Higher Tier - Grades 6-7

Higher Secure B

A second secure Higher paper for mixed revision and timed practice.

27 questions - 80 marks - calculator allowed

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Questions

Question 1 [1 marks]

Indices and Standard Form

Work out the value of 3^4.

Question 2 [2 marks]

Angles and Geometrical Reasoning

Draw a straight line segment AB of length 7 cm. Using a ruler and compasses only, construct the perpendicular bisector of AB. You must show all your construction lines.

Question 3 [2 marks]

Ratio and Proportion

P is directly proportional to Q. When Q = 5, P = 30. Find the value of P when Q = 9.

Question 4 [4 marks]

Area, Volume and Measures

The exchange rate is £1 sterling (GBP) = 1.16 euros (EUR).

Nadia changes £250 into euros. Work out how many euros she receives.

On the way home Nadia changes 87 euros back into £, using the same exchange rate. Work out how many £ she receives.

Question 5 [1 marks]

Graphs and Coordinates

The graph of y = g(x) is obtained from y = f(x) by first reflecting in the y-axis and then translating up by 6. Write g(x) in terms of f.

Question 6 [1 marks]

Statistics and Probability

A bag contains 3 blue and 9 red counters. One counter is chosen at random. Write down the probability it is red.

Question 7 [2 marks]

Graphs and Coordinates

Let f(x) = x^2 and g(x) = x + 1. Find (g o f)(3).

Question 8 [2 marks]

Quadratics

Factorise x^2 - 13x + 40

Question 9 [2 marks]

Number and Calculation

A plumber charges a call-out fee of £45 plus £30 per hour worked. Show that the total cost of a job lasting 2.5 hours is £120.

Question 10 [2 marks]

Statistics and Probability

A scatter graph shows the number of gym sessions attended per month (x) and the kilograms of weight lost that month (y), for data collected between 1 and 12 sessions per month. Aisha uses the line of best fit to estimate the weight lost by someone who attends 40 sessions in a month. Explain why this estimate would not be reliable.

Question 11 [2 marks]

Linear Algebra

Solve 2x + 5 = -3

Question 12 [2 marks]

Quadratics

Solve x^2 + 9x = 0

Question 13 [2 marks]

Linear Algebra

Anjali has p pounds. She has more than £15 but no more than £40. Write down an inequality, in terms of p, to show this information.

Question 14 [2 marks]

Area, Volume and Measures

Convert 0.003 km^2 into square metres (m^2).

Question 15 [3 marks]

Sequences

Here are the first four terms of a sequence. 1, 1.5, 2, 2.5

Write down the next term of the sequence.

Find an expression, in terms of n, for the nth term of the sequence.

Question 16 [3 marks]

Pythagoras and Trigonometry

In triangle LMN, angle L = 115 degrees, MN = 18 cm (opposite angle L), and LN = 9 cm (opposite angle M). Calculate the size of angle N. Give your answer correct to 1 decimal place.

Question 17 [3 marks]

Angles and Geometrical Reasoning

Enlarge triangle T with vertices (2, 1), (4, 1) and (2, 3) by a scale factor of -2 centre (1, 1). Find the coordinates of the image vertices.

Question 18 [5 marks]

Statistics and Probability

In a class of 30 students, 14 study French, 16 study Spanish and 5 study both French and Spanish. Every student studies French, Spanish, both, or neither.

Work out the number of students who study only French.

Work out the number of students who study neither French nor Spanish.

A student is picked at random from the class. Find the probability that the student studies both French and Spanish. Give your answer as a fraction in its simplest form.

Question 19 [5 marks]

Ratio and Proportion

A metal rod is stretched so that its length increases in direct proportion to the temperature increase. At 20 degrees C the rod is 1.25 m long. At 80 degrees C the rod is 1.253 m long. Assuming linear proportional change, find an expression for the length L (in metres) as a function of temperature t in degrees C, of the form L = m t + c. Give m and c to 6 decimal places.

Question 20 [6 marks]

Linear Algebra

A rectangle has length (2x + 5) cm and width (x + 3) cm. The perimeter of the rectangle is 46 cm.

Show that 6x + 16 = 46

Solve the equation to find the value of x.

Work out the length and the width of the rectangle.

Question 21 [7 marks]

Fractions, Decimals and Percentages

Deepa and her brother each invest £2000 for 5 years, in different types of account.

Account A pays compound interest at a rate of 2.5% per year. Deepa invests £2000 in Account A. Calculate the total amount in Account A after 5 years. Give your answer to the nearest penny.

Account B pays simple interest at a rate of 3% per year. Deepa's brother invests £2000 in Account B. Calculate the total amount in Account B after 5 years.

State which account gives the greater return after 5 years, and work out by how much.

Question 22 [1 marks]

Graphs and Coordinates

The graph of y = f(x) is transformed to give the graph of y = f(x) + 4. Which of the following correctly describes this transformation?

Question 23 [1 marks]

Quadratics

The equations y = x^2 - 1 and y = 2x + 2 are solved simultaneously. Which one of these points is a solution to both equations?

Question 24 [5 marks]

Ratio and Proportion

A metal alloy is made by mixing 300 g of metal A, which has a density of 8.4 g/cm^3, with 500 g of metal B, which has a density of 7.2 g/cm^3. Assuming there is no change in total volume when the metals are mixed, calculate the density of the alloy. Give your answer to 3 significant figures.

Question 25 [5 marks]

Graphs and Coordinates

Consider the family of cubics y = x^3 + kx^2 where k is a real constant. (a) Show that x = 0 is always a root. (b) For k = -3 find the other two roots and classify the nature of the turning points of the cubic (local max, local min or point of inflection).

For k = -3 find the other two roots. Then, by working out y at x = -1, 0, 1, 2, 3 and 4 and comparing neighbouring values, say whether the graph has a local maximum or a local minimum at x = 0 and at x = 2.

Question 26 [5 marks]

Number and Calculation

A padlock code has 4 digits. Each digit can be any number from 0 to 9.

The digits may be repeated. Work out the number of different codes possible.

The first digit of the code cannot be 0, although the other three digits may still be any digit from 0 to 9 and may repeat. Work out the number of different codes now possible.

Question 27 [4 marks]

Graphs and Coordinates

A curve y = r(x) passes through the point (8, 3).

After the transformation y = r(x) + c, the image of this point is (8, -5). Find the value of c.

The curve y = r(kx) has an image point (2, 3) that corresponds to (8, 3) on y = r(x). Find the value of k.

Model solutions

Mark scheme for Question 1 [1 mark]
Question 1[1 mark]
Answer or workingMarks
81B1cao
Mark scheme for Question 2 [2 marks]
Question 2[2 marks]
Answer or workingMarks
two pairs of intersecting arcs of equal radius (radius greater than half of AB), one pair centred on A and one pair centred on BC1
correct straight line drawn through both points of intersection, extending beyond AB, within 2 mm and 2 degrees of the true perpendicular bisectorA1
Final answer: A straight line perpendicular to AB, passing through its midpoint, 3.5 cm from both A and B.
Mark scheme for Question 3 [2 marks]
Question 3[2 marks]
Answer or workingMarks
finds the constant of proportionality k = 30/5 (=6), or uses a valid scale factor methodM1
P = 54A1cao
Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or workingMarks
250 x 1.16M1
290 (euros)A1cao
87 / 1.16M1
75 (pounds)A1cao
Final answer: 290 euros; £75
Mark scheme for Question 5 [1 mark]
Question 5[1 mark]
Answer or workingMarks
g(x) = f(-x) + 6B1cao
Mark scheme for Question 6 [1 mark]
Question 6[1 mark]
Answer or workingMarks
9/12 or 3/4B1cao
Final answer: 3/4
Mark scheme for Question 7 [2 marks]
Question 7[2 marks]
Answer or workingMarks
compute f(3)=9 then g(9)=9+1 or equivalentM1
10A1cao
Mark scheme for Question 8 [2 marks]
Question 8[2 marks]
Answer or workingMarks
attempts a factor pair of 40 that sums to -13, e.g. -8 and -5M1
(x - 8)(x - 5)A1cao
Mark scheme for Question 9 [2 marks]
Question 9[2 marks]
Answer or workingMarks
30 x 2.5 = 75 seenM1oe
45 + 75 = 120 shown correctlyA1cso
Final answer: £120 (shown). Working check: 30 x 2.5 = 75; 45 + 75 = 120.
Mark scheme for Question 10 [2 marks]
Question 10[2 marks]
Answer or workingMarks
40 sessions is outside the range of data collected (1 to 12), so this is extrapolationB1
the same trend/rate of weight loss may not continue that far beyond the data, e.g. there could be a natural limit to how much weight can be lost, or attending 40 sessions a month may not be realisticB1oe
Final answer: The estimate is unreliable because 40 sessions is far outside the data collected (1 to 12 sessions), so it involves extrapolation and the same trend may not continue.
Mark scheme for Question 11 [2 marks]
Question 11[2 marks]
Answer or workingMarks
2x = -8M1oe
x = -4A1cao
Mark scheme for Question 12 [2 marks]
Question 12[2 marks]
Answer or workingMarks
x(x + 9) = 0M1oe
x = 0 and x = -9A1cao
Final answer: x = 0 or x = -9
Mark scheme for Question 13 [2 marks]
Question 13[2 marks]
Answer or workingMarks
15 < p oe or p <= 40 oe, one boundary correctly representedM1
15 < p <= 40 oe cao, both boundaries correct with correct strict/non-strict signsA1
Final answer: 15 < p <= 40
Mark scheme for Question 14 [2 marks]
Question 14[2 marks]
Answer or workingMarks
use 1 km^2 = 1 000 000 m^2, multiply 0.003 by 1 000 000M1oe
3000 m^2A1cao
Mark scheme for Question 15 [3 marks]
Question 15[3 marks]
Answer or workingMarks
3B1cao
common difference of 0.5 identified, e.g. 0.5n + ...M1oe
0.5n + 0.5 oe cao, e.g. (n + 1)/2A1
Final answer: 3 | 0.5n + 0.5 (oe)
Mark scheme for Question 16 [3 marks]
Question 16[3 marks]
Answer or workingMarks
sin(M)/9 = sin(115)/18M1oe
angle M = awrt 26.9 (degrees), dep on correct sine rule setupM1
angle N = 180 - 115 - their M = awrt 38.1 (degrees)A1ft
Final answer: angle N = 38.1 degrees (1 d.p.)
Mark scheme for Question 17 [3 marks]
Question 17[3 marks]
Answer or workingMarks
subtract centre (1,1) then multiply by -2 for each coordinate (method shown)M1oe
image of (2,1) is (-1,1)A1cao
images of other vertices: (4,1) -> (-5,1) and (2,3) -> (-1,-3)A1cao
Final answer: Images: (2,1) -> (-1,1); (4,1) -> (-5,1); (2,3) -> (-1,-3).
Mark scheme for Question 18 [5 marks]
Question 18[5 marks]
Answer or workingMarks
9B1cao
30 - (9 + 11 + 5)M1oe
5A1cao
5/30M1oe
1/6A1cao
Final answer: 9 | 5 | 1/6
Mark scheme for Question 19 [5 marks]
Question 19[5 marks]
Answer or workingMarks
use two points to find gradient m = (1.253 - 1.25) / (80 - 20)M1
m = 0.000050 cao to 6 dpA1
substitute one point to find c using L = m t + cM1
c = 1.249000 cao to 6 dpA1
final correct expression L = 0.000050 t + 1.249000A1cao
Final answer: L = 0.000050 t + 1.249000. Working: m = (1.253 - 1.25)/(80 - 20) = 0.003/60 = 0.00005 = 0.000050 to 6 dp. Then c = 1.25 - 0.00005*20 = 1.25 - 0.001 = 1.249000. Check: at t = 80, L = 0.00005*80 + 1.249 = 0.004 + 1.249 = 1.253.
Mark scheme for Question 20 [6 marks]
Question 20[6 marks]
Answer or workingMarks
2[(2x + 5) + (x + 3)] = 46 oe (correct perimeter expression set equal to 46)M1
correctly expands and simplifies to 6x + 16 = 46A1dep
6x = 30M1oe
x = 5A1cao
substitutes x = 5 into both expressions (ft their x)M1
length = 15 cm and width = 8 cm (both requiredA1cao
Final answer: 6x + 16 = 46 (shown) | x = 5 | length = 15 cm, width = 8 cm
Mark scheme for Question 21 [7 marks]
Question 21[7 marks]
Answer or workingMarks
1.025^5 (= 1.131408212890625)M1
2000 x their 1.131408212890625M1
2262.82 (pounds)A1cao
2000 x 0.03 x 5 (= 300)M1
2300 (pounds)A1cao
Account B ft from (a) and (b)B1
37.18 (pounds) ft, correct differenceB1
Final answer: £2262.82 | £2300 | Account B, by £37.18
Mark scheme for Question 22 [1 mark]
Question 22[1 mark]
Answer or workingMarks
BB1
Mark scheme for Question 23 [1 mark]
Question 23[1 mark]
Answer or workingMarks
B) (3, 8) selectedB1
Final answer: B) (3, 8)
Mark scheme for Question 24 [5 marks]
Question 24[5 marks]
Answer or workingMarks
volume of A = 300 / 8.4 oe, awrt 35.7M1
volume of B = 500 / 7.2 oe, awrt 69.4M1
total volume = their vol A + their vol B (ft), awrt 105M1
density of alloy = 800 / their total volumeM1ft
7.61 (g/cm^3)A1awrt
Final answer: 7.61 g/cm^3 (3 s.f.)
Mark scheme for Question 25 [5 marks]
Question 25[5 marks]
Answer or workingMarks
substitute x = 0 to give y = 0 thereby showing x=0 is a rootB1
factor y = x^3 - 3x^2 as x^2(x - 3) or equivalent to find rootsM1
identify roots x = 0 (double), x = 3 or explicitly x = 0,0,3M1
work out values e.g. y(-1) = -4, y(0) = 0, y(1) = -2, y(2) = -4, y(3) = 0 and compare neighbouring values around x = 0 and x = 2M1
x = 0 is a local maximum (0 is higher than the neighbouring values -4 and -2) and x = 2 is a local minimum (-4 is lower than the neighbouring values -2 and 0)A1cao
Final answer: Substitute x = 0 gives y = 0, so x = 0 is a root. | Other roots: x = 0 (double root) and x = 3. Using y(-1) = -4, y(0) = 0, y(1) = -2, y(2) = -4, y(3) = 0: x = 0 is a local maximum (0 is higher than the values on either side, -4 and -2) and x = 2 is a local minimum (-4 is lower than the values on either side, -2 and 0).
Mark scheme for Question 26 [5 marks]
Question 26[5 marks]
Answer or workingMarks
10 * 10 * 10 * 10 oe seenM1
10000A1cao
identifies 9 choices for the first digitM1
(dep) 9 * 10 * 10 * 10M1oe
9000A1cao
Final answer: 10000 | 9000
Mark scheme for Question 27 [4 marks]
Question 27[4 marks]
Answer or workingMarks
3 + c = -5M1oe
c = -8A1cao
8 / k = 2M1oe
k = 4A1cao
Final answer: c = -8; k = 4