8 higher-order questions on the Class 10 syllabus, set out like a 25-mark, 1-hour paper. Try it in one sitting, then open each worked solution to mark your work.
25 marks
1 hour
8 questions
No calculator
Our own practice paper, not a CBSE paper. CBSE has not yet published a sample paper, blueprint or Class 10 syllabus for Mathematics Advanced, so we wrote these questions ourselves on the current Class 10 chapters, at the higher-order (HOTS) level CBSE says the paper will have. What CBSE has said about the paper.
Instructions
All questions are compulsory. Marks are shown with each question.
Calculators are not allowed. Leave answers in surd form or in terms of π unless a question says otherwise.
Show your working: in the solutions, marks in brackets show where the method marks go.
Time: 1 hour. Try the whole paper before you open any solution.
Question 1
Real Numbers2 marks
The HCF of two positive integers is 24 and their product is 6912. Find every possible pair of such integers.
Worked solution
Answer: 24 and 288, or 72 and 96
Write the numbers as 24m and 24n, where m and n have no common factor (otherwise the HCF would be bigger than 24).
24m × 24n = 6912, so 576mn = 6912 and mn = 12. [1]
Factor pairs of 12 with m < n: (1, 12), (2, 6), (3, 4). The pair (2, 6) shares the factor 2, so it is rejected.
So the numbers are 24 and 288, or 72 and 96. [1]
The step most often missed: rejecting (2, 6). It gives 48 and 144, whose HCF is 48, not 24.
Question 2
Polynomials2 marks
α and β are the zeros of the polynomial 2x² − 6x + 3. Without finding α and β, find a quadratic polynomial with integer coefficients whose zeros are α² and β².
Worked solution
Answer: 4x² − 24x + 9 (or any non-zero multiple)
From the coefficients: α + β = 6/2 = 3 and αβ = 3/2. [½]
Sum of the new zeros: α² + β² = (α + β)² − 2αβ = 9 − 3 = 6. [½]
Product of the new zeros: α²β² = (αβ)² = 9/4. [½]
A polynomial is x² − 6x + 9/4; multiplying by 4 gives 4x² − 24x + 9. [½]
Question 3
Probability2 marks
Two fair dice, each numbered 1 to 6, are thrown together. Find the probability that the product of the two numbers is a perfect square.
Worked solution
Answer: 2/9
There are 6 × 6 = 36 equally likely outcomes.
Products that are perfect squares (1, 4, 9, 16, 25 or 36): (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 4) and (4, 1). That is 8 outcomes. [1]
Probability = 8/36 = 2/9. [1]
Check the non-doubles carefully: 2 × 8 or 3 × 12 would be squares, but no die shows 8 or 12, so (1, 4) and (4, 1) are the only ones.
Question 4
Coordinate Geometry3 marks
A(3, 0), B(6, 4) and C(−1, 3) are three points.
Show that triangle ABC is right-angled and isosceles.
Find the point D such that ABDC is a square, and find the area of the square.
Worked solution
Answer: (a) AB = AC = 5, BC = 5√2, AB² + AC² = BC²; (b) D(2, 7), area 25 square units
AB = AC = 5, so the triangle is isosceles, and AB² + AC² = 50 = BC², so by the converse of Pythagoras the angle at A is 90°. [1]
(b) In square ABDC, BC is a diagonal, so the diagonals AD and BC bisect each other. The midpoint of BC is (5/2, 7/2), so D = (2 × 5/2 − 3, 2 × 7/2 − 0) = (2, 7). Area = AB² = 25 square units. [1]
Check: BD² = 16 + 9 = 25 and CD² = 9 + 16 = 25, so all four sides are 5.
Question 5
Arithmetic Progressions3 marks
The sum of the first n terms of a sequence is Sn = 3n² + 5n for every natural number n.
Show that the sequence is an AP and find its nth term.
Which term of the AP is 152?
Is 247 a term of this AP? Give a reason.
Worked solution
Answer: (a) aₙ = 6n + 2, d = 6; (b) 25th term; (c) no, n would be 245/6
(a) an = Sn − Sn−1 = (3n² + 5n) − (3(n − 1)² + 5(n − 1)) = 6n + 2 for n ≥ 2, and a1 = S1 = 8 also fits. So an = 6n + 2. Consecutive terms differ by 6 each time, so it is an AP with d = 6. [1]
(b) 6n + 2 = 152 gives n = 25: the 25th term. [1]
(c) 6n + 2 = 247 gives n = 245/6, which is not a natural number, so 247 is not a term. [1]
Do not stop at n ≥ 2: check that the formula also gives a1 = S1.
Question 6
Introduction to Trigonometry3 marks
θ is an acute angle and sec θ + tan θ = p. Prove that sin θ = (p² − 1)/(p² + 1).
Worked solution
Answer: Proof (sec θ − tan θ = 1/p)
Since sec²θ − tan²θ = 1, (sec θ + tan θ)(sec θ − tan θ) = 1, so sec θ − tan θ = 1/p. [1]
Adding and subtracting: sec θ = (p + 1/p)/2 = (p² + 1)/(2p) and tan θ = (p − 1/p)/2 = (p² − 1)/(2p). [1]
The key idea is the factorisation of sec²θ − tan²θ = 1. Squaring sec θ + tan θ = p straight away also works but takes much longer.
Question 7
Some Applications of Trigonometry5 marks
A man standing on top of a vertical cliff 50√3 m high sees a boat sailing straight towards the foot of the cliff at a steady speed. The angle of depression of the boat changes from 30° to 60° in 10 minutes.
Find the distance of the boat from the foot of the cliff at each sighting.
Find the speed of the boat in metres per minute.
How much longer will the boat take to reach the foot of the cliff?
Show that the answer to part (c) would be the same for a cliff of any height.
Worked solution
Answer: (a) 150 m and 50 m; (b) 10 m/min; (c) 5 minutes; (d) remaining distance is always half the distance covered
Let the cliff be AB with A at the top, and let the boat be at C and then at D on the ground.
(a) The angle of depression equals the angle of elevation from the boat (alternate angles). At 30°: tan 30° = AB/BC, so BC = 50√3 ÷ (1/√3) = 50 × 3 = 150 m. At 60°: BD = 50√3 ÷ √3 = 50 m. [2]
(b) The boat covers 150 − 50 = 100 m in 10 minutes: speed = 10 m per minute. [1]
(c) It still has 50 m to go: 50 ÷ 10 = 5 minutes. [1]
(d) For height h: BC = h√3 and BD = h/√3. Distance covered = h√3 − h/√3 = 2h/√3, distance left = h/√3, which is half the distance covered. At a steady speed it takes half the time, 5 minutes, whatever h is. [1]
Question 8
Surface Areas and Volumes5 marks
A sphere of radius r fits exactly inside a closed right circular cylinder: it touches the top, the bottom and the curved surface. A cone has the same base and the same height as the cylinder.
Show that the curved surface area of the cylinder equals the surface area of the sphere.
Show that the volumes of the cylinder, the sphere and the cone are in the ratio 3 : 2 : 1.
If r = 3 cm, find the volume of the space inside the cylinder but outside the sphere. Leave your answer in terms of π.