags, and the selling price per bag. We can then work out the amount of money she could make if she sells all the bags, and from this work out the percentage profit that is made by comparing with the initial outlay.

Question 4 Estimations

 

This is a question about finding an estimated answer to a question. Here we are given accurate data about a cycle race across America, which includes the distance of the race, and the average speed that the cyclist will be cycling each day, and for how many hours. We are asked to estimate the total number of days it will take the cyclist to complete the race. To do this, we need to round up the values given, so that the distance is rounded from 3069.25 miles to 3000miles, and the average speed from 15.12 miles per hour to 15 miles per hour. Using the information given, if he cycles for 8 hours a day, we can estimate he will travel 8 x 15 =120 miles per day. We then divide the total distance to travel by the miles travelled per day 3000/120  which gives us 25 days.

In the second part of the question we are told that he trains and his average cycling speed improves to 16.72 miles per hour, and asked how this will affect our answer to the first part of the question. Since the speed increases, we know that he will be covering more distance every day, which will reduce the number of days he will take to complete the race.

Question 5 Front Elevations.

This is a question in which we are given a 3D drawing of a pyramid and asked to draw the front elevation of the pyramid, and also work out the surface area of the pyramid.

To draw the front elevation, we are given a square grid, and it is necessary to view the pyramid directly from in front. To do this we must try and take suitable measurements and transfer them on to the square grid. The first measurement we take is the height of the pyramid, followed by the width of the base of the pyramid. We then transfer these measurements to the grid, join them up, and that gives us the front elevation.

To work out the surface area, we need to imagine the net of the shape so that we can view it in 2D. A series of triangles and a square are created by doing this, from which it is simply a case of working out the area of each separately and adding together to give the total surface area of the shape.

For more questions and practice on this topic click here

Question 6 Coordinates

Here we are given a pattern made up of identical squares which is drawn on a Grid. We are given two coordinates on the pattern, and asked to find the coordinate of a given point C in the pattern.

We can see that the distance covered in the x direction is the difference between the x coordinates of A and B and is 38 – 6 = 32. If we divide this number by 4 we get the side length of the squares = 32/4 = 8. We can now find the x coordinate of C by adding two lots of 8 to the x coordinate of A to give

6 + 2×8 = 22

The y coordinate of C is given by subtracting two lots of 8 from the y coordinate of B to give

36 – (2 x 8) = 20

combining these gives us the coordinate of C:

(22,20)

 

Question 7 Transformations

 

Here  shape T is transformed to a new position S after first undergoing two reflections. We are asked to describe the single transformation that will take us from T to S in one go.

First we have to reflect shape T in the line x = – 1  to give shape R, this is achieved by drawing the line x = -1 which is a vertical line through the point x = -1. We then reflect shape T by taking each point on the shape, drawing a line from it to the line of reflection and beyond to an equal distance on the other side. Shape R is found by joining the new points together. Shape S is found by reflecting shape R in the line y = -2, which we achieve in a similar manner.

Now we have shape T and shape S on the same axes, we can see that the transformation has to be a rotation because of the way the shape is orientated. To describe a rotation fully, we need to give the centre of rotation, the angle of rotation, and the direction of rotation….i.e clockwise or anticlockwise. Tracing paper placed on the axes allows us to achieve this. We place it over the axes, draw round shape T, and then place our pencil at a suitable point to act as a pivot, to rotate the paper so that shape T exactly covers shape S. In reality, this is a process of trial and error, and several attempts may be required before we have found the correct pivot point. This pivot point when correctly found becomes our centre of rotation.

Question 8 Ratios

This Question is about ratios, and we are asked to work out the area of a triangle given its perimeter, the ratio of sides, and the fact that it is a right angled triangle.

The first thing we can do is find the actual lengths of all the sides given the perimeter and the ratio of the sides. To do this we need to add the numbers in the ratio together and equate the number of parts to the perimeter:

3 + 4 + 5 = 12 parts

12 parts = 72cm

We then divide the perimeter by this number pf parts to give the value of one part

1 part = 72/12 = 6

now we multiply each number in the ratio by 6 to give the lengths of the sides

18cm, 24cm, 30cm

Having got these values we know the longest side is the hypotenuse and the other two are the base and the height.

As a result we can use the formula Area = Base x Height

Giving Area = 24 x 18 = 216cm squared (note, it doesn’t matter which way round the base and the height are!)

Question 9 Indices

Here we are given three questions regarding indices. The first one has a fractional indices, and we need to know that when the indices is a fraction, the denominator becomes a root. In this case the denominator is a 2, which means we are finding

36

which equals 6.

 

Part b) asks us to work out the value of

230

, and we need to know that anything to the power zero is equal to one. So the answer in this case is 1.

part c) asks us to find the value of

2723

. Since the denominator is 3, we can first work out the cube root of 27, which equals 3. Next, because the numerator of the fraction is 2, we square the 3, this gives us 9. Finally, we take into account the minus sign, which tells us to take the reciprocal of the answer we have calculated. This gives us

19

Question 10 Box Plots

This is a question in which we are given a table which gives us information about heights of 80 girls. This information comprises of the lowest, greatest, lower and upper quartile, median values. We are asked to represent this information on a box plot, and to do this we draw vertical lines at all the points given, and then draw a box around the values which make up the lower and upper quartiles.

The second part of the question asks us to estimate how many girls have heights between 133cm and 157cm. This is easy to do, because the values that have been given are the lowest recorded, and the upper quartile. Since the upper quartile represents the value below which 75% of all the data will lie, we know that 75% of the girls fall into the required group. Hence we can just take 75% of the total number (80), which gives us 60.

 

 

 

Question 11 Angles in Circles

 

This is a question in which we are asked to find an angle in a shape comprising of a circle and a tangent and extended diameter. The question asks us to find the angle ACB in terms of x, which is the size of angle ABC, and is the angle which is formed between the chord of the circle and a radius.

The first thing to recognise here is that OB is the radius of a circle, and the angle it forms with the tangent is a right angle. We also need to spot that if we draw the line OB, it creates an isoceles triangle AOB, and from that we know that angle BAO is also x. We can then use the rule that an angle at the circumference is half that at the centre to give us that BOC is 2x. Since we also know that OBC is equal to 90 degrees, because it is the angle between a tangent and a radius, we can find ABC because it is in the same triangle as OBC and BOC.

ABC = 180  (BOC + OBC)Since BOC = 2x, and OBC = 90oABC = 180o  (2x + 90o)ABC = 90o  2x

Question 12 Proofs

Here we  have a proofs question in which we are asked to prove that the square of an odd number is always one more than a multiple of 4.

In order to do this we first define an odd number as follows:

If n is an integer then 2n is always an even number, hence an odd number is given by:2n + 1If we square this we get:(2n + 1)2= 4n2+4n + 1We can then factorise this giving:4(n2+n) + 1Since the 4 times the brackets must be a multiple of 4, what we have hereis a multiple of 4 with 1 added to it. 

Question 13 Surds

 

 

 

 

Here we have a  question about surds, in which we are asked to multiply out brackets containing surds, and then simplify the answer.

We first start by simplifying the8 and 18

 

We add these to give 22 + 32 = 52

 

Putting this in place of the bracket gives us:5×52 = 510This tells us that a = 5 and completes the question

 

Question 14 Inverse Proportionality

 

This is a question on proportionality, in which we are given two proportional relationships. The first is an inverse proportionality between y and d squared, the second is a direct proportionality between d and x squared.

Here we are asked to combine the two proportionalities  to form a relationship between y and x.

In order to achieve this we have to form equations which include a constant of proportionality between y and d, and then d and x. The constant of proportionality is then found in both cases by applying the boundary conditions, which are, in the case of y and d, that d = 10, when y = 4. In the case of x and d, we have x = 2 when d = 24. Once we have calculated the constant of proportionality in each case, we have created two equations that link y and d, and x and d. We can now use the first equation to find d in terms of y, and substitute for d in the equation which links d and x. Now that we have done that, we have created the required equation linking y and x, which we then simplify and make y the subject of to give us our required answer.

Question 15 Difference of Two Squares

 

This is a question about using the difference of two squares rule, which is as follows:

x2  y2 = (x + y)(x  y)in this case we can directly replace x and y by a and b to give:(a + b)(a  b)

For the second part of the question, we can let our a from the first part of the question be the first bracket, and our b be the second bracket:

a = (x2+4),     b = (x22)Substituting into the formula gives us:( (x2+4) + (x22))( (x2+4)   (x22))This simplifies to:(2x2 + 2)(6) = 12x2 + 12Which is the simplified answer as required.

 

For more help on this check out our other videos on completing the square at  Garry Teaches Maths

Question 16 Ratio and Probability

 

 

 

 

 

 

Here we have a question which combines ratio and probability, in which we have a bag containing red, blue, and purple counters. We aretold the ratio of red counters to blue counters in the bag, and told the probability that we get a purple out is 0.2. From this we are asked to work out the probability that we pick out a red counter.

The first step is to realise that if we know the probability of getting a purple is 0.2, then the probability of getting either red or blue must be 1 – 0.2 = 0.8

From here we can split the 0.8 into the ratio of 3 : 17 and this will give us the probability of getting a red:

Total number of parts = 3 + 17 = 2020 parts = 0.81 part = 0.8 ÷ 20 = 0.04probability of getting a red = 3 Parts = 3 × 0.04 = 0.12

Question 17 Algebraic Fractions

 

Here we have a question which involves an algebraic fraction, which we are asked to simplify fully. To achieve this it is first necessary to factorise both top and bottom of the fraction. Once we have done this, we can look for common factors on both the top and bottom of the fraction, which we can then cancel to leave us with a simplified fraction:

Factorise the top3x2  8x  3 = (3x + 1)(x  3)Factorise the bottom2x2  6x = 2x(x  3)If we write the fraction in its factorised form we get:(3x + 1)(x  3)2x(x  3)We can see that the (x  3) is on the top and the bottom, and will cancel.This leaves:3x + 12xWhich is the simplest form of the fraction

For more help on quadratics see our other posts here

Question 18 Sine Graphs

 

This question shows us a sine graph

y = sin(xo)  in the range   180  x 180

, and we are told to sketch the graph of

y = sin xo   2

In order to achieve this we need to recognise that this is a simple graph transformation of the form

f(x) is transformed to f(x) 2

. A graph transformation of this type is simply a transformation of – 2 in the y axis, so all that is required is to move the graph shown down by two in the y direction. This is easily achieved by picking several points on the original graph, moving each one of them down by two, and then joining up the resulting points.

Question 19 Gradients of Perpendicular Lines

 

This question is about perpendicular lines, and asks us to calculate the value of b in terms of a, given that (a,b) is a coordinate on the line PQ, and the equation of a line perpendicular to PQ

The first piece of knowledge that we have to use is that the gradients of two line which are perpendicular to each other have the negative reciprocal relationship:

 

m = 1n Where the two lines have gradients m and n respectively 

 

This allows us to calculate the gradient of PQ as follows:

 

Rearrange the equation of the line perpendicular to PQ into the form y = mx + cWhere m is the gradient and c is the y interceptStarting with3x + 2y = 72y = 7  3xy  = 72 3x2  changing the order givesy =  3x2 + 72from which we can see that the gradient of the line perpendicular to PQis  32If we now take the negative reciprocal of this we getGradient of PQ = 23

 

We then use a second relationship to do with gradient of a line between twopoints, this is that if a line is between the points (x1,y1) and (x2, y2),then the gradient between them is:Gradient = y2  y1x2  x1If we apply this to the points on the line PQ and equate it to the gradient of the lineas calculated previously, we get the equation:

 

b  4a  3 = 23   If we multiply by (a  3) and then by 3 we get3(b  4)  = 2(a  3)   which gives3b  12 = 2a  6 we then add 12 to both sides3b = 2a + 6   and then divide both sides by 3 to give:b = 2a + 63Which is our answer putting b in terms of a, as required

Question 20 Inequalities

 

This is a question about inequalities, in which we are asked to find all the possible values of an integer n. To find the acceptable values of n, we are given two inequalities which need to be solved separately. The first of the inequalities is a linear one, and the second is a non linear quadratic inequality.

 

If we turn our attention first to the linear inequality we solve it as follows:

3n + 2  14     Subtract 2 from both sides3n  12     Divide by 3n  4So from the first one we know that n must be an integer less than or equal to 4

 

Now we look at the second, non linear inequality:

6nn2 + 5 > 1        We multiply both sides by (n2 + 5) to give6n > n2 + 5        Now subtract 6n so that the left hand side is zero0 >n2 6n + 5   which looks neater if we change the order giving:n2 6n + 5 < 0

 

Next we need to factorise the quadratic in the inequality

 

(n  5)(n  1) < 0

 

If we were to plot the graph of

y = (n  5)(n 1)

, we would see that it would be a quadratic shape graph which crossed the horizontal axis at the points given by

(n 5)(n  1) = 0These are the points n = 5 and n = 1, If we consider the shape of the graphwe can see that it will go below the horizontal axis between these points. i.e.the range of values which satisfy our inequality are in the range     1 < n < 5, if we consider the integer values of n whichsatisfy both our inequalities, these are:2, 3, and 4

 

Which brings us to the end of our post. If you feel after watching these videos and reading the descriptions that you need more help, then please contact us for help in finding a tutor here. Thanks for watching and reading, we wish you every success and hope you have found our post helpful!

 

 

 

 

 

 

 

 

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