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class 9 ganita manjari chapter 10
📖 Textbook Reference: 10.1.1 Average of Averages
When does Method 1 (adding averages and dividing by the number of averages) work and when does it not work? Let us find out.
In the cycling scenario, Jaspreet cycled 5 days every week for 3 weeks. This means that all three collections are of the same size, that is 5.
Thus, in the general form, the average becomes: $\frac{ap+bp+cp}{p+p+p} = \frac{(a+b+c)p}{3p} = \frac{a+b+c}{3}$.
We see that when the sizes of the collections are the same, to find the average of the combined collection, we can add the averages of the collections and divide the sum by the number of collections.
Jaspreet recently learnt cycling. She tracked how much time she cycled on weekdays over the last 3 weeks. The weekly averages for Week 1, Week 2, and Week 3 are 12.8, 15.8, and 18 minutes respectively (each calculated over 5 weekdays). Find the mean time spent cycling per weekday over the last 3 weeks.
1
When combining collections that have the exact same size, the weighted average formula simplifies beautifully.
2
If all weights $p$ are equal, then $\frac{ap + bp + cp}{p + p + p} = \frac{(a+b+c)p}{3p} = \frac{a+b+c}{3}$. This means you can simply add the averages and divide by the number of collections!
1
Try Method 1: Simply add the three weekly averages and divide by 3.
2
Try Method 2: Multiply each weekly average by 5, add them up, and divide by 15 (the total number of days).
3
Compare the results. You will find they are identical!
!
Assuming this works everywhere: The biggest mistake students make after learning this shortcut is applying it to problems like Example 1 (the badminton players). This shortcut *only* works if the number of items in each group is identical.
Method 1 (The Shortcut)
Since each week has exactly 5 days, we can average the averages directly.
$\text{Mean} = \frac{12.8 + 15.8 + 18}{3}$
$= \frac{46.6}{3}$
$= 15.533...$
Method 2 (Standard Weighted Average)
$\text{Mean} = \frac{(12.8 \times 5) + (15.8 \times 5) + (18 \times 5)}{5 + 5 + 5}$
$= \frac{64 + 79 + 90}{15}$
$= \frac{233}{15}$
$= 15.533...$
Ans: The mean time spent cycling is 15.53 minutes.
A school has three sections of Class 9: A, B, and C. Every section has exactly 40 students. The average math scores for sections A, B, and C are 75, 82, and 89 respectively. Find the combined average score of all Class 9 students.
1
Because all three sections have the exact same number of students (40), the "weights" are equal.
2
This allows you to bypass the long weighted average formula and use the "Equal Weights Shortcut".
1
You don't need to multiply the scores by 40. Just add the three averages together and divide by 3.
!
Doing it the long way: Calculating $\frac{75(40) + 82(40) + 89(40)}{120}$ will give you the correct answer, but it wastes precious time during an exam! Spotting equal weights saves time.
Step-by-Step Calculation
Since group sizes are equal ($n=40$ for all):
$\text{Combined Average} = \frac{75 + 82 + 89}{3}$
$= \frac{246}{3}$
$= 82$
Ans: The combined average score is 82.
In the weighted average formula $\frac{an+bm+cp}{n+m+p}$, if $n = m = p$, what does the expression simplify to?
A.
$$\frac{a + b + c}{n + m + p}$$
B.
$$\frac{a + b + c}{3}$$
C.
$$a + b + c$$
D.
$$\frac{an + bm + cp}{3}$$
1
This tests your ability to factor out variables algebraically. Since $n, m,$ and $p$ are exactly the same number, you can replace them all with a single variable, say $n$.
1
Substitute $n$ for $m$ and $p$: $\frac{an + bn + cn}{n + n + n}$.
2
Factor out the $n$ from the numerator and the denominator, then cancel it out.
!
Option A leaves the variables in the denominator, which isn't fully simplified. Option D leaves them in the numerator.
Formula Simplification
Given: $n = m = p$. Substitute $n$ everywhere:
$\frac{an + bn + cn}{n + n + n}$
$= \frac{n(a + b + c)}{3n}$
Cancel out the $n$ on top and bottom:
$= \frac{a + b + c}{3}$
Ans: Option (B)
Why does simply averaging the weekly averages (Method 1) work correctly for Jaspreet's cycling data, but not for the badminton academy in Example 1?
A. Because Jaspreet cycled for an odd number of weeks.
B. Because decimal numbers are easier to average directly.
C. Because the sample size (5 days) is identical for each week.
D. Because time is a continuous variable, whereas heights are not.
1
The critical factor in weighted averages is the "weight" (the size of the collection).
1
In the badminton academy, the groups were 8 and 3 (unequal). In Jaspreet's data, the groups are 5, 5, and 5.
!
Don't get distracted by the variable types (decimals, odd numbers, continuous time). Statistical rules depend on sample sizes, not variable formats.
Logical Explanation
Jaspreet cycled exactly 5 weekdays every single week. Since each weekly average represents the exact same amount of data (same weight), they influence the final average equally.
Therefore, a simple arithmetic mean of the averages works perfectly here, unlike the badminton example where one group heavily outweighed the other.
Ans: Option (C)
