Higher Tier - Grades 4-9

IGCSE Maths Higher Short Paper B

Covers Functions and Rates of Change, Sequences and Graphs, Angles, Polygons and Circle Theorems and 3 more.

8 questions - 40 marks - calculator allowed

Questions

Angles, Polygons and Circle Theorems

In a triangle, the angles are 2x degrees, 3x degrees and 55 degrees.

Find the value of x.

Mensuration and Vectors

A prism has cross-sectional area 22 cm^2 and length 16 cm.

Find the volume of the prism.

Sequences and Graphs

The straight line L has equation y = 2x - 7.

Find the equation of the line parallel to L that passes through (3, 4).

Pythagoras and Trigonometry

A triangular field has vertices A, B and C.

Angle BAC = 55 degrees, angle ABC = 72 degrees, and side BC = 84 m.

Find the length of side AC, to the nearest metre.

Angles, Polygons and Circle Theorems

ABCD is a cyclic quadrilateral. The side DC is extended to a point E, so angle BCE is an exterior angle.

Angle DAB = 112 degrees. Find angle BCD and angle BCE.

Functions and Rates of Change

The displacement, s metres, of a particle at time t seconds is given by s = t^3 - 6t^2 + 9t.

Find the particle's velocity when t = 4, and find the value(s) of t at which the particle is momentarily at rest.

Statistics and Probability

A box contains 5 milk chocolates and 7 dark chocolates. Two chocolates are taken at random without replacement.

Find the probability that both chocolates are dark, given that at least one of the two chocolates is dark.

Mensuration and Vectors

u = (8, 6) and w = (-2, 5), given as column vectors.

Find v = 2u - w as a column vector, then find the magnitude of v, giving your answer to 1 decimal place.

Mark scheme for Question 1 [4 marks]
Question 1[4 marks]
Answer or working	Marks
using angles in a triangle sum to 180 degrees	M1
2x + 3x + 55 = 180	M1
5x = 125	M1
x = 25	A1
Final answer: 25

Mark scheme for Question 2 [4 marks]
Question 2[4 marks]
Answer or working	Marks
volume = cross-sectional area x length	M1
22 x 16	M1
352	A1
cm^3	B1
Final answer: 352 cm^3

Mark scheme for Question 3 [4 marks]
Question 3[4 marks]
Answer or working	Marks
identifying gradient 2 from a parallel line	M1
using y = 2x + c	M1
substituting (3, 4) to get 4 = 6 + c	M1
y = 2x - 2	A1

Mark scheme for Question 4 [5 marks]
Question 4[5 marks]
Answer or working	Marks
finding angle ACB = 180 - 55 - 72 = 53 degrees	M1
using the sine rule AC/sin B = BC/sin A	M1
AC = 84 x sin 72 / sin 55	M1
97.5 m before rounding	A1
98 m (nearest metre)	A1
Final answer: 98 m

Mark scheme for Question 5 [5 marks]
Question 5[5 marks]
Answer or working	Marks
using opposite angles in a cyclic quadrilateral sum to 180 degrees	M1
angle BCD = 180 - 112 = 68 degrees	M1
using angles BCD and BCE lie on the straight line DCE, summing to 180 degrees	M1
angle BCE = 112 degrees	A1
noting angle BCE equals angle DAB, the exterior angle of a cyclic quadrilateral equals the interior opposite angle	B1
Final answer: angle BCD = 68 degrees, angle BCE = 112 degrees

Mark scheme for Question 6 [6 marks]
Question 6[6 marks]
Answer or working	Marks
differentiating to get v = 3t^2 - 12t + 9	M1
substituting t = 4 into v	M1
velocity = 9 m/s	A1
setting v = 0 and dividing by 3 to get t^2 - 4t + 3 = 0	M1
factorising to (t - 1)(t - 3) = 0	M1
t = 1 and t = 3 seconds	A1
Final answer: velocity = 9 m/s; t = 1 and t = 3 seconds

Mark scheme for Question 7 [6 marks]
Question 7[6 marks]
Answer or working	Marks
P(both milk) = 5/12 x 4/11 = 20/132	M1
P(at least one dark) = 1 - 20/132 = 112/132	M1
P(both dark) = 7/12 x 6/11 = 42/132	M1
using conditional probability P(both dark \| at least one dark) = P(both dark)/P(at least one dark)	M1
42/132 divided by 112/132 = 42/112	M1
3/8	A1