Class 9 Maths Chapter 10 End-of-Chapter Exercises Solutions

Runs scored in the first over = 6

Runs scored in the second over = 12

Using the combined average formula:

where a, b are the averages and n, m are the respective numbers of overs.

Run rate = 6 × 1 + 12 × 11 + 1

Runs scored in the first 19 overs:

Runs scored in the 20th over = 12.

Run rate = 6 × 19 + 12 × 119 + 1

Let the average prices collected by Yusuf, Srikanth, Kashvi, Prasanna and Gracy be a_y, a_s, a_k, a_p, a_g, respectively.

The numbers of prices collected are Yusuf = 6, Srikanth = 4, Kashvi = 5, Prasanna = 4, and Gracy = 4.

6 + 4 + 5 + 4 + 4 = 23

Using the weighted mean formula:

If their average prices are a_n and a_f, respectively:

Yes, both expressions give the same average price because they combine the same 23 prices. The first method groups the prices by friend, while the second groups them by shuttlecock type.

Initial shares = 25, initial average price = ₹150, additional shares = 10, and price of each additional share = ₹30.

Using the weighted mean formula:

Let the number of additional shares be x.

70 = 5 × 150 + x × 305 + x

A pool 30 hastas long is divided into sections of lengths 4, 5, 6, 7, 8 hastas, with respective depths 9, 7, 7, 3, 2 hastas.

Mean depth = Sum of (section length × depth)Total length

Mean depth = 4 × 9 + 5 × 7 + 6 × 7 + 7 × 3 + 8 × 24 + 5 + 6 + 7 + 8

Suvarna already owns 1 g of gold at an average price of ₹15k per gram. Here, k means thousand rupees.

Mark these values on the existing number line from 0 to 30, in thousands of rupees per gram. The original point O is at 15, so scenario (vi) remains at O.

The weighted mean is:

x̄ = w₁x₁ + w₂x₂ + … + wₙxₙw₁ + w₂ + … + wₙ

If all weights are doubled:

x̄_new = 2w₁x₁ + 2w₂x₂ + … + 2wₙxₙ2w₁ + 2w₂ + … + 2wₙ

The factor of 2 cancels out from the numerator and denominator.

The supplied stacked column chart shows cumulative objects currently orbiting Earth, by year of launch separation, as of April 24, 2026.

From the chart, the total in 2003 was approximately 10,000 objects.

The graph reaches approximately 20,000 objects around 2021.

Reading the 2025 bar approximately from the chart:

Total objects:

Payload share:

Other-object share:

The infographic gives these percentages:

Haryana = 90%, Uttarakhand = 78%, Arunachal Pradesh = 68%, Punjab = 98%, Maharashtra = 93%, and Telangana = 74%.

(a) Haryana has a higher percentage than Uttarakhand, but this does not prove that more schools have playgrounds in Haryana because the total number of schools in each state is not given. Not valid.

(b) Arunachal Pradesh has approximately 68% of schools with playgrounds.

Since 68% is close to 66.67%, valid.

(c) Punjab has the highest percentage on the map, but percentages alone do not tell us which state has the highest actual number of schools with playgrounds. Not valid.

(d) Maharashtra has more schools than Telangana, and its playground percentage is also higher (93% versus 74%). Therefore, it must have more schools with playgrounds. Valid.

No. We need the total number of schools in each state and Union Territory.

Nationwide percentage = (Total schools with playgrounds across India / Total schools across India) × 100%

The state percentages must be combined using the number of schools in each state as weights. A simple average of the state percentages would not give the correct nationwide percentage.

Play A and Play B each have approximately 72% combined four-star and five-star ratings. Play A has a larger share of five-star ratings.

Yes, I would change my decision to Play B.

The ordinary stacked bar chart shows actual numbers of ratings. Play B has received considerably more ratings than Play A, while their combined proportions of four-star and five-star ratings are approximately equal. Thus, Play B has a larger number of positive ratings.

A stacked bar chart is more suitable because it shows both the actual time spent in each segment and the total time taken.

Athlete 1:

1 h 08 min + 5 h + 3 h 15 min = 9 h 23 min

Athlete 2:

1 h 05 min + 5 h 10 min + 3 h 35 min = 9 h 50 min

Athlete 3:

1 h 22 min + 5 h 55 min + 3 h 50 min = 11 h 07 min

Athlete 1 has the shortest total time and finishes first.

(a) Who finished first? Cannot be determined because all bars are scaled to 100%, hiding total durations.

(b) Approximately what fraction of Athlete 1's race time was spent cycling?

Total time:

Cycling time:

This fraction can be read from a 100% stacked bar chart.

(c) Which athlete took the longest for running? Cannot be determined from the 100% stacked chart alone, because it compares proportions rather than actual running durations. From the table, Athlete 3 took the longest for running, at 3 h 50 min.

The supplied 100% stacked bar chart shows the distribution of persons with disabilities by age group and type of disability in India, according to Census 2011.

Observations:

1. The proportion of disability types differs across the four age groups.

2. Seeing-related disabilities appear to account for a larger share in the 60+ age group than in younger groups.

3. Movement-related disabilities form a substantial share across age groups, particularly among adults.

4. Speech-related disabilities appear to make up a larger share in the 0–19 group than in the oldest group.

What I wonder:

The chart compares percentage distributions within age groups. It does not show which group contains the greatest total number of persons with disabilities. Actual population counts would be needed.

The textbook asks students to complete at least one project. The following are model examples only; students should replace sample data with their own observations.

Each person's total must equal 24 hours.

Draw a stacked bar chart with one bar per person and three segments for the body states. Use a consistent scale from 0 to 24 hours.

For the student, the segment boundaries are:

Observation: The chart compares how different people divide their day among the three activities.

Identify expense categories, collect expenditure for at least three months, and calculate each category's percentage share.

Percentage share = (Expenditure in a category / Total monthly expenditure) × 100

Illustrative sample data:

For Month 1, the percentage spent on food is

Calculate the other shares similarly and create a 100% stacked bar chart with one bar for each month.

Observations: Food and housing account for a large share of the illustrative budget. Housing remains constant in amount, but its percentage share decreases when the total monthly expenditure increases.

This is an illustrative model for a clothing-store shopping experience. Actual projects should collect ratings from at least 10 people.

Product quality and price receive higher weights. Variety and staff service receive slightly lower weights.

Ratings use a scale of 1–5, where 5 means very good.

For Person 1:

Rating = 3(4) + 3(4) + 2(5) + 2(4)3 + 3 + 2 + 2

Using the same formula for all 10 people:

Overall average = 4.2 + 4.1 + 4.3 + 3.4 + 4.2 + 4.0 + 3.3 + 3.7 + 4.8 + 4.210

Draw four horizontal bars, each representing 100%, and divide each bar according to these percentages.

Most people gave four-star or five-star ratings. Quality and variety have the same rating distribution, while staff service received the highest proportion of four-star ratings. The overall average rating of 4.02 out of 5 suggests a positive experience among this sample.

Since families have different numbers of members, use the weighted mean formula rather than simply averaging the family averages.

Illustrative data:

The simple average of the family averages is:

This is not correct for all family members because families have different sizes.

The weighted average is:

x̄ = w₁x₁ + w₂x₂ + … + wₙxₙw₁ + w₂ + … + wₙ

If each weight increases by 1, the new weighted average becomes:

x̄_new = [(w₁ + 1)x₁ + (w₂ + 1)x₂ + … + (wₙ + 1)xₙ] / [(w₁ + 1) + (w₂ + 1) + … + (wₙ + 1)]

x̄_new = w₁x₁ + … + wₙxₙ + (x₁ + … + xₙ)w₁ + w₂ + … + wₙ + n

Example:

Consider values 10 and 20, with weights 1 and 3.

x̄ = 1 × 10 + 3 × 201 + 3

After increasing each weight by 1, the new weights are 2 and 4.

x̄_new = 2 × 10 + 4 × 202 + 4

The ordinary arithmetic mean is:

The weighted average has moved from 17.5 towards 15.

Observation: Adding the same positive constant to every weight moves the weighted average towards the ordinary arithmetic mean of the values. It does not necessarily increase or decrease the weighted average in every case; the direction depends on whether the original weighted average is above or below the ordinary mean.

If the original weighted average already equals the ordinary mean, it remains unchanged.

Multiplying all weights by a common factor is different: it leaves the weighted average unchanged.

The initial percentage shares are cows = 60%, sheep = 25%, and chickens = 15%.

(i) All three percentage shares decrease.

Impossible because their total must remain 100%.

(ii) All three percentage shares increase.

Impossible because their total would exceed 100%.

(iii) All three shares remain the same.

Possible. If the populations start at 60, 25, 15, they could decrease to 48, 20, 12. Every population decreases, but the shares remain 60%, 25%, 15%.

(iv) Cows' share decreases, sheep's share increases, and chickens' share remains the same.

Possible. For example, populations could change from 600, 250, 150 to 440, 240, 120. All populations decrease, while the shares become:

(v) Cows' and sheep's shares increase, while chickens' share decreases.

Possible. Populations could change from 600, 250, 150 to 520, 240, 40. All populations decrease, while the new shares are: