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FREE GCSE MATHS

HIGHER-LEVEL SAMPLE QUESTIONS

PRACTISE FOR THE 2017 GCSE MATHS PAPERS

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Question 1

Question 2

FREE GCSE SAMPLE QUESTIONS: PRACTISE FOR THE 2017 GCSE MATHS PAPERS

Below you will find sample questions and answers replicating some of the major content

areas that appear on the GCSE higher tier Maths exam. For more information about the

GCSEs, visit our GCSE Past Papers page.

GCSE MATHS HIGHER LEVEL:

NON-CALCULATOR QUESTIONS

The following questions are examples of a questions likely to appear on a GCSE Maths

higher level exam. Note, that the use of calculators would NOT be permitted for such

questions.

There are 11 cards face down on a table.

There are y red cards. All the other cards are black.

Lucy picks, at random, two cards from the table.

Find an expression, in terms of y, for the probability that Lucy picks one of each card. (5)

Show that 6 + 5 − √2

6 − √18 can be written as a fraction in its simplest form. (5)

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Question 3

Question 4

In a cafeteria, a meal usually costs £16.

If you have a student card, you can buy three meals for just £38.40.

What is the percentage reduction in cost of buying three meals if you have a student card? (3)

GCSE MATHS HIGHER LEVEL:

CALCULATOR QUESTIONS

The following questions are examples of a questions likely to appear on a GCSE Maths

higher level exam. Note, that the use of calculators is permitted for these questions.

In a Sports College, there are 80 students.

15 students train in both track and field events.

32 students train only in track events.

The total number of students who train in field events is 41.

a) Complete the Venn Diagram for this information. (3)

One student is picked at random to win a scholarship.

b) What is the probability it is a student who trains in neither track nor field? (1)

Track Field



Question 5

Question 6

The number of users on a new website increases as a quadratic sequence.

4 10 18 28

a) Find an expression, in terms of n, for the nth term of this sequence. (3)

The nth term of a different sequence is 2𝑛 − 6.

b) Show that 45 is not a term of this sequence. (2)

Below are the equations of four straight lines.

Line A 𝑦 = −3𝑥 + 2

Line B 𝑦 + 3𝑥 = 0

Line C 𝑦 − 5 = −3𝑥

Line D 𝑦 − 3𝑥 – 4 = 0

Three of the lines are parallel. Write down the equation of the line which is not parallel to the other three.



Answers & Explanations

Question Answer Notes

1. 11𝑦 − 𝑦2

55

P1 for 𝑦

11 or

11 − 𝑦

11 or

𝑦 − 1

10 or

11 − 𝑦

10 or

10 −𝑦

10 (seen on diagram or in a

calculation)

P1 for 𝑦

11×

11 − 𝑦

10 or

11 − 𝑦

11 ×

𝑦

10 for

𝑦

11 ×

𝑦 − 1

10+

11 − 𝑦

11×

10 −𝑦

10

P1 for 𝑦

11×

11 − 𝑦

10 +

11 − 𝑦

11 ×

𝑦

10 for 1 − (

𝑦

11 ×

𝑦 − 1

10+

11 − 𝑦

11×

10 −𝑦

10)

P1 for beginning to process the algebra

A1 11𝑦 − 𝑦2

55 oe

For Lucy to pick one of each colour, she either has to choose a red then a black or a black then a red. You must consider both of these options when calculating the probability. Remember that in probability: AND means × OR means + So, to find probability of a red AND then a black, you must multiply the probabilities. Use the probability formula to correctly write your expressions:

Probability = Number of Successful Outcomes

Total Number of Outcomes

In this question, as there are 11 cards in total, there are 11 total outcomes at the start. Red: There are y red cards, so the probability is

𝒚

𝟏𝟏

Black: You are told the rest of the cards are black. If there are y red cards, then to find the rest, you subtract from the total. 11 − 𝑦 must be the total of black cards. Once, you have picked a red card, however, the total number of cards reduces to 10. So,

the probability of picking the black card second is 𝟏𝟏 − 𝒚

𝟏𝟎.

Now, multiply: 𝑦

11×

11 − 𝑦

10.

Now, consider the second option, which was picking a black card and then a red. Notice, the denominators have switched as there are always 11 cards at the start and then 10 after she has made her first selection.

So, 11 − 𝑦

11 ×

𝑦

10.

Finally, you must add the probabilities to find the total probability of red then black OR black then red.



𝑦

11×

11 − 𝑦

10 +

11 − 𝑦

11 ×

𝑦

10.

Start to simplify each pair of fractions: 11𝑦 − 𝑦2

110+

11𝑦 − 𝑦2

110

You should notice this is the same as 2 (11𝑦 – 𝑦2

110).

Divide numerator and denominator by 2 to get the final answer 𝟏𝟏𝒚 – 𝒚𝟐

𝟓𝟓.

Alternative Method: Another option is to consider what you do not want to happen and then subtract the answer from 1 to find the probability that you do want. If Lucy chooses one of each colour, then the opposite outcome would be choosing two colours that are the same—red AND red OR black AND black. Calculate the probabilities in the same way, remembering the AND and OR rules.

The probability you don't want is 𝑦

11 ×

𝑦 – 1

10+

11 – 𝑦

11×

10 −𝑦

10.

So now, subtract from 1 to find the probability you do want: 1 − (𝑦

11 ×

𝑦 − 1

10+

11 − 𝑦

11×

10 −𝑦

10).

Simplifying, this becomes: 1 − (𝑦2 −𝑦

110 +

110 −21𝑦 + 𝑦2

110)

Add the fractions to get: 1 − (2𝑦2 −22𝑦+110

110)

Divide top and bottom by 2 to get: 1 − (𝑦2 −11𝑦+55

55)

Write 1 as a fraction with the same denominator: 55

55.

Finally, do the subtraction: 55

55 − (

𝑦2 −11𝑦+55

55) =

𝟏𝟏𝒚 – 𝒚𝟐

𝟓𝟓.




2.

𝟒𝟒 + 𝟑√𝟐

𝟔

M1 for intention to multiply numerator and denominator by 6

+ √18 or

6 + 3√2.

M1 for correct expansion of either (5 – √2) (6 + 3√2 ∗)

Or (6 − 3√2 ∗) (6 + 3√2 ∗), at least 3 terms correct or 4 correct terms ignoring signs. * using √18 or 3√2.

M1 for fully correct working leading to 8 +3√2

6.

M1 for writing 6 as 36

6 oe, leading both denominators to match.

A1 for fully correct working leading to 𝟒𝟒 +𝟑√𝟐

𝟔.

First, simplify the fraction by rationalising the denominator, before adding the 6.

√18 can first be simplified to √9 ×2, which is 3√2. If you forget to do this step, it can be done after you have rationalised the denominator. Multiply the top and bottom by the conjugate partner of the denominator to simplify.

5 − √2

6 − 3√2 ×

6 + 3√2

6 + 3√2

As the denominator is a conjugate pair, it can be written as the difference of two squares, using this formula: (a – b) (a + b) = a2 – b2.

So, (6 − 3√2) (6 + 3√2) = 62 − (3√2)2

= 36 − (√18)2 = 36 − 18 = 18.

Expand the numerator* and then simplify to get:

30 + 15√2 − 6√2 − 6 = 𝟐𝟒 + 𝟗√𝟐. Rewriting the whole fraction: 24 +9√2

18

Simplify further by dividing by the highest common factor, which is 3, ensuring you divide each term.

8 + 3√2

6



Now that you have the simplified fraction, it is possible to add the 6 from the question. To add fractions, you need common denominators, so change 6 to have a denominator of 6 before adding. 6

1 =

36

6

So, 36

6 +

8 +3√2

6=

44 +3√2

6.

Check that this fraction cannot be simplified further before submitting your answer, as the question specifies it must be in its most simplified form. As 44, 3 and 6 have no common factors, this fraction is in its simplest form and can be stated as the answer.

𝟒𝟒 + 𝟑√𝟐

𝟔

Solving Tip: If you need help expanding double brackets, use the acronym FOIL. This stands for First, Outer, Inner, Last, and it denotes the order of multiplying in a double bracket. See example: (a + b) (c + d) F: Multiply the first term in each bracket, a with c, to get ac. O: Multiply the outer two terms, a with d, to get ad. I: Multiply the inside two terms, b with c, to get bc. L: Multiply the last term in each bracket, b with d, to get bd. Putting this together gives: (a + b) (c + d)= ac + ad + bc + bd




3.

20%

M1 for 48 – 38.40 (=9.6) M1 for "9.6" ÷ 48 ×100 A1 for 20

First, calculate the cost of buying three meals without a student card: 16 × 3 = 48. With the student card, the three meals only cost £38.40. Calculate how much money you would be saving: 48 − 38.40 = £9.60. You need to convert the actual reduction in price into a percentage reduction in price. The formula for this is: The original price would be £48, i.e. the price before the reduction.

Therefore, 9.60

48 ×100.

As this is a non-calculator question, you need to simplify the fraction as much as you can by dividing by common factors, e.g. 2. One route is: 9.60

48 =

4.80

24 =

1.20

6 =

0.60

3 =

6

30 =

1

5

÷ 2 ÷ 4 ÷ 2 x 10 ÷ 6 Now that it is a simpler fraction, multiply by 100 to give 20%, which is the answer.

Percentage Reduction = Actual reduction

Original price ×100




4(a)

(A3 for fully correct diagram) B1 for 32 and 15 in correct positions M1 for 41 – 15 (= 26) or 80 – 41 – 32 (= 7) A1 correct diagram

(b)

7

80

B1 for 7

80 or ft from Venn diagram for "

7

80"

a) Start to fill in the Venn Diagram in the order the question gives you information.

15 students train in both track and field, which means 15 needs to go in the intersection, as it is the only place which is in both the circle for track and the circle for field.

Next, you are told that there are 32 students who train in track only. Those students go in the part of the track circle that is not in the intersection. There are 41 students altogether who train in field events, which means there needs to be 41 students in that entire circle. The intersection part of the circle already contains 15 students, so the rest of the circle must contain 26 students. (41 − 15 = 26).

Track Field

32 15 26

7

Track Field

15

Track Field

32 15

Track Field

32 15 26



At this stage, it is crucial to check you have displayed all the data. There are 80 students in the College altogether, so add up the numbers in the circles to see if all students are represented.

32 + 15 + 26 = 73 There are some students missing. This means those students must train in something other than track or field. Therefore, they appear in the box but outside of the circles. To work out how many students go outside the circles, subtract the current total from 80. 80 − 73 = 7 Now, finish off the Venn Diagram.

b) This question is asking about those students who train in neither track nor field. Those students are the ones outside the circle, so there are seven of them. If you

pick a student at random out of the whole college, then the probability is 7

80 , as

there are seven students that give the correct outcome out of 80 possible students who could be picked.

So, the correct answer is 𝟕

𝟖𝟎.

Track Field

32 15 26

7




5.(a)

𝑛2 + 3𝑛

M1 for correct deduction from differences Eg 2nd difference of 2 implies 1𝑛2 or 12, 22, 32 M1 for 12, 22, 32 linked with 3, 6, 9 A1 for 𝑛2 + 3𝑛 oe

(b)

Explanation

C1 explanation that 51 is not a power of 2. Or all powers of 2 are even numbers. Even – even = even.

a) A quadratic sequence has a common second difference. Firstly, find the first line of

differences between the terms by subtracting.

4 10 18 28 (+ 6) (+ 8) (+ 10) Next, find the second differences by subtracting the terms in the first difference.

4 10 18 28 (+6) (+8) (+10) (+2) ( +2) There is a common second difference of 2. This means that the coefficient of 𝑛2 in the sequence is 1. This is because you halve the second difference to find the coefficient. So, the sequence begins as 1𝑛2 or 𝑛2. It is then important to match up the part of the sequence formula you have with the actual sequence to see if it fits.

𝑛2

1

4

9

16

Original Sequence

4

10

18

28

Difference

+3

+6

+9

+12



The difference between the two sequences is, in fact, the three times table. This has the formula 3n, as each position in the three times table is 3 x the position number. Position 1: 3 x 1 = 3 Position 2: 3 x 2 = 6 Position 3: 3 x 3 = 9 Position n: 3 x n = 3n Therefore, to match 𝑛2 to our sequence, you simply need to add 3n.

The final formula for the nth term is 𝒏𝟐 + 𝟑𝒏. Solving Tip: Once you have found the nth term for your sequence, it is good practice to always test it out to see if it works. To do this, substitute 1, 2, 3, and 4 in as n to see if you generate the first four terms of the sequence.

n Substitution Value 1 12 + (3 ×1) = 1 + 3 4 2 22 + (3 ×2) = 4 + 6 10 3 32 + (3 ×3) = 9 + 9 18 4 42 + (3 ×4) = 16 + 12 28

You can now be sure that your formula is correct as it gives you the sequence from the question.

b) If 45 were a term in this sequence, then you could write:

2𝑛 − 6 = 45. Start to solve this equation to find n, the position in the sequence where you would get the number 45. 2𝑛 = 51 // added 6 to both sides This means that 2 to some power would give you 51. This is impossible as 51 is not a power of 2. Furthermore, all the powers of 2 are even because each term is double the previous term. Write out the powers of 2 to see this more clearly: 21 = 2 22 = 4 23 = 8 24 = 16 25 = 32 26 = 64 You can see that 51 does not appear in this sequence, and you can therefore conclude that 45 was not a term in the sequence of 2𝑛 − 6.



Solving Tip: If you are struggling to begin solving a proof question, there are a couple of helpful options to get you started:

(i) Try out some numbers in the formula to see if any of them give the value you want or to see if you can eliminate a value because it will not appear in the sequence.

(ii) If you are trying to disprove a situation, start by trying to prove it, for example by trying to solve the equation. You should then get to a point where the equation is impossible to solve, thereby proving that the original statement could not have been true.




6

Line D

M1 for rearranging 3 lines into y = mx + c form. A1 for correct answer.

Lines are said to be parallel if their gradients are equal. This means they have the same slope and that walking up them would be like walking up an identical hill. To find the gradient of a straight line, you must find the coefficient of x when the line is in the form:

y = mx + c.

Line A is already in this form, and therefore you can see that the coefficient of x (the number attached to the x) is −3. This means the gradient of Line A is − 3. Rearrange Line B so that the x term is on the other side of the equal sign: y = −3x // subtract 3x from both sides Rearrange Line C so that y is on its own: y = −3x + 5 // add 5 to both sides Finally, rearrange Line D so that y is on its own:

y = 3x + 4 // subtract 3x and 4 from both sides You can now read off the other three coefficients of x. Line B: The coefficient is −3, which means the gradient is −3. Line C: The coefficient is −3, which means the gradient is −3. Line D: The coefficient is 3, which means the gradient is 3. This means that lines A, B, and C are all parallel. Therefore, the line which is not parallel is Line D.



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