(i) Adding 12 runs in the second over brings the total to 6 + 12 = 18 runs across 2 overs. Dividing 18 by 2 gives a current run rate of 9. (ii) A run rate of 6 after 19 overs means 19 × 6 = 114 runs were scored. Adding 12 runs gives 126 runs in 20 overs. Dividing 126 by 20 gives an updated run rateRead more
(i) Adding 12 runs in the second over brings the total to 6 + 12 = 18 runs across 2 overs. Dividing 18 by 2 gives a current run rate of 9.
(ii) A run rate of 6 after 19 overs means 19 × 6 = 114 runs were scored. Adding 12 runs gives 126 runs in 20 overs. Dividing 126 by 20 gives an updated run rate of 6.3.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
(i) Mixing yields [(1 × 34) + (2 × 0.001)] / 3 ≈ 11.33% salinity. (ii) Yes, because groundwater salinity (0.01%) lies strictly between Dead Sea water (34%) and drinking water (0.001%). Solving (34 + 0.001x) / (1 + x) = 0.01 gives x ≈ 3776.7 litres of drinking water. (iii) No, it is impossible becausRead more
(ii) Yes, because groundwater salinity (0.01%) lies strictly between Dead Sea water (34%) and drinking water (0.001%). Solving (34 + 0.001x) / (1 + x) = 0.01 gives x ≈ 3776.7 litres of drinking water.
(iii) No, it is impossible because combining two liquids cannot produce a salinity lower than both components (34% and 0.01%).
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
The simple midpoint of 16.5 kg and 13.8 kg is 15.15 kg. Since the combined average (14.925 kg) is closer to the females' average, female langurs outnumber males. (i) Expressions (a) and (b) are correct since x + y = 60. (ii) Solving gives 25 males and 35 females. (iii) New female average is (483 + 1Read more
The simple midpoint of 16.5 kg and 13.8 kg is 15.15 kg. Since the combined average (14.925 kg) is closer to the females’ average, female langurs outnumber males. (i) Expressions (a) and (b) are correct since x + y = 60. (ii) Solving gives 25 males and 35 females. (iii) New female average is (483 + 15.2) / 36 ≈ 13.839 kg. (iv) Releasing two males weighing 33 kg gives (412.5 − 33) / 23 = 16.5 kg. (v) If one male loses 1 kg, the new male average becomes (412.5 − 1) / 25 = 16.46 kg.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
In cricket, the run rate is the average number of runs scored per over. In a T20 match, a team scored 6 runs in the first over making the run rate 6. (i) In the second over they scored 12 runs. What is the run rate now? (ii) After Over 19, their run rate was 6. What is the run rate after 20 overs given the team made 12 runs in the last over?
(i) Adding 12 runs in the second over brings the total to 6 + 12 = 18 runs across 2 overs. Dividing 18 by 2 gives a current run rate of 9. (ii) A run rate of 6 after 19 overs means 19 × 6 = 114 runs were scored. Adding 12 runs gives 126 runs in 20 overs. Dividing 126 by 20 gives an updated run rateRead more
(i) Adding 12 runs in the second over brings the total to 6 + 12 = 18 runs across 2 overs. Dividing 18 by 2 gives a current run rate of 9.
(ii) A run rate of 6 after 19 overs means 19 × 6 = 114 runs were scored. Adding 12 runs gives 126 runs in 20 overs. Dividing 126 by 20 gives an updated run rate of 6.3.
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 lessDorjee has collected 1 litre of water from the Dead Sea! Using the information in the table, answer the following questions. A calculator can be used if necessary. (i) What is the salinity of the mixture if he mixes 1 litre of water from the Dead Sea with 2 litres of purified drinking water? (ii) Is it possible to mix water from the Dead Sea and purified drinking water to get a mixture with the salinity of groundwater? Why/Why not? What quantity of purified drinking water should Dorjee mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of groundwater? (iii) Is it possible to mix water from the Dead Sea and groundwater to get a mixture with the salinity of purified drinking water? Why/Why not? What quantity of groundwater should he mix with 1 litre of water from the Dead Sea to get a mixture having the salinity of purified drinking water?
(i) Mixing yields [(1 × 34) + (2 × 0.001)] / 3 ≈ 11.33% salinity. (ii) Yes, because groundwater salinity (0.01%) lies strictly between Dead Sea water (34%) and drinking water (0.001%). Solving (34 + 0.001x) / (1 + x) = 0.01 gives x ≈ 3776.7 litres of drinking water. (iii) No, it is impossible becausRead more
(i) Mixing yields [(1 × 34) + (2 × 0.001)] / 3 ≈ 11.33% salinity.
(ii) Yes, because groundwater salinity (0.01%) lies strictly between Dead Sea water (34%) and drinking water (0.001%). Solving (34 + 0.001x) / (1 + x) = 0.01 gives x ≈ 3776.7 litres of drinking water.
(iii) No, it is impossible because combining two liquids cannot produce a salinity lower than both components (34% and 0.01%).
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 lessThe following table shows the weight data of langurs in an animal facility. Without doing any computations, can you tell whether there are more male langurs or more female langurs? Or are they equal in number? (i) Which of the following expression(s) describes the given scenario? (a) (16.5x + 13.8y) / (x + y) = 14.925 (b) (16.5x + 13.8y) / 60 = 14.925 (c) (16.5x + 13.8y) / 2 = 14.925 (d) (16.5x + 13.8y) / (16.5 + 13.8) = 14.925 (ii) Find out how many male langurs are present. (iii) A female langur weighing 15.2 kg is admitted to the facility. What is the average weight of the female langurs after this? (iv) Two male langurs weighing 16.9 kg and 16.1 kg are released from the facility. What is the average weight of the male langurs after this? (v) Now, suppose one of the male langurs lost 1 kg of weight. What is the average weight of all the male langurs after this?
The simple midpoint of 16.5 kg and 13.8 kg is 15.15 kg. Since the combined average (14.925 kg) is closer to the females' average, female langurs outnumber males. (i) Expressions (a) and (b) are correct since x + y = 60. (ii) Solving gives 25 males and 35 females. (iii) New female average is (483 + 1Read more
The simple midpoint of 16.5 kg and 13.8 kg is 15.15 kg. Since the combined average (14.925 kg) is closer to the females’ average, female langurs outnumber males. (i) Expressions (a) and (b) are correct since x + y = 60. (ii) Solving gives 25 males and 35 females. (iii) New female average is (483 + 15.2) / 36 ≈ 13.839 kg. (iv) Releasing two males weighing 33 kg gives (412.5 − 33) / 23 = 16.5 kg. (v) If one male loses 1 kg, the new male average becomes (412.5 − 1) / 25 = 16.46 kg.
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 lessA restaurant collected ratings from 10 customers on a scale of 1 to 5. The resulting data is shown in the table below. What is the average rating if the metrics are combined with the weights food : ambience : service = 6 : 5 : 4?
The mean rating for each metric out of 10 customers is: Food = [(5 × 5) + (4 × 3) + (3 × 2)] / 10 = 43 / 10 = 4.3; Ambience = [(4 × 4) + (3 × 5) + (2 × 1)] / 10 = 33 / 10 = 3.3; Service = [(5 × 1) + (4 × 2) + (3 × 2) + (2 × 4) + (1 × 1)] / 10 = 28 / 10 = 2.8. Combining these using weights 6:5:4 yielRead more
The mean rating for each metric out of 10 customers is: Food = [(5 × 5) + (4 × 3) + (3 × 2)] / 10 = 43 / 10 = 4.3; Ambience = [(4 × 4) + (3 × 5) + (2 × 1)] / 10 = 33 / 10 = 3.3; Service = [(5 × 1) + (4 × 2) + (3 × 2) + (2 × 4) + (1 × 1)] / 10 = 28 / 10 = 2.8. Combining these using weights 6:5:4 yields [(4.3 × 6) + (3.3 × 5) + (2.8 × 4)] / 15 = 53.5 / 15 ≈ 3.57.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 10 How Quantities Combine Understanding Data Question Answer (2026-27)
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See lessCompute gcd(2587392, 157656) using the improved version of Euclid’s algorithm.
Successive divisions give: 2587392 = 16 × 157656 + 64896 ⇒ gcd(157656, 64896); 157656 = 2 × 64896 + 27864 ⇒ gcd(64896, 27864); 64896 = 2 × 27864 + 9168 ⇒ gcd(27864, 9168); 27864 = 3 × 9168 + 360 ⇒ gcd(9168, 360); 9168 = 25 × 360 + 168 ⇒ gcd(360, 168); 360 = 2 × 168 + 24 ⇒ gcd(168, 24); 168 = 7 × 24Read more
Successive divisions give:
2587392 = 16 × 157656 + 64896 ⇒ gcd(157656, 64896);
157656 = 2 × 64896 + 27864 ⇒ gcd(64896, 27864);
64896 = 2 × 27864 + 9168 ⇒ gcd(27864, 9168);
27864 = 3 × 9168 + 360 ⇒ gcd(9168, 360);
9168 = 25 × 360 + 168 ⇒ gcd(360, 168);
360 = 2 × 168 + 24 ⇒ gcd(168, 24);
168 = 7 × 24 + 0 ⇒ gcd(24, 0).
Thus, the GCD is 24.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 11 The World of Algorithms Question Answer (2026-27)
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See less