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Average - Formula
In this article, we will learn about Average as per the requirements of Primary 5 Math level.
The learning objectives are:
- Find the average of a set of data
- Find the total given the average
- Find the number of data given the average
Average
To find the average, we divide the total by the number of data.
Formula For Average
\(\begin{align} \text{Average} = \frac{\text{Total Number or Amount}}{\text{Number Of Data}} \end{align} \)
1. Find the average of a set of data
Question 1:
Andy has \(2\) toy cars. Ben has \(7\) toy cars. Clement has \(3\) toy cars. What is the average number of toy cars each of them have?
Solution:
Total number of toy cars the three boys have
\(= 2 + 7 + 3 \\ = 12\)
\(\begin{align*} \text{Average} &= \frac {\text{Total numbe of toy cars}}{\text{Total number of boys}} \\ &= \frac {12}{3} \\ &= 4 \end{align*}\)
The average number of toy cars each boy has is \(4\).
Answer:
\(4\) toy cars
Question 2:
The table shows the results of Yong Jun's jumps in a high jump event. Find the average height of his jumps.
| Jump | Height |
|---|---|
| 1st jump | \(87 \text{ cm}\) |
| 2nd jump | \(92 \text{ cm}\) |
| 3rd jump | \(85 \text{ cm}\) |
Solution:
Total number of jumps \(= 3\)
Total height in \(3\) jumps\(\begin{align*} \\ &= 87 \text{ cm} + 92 \text{ cm} + 85 \text{ cm} \\ &= 264 \text{ cm} \end{align*}\)
Average height of jump \(\begin{align*} \\ &= 264 \text{ cm} \div 3 \\ &= 88 \text{ cm} \end{align*}\)
Answer:
\(88 \text{ cm}\)
Question 3:
Sean scored the following marks for the various subjects in his mid-year examination.
| English | Mathematics | Science | Mother Tounge | |
|---|---|---|---|---|
| Marks | \(78\) | \(92\) | \(87\) | \(83\) |
What was Sean’s average mark for the \(4\) subjects?
Solution:
Total number of subjects \(= 4\)
Total marks obtained in all \(4\) subjects
\(\begin{align} &= 78 + 92 + 87 + 83 \\ &= 340 \end{align}\)
Average mark
\(\begin{align} &= 340 \div 4 \\ &= 85 \end{align} \)
Answer:
\(85\) marks
Question 4:
A website had an average of \(392\) views per day over a \(4\text{ day}\) period. There were another \(1450\) views on the \(5^\text{th}\) day. What was the average number of views per day for the \(5\) days? Give your answer to the nearest whole number.
Solution:
Average number of views per day over a \(4\text{ day}\) period \(= 392\) hits
Total number of views over a \(4 \text{ day}\) period
\(= 4 \times 392 \\ = 1568 \)
Total number of views for \(5\) days
\(= 1568 + 1450 \\ = 3018 \)
Average number of views per day for the \(5\) days
\(\begin{align} &= 3018 \div 5 \\ &= 603.6 \\ &≈ 604 \end{align}\)
Answer:
\(604\) views
Question 5:
Wendy baked \(45, 21, 24\) and \(57\) cupcakes for her bakery shop from Monday to Thursday respectively. She did not bake any cupcakes for the rest of the week. What was the average number of cupcakes she baked per day in that week?
Solution:
Total number of cupcakes baked in the week
\(\begin{align} &= 45 + 21 + 24 + 57 \\ &= 147 \end{align}\)
Average number of cupcakes she baked per day in that week
\(\begin{align} &= 147 \div 7 \\ &= 21 \end{align}\)
Answer:
\(21\) cupcakes
2. Find the total given the average
\(\text{Total Number or Amount} = \text{Average} \times \text{Number Of Data}\)
Question 1:
The average length of \(4\) pieces of wire is \(14.4 \text{ m}\). Find their total length.
Solution:
Average length of \(4\) pieces \(=14.4 \text{ m}\)
Total length of \(4\) pieces
\(\begin{align} &= 4 × 14.4 \text{ m} \\ &= 57.6 \text{ m} \end{align}\)
Answer:
\(57.6 \text{ m} \)
Question 2:
The average mass of John and Sam is \(48 \text{ kg}\). The ratio of John's mass to Sam's mass is \(5 : 3\). Find John's mass.
Solution:
\(\begin{align} \text{John} &: \text{Sam} \\ \text{5} &: \text{3} \end{align}\)
Total units of John’s mass and Sam’s mass
\(= 5 \text{ units} + 3 \text{ units} \\ = 8 \text{ units}\)
Total mass of John and Sam
\(= 2 \times 48 \text{ kg} \\ = 96 \text{ kg}\)
\(\begin{align} 8 \text{ units} &= 96 \text{ kg} \\ 1 \text{ unit} &= 96 \text{ kg} \div 8 \\ &= 12 \text{ kg} \end{align}\)
John’s mass
\(= 5 \text{ units} \\ = 5 \times 12 \text{ kg} \\ = 60 \text{ kg}\)
Answer:
\(60 \text{ kg}\)
Question 3:
Heather's results slip was torn. However, she remembered that her average for the \(4\) subjects was \(68.75\) marks.
What could be the smallest possible difference between her Chinese and English marks?
Solution
Average marks of \(4\) subjects \(= 68.75\) marks
Total marks for \(4\) subjects
\(= 4 \times 68.75 \\ = 275\)
Total marks for Science and Mathematics
\(= 69 + 80 \\ = 149\)
Total marks for Chinese and English
\(\begin{align} &= 275 − 149 \\ &= 126 \end{align}\)
To get the smallest possible difference between her Chinese and English marks, either the Chinese marks has to be as high as possible or the English marks has to be as low as possible.
Highest possible Chinese marks \(= 59\)
English marks
\(\begin{align} &= 126 − 59 \\ &= 67 \end{align} \)
However, according to her results slip, the tens place of her English marks is \(7\).
Hence, it is not possible for \(67\) to be her English marks.
Lowest possible English marks \(= 70\)
Chinese marks
\(\begin{align} &= 126 − 70 \\ &= 56 \end{align}\)
Difference between her Chinese and English marks
\(\begin{align} &= 70 − 56 \\ &= 14 \end{align}\)
Hence, the smallest possible difference between her Chinese and English marks \(= 14\)
Answer:
\(14\)
3. Find the number of data given the average
\(\begin{align} \text{Number Of Data} = \frac{\text{Total Number or Amount}}{\text{Average}} \end{align}\)
Question 1:
Ronald bought some T-shirts for a total of \($207\). The average cost of a T-shirt was \($23\). How many T-shirts did he buy?
Solution:
Number of T-shirts bought
\(= $207 \div $23 \\ = 9\)
Answer:
\(9\) T-shirts
Question 2:
Jerry wanted to buy a tablet that cost \($312\). On average, he saved \($8\) a day. How many days did Jerry take to save enough to buy the tablet?
Solution:
Number of days
\(= $312 ÷ $8 \\ = 39\)
Answer:
\(39\) days
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