IGCSE Mathematics Paper 2 · Non-calculator practice papers
Extended · 100 marks · 2h. Each generated paper follows the 0580 Paper 2 · Non-calculator format, with the same number and style of questions as the real exam, and comes with a Cambridge-style mark scheme.
The lengths, L mm, of 80 leaves are summarised in the table.
Length (L mm)
0 < L ≤ 20
20 < L ≤ 40
40 < L ≤ 60
60 < L ≤ 80
80 < L ≤ 100
Frequency
8
p
q
r
12
Table 1.1
A cumulative frequency polygon is drawn: the cumulative frequencies are plotted at the upper class boundaries and the points are joined with straight lines. From the polygon, the median is 50 mm and the upper quartile is 72 mm. For n leaves, use the n2th value for the median, the n4th value for the lower quartile and the 3n4th value for the upper quartile.
(a)
Find the values of p, q and r.
[5]
(b)
Find the interquartile range.
[2]
(c)
More leaves are added to the original 80: m leaves with 0 < L ≤ 20 and k leaves with 80 < L ≤ 100. From the new polygon, the median is still 50 mm and the upper quartile is now 76 mm. Find m and k.
[5]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
p = 24, q = 16, r = 20
p + q + r = 60 B1 median in 40 < L ≤ 60: 8 + p + q2 = 40 M1 upper quartile in 60 < L ≤ 80: 60 + 20 × 60 − (8 + p + q)r = 72 M1 uses p + q = 60 − r to get r − 8r = 0.6, r = 20 A1 p = 24, q = 16 A1
5
(b)
42 mm (lower quartile 30 mm)
lower quartile: 20 + 20 × 1224 = 30 M1 42 A1
2
(c)
m = 8, k = 8
new cumulative frequencies 8 + m, 32 + m, 48 + m, 68 + m, 80 + m + k B1 median at 80 + m + k2 = 32 + m + 8 (halfway through the 40–60 class), giving k = m M1 upper quartile at 3(80 + 2m)4 = 60 + 1.5m, in the 60–80 class M1 60 + 20 × 60 + 1.5m − (48 + m)20 = 76, i.e. 72 + 0.5m = 76 M1 m = 8, k = 8 A1
5
Question 2Structured · 13 marksA* challenge
Seven positive integers have a mean of 10 and a median of 9. The list has exactly one mode, which is 12.
(a)
Find the greatest possible range of the seven integers.
[4]
(b)
Find the least possible range.
[3]
(c)
Prove that only one list of seven integers has the range in part (b).
[3]
(d)
An eighth positive integer, k, is added to the list in part (b). The mean and the median of the eight integers are now equal. Find all the possible values of k.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
30 (1, 2, 3, 9, 12, 12, 31)
total 70, fourth value 9, 12 at least twice above the median B1 the three values below 9 must be different (a repeat would tie with 12) M1 smallest values 1, 2, 3, largest 70 − 6 − 9 − 24 = 31 M1 30 A1
4
(b)
4 (8, 8, 9, 9, 12, 12, 12)
if 12 occurs only twice, the values below 9 are distinct, so the range is at least 10 (e.g. smallest values total at most 6 + 7 + 8 = 21, largest at least 70 − 21 − 9 − 24 = 16) M1 so 12 occurs three times and the three smallest values total 25 M1 4 A1
3
(c)
smallest is 8, so the other two total 17 with each ≤ 9: 8 and 9 only
a + b + c = 25 with a ≤ b ≤ c ≤ 9 and the largest value 12 forces a = 8 for range 4 M1 b + c = 17 with b, c ≤ 9 gives only b = 8, c = 9 M1 (a = 7 would need 9, 9 and three 9s ties the mode) so 8, 8, 9, 9, 12, 12, 12 is the only list A1
3
(d)
k = 2 or k = 14
mean = 70 + k8; median 9 if k ≤ 9, 9 + k2 if 9 ≤ k ≤ 12, 10.5 if k ≥ 12 M1 solves each case: 70 + k8 = 9 gives k = 2; 70 + k8 = 9 + k2 gives k = 343 (rejected); 70 + k8 = 10.5 gives k = 14 M1 k = 2 or k = 14 A1
3
Question 3Structured · 5 marks
The vertices of triangle ABC are A(−3, −1), B(0, 7) and C(3, −1).
(a)
Show that the triangle is isosceles.
[3]
(b)
Find the area of the triangle.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
AB = BC
AB = √(32 + 82) M1 BC = √(32 + 82) M1 AB = BC so isosceles A1
3
(b)
24
½ × 6 × 8 M1 24 A1
2
Question 4Structured · 7 marks
A bag contains 7 red counters and 4 blue counters. Amira takes a counter at random, notes its colour and does not replace it. She then takes a second counter at random.
Fig. 4.1
(a)
Complete the tree diagram.
[2]
(b)
Find the probability that both counters are red.
[2]
(c)
Find the probability that the counters are different colours.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
Lower branches: 7/10 and 3/10
7/10 B1 3/10 B1
2
(b)
2155
711 × 610M1 2155A1
2
(c)
2855
711 × 410M1 411 × 710 added M1 2855A1
3
Question 5Structured · 4 marks
Triangle ABC is similar to triangle PQR. AB = 3 cm, BC = 4 cm and PQ = 7.5 cm.
(a)
Calculate QR.
[2]
(b)
The area of triangle ABC is 22 cm². Calculate the area of triangle PQR.
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in