Higher Tier - Year 11

Year 11 Paper 4: Algebra and Trigonometry

Covers linear algebra, quadratics, graphs and coordinates, sequences, Pythagoras' theorem and trigonometry, and statistics and probability.

25 questions - 60 marks - calculator allowed

Year 11 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 11, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.

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Questions

Question 1 [1 marks]

Quadratics

When you fully expand three linear brackets, each containing a term in x, such as (x + a)(x + b)(x + c), what is the highest power of x in the fully simplified answer?

Question 2 [1 marks]

Graphs and Coordinates

Which of these lines is parallel to the line with equation y = 4x - 1?

Question 3 [1 marks]

Pythagoras and Trigonometry

The diagram shows a right-angled triangle with the right angle marked. The sides are labelled p, q and r, where r is the side opposite the right angle. Diagram: right-angled triangle, right angle shown at the bottom-left corner, side p along the base, side q up the vertical side, side r as the slanted side joining the two ends.

Question 4 [3 marks]

Sequences

The nth term of a sequence is 6n - 4.

Work out the first three terms of the sequence.

Work out the 20th term of the sequence.

Question 5 [1 marks]

Statistics and Probability

Using the same table as Questions 7 and 8: Time, t (minutes): 0 < t <= 20, 20 < t <= 40, 40 < t <= 60, 60 < t <= 80, 80 < t <= 100 Frequency (f): 4, 9, 11, 4, 2 Write down the midpoint of the class 20 < t <= 40.

Question 6 [1 marks]

Pythagoras and Trigonometry

Write down the exact value of sin 30 degrees.

Question 7 [1 marks]

Statistics and Probability

A school recorded the time, t minutes, spent on homework by 30 pupils in one evening. The table shows the results. Time, t (minutes): 0 < t <= 20, 20 < t <= 40, 40 < t <= 60, 60 < t <= 80, 80 < t <= 100 Frequency (f): 4, 9, 11, 4, 2 Total number of pupils = 30 Write down the modal class.

Question 8 [1 marks]

Graphs and Coordinates

The graphs of two straight lines intersect at the point (-2, 5). Which of the following is the solution to the pair of simultaneous equations represented by the lines?

Question 9 [2 marks]

Statistics and Probability

The box plot shows the scores, out of 40, of 32 students in a spelling test.

Write down the median score.

Work out the range of scores.

Question 10 [2 marks]

Linear Algebra

Solve 2x + 5 = 17 Show your working.

Question 11 [2 marks]

Pythagoras and Trigonometry

A cuboid measures 3 m by 4 m by 12 m. Work out the length of the space diagonal from one corner to the opposite corner.

Question 12 [2 marks]

Quadratics

Factorise x^2 - 11x + 28

Question 13 [3 marks]

Linear Algebra

The circumference of a circle of radius r is given by the formula C = 2 * pi * r

Rearrange the formula to make r the subject.

A circular pond has a circumference of 18.84 metres. Work out the radius of the pond. Give your answer correct to 1 decimal place.

Question 14 [2 marks]

Graphs and Coordinates

The point M(-3, k) lies directly above the point N(-3, -2). The distance MN is 9 units. Work out the value of k.

Question 15 [3 marks]

Linear Algebra

Solve 6/(x + 2) = 3

Question 16 [3 marks]

Graphs and Coordinates

A straight line has gradient 5 and passes through the points (3, k) and (5, 16). Find the value of k.

Question 17 [4 marks]

Statistics and Probability

Two fair, six-sided dice are rolled and the scores are added together.

Show that there are 6 ways of getting a total of 7.

Find the probability that the total is at least 10.

Question 18 [3 marks]

Linear Algebra

Solve 3(2y - 1) = 4y + 5.

Question 19 [4 marks]

Statistics and Probability

A cafe is choosing between two vending machines. Each machine was tested with 30 customers, and the waiting time for a drink, w seconds, was recorded for each customer. Calculate an estimate of the mean waiting time for each machine, and use your answers to advise the cafe which machine has the shorter typical waiting time. You must show your working.

Question 20 [3 marks]

Linear Algebra

Show that (x + 4)(2x - 1) = 2x^2 + 7x - 4. You must show your steps.

Question 21 [1 marks]

Graphs and Coordinates

The graph of y = g(x) is transformed to give the graph of y = g(x - 5). Which of the following correctly describes this transformation?

Question 22 [4 marks]

Linear Algebra

Prove algebraically that the sum of any four consecutive integers is always even, but is never a multiple of 4.

Question 23 [3 marks]

Statistics and Probability

A bag contains 8 counters, which are either green or red. Two counters are taken at random from the bag, one after another, without replacement. The probability that both counters are green is 3/14.

Show that there are 4 green counters in the bag.

Question 24 [4 marks]

Linear Algebra

The total surface area of a closed cylinder is given by the formula A = 2 x pi x r^2 + 2 x pi x r x h, where r is the radius and h is the height. A cylinder has radius r = 3 cm and total surface area A = 66 pi cm^2. Work out the value of h.

Question 25 [5 marks]

Linear Algebra

Stretch question. Prove that n^5 - n is divisible by 30 for any integer n.

Model solutions

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or working	Marks
C selected	B1cao
Final answer: C (x^3)

Mark scheme for Question 2 [1 mark]
Question 2[1 mark]
Answer or working	Marks
A	B1cao
Final answer: A) y = 4x + 5

Mark scheme for Question 3 [1 mark]
Question 3[1 mark]
Answer or working	Marks
r stated as the hypotenuse	B1cao
Final answer: C) r

Mark scheme for Question 4 [3 marks]
Question 4[3 marks]
Answer or working	Marks
at least one correct substitution, e.g. n=1 gives 2	M1
2, 8, 14 all correct	A1cao
116	B1cao
Final answer: 2, 8, 14 \| 116

Mark scheme for Question 5 [1 mark]
Question 5[1 mark]
Answer or working	Marks
30	B1cao

Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or working	Marks
1/2	B1cao

Mark scheme for Question 7 [1 mark]
Question 7[1 mark]
Answer or working	Marks
40 < t <= 60	B1cao

Mark scheme for Question 8 [1 mark]
Question 8[1 mark]
Answer or working	Marks
B	B1
Final answer: B) x = -2, y = 5

Mark scheme for Question 9 [2 marks]
Question 9[2 marks]
Answer or working	Marks
28	B1cao
30	B1cao
Final answer: 28 \| 30

Mark scheme for Question 10 [2 marks]
Question 10[2 marks]
Answer or working	Marks
2x = 12	M1oe
x = 6	A1cao

Mark scheme for Question 11 [2 marks]
Question 11[2 marks]
Answer or working	Marks
use 3D Pythagoras: sqrt(3^2 + 4^2 + 12^2) or equivalent method	M1
13 m	A1cao

Mark scheme for Question 12 [2 marks]
Question 12[2 marks]
Answer or working	Marks
correct factor pair of 28 that sums to -11 identified (-4 and -7)	M1
(x - 4)(x - 7)	A1cao

Mark scheme for Question 13 [3 marks]
Question 13[3 marks]
Answer or working	Marks
r = C/(2 * pi)	B1oe
substitutes C = 18.84 into r = C/(2 * pi), ft from part a	M1
awrt 3.0 (m)	A1
Final answer: r = C/(2 * pi) \| r = 3.0 m (1 dp)

Mark scheme for Question 14 [2 marks]
Question 14[2 marks]
Answer or working	Marks
set up k - (-2) = 9 or equivalent equation for k, shown	M1
7	A1cao

Mark scheme for Question 15 [3 marks]
Question 15[3 marks]
Answer or working	Marks
multiplies both sides by (x + 2): 6 = 3(x + 2)	M1
expands and rearranges: 3x = 0	M1
x = 0	A1cao

Mark scheme for Question 16 [3 marks]
Question 16[3 marks]
Answer or working	Marks
(16 - k) / (5 - 3) = 5	M1oe
16 - k = 10	M1dep
k = 6	A1cao

Mark scheme for Question 17 [4 marks]
Question 17[4 marks]
Answer or working	Marks
at least 4 correct pairs listed, e.g. (1,6), (2,5), (3,4), (4,3)	M1
all six pairs listed with no repeats: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)	A1cso
identifies 6 outcomes: (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)	M1
1/6	A1cao
Final answer: 6 ways (shown) \| 1/6

Mark scheme for Question 18 [3 marks]
Question 18[3 marks]
Answer or working	Marks
expand and collect terms, 6y - 3 = 4y + 5 or equivalent	M1
isolate y, 6y - 4y = 5 + 3	M1
y = 4	A1cao
Final answer: 4

Mark scheme for Question 19 [4 marks]
Question 19[4 marks]
Answer or working	Marks
estimate for Machine A: sum(f x midpoint) = 410, mean = 410 / 30 = 13.67 awrt	M1oe
estimate for Machine B: sum(f x midpoint) = 490, mean = 490 / 30 = 16.33 awrt	M1oe
both means correct, awrt 13.67 and awrt 16.33	A1
correct conclusion, Machine A, with valid reason based on the two calculated means (ft their means)	B1
Final answer: Machine A: 13.67 seconds; Machine B: 16.33 seconds; recommend Machine A, as it has the shorter mean waiting time.

Mark scheme for Question 20 [3 marks]
Question 20[3 marks]
Answer or working	Marks
expand brackets correctly: x2x, x(-1), 42x, 4(-1) shown	C1
combine the four terms to form 2x^2 + 7x - 4	C1
final statement equal to 2x^2 + 7x - 4 with clear workings	C1cso
Final answer: 2x^2 + 7x - 4

Mark scheme for Question 21 [1 mark]
Question 21[1 mark]
Answer or working	Marks
A	B1

Mark scheme for Question 22 [4 marks]
Question 22[4 marks]
Answer or working	Marks
Four consecutive integers n, n+1, n+2, n+3 summed and simplified to 4n + 6	M1
Factorised to 2(2n + 3), showing the sum is even (has factor 2)	A1
2n + 3 identified as odd (2n is even, plus 3 is odd)	M1
Conclusion: 2(2n+3) is even but, since 2n+3 is odd, it cannot be a multiple of 4	A1cso
Final answer: n + (n+1) + (n+2) + (n+3) = 4n + 6 = 2(2n+3). This is even, but since 2n+3 is odd, the sum is never a multiple of 4.

Mark scheme for Question 23 [3 marks]
Question 23[3 marks]
Answer or working	Marks
g/8 x (g-1)/7 = 3/14 formed, where g is the number of green counters	M1
rearranges to g(g-1) = 12 (oe g^2 - g - 12 = 0)	M1
solves to g = 4, rejecting g = -3	A1cso
Final answer: g = 4

Mark scheme for Question 24 [4 marks]
Question 24[4 marks]
Answer or working	Marks
2 x pi x 3^2 (= 18 pi) seen	M1oe
18 pi + 2 x pi x 3 x h = 66 pi oe, correct equation formed	M1
18 + 6h = 66 seen (equation divided through by pi), or 6h = 48 seen	dM1
h = 8 (cm)	A1cao
Final answer: h = 8 cm

Mark scheme for Question 25 [5 marks]
Question 25[5 marks]
Answer or working	Marks
factorise n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = n(n - 1)(n + 1)(n^2 + 1)	M1
state among n - 1, n, n + 1 one is even, so there is a factor 2	M1
state among n - 1, n, n + 1 one is multiple of 3, so there is a factor 3	M1
show divisibility by 5 by checking n mod 5 = 0, 1, 2, 3, 4 (or use Fermat): in each case n^5 - n is a multiple of 5	M1
conclude n^5 - n is divisible by 2, 3 and 5 and hence by 30	A1cao
Final answer: n^5 - n = n(n - 1)(n + 1)(n^2 + 1) is divisible by 2 and 3 from the three consecutive factors, and by 5 by case or Fermat argument; therefore divisible by 30.