NCERT Solutions for Class 9 Maths Chapter 10 — End-of-Chapter Exercises#

$$\text{Run rate} = \dfrac{6 + 12}{2}$$

$$\text{Run rate} = 9$$

$$\text{Runs after 19 overs} = 6 \times 19 = 114$$$$\text{Run rate after 20 overs} = \dfrac{114 + 12}{20}$$

$$\text{Run rate after 20 overs} = \dfrac{126}{20}$$

$$\text{Run rate after 20 overs} = 6.3$$

(One big over raises the rate a lot early in the innings, but only a little after 19 overs.)

The friends collected different numbers of prices (6, 4, 5, 4, 4 — a total of 23), so each average is weighted by its count:

$$\text{Combined average} = \dfrac{6a_y + 4a_s + 5a_k + 4a_p + 4a_g}{23}$$

With the data ($a_y \approx 152.83$, $a_s = 123.25$, $a_k = 153.2$, $a_p = 224.25$, $a_g = 247.5$), this gives $\dfrac{4063}{23} \approx$ ₹176.65.

Counting the types: there are 13 nylon and 10 feather prices. So:

$$\text{Average} = \dfrac{13a_n + 10a_f}{23}$$

With $a_n = \dfrac{1429}{13} \approx 109.92$ and $a_f = \dfrac{2634}{10} = 263.4$, this is again $\dfrac{4063}{23} \approx$ ₹176.65.

Yes, the two answers are equal. Both are the total of all 23 prices divided by 23 — we have just grouped the same prices in two different ways. (But the simple average $\dfrac{a_n + a_f}{2} \approx ₹186.66$ would be wrong, because there are more nylon than feather prices.)

$$\text{Average} = \dfrac{25 \times 150 + 10 \times 30}{25 + 10}$$

$$\text{Average} = \dfrac{3750 + 300}{35}$$

$$\text{Average} = \dfrac{4050}{35}$$

$$\text{Average} \approx ₹115.71$$

Let her buy $k$ shares. The new average is ₹70, so:

$$\dfrac{5 \times 150 + 30k}{5 + k} = 70$$

$$\Rightarrow 750 + 30k = 350 + 70k$$

$$\Rightarrow 400 = 40k$$

$$\Rightarrow k = 10$$

∴ Vaishnavi bought 10 shares.

Each depth is weighted by the length of its section:

$$\text{Mean depth} = \dfrac{4 \times 9 + 5 \times 7 + 6 \times 7 + 7 \times 3 + 8 \times 2}{30}$$

$$\text{Mean depth} = \dfrac{36 + 35 + 42 + 21 + 16}{30}$$

$$\text{Mean depth} = \dfrac{150}{30}$$

$$\text{Mean depth} = 5 \text{ hastas}$$

∴ The mean depth is 5 hastas (about 90 inches).

After each purchase, the new average price (in ₹k per gram) is:

Scenario	Average price	Where it lies
(i) 1g at ₹30k	$\dfrac{15 + 30}{2} = 22.5$	halfway between 15 and 30
(ii) 2g at ₹30k	$\dfrac{15 + 60}{3} = 25$	two-thirds of the way to 30
(iii) 1g at ₹10k	$\dfrac{15 + 10}{2} = 12.5$	halfway between 10 and 15
(iv) 10g at ₹10k	$\dfrac{15 + 100}{11} \approx 10.45$	very close to 10
(v) 0.5g at ₹30k	$\dfrac{15 + 15}{1.5} = 20$	one-third of the way to 30
(vi) 0.5g at ₹15k	$\dfrac{15 + 7.5}{1.5} = 15$	stays at O

The more gold she buys, the closer her average moves towards the new price; buying at the same price (vi) leaves it unchanged.

It does not change. For example, values 60, 70 with weights 1, 3 give $\dfrac{60 + 210}{4} = 67.5$; with weights 2, 6 they give $\dfrac{120 + 420}{8} = 67.5$.

Algebra: doubling every weight doubles the numerator and the denominator:

$$\dfrac{2w_1x_1 + 2w_2x_2 + \cdots}{2w_1 + 2w_2 + \cdots} = \dfrac{2(w_1x_1 + w_2x_2 + \cdots)}{2(w_1 + w_2 + \cdots)}$$

$$\dfrac{2w_1x_1 + 2w_2x_2 + \cdots}{2w_1 + 2w_2 + \cdots} = \dfrac{w_1x_1 + w_2x_2 + \cdots}{w_1 + w_2 + \cdots}$$

The 2s cancel. Only the ratio of the weights matters — multiplying all of them by the same number leaves the weighted average unchanged.

The 2003 bar reaches about 10,000 objects. Double this is 20,000, which the bars first reach in about 2021.

The 2025 bar is about 33,000 tall, with the payload part up to about 17,000:

$$\text{Payload share} \approx \dfrac{17000}{33000} \approx 52\%$$

$$\text{Other objects} \approx 33000 - 17000 = 16000, \text{ share} \approx 48\%$$

• Numbers: payload objects grew slowly for decades, then shot up after about 2019 (large satellite constellations). Other objects (debris, rocket stages) grew steadily, with sudden jumps around 2007 and 2009 (debris from a satellite collision and a missile test).
• Shares: for most of the period, other objects made up the large majority (about 75–85%). In the last few years the payload share rose sharply, to about half by 2025.

Statement	Correct?	Reason
(a) More schools have a playground in Haryana compared to Uttarakhand.	No	The map shows percentages (90% vs 78%), not numbers of schools. Without the number of schools in each state, we cannot compare counts.
(b) Approximately every 2 out of 3 schools in Arunachal Pradesh have a playground.	Yes	68% ≈ $\dfrac{2}{3}$ (66.7%).
(c) Punjab has the highest number of schools with a playground.	No	98% is a high percentage, but it says nothing about the number of schools. (Dadra & Nagar Haveli and Daman & Diu even has a higher percentage, 99.8%.)
(d) Suppose it is given that Maharashtra has more schools than Telangana. Then the number of schools having a playground is more in Maharashtra.	Yes	Maharashtra has both more schools and a higher percentage (93% vs 74%), so 93% of a larger number is more than 74% of a smaller one.

∴ The correct inferences are (b) and (d).

No. The nation-wide percentage is a weighted average of the state percentages, and the weights are the numbers of schools in each state. We need the number of schools in each state (and union territory):

$$\text{National \%} = \dfrac{\sum (\text{schools in state} \times \text{state \%})}{\text{total schools}}$$

From the 100% stacked chart, about 70–72% of the ratings for Play A and Play B are 4 or 5 stars, with few low ratings. Play C has a large share of 5-star ratings (about 40%) but also about a third 1-star ratings — opinions are split.

So, from shares alone, Play A or Play B is the safer choice (Play A has a slightly larger 5-star share).

The stacked chart shows the number of ratings: about 920 for Play B, about 380 for Play C, but only about 210 for Play A. Play B’s good share is based on far more viewers, so it is more reliable.

Many would now choose Play B — similar high share of good ratings, backed by many more people. (A reasonable answer can differ, as long as it uses both the shares and the totals.)

A stacked bar chart. The race is decided by the total time, and a stacked bar shows each athlete’s total (Athlete 1: 9 h 23 min, Athlete 2: 9 h 50 min, Athlete 3: 11 h 7 min) while still showing the three segments. A 100% stacked chart would make every bar the same length and hide who was fastest.

• (a) No. The 100% chart shows only shares, not total times.
• (b) Yes. The cycling piece of Athlete 1’s bar shows its share directly: $\dfrac{300}{563} \approx 0.53$, a little over half.
• (c) No. A larger running share does not mean a longer running time, because the totals differ.

∴ Only (b) can be answered.

This is an open question; here are some points you might notice.

What we notice:

• Movement disability takes a bigger share as age increases — it is largest in the 60+ group (about a quarter).
• Seeing disability is a large share (about 18–26%) in every age group.
• Mental retardation and multiple disability take larger shares in the 0–19 group than in middle age.
• Speech disability takes a small share, smallest in the 60+ group.

What we wonder:

• Why does movement disability rise so much with age? (Falls, joint problems, strokes?)
• The chart shows shares, not numbers — are there more disabled people in older age groups or not?
• Which disabilities could be prevented or treated early?

Inference: the needs of disabled people change with age, so support services (schools, hospitals, public buildings) should plan for different types of disability in different age groups.

This is a project, so your data will be your own. Steps:

1. Collect the data in a table: one row per person (or per month), one column per category. For body-states, each row must add up to 24 hours.
2. Choose the chart. For (i), a stacked bar chart (equal totals of 24 hours, so stacked and 100% stacked look the same). For (ii), a 100% stacked bar chart, so that months with different total spending can be compared by share.
3. Draw the bars: convert each part to a percentage of its row total, mark the pieces in the same order and colour on every bar, and add a legend.
4. Write observations, for example “Food is the largest share every month” or “Grandmother spends the most time lying down”.

Aspect	Why it matters	Weight
Punctuality	Passengers plan their day around it	3
Safety of driving	Most important for the passenger	4
Cleanliness	Comfort and health	2
Seat comfort	Long journeys	2
Staff behaviour	Overall experience	1

Each person rates every aspect from 1 to 5, and their overall rating is the weighted mean, for example $\dfrac{3 \times 4 + 4 \times 5 + 2 \times 3 + 2 \times 3 + 1 \times 4}{12} = 4$. The group then averages the overall ratings of all 10 people, draws a 100% stacked bar chart of the ratings for each aspect, and writes what the scheme captures well (for example safety) and what it misses (for example cost of the ticket).

Families have different sizes, so a simple average of the family averages would be wrong. Each family’s average age must be weighted by its number of members:

$$\text{Average age of all family members} = \dfrac{\sum (\text{family size} \times \text{family's average age})}{\sum \text{family sizes}}$$

For example, a family of 4 with average age 30 and a family of 6 with average age 25 give $\dfrac{4 \times 30 + 6 \times 25}{10} = 27$ years, not 27.5. Each numerator term is just that family’s total age, so the class adds up all the total ages and divides by the total number of people.