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CE.M35 Maths: Scholarship Challenge Paper

ISEB COMMON ENTRANCE CE.M35 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Time allowed: 60 minutes. Answer ALL 18 questions. Write your answers in the spaces provided. You ARE allowed to use a calculator. You must show all your working; marks are available for a correct method as well as for the final answer, and method marks are awarded generously throughout. The number of marks for each question, or part-question, is shown in brackets. This is a Scholarship Challenge Paper: a step up from the Additional (Level 3) practice papers, requiring multi-step problem solving, proof-style algebraic reasoning, harder number and algebra, and geometry with justification. Topics include: surds, simultaneous equations (linear and quadratic), polygon angle facts, algebraic proof, ratio and percentage problems, quadratic sequences, similar shapes, compound interest and exponential growth, 3-D Pythagoras, combined inequalities, indices and standard form, forming and solving equations from a word problem, right-angled trigonometry, and optimisation by completing the square.

Simplify fully, showing your working: √200 - √50 + √18 (2 marks)

(Total for Question 1 is 2 marks)

A line has equation y = x + 1. A curve has equation y = x² - 5. The line and the curve intersect at two points.

(a)Show that, at the points of intersection, x satisfies the equation x² - x - 6 = 0. (1 mark)(1)

(b)Hence solve the equation to find the coordinates of both points of intersection. (2 marks)(2)

(c)State which of the two points of intersection satisfies x > 0. (1 mark)(1)

(Total for Question 2 is 4 marks)

A regular polygon has an interior angle of 156 degrees.

(a)Find the number of sides of the polygon, showing your method. (2 marks)(2)

(b)All the diagonals from one vertex of this polygon are drawn, dividing it into triangles. State how many triangles are formed, and justify your answer using the formula for the sum of interior angles of a polygon. (2 marks)(2)

(Total for Question 3 is 4 marks)

Prove that the sum of any three consecutive integers is always a multiple of 3. (3 marks)

(Total for Question 4 is 3 marks)

Solve the equation (2x - 1)/3 - (x + 2)/4 = 1, giving your answer as a fraction in its simplest form. (3 marks)

(Total for Question 5 is 3 marks)

A cafe mixes Kenyan and Colombian coffee beans in the ratio 5:3 by weight to make a house blend. Kenyan beans cost £14.40 per kg and Colombian beans cost £11.20 per kg.

(a)Find the cost of 1 kg of the house blend. (2 marks)(2)

(b)The cafe sells the house blend in 250 g bags for £4.75 each. Find the percentage profit made on each bag, giving your answer to 1 decimal place. (3 marks)(3)

(Total for Question 6 is 5 marks)

The nth term of a sequence is given by T(n) = n² + n.

(a)Find T(10). (1 mark)(1)

(b)Show that T(n) can be written as n(n + 1), and hence explain why T(n) is always even for every positive integer n. (2 marks)(2)

(c)Find the positive value of n for which T(n) = 132. (2 marks)(2)

(Total for Question 7 is 5 marks)

Triangle ABC is similar to triangle DEF, with AB corresponding to DE. AB = 6 cm, DE = 9 cm. The area of triangle ABC is 24 cm^2.

(a)Find the linear scale factor from triangle ABC to triangle DEF. (1 mark)(1)

(b)Find the area of triangle DEF. (2 marks)(2)

(Total for Question 8 is 3 marks)

Priya invests £2,500 in a savings account paying compound interest at 3.5% per annum.

(a)Find the value of her investment after 4 years, giving your answer to the nearest penny. (2 marks)(2)

(b)A separate investment of £4,000 grows at 2% per annum compound interest. Find the smallest integer number of years, y, after which this investment first exceeds £4,500. Show your working. (3 marks)(3)

(Total for Question 9 is 5 marks)

Prove algebraically that the difference between the squares of any two consecutive odd numbers is always a multiple of 8. (4 marks)

(Total for Question 10 is 4 marks)

Aisha cycles from her house to the library, a distance of 18 km, at an average speed of 15 km/h. She stops at the library for 45 minutes, then cycles back home along the same route at an average speed of 12 km/h.

(a)Find the total time, in hours, that Aisha spends cycling (not including the stop). (2 marks)(2)

(b)Find Aisha's average speed for the whole journey away from home, including the time she spends stopped at the library, giving your answer in km/h to 1 decimal place. (2 marks)(2)

(Total for Question 11 is 4 marks)

A bag contains 5 red counters and 7 blue counters. Two counters are drawn at random from the bag, one after the other, without replacement.

(a)Find the probability that both counters are red. (2 marks)(2)

(b)Find the probability that the two counters are different colours. (2 marks)(2)

(c)Given that the two counters drawn are different colours, find the probability that the first counter drawn was red. (1 mark)(1)

(Total for Question 12 is 5 marks)

A cuboid has a rectangular base measuring 8 cm by 6 cm, and a height of 10 cm.

(a)Find the length of the diagonal of the base of the cuboid. (1 mark)(1)

(b)Find the length of the space diagonal of the cuboid (the diagonal from one bottom corner to the opposite top corner), giving your answer to 3 significant figures. (2 marks)(2)

(c)Explain why a single straight rod of length 15 cm could not fit inside the cuboid without bending. (1 mark)(1)

(Total for Question 13 is 4 marks)

Find all integer values of x that satisfy both of the following inequalities, showing your working clearly: 3x - 7 < 11 and 2x + 5 ≥ x - 1 (3 marks)

(Total for Question 14 is 3 marks)

Simplify and evaluate the following.

(a)Write 2⁵ x 2^-2 / 2³ as a single power of 2, and hence evaluate it. (2 marks)(2)

(b)A number is given in standard form as 4.5 x 10^7. A second number is 3 x 10^4. Find, in standard form, the value of the first number divided by the second number. (2 marks)(2)

(Total for Question 15 is 4 marks)

Tickets for a school concert cost £6 for adults and £3.50 for children. A total of 140 tickets were sold, raising £715 in total. Let x be the number of adult tickets sold.

(a)Form an equation in x, and show that it simplifies to 2.5x = 225. (2 marks)(2)

(b)Hence find the number of adult tickets and the number of child tickets sold. (3 marks)(3)

(Total for Question 16 is 5 marks)

A ladder of length 5 m leans against a vertical wall, with its foot on horizontal ground 3 m from the base of the wall.

(a)Find the angle the ladder makes with the ground, giving your answer to 1 decimal place. (2 marks)(2)

(b)The foot of the ladder is then moved further from the wall, to a point 4 m from the base of the wall, with the top of the ladder still touching the wall. Find the new height reached by the ladder on the wall, and hence find the vertical distance the top of the ladder has slipped down the wall. (2 marks)(2)

(Total for Question 17 is 4 marks)

A farmer has 60 m of fencing available. She wants to enclose a rectangular pen using fencing on three sides, with the fourth side formed by an existing straight wall (so no fencing is needed there). Let x metres be the length of each of the two sides perpendicular to the wall, and let y metres be the length of the side parallel to the wall.

Figure (to be drawn): Rectangular pen against a straight wall: two sides of length x perpendicular to the wall, one side of length y parallel to the wall opposite the wall, the wall forms the fourth (unfenced) side. Not to scale.

(a)Show that y = 60 - 2x, and hence show that the area, A m², enclosed by the pen is given by A = 60x - 2x^2. (2 marks)(2)

(b)By writing A in the form A = -2(x - a)² + b, find the values of a and b. (2 marks)(2)

(c)Hence find the maximum possible area of the pen, and state the corresponding value of y. Justify why this value of x gives a maximum area, not a minimum. (3 marks)(3)

(d)The farmer decides she also needs a gate of width 2 m somewhere along one of the fenced sides; the gate does not change the total length of fencing used for that side, it simply interrupts it. Explain why your answer to part (c) is unaffected by the addition of the gate. (1 mark)(1)

(Total for Question 18 is 8 marks)

Mark scheme · CE.M35 Maths: Scholarship Challenge Paper

Question 1

M1 simplifies at least two of the three surds correctly: √200 = 10sqrt(2), √50 = 5sqrt(2), √18 = 3sqrt(2)
A1 8sqrt(2) cao
Answer: 8sqrt(2)

Question 2

(a) B1 sets x + 1 = x² - 5 and rearranges to x² - x - 6 = 0 (shown)
(a) Answer: x² - x - 6 = 0 (shown)
(b) M1 factorises (x-3)(x+2) = 0, or uses the quadratic formula, to find x = 3 and x = -2
(b) A1 both points correctly stated: (3, 4) and (-2, -1)
(b) Answer: (3, 4) and (-2, -1)
(c) B1 (3, 4) cao
(c) Answer: (3, 4)

Question 3

(a) M1 exterior angle = 180 - 156 = 24 degrees, and number of sides = 360 / 24
(a) A1 15 cao
(a) Answer: 15 sides
(b) B1 13 triangles
(b) B1 valid justification: diagonals drawn from one vertex of an n-sided polygon always create (n - 2) triangles, consistent with the interior angle sum formula (n - 2) x 180 degrees; here n = 15, so 15 - 2 = 13
(b) Answer: 13 triangles

Question 4

B1 lets the three consecutive integers be n, n+1 and n+2, for integer n
M1 sums to n + (n+1) + (n+2) = 3n + 3
A1 = 3(n+1), which is a multiple of 3 for every integer n (conclusion stated)
Answer: 3n + 3 = 3(n+1), a multiple of 3 for all integers n

Question 5

M1 multiplies every term by 12 (the LCM of 3 and 4) to obtain 4(2x-1) - 3(x+2) = 12
M1 expands and simplifies to 5x - 10 = 12
A1 x = 22/5 oe
Answer: x = 22/5

Question 6

(a) M1 (5 x 14.40 + 3 x 11.20) / (5 + 3)
(a) A1 £13.20 per kg cao
(a) Answer: £13.20 per kg
(b) M1 cost of 250 g bag = 0.25 x 13.20 = £3.30 (ft from part a)
(b) M1 profit = 4.75 - 3.30 = £1.45, and percentage profit = (1.45 / 3.30) x 100
(b) A1 43.9% awrt
(b) Answer: 43.9%

Question 7

(a) B1 110 cao
(a) Answer: 110
(b) M1 factorises n² + n = n(n+1)
(b) A1 correct explanation: n and n+1 are consecutive integers so exactly one of them is even, and an even number multiplied by any integer is even, so T(n) is always even
(b) Answer: n(n+1); always even since exactly one of two consecutive integers is always even
(c) M1 forms n² + n - 132 = 0 and factorises to (n-11)(n+12) = 0, or uses the quadratic formula
(c) A1 n = 11, rejecting n = -12 since n must be positive
(c) Answer: n = 11

Question 8

(a) B1 1.5 oe (3/2)
(a) Answer: 1.5 (3/2)
(b) M1 area scale factor = (linear scale factor)² = 1.5² = 2.25, and area of DEF = 24 x 2.25
(b) A1 54 cm² cao
(b) Answer: 54 cm²

Question 9

(a) M1 2500 x 1.035⁴ (or an equivalent year-by-year method using a multiplier of 1.035)
(a) A1 £2868.81 awrt (nearest penny)
(a) Answer: £2868.81
(b) M1 sets up the inequality 4000 x 1.02^y > 4500, or equivalent
(b) M1 valid trial or logarithmic method showing year 5 fails (4000 x 1.02⁵ = £4416.32) and year 6 succeeds (4000 x 1.02⁶ = £4504.65)
(b) A1 y = 6 cao
(b) Answer: y = 6

Question 10

B1 lets the two consecutive odd numbers be 2n+1 and 2n+3, for integer n
M1 uses the difference of two squares: (2n+3)² - (2n+1)² = [(2n+3)-(2n+1)] x [(2n+3)+(2n+1)]
A1 simplifies to 2 x (4n+4) = 8n + 8
B1 = 8(n+1), which is a multiple of 8 for every integer n (conclusion stated)
Answer: 8n + 8 = 8(n+1), a multiple of 8 for all integers n

Question 11

(a) M1 time out = 18/15 = 1.2 hours and time back = 18/12 = 1.5 hours
(a) A1 total cycling time = 2.7 hours cao
(a) Answer: 2.7 hours
(b) M1 total distance = 36 km and total time away = 2.7 + 0.75 = 3.45 hours (45 minutes = 0.75 hours), then 36/3.45
(b) A1 10.4 km/h awrt
(b) Answer: 10.4 km/h

Question 12

(a) M1 5/12 x 4/11
(a) A1 5/33 oe
(a) Answer: 5/33
(b) M1 (5/12 x 7/11) + (7/12 x 5/11)
(b) A1 35/66 oe
(b) Answer: 35/66
(c) B1 1/2 cao, from (35/132) / (70/132)
(c) Answer: 1/2

Question 13

(a) B1 10 cm cao
(a) Answer: 10 cm
(b) M1 uses their base diagonal: √10² + 10², or equivalently √8²+6²+10²
(b) A1 14.1 cm awrt (3sf)
(b) Answer: 14.1 cm
(c) B1 correct explanation: the space diagonal (about 14.1 cm) is the longest straight-line distance available inside the cuboid, and 15 cm is greater than this, so a 15 cm rod cannot fit without bending
(c) Answer: 15 cm > 14.1 cm, the maximum straight-line distance inside the cuboid, so the rod cannot fit

Question 14

M1 solves both inequalities: 3x - 7 < 11 gives x < 6, and 2x + 5 ≥ x - 1 gives x ≥ -6
A1 combines to give -6 ≤ x < 6
B1 correctly lists all 12 integer values: -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5
Answer: -6 ≤ x < 6; integers -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 (12 values)

Question 15

(a) M1 combines indices: 2^5-2-3 = 2⁰
(a) A1 = 1 cao
(a) Answer: 2⁰ = 1
(b) M1 4.5 / 3 = 1.5 and 10^7-4 = 10³
(b) A1 1.5 x 10³ cao
(b) Answer: 1.5 x 10³

Question 16

(a) M1 forms the equation 6x + 3.50(140 - x) = 715
(a) A1 correctly expands and simplifies to 2.5x = 225 (shown)
(a) Answer: 2.5x = 225 (shown)
(b) M1 x = 225 / 2.5
(b) A1 90 adult tickets cao
(b) B1 50 child tickets (140 - 90), ft
(b) Answer: 90 adult tickets, 50 child tickets

Question 17

(a) M1 cos(angle) = 3/5 (= 0.6)
(a) A1 53.1 degrees awrt (1dp)
(a) Answer: 53.1 degrees
(b) M1 finds both heights using Pythagoras: original height √5²-3² = 4 m, new height √5²-4² = 3 m
(b) A1 slip = 4 - 3 = 1 m cao
(b) Answer: new height 3 m; the ladder slips down by 1 m

Question 18

(a) M1 total fencing 2x + y = 60 gives y = 60 - 2x, and area A = x times y = x(60-2x)
(a) A1 = 60x - 2x² (shown)
(a) Answer: A = 60x - 2x² (shown)
(b) M1 takes out a factor of -2 and completes the square: A = -2(x²-30x) = -2[(x-15)²-225]
(b) A1 a = 15, b = 450 (both correct)
(b) Answer: a = 15, b = 450
(c) M1 maximum A = 450 m² when x = 15 (ft from part b)
(c) A1 y = 60 - 2(15) = 30 m
(c) B1 valid justification: since (x-15)² ≥ 0 for all x, -2(x-15)² ≤ 0, so A = 450 - 2(x-15)² ≤ 450 always; the coefficient of the squared term is negative, so moving x away from 15 in either direction decreases A, confirming a maximum
(c) Answer: maximum area 450 m² when x = 15 m and y = 30 m
(d) B1 correct explanation: the total length of fencing used (2x + y = 60) is unchanged by adding a gate within an existing side, so the equation, the area formula, and the maximum area found in part (c) are all unaffected
(d) Answer: the total fencing length is unchanged by the gate, so the model and maximum area still hold

Maths: Scholarship Challenge Paper - Worksheets, Questions and Revision