NCERT Class 9 Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data PDF Download

Chapter 10: How Quantities Combine: Understanding Data

10.1 Combining Things

In earlier grades, we looked at how the average conveys the 'centre' of given data. In this chapter, we shall extend the concept of average to a more general setting called 'weighted average'.

10.1.1 Average of Averages

Example 1: In a badminton academy, there is a group of 11 trainees - 8 seniors and 3 juniors. Their heights and average heights (in cm) are given in the table below. Find the average height of the whole group.

Group	Heights (in cm)	Average Height (in cm)
Seniors	165, 169, 164, 167, 170, 159, 164, 166	165.5
Juniors	146, 149, 153	149.33

Two students calculate the average height of the whole group in two ways:

Method 1

\[ \frac{165.5 + 149.33}{2} = \frac{314.83}{2} = 157.415 \]

Method 2

\[ \frac{165 + 169 + 164 + 167 + 170 + 159 + 164 + 166 + 146 + 149 + 153}{11} = \frac{1772}{11} = 161.09 \]

Whose calculation gives the correct average height of the whole group? The average height of the whole group is the sum of the heights of all eleven members divided by 11, as in Method 2.

Think and Reflect

Why did Method 1 not work?

Teacher's Note

Method 1 fails because it treats the two averages equally, but the groups have different sizes. The seniors group is much larger (8 people) than the juniors group (3 people), so the seniors' average should have more influence on the final answer. Always count how many values went into each average - this count matters when you combine them.

Method 1 does not work because it treats the average heights of seniors and of juniors equally - but there are many more seniors than juniors!

Shreyas calculated it differently as

\[ \frac{(165.5 \times 8) + (149.33 \times 3)}{8 + 3} = \frac{1324 + 448}{11} = \frac{1772}{11} = 161.09 \]

Do you understand why this also works? Can you see why \( 165.5 \times 8 \) gives the sum of the heights of all the seniors and \( 149.33 \times 3 \) gives the sum of the heights of all the juniors?

This phenomenon is explained below with a generalisation.

Suppose we have two collections of data - Collection 1 and Collection 2.

Suppose the data in Collection 1, having \( n \) values, is \( x_1, x_2, x_3, \ldots, x_n \).

The average of Collection 1 is

\[ \frac{x_1 + x_2 + x_3 + \ldots + x_n}{n} = a. \]

Thus the sum of the data in Collection 1 is

\[ x_1 + x_2 + x_3 + \ldots + x_n = an. \]

Suppose the data in Collection 2, having \( m \) values, is \( y_1, y_2, y_3, \ldots, y_m \).

The average of Collection 2 is

\[ \frac{y_1 + y_2 + y_3 + \ldots + y_m}{m} = b. \]

Thus the sum of the data in Collection 2 is

\[ y_1 + y_2 + y_3 + \ldots + y_m = bm. \]

The average of the combined data is therefore

\[ \frac{(\text{Sum of values in Collection 1}) + (\text{Sum of values in Collection 2})}{(\text{Number of values in Collection 1}) + (\text{Number of values in Collection 2})} = \frac{an + bm}{n + m}. \]

Example 2: Jaspreet recently learnt cycling. She has explored different routes in her town. She tracked how much time she cycled on weekdays over the last 3 weeks. Find the mean time spent cycling per weekday over the last 3 weeks.

Week	Day 1	Day 2	Day 3	Day 4	Day 5	Weekly Average
Week 1	10	13	10	15	16	12.8
Week 2	14	10	20	17	18	15.8
Week 3	15	18	14	20	23	18

Two ways of calculating this are given here.

Method 1

\[ \frac{(12.8 + 15.8 + 18)}{3} = \frac{46.63}{3} = 15.53 \]

Method 2

\[ \frac{(12.8 \times 5) + (15.8 \times 5) + (18 \times 5)}{5 + 5 + 5} = \frac{(64 + 79 + 90)}{15} = 15.53 \]

The average time spent cycling by Jaspreet on weekdays is 15.53 minutes.

We saw earlier how Method 1 (badminton example) did not produce the correct value. Why does Method 1 give the correct answer in this case? When does Method 1 work and when does it not work? Let us find out.

In general, the average of data in three collections is given as

\[ \frac{(\text{Sum of values in Collection 1}) + (\text{Sum of values in Collection 2}) + (\text{Sum of values in Collection 3})}{(\text{No. of values in Collection 1}) + (\text{No. of values in Collection 2}) + (\text{No. of values in Collection 3})} = \frac{ap + bq + cr}{p + q + r}, \]

where \( a, b, c \) are the averages of each collection, and \( p, q, r \) are the sizes of the respective collections.

Teacher's Note

The key insight is that the group sizes are the weights. When all groups have equal size, you can just average the averages. But when groups are unequal, multiply each average by its group size first, add those products, then divide by the total number of items. This is called a weighted average, and you will use it many times in exams.

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,

\[ \frac{ap + bp + cp}{p + p + p} = \frac{(a + b + c)p}{3p} = \frac{a + b + c}{3} \]

(where \( p \) is the size of the collections).

This is the same as adding the 3 weekly averages and dividing the sum by 3. Here, each week has 5 days, so each average comes from the same number of days; that is why treating each week's average equally works. If the weeks had different numbers of cycling days, this method would no longer be correct.

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.

Key Points

• When combining data from different groups, the average of the combined data is the total sum of all values divided by the total count of all values.
• If you know the average and size of each group, use the formula \( \frac{ap + bq + cr}{p + q + r} \) where \( a, b, c \) are the averages and \( p, q, r \) are the group sizes.
• You can only add the averages and divide by the number of groups if all groups have the same size; otherwise, you must use the weighted average formula.

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