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Observe the following infographic. (i) Identify the correct inference(s) from the statements given. (a) More schools have a playground in Haryana compared to Uttarakhand. (b) Approximately every 2 out of 3 schools in Arunachal Pradesh have a playground. (c) Punjab has the highest number of schools with a playground. (d) Suppose it is given that Maharashtra has more schools than Telangana. Then the number of schools having a playground is more in Maharashtra. (ii) Using the information given, can we find the nation-wide percentage of schools with a playground? If not, what additional information is needed?
(i) Statement (d) is the only correct inference. Since Maharashtra has a higher percentage (93%) than Telangana (74%) and also has more total schools, it must have more playground-equipped schools. Statements (a), (b) and (c) are incorrect. (ii) No, simple state averages are invalid. We need the totRead more
(i) Statement (d) is the only correct inference. Since Maharashtra has a higher percentage (93%) than Telangana (74%) and also has more total schools, it must have more playground-equipped schools. Statements (a), (b) and (c) are incorrect. (ii) No, simple state averages are invalid. We need the total number of schools in each state to calculate the correct weighted average across India.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-10/
See lessDecision Dilemma: (i) Which of the plays would you choose to watch based on the following chart? (ii) The corresponding stacked bar chart is shown below. Would you change your decision after looking at this chart? Why/Why not? Discuss.
(i) Play A or Play B would be preferred because combined 4-star and 5-star ratings account for over 70% of responses, whereas Play C has a huge 1-star share. (ii) Yes, the decision favors Play B. The stacked bar chart shows Play A has only about 200 total reviews, whereas Play B has roughly 850 reviRead more
(i) Play A or Play B would be preferred because combined 4-star and 5-star ratings account for over 70% of responses, whereas Play C has a huge 1-star share. (ii) Yes, the decision favors Play B. The stacked bar chart shows Play A has only about 200 total reviews, whereas Play B has roughly 850 reviews. A larger sample size makes Play B’s high rating far more dependable and proven.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-10/
See lessAnswer the following questions based on the graph. (i) In 2003, approximately how many total objects were found orbiting Earth in space? In what year did this number double? (ii) Find the approximate number and share of payload objects and other objects in the year 2025. (iii) What can you say about the number of payload objects over time and the number of other objects over time? What can you say about the share of payload objects over time and the share of other objects over time?
(i) In 2003, total orbiting objects were roughly 9,500, which doubled to about 19,000 around 2021. (ii) In 2025, out of approximately 33,000 total objects, payloads comprised about 18,000 (around 55%) and other debris objects accounted for roughly 15,000 (around 45%). (iii) Both numbers grew steadilRead more
(i) In 2003, total orbiting objects were roughly 9,500, which doubled to about 19,000 around 2021. (ii) In 2025, out of approximately 33,000 total objects, payloads comprised about 18,000 (around 55%) and other debris objects accounted for roughly 15,000 (around 45%). (iii) Both numbers grew steadily over time, but after 2019 payload counts surged dramatically, significantly increasing the relative percentage share of active payloads compared to space debris.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-10/
See less(Pṛthūdakasvāmī, commentary on Brahmagupta’s Brāhmasphuṭasiddhānta, c. 864 CE) A “hasta” (meaning “forearm”) refers to a unit of length measuring around 18 inches. A pool 30 hastas long is dug to different depths along its length. It is divided into 5 sections having lengths 4, 5, 6, 7 and 8 hastas and is dug to depths of 9, 7, 7, 3 and 2 hastas, respectively. Find the mean depth of the pool.
To find the mean depth, we weight each section's depth by its length. The total volume-cross product equals (4 × 9) + (5 × 7) + (6 × 7) + (7 × 3) + (8 × 2) = 36 + 35 + 42 + 21 + 16 = 150. Dividing this total by the full length of the pool, which is 4 + 5 + 6 + 7 + 8 = 30 hastas, gives a mean depth oRead more
To find the mean depth, we weight each section’s depth by its length. The total volume-cross product equals (4 × 9) + (5 × 7) + (6 × 7) + (7 × 3) + (8 × 2) = 36 + 35 + 42 + 21 + 16 = 150. Dividing this total by the full length of the pool, which is 4 + 5 + 6 + 7 + 8 = 30 hastas, gives a mean depth of exactly 5 hastas.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-10/
See lessGiven some data with corresponding weights, how would the weighted average change if all the weights are doubled? If required, experiment with some data. What do you observe? Justify your answer using algebra.
The weighted average remains completely unchanged. Algebraically, if each weight wᵢ is doubled to 2wᵢ, the new mean is Σ(2wᵢxᵢ)/Σ(2wᵢ). Factoring out the common factor of 2 gives [2Σ(wᵢxᵢ)]/[2Σwᵢ]. The factor 2 cancels from both numerator and denominator, leaving the original weighted mean formula iRead more
The weighted average remains completely unchanged. Algebraically, if each weight wᵢ is doubled to 2wᵢ, the new mean is Σ(2wᵢxᵢ)/Σ(2wᵢ). Factoring out the common factor of 2 gives [2Σ(wᵢxᵢ)]/[2Σwᵢ]. The factor 2 cancels from both numerator and denominator, leaving the original weighted mean formula intact. Scaling all weights by any constant preserves relative weights.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-10/
See less