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Number problems and reasoning - Exam Questions.docx

Built from the lesson script on 28 September 2026.

EDEXCEL GCSE MATHS · FOUNDATION & HIGHER

Exam Practice: Number Problems and Reasoning

Number problems and reasoning · Number · Lesson 1 of 7 · 14 marks · 25 minutes

Name Date

Answer all questions. Show your working. Questions 1 to 4 are non-calculator.

Question 1 NON-CALCULATOR [2 marks]

Here are three number cards: 2, 5 and 7. List all the three-digit numbers that can be made using each card once.

Question 2 NON-CALCULATOR [2 marks]

Sam has 5 shirts and 4 ties. He chooses one shirt and one tie. How many different combinations of shirt and tie can he choose?

Question 3 NON-CALCULATOR · HIGHER [3 marks]

There are 12 students in a chess club. Each student plays every other student once. How many games are played?

Question 4 NON-CALCULATOR · HIGHER [3 marks]

A security code is 2 letters followed by 3 digits. The letters are chosen from the 26 letters of the alphabet and cannot be repeated. The digits are chosen from 0 to 9 and can be repeated. How many different codes are possible?

Question 5 CALCULATOR · HIGHER [4 marks]

How many odd numbers between 300 and 600 have three different digits?

 


Answers

Check your answer only once you have written one.

Question 1 [2 marks]

257, 275, 527, 572, 725, 752

▸ All 6 numbers listed, with no repeats and no extras B2

▸ At least 4 correct numbers listed B1

Question 2 [2 marks]

5 × 4 = 20 combinations.

▸ 5 × 4 seen, or a systematic list of at least 8 correct combinations M1

▸ 20 A1

Question 3 [3 marks]

Each student plays 11 others: 12 × 11 = 132. Each game has been counted twice, so 132 ÷ 2 = 66 games.

▸ 12 × 11 or 132 seen M1

▸ Dividing by 2 M1

▸ 66 A1

Question 4 [3 marks]

Letters: 26 × 25 = 650. Digits: 10 × 10 × 10 = 1000. Total 650 × 1000 = 650 000 codes.

▸ 26 × 25 or 650 P1

▸ 10 × 10 × 10 or 1000 P1

▸ 650 000 A1

Question 5 [4 marks]

The first digit is 3, 4 or 5. If it is 3 or 5 (odd), the last digit has 4 odd choices left and the middle digit has 8 choices: 2 × 4 × 8 = 64. If it is 4, the last digit has 5 odd choices and the middle 8: 1 × 5 × 8 = 40. Total 64 + 40 = 104.

▸ Splits into cases by first digit (odd first digit and 4 as first digit) P1

▸ 2 × 4 × 8 or 64 P1

▸ 1 × 5 × 8 or 40 P1

▸ 104 A1