Higher Grades 7-9 Mini Test
A 45-minute, 40-mark test using only authored grades 7-9 questions, focused on six topics with enough top-grade material for an exact test.
Questions
Question 1 [3 marks]
Fractions, Decimals and Percentages
Convert 0.58333... (3 recurring) to a fraction in its simplest form.
Question 2 [5 marks]
Linear Algebra
Solve the inequality (x+1)/(x-2) >= 0. State clearly any values excluded from the solution and give the final solution as a single inequality or union of intervals.
Question 3 [1 marks]
Pythagoras and Trigonometry
In triangle ABC, angle A and sides a and b are given (side a is opposite angle A). Which condition guarantees that the ambiguous case (two possible triangles) cannot occur?
Question 4 [3 marks]
Angles and Geometrical Reasoning
O is the centre of a circle. A, B and C are points on the circumference, with C on the major arc AB. Angle AOB = 148 degrees. Diagram: circle, centre O, A and B on the circumference with angle AOB = 148 degrees marked at the centre, C a separate point on the major arc AB, joined to A and B. Calculate the size of angle ACB, giving a reason for your answer.
Question 5 [3 marks]
Indices and Standard Form
The volume of a cube is 64x^12 cm^3. Find an expression, in terms of x, for the length of one edge of the cube. Give your answer in its simplest form.
Question 6 [4 marks]
Pythagoras and Trigonometry
Triangle PQR has PQ = 11 cm, QR = 6 cm and PR = 14 cm.
Calculate the size of angle PQR. Give your answer correct to 1 decimal place.
Hence, without further calculation, state whether triangle PQR is acute-angled, right-angled or obtuse-angled. Give a reason for your answer.
Question 7 [5 marks]
Ratio and Proportion
y is directly proportional to x and inversely proportional to z. When x = 4 and z = 2, y = 10.
Find a formula for y in terms of x and z.
Work out the value of y when x = 9 and z = 3.
Question 8 [4 marks]
Linear Algebra
A machine applies f(x) = kx + 4. Applying the machine twice to the input 1 gives 49, i.e. f(f(1)) = 49. Find all possible values of k.
Question 9 [6 marks]
Ratio and Proportion
The signal strength, S bars, received from a phone mast is inversely proportional to the square of the distance, d km, from the mast. When d = 2, S = 63.
Find a formula for S in terms of d.
Work out the signal strength at a distance of 6 km.
Priya says: 'If you treble the distance from the mast, the signal strength becomes a third of what it was.' Using your answers to parts (a) and (b), or otherwise, explain whether Priya is correct.
Question 10 [6 marks]
Pythagoras and Trigonometry
The diagram shows the triangular cross-section of a roof truss, made from two right-angled triangles PQR and QRS sharing the vertical ridge support QR. The horizontal base PS is divided at R into PR = 12 m and RS = 8 m. The angle between PQ and the base at P is 50 degrees. Diagram: triangular roof cross-section, horizontal base PS with R between P and S so that PR = 12 m and RS = 8 m, vertical ridge support QR drawn from R up to the apex Q, angle at P (between PQ and PR) = 50 degrees.
Calculate the height of the ridge support, QR. Give your answer correct to 3 significant figures.
Calculate the size of the angle between QS and the base at S (angle QSR). Give your answer correct to 1 decimal place.
Model solutions
| Question 1[3 marks] | |
|---|---|
| Answer or working | Marks |
| x = 0.58333..., 100x = 58.333... and 1000x = 583.333... all written (or equivalent valid method) | M1 |
| subtracts correctly to get 900x = 525, dependent on M1 | dM1 |
| 7/12 cao (from 525/900 simplified) | A1 |
| Final answer: 7/12 | |
| Question 2[5 marks] | |
|---|---|
| Answer or working | Marks |
| identify critical points x = -1 (numerator zero) and x = 2 (denominator zero) and state x = 2 is excluded | M1 |
| write the sign intervals: (-infinity, -1), {-1}, (-1,2), (2, infinity) or equivalent interval partition | M1 |
| test signs or use multiplicative signs to determine where (x+1)(x-2) is >= 0 | M1 |
| decide whether x = -1 is included since numerator zero gives value 0, and x = 2 is excluded | M1 |
| final solution: (-infinity, -1] U (2, infinity) cao, with x = 2 excluded | A1 |
| Final answer: x <= -1 or x > 2 (i.e. (-infinity, -1] U (2, infinity)), with x = 2 excluded | |
| Question 3[1 mark] | |
|---|---|
| Answer or working | Marks |
| B | B1oe |
| Final answer: B) angle A is acute and a >= b | |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| 148 / 2 oe, using angle at centre = 2 x angle at circumference | M1 |
| 74 degrees | A1cao |
| reason: the angle at the centre is twice the angle at the circumference, standing on the same arc AB | B1 |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| cube root of 64 = 4 seen | M1 |
| cube root of x^12 = x^4 seen (12 / 3) | M1 |
| 4x^4 oe | A1cao |
| Final answer: 4x^4 cm | |
| Question 6[4 marks] | |
|---|---|
| Answer or working | Marks |
| correct substitution, e.g. cos(PQR) = (11^2+6^2-14^2)/(2*11*6) | M1 |
| cos(PQR) = -39/132 oe (awrt -0.295) | A1 |
| angle PQR = 107.2 degrees (awrt 107.1-107.2) | A1 |
| obtuse-angled (ft their angle from part (a)), because angle PQR (awrt 107 degrees) is greater than 90 degrees, oe (equivalently, because cos(PQR) is negative) | B1 |
| Final answer: Angle PQR = 107.2 degrees (1 d.p.) | Obtuse-angled (angle PQR > 90 degrees) | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| y = kx/z oe seen | M1 |
| 10 = k x 4/2 (= 2k) solved, dependent on the first M1 | dM1 |
| y = 5x/z | A1oe |
| substitutes x = 9, z = 3 into their formula | M1 |
| y = 15 | A1cao |
| Final answer: y = 5x/z | y = 15 | |
| Question 8[4 marks] | |
|---|---|
| Answer or working | Marks |
| substitute: f(1) = k + 4, then f(f(1)) = k(k + 4) + 4 | M1 |
| form quadratic: k^2 + 4k + 4 = 49 -> k^2 + 4k - 45 = 0 | M1 |
| factor or use formula, e.g. (k + 9)(k - 5) = 0 | M1 |
| k = 5 or k = -9 | A1cao |
| Final answer: k = 5 or k = -9. | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| finds k = 63*2^2 (= 63*4) | M1oe |
| S = 252/d^2 | A1oe |
| substitutes d = 6 into their formula | M1 |
| 7 (bars) | A1cao |
| compares S at d = 2 (63) with S at d = 6 (7) and notes that 7 is a ninth of 63, not a third (dep on a correct or ft value from part b) | B1 |
| concludes Priya is incorrect, because S is inversely proportional to d^2 (not d), so trebling d divides S by 9, not 3 | B1oe |
| Final answer: S = 252/d^2 | 7 bars | Priya is incorrect. Trebling d from 2 to 6 changes S from 63 to 7, which is a ninth of the original value, not a third, because S is inversely proportional to d^2, not d. | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| tan(50) = QR / 12 | M1oe |
| QR = 12 x tan(50) | M1 |
| awrt 14.3 m | A1 |
| tan(QSR) = (their QR) / 8 | M1oe |
| QSR = tan^-1((their QR) / 8) | M1 |
| awrt 60.8 degrees | A1 |
| Final answer: 14.3 m (3 s.f.) | 60.8 degrees (1 d.p.) | |