Problem Solving

 
8.3  Problem Solving
 

EXAMPLE:

The graph below shows the medal collection for sports houses in a school.

Based on the graph,
a) State the ratio of medal collection for red, blue and yellow houses.
b) Determine the range, median and mean for the medal collected.

SOLUTION:
a) State the ratio of medal collection for red, blue and yellow houses.

Red house's medal collection: 14
Blue house's medal collection: 42
Yellow house's medal collection: 28

Ratio = Red : Blue : Yellow

\(\text{Ratio}=14:42:28\)

Simplify and we will get

\(=14\color{red}{\div7}:42\color{red}{\div7}:28\color{red}{\div7}\\=2:6:4\)

b) Determine the range, median and mean for the medal collected.
Range 
= Maximum Value - Minimum Value
Range = 42 -14 = 28

Median = Middle Value
By sorting in ascending order, we will get 14, 28 and 42.
Median = 28

Min = Total of medals \(\div\) Number of houses
\(=\frac{42+28+14}{3}\\=\frac{84}{3}\)

\(\text{Mean}=28\)

 

 

 

Problem Solving

 
8.3  Problem Solving
 

EXAMPLE:

The graph below shows the medal collection for sports houses in a school.

Based on the graph,
a) State the ratio of medal collection for red, blue and yellow houses.
b) Determine the range, median and mean for the medal collected.

SOLUTION:
a) State the ratio of medal collection for red, blue and yellow houses.

Red house's medal collection: 14
Blue house's medal collection: 42
Yellow house's medal collection: 28

Ratio = Red : Blue : Yellow

\(\text{Ratio}=14:42:28\)

Simplify and we will get

\(=14\color{red}{\div7}:42\color{red}{\div7}:28\color{red}{\div7}\\=2:6:4\)

b) Determine the range, median and mean for the medal collected.
Range 
= Maximum Value - Minimum Value
Range = 42 -14 = 28

Median = Middle Value
By sorting in ascending order, we will get 14, 28 and 42.
Median = 28

Min = Total of medals \(\div\) Number of houses
\(=\frac{42+28+14}{3}\\=\frac{84}{3}\)

\(\text{Mean}=28\)