Skip to main content
Links to the University of Cambridge website
Links to the NRICH website Home page
Nurturing young mathematicians: teacher webinars
30 April (Primary)
,
1 May (Secondary)
menu
search
Teachers
expand_more
Early years
Primary
Secondary
Post-16
Events
Professional development
Students
expand_more
Primary
Secondary
Post-16
Parents
expand_more
Early Years
Primary
Secondary
Post-16
Problem-solving Schools
About NRICH
expand_more
About us
Impact stories
Support us
Our funders
Contact us
search
Site search
search
Or search by topic
Number and algebra
The Number System and Place Value
Calculations and Numerical Methods
Fractions, Decimals, Percentages, Ratio and Proportion
Properties of Numbers
Patterns, Sequences and Structure
Algebraic expressions, equations and formulae
Coordinates, Functions and Graphs
Geometry and measure
Angles, Polygons, and Geometrical Proof
3D Geometry, Shape and Space
Measuring and calculating with units
Transformations and constructions
Pythagoras and Trigonometry
Vectors and Matrices
Probability and statistics
Handling, Processing and Representing Data
Probability
Working mathematically
Thinking mathematically
Mathematical mindsets
For younger learners
Early Years Foundation Stage
Advanced mathematics
Decision Mathematics and Combinatorics
Advanced Probability and Statistics
Mechanics
Calculus
Choose the Mean
Age
11 to 14
Short
Challenge Level
Secondary curriculum
Problem
Solutions
For the mode to be 9, at least two of the numbers must be 9s.
The median and the mean are both smaller than 9, so the other numbers must be smaller than 9.
_ _ 9 9
The median will be halfway between 9 and the next largest number, so if the median is 8, then the next largest number must be 7.
_ 7 9 9
Then 9, 9, 7 and the fourth number must have a mean of 7, so, since there are 4 of them, they must have a total of 7$\times$4 = 28.
So the numbers are:
3 7 9 9
You can find more short problems, arranged by curriculum topic, in our
short problems collection
.