Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.3 Solutions How Quantities Combine: Understanding Data
The NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.3 – How Quantities Combine: Understanding Data help students develop a better understanding of how data can be represented, compared, and interpreted using different mathematical methods. This exercise focuses on applying concepts such as averages, proportions, and data representation to practical situations, making it easier for students to connect mathematical ideas with real-life examples.
The solutions are explained in a clear, step-by-step manner to help students understand the reasoning behind each answer and solve similar questions confidently. Students can also refer to the Solutions PDF for convenient practice, revision, and exam preparation.
NCERT Class 9 Maths Ganita Manjari Chapter 10 Exercise Set 10.3
1. A stationery shop owner made ₹8000 selling books, of which 30% is the profit amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the percentage of profit on the total sales?
Answer:
Profit from selling books:
30% of ₹8000
= 30/100 × 8000
= ₹2400
Profit from selling book covers:
50% of ₹1000
= 50/100 × 1000
= ₹500
Total profit:
₹2400 + ₹500 = ₹2900
Total sales:
₹8000 + ₹1000 = ₹9000
Percentage profit on total sales:
(2900 / 9000) × 100
= 32.22%
Therefore, the percentage of profit on the total sales is approximately 32.22%.
2. A white stork’s migration is tracked. The average daily distance travelled, calculated over 20 days, is 44.5 km. On the 21st day, it flew 55 km. What is the average daily distance travelled over these 21 days? Make a guess before you calculate.
Answer:
Guess: Since the stork travelled 55 km on the 21st day, which is more than the previous average of 44.5 km, the new average should be slightly more than 44.5 km.
Total distance travelled in the first 20 days:
44.5 × 20 = 890 km
Distance travelled on the 21st day:
55 km
Total distance travelled in 21 days:
890 + 55 = 945 km
Average daily distance:
945 ÷ 21 = 45 km
Therefore, the average daily distance travelled over 21 days is 45 km.
3. A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar. What are the concentrations of salt and sugar in the mixture?
Answer:
Salt in the first solution:
5% of 600 mL
= 5/100 × 600
= 30 mL
Sugar in the second solution:
8% of 300 mL
= 8/100 × 300
= 24 mL
Total volume of the mixture:
600 + 300 = 900 mL
Salt concentration in the mixture:
(30 / 900) × 100 = 3.33%
Sugar concentration in the mixture:
(24 / 900) × 100 = 2.67%
Therefore, the mixture contains 3.33% salt and 2.67% sugar.
4. At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water) was 8%. Many customers complained that it was too spicy. What quantity of regular water should be mixed into the 10 litres of pani so that the spice level is reduced to (3/4)ᵗʰ of the original concentration?
Answer:
Original concentration of spice = 8%
Required concentration = 3/4 of 8%
= 3/4 × 8%
= 6%
Amount of spice in 10 litres of pani:
8% of 10
= 8/100 × 10
= 0.8 litre
Let the quantity of water added be x litres.
After adding water, total volume:
10 + x litres
Since the amount of spice remains unchanged:
0.8 / (10 + x) × 100 = 6
0.8 / (10 + x) = 0.06
0.8 = 0.06(10 + x)
0.8 = 0.6 + 0.06x
0.2 = 0.06x
x = 0.2 / 0.06
= 10/3
= 3.33 litres
Therefore, approximately 3.33 litres of regular water should be added.
5. A physical fitness evaluation is being undertaken. The final marks are calculated by combining the marks for strength, flexibility, and agility in the ratio 4 : 5 : 6. Rashi has scored 60, 65, and 70. Keerthi has scored 55, 65, and 75 respectively.
(i) Find out whose total is more without calculating.
(ii) What are their final marks?
Answer:
Solution:
The marks for strength, flexibility, and agility are combined in the ratio 4 : 5 : 6.
(i) Find out whose total is more without calculating.
Both Rashi and Keerthi have the same score in flexibility: 65.
Rashi scores 5 marks more in strength, while Keerthi scores 5 marks more in agility.
Since agility has a higher weight (6) than strength (4), Keerthi’s advantage contributes more to the final score.
Therefore, Keerthi’s total is higher.
(ii) What are their final marks?
For Rashi:
Final marks
= (60 × 4 + 65 × 5 + 70 × 6) / (4 + 5 + 6)
= (240 + 325 + 420) / 15
= 985 / 15
= 65.67
For Keerthi:
Final marks
= (55 × 4 + 65 × 5 + 75 × 6) / 15
= (220 + 325 + 450) / 15
= 995 / 15
= 66.33
Therefore, Rashi’s final marks are 65.67 and Keerthi’s final marks are 66.33.
6. A 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?
Answer:
First, calculate the average rating for each metric.
Food
Average rating
= (5 × 5 + 4 × 3 + 3 × 2 + 2 × 0 + 1 × 0) / 10
= (25 + 12 + 6) / 10
= 43 / 10
= 4.3
Ambience
Average rating
= (5 × 0 + 4 × 4 + 3 × 5 + 2 × 1 + 1 × 0) / 10
= (16 + 15 + 2) / 10
= 33 / 10
= 3.3
Service
Average rating
= (5 × 1 + 4 × 2 + 3 × 2 + 2 × 4 + 1 × 1) / 10
= (5 + 8 + 6 + 8 + 1) / 10
= 28 / 10
= 2.8
Now, the metrics are combined in the ratio:
Food : Ambience : Service = 6 : 5 : 4
Overall average rating:
= (4.3 × 6 + 3.3 × 5 + 2.8 × 4) / (6 + 5 + 4)
= (25.8 + 16.5 + 11.2) / 15
= 53.5 / 15
≈ 3.57
Therefore, the restaurant’s average rating is approximately 3.57 out of 5.
7. The 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?
Answer:
Solution:
The average weight of male langurs is 16.5 kg, while that of female langurs is 13.8 kg. The overall average, 14.925 kg, is closer to 13.8 kg than to 16.5 kg.
Therefore, there are more female langurs than male langurs.
(i) Which expression describes the given scenario?
Let:
x = number of male langurs
y = number of female langurs
Since the combined average weight is 14.925 kg,
(16.5x + 13.8y) / (x + y) = 14.925
Therefore, the correct expression is:
(a) (16.5x + 13.8y)/(x + y) = 14.925
(ii) Find the number of male langurs.
Total number of langurs = 60
Therefore,
x + y = 60
So,
y = 60 − x
Using the weighted-average equation:
(16.5x + 13.8y) / 60 = 14.925
Substituting y = 60 − x:
16.5x + 13.8(60 − x) = 14.925 × 60
16.5x + 828 − 13.8x = 895.5
2.7x = 67.5
x = 25
Therefore, there are 25 male langurs and 35 female langurs.
(iii) A female langur weighing 15.2 kg is admitted. Find the new average weight of the female langurs.
Number of female langurs = 35
Their total weight:
35 × 13.8 = 483 kg
After adding the new langur:
New total weight = 483 + 15.2 = 498.2 kg
New number of female langurs = 36
New average:
498.2 / 36 ≈ 13.84 kg
Therefore, the new average weight of the female langurs is approximately 13.84 kg.
(iv) Two male langurs weighing 16.9 kg and 16.1 kg are released. Find the new average weight.
Original total weight of male langurs:
25 × 16.5 = 412.5 kg
Weight of the two released langurs:
16.9 + 16.1 = 33 kg
Remaining total weight:
412.5 − 33 = 379.5 kg
Remaining number of male langurs:
25 − 2 = 23
New average:
379.5 / 23 = 16.5 kg
Therefore, the average weight of the remaining male langurs is 16.5 kg.
(v) One of the male langurs loses 1 kg. Find the new average.
After part (iv), there are 23 male langurs with a total weight of 379.5 kg.
If one langur loses 1 kg:
New total weight = 379.5 − 1 = 378.5 kg
New average:
378.5 / 23 ≈ 16.46 kg
Therefore, the new average weight of the male langurs is approximately 16.46 kg.
8. Dorjee 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?
Answer:
Solution:
From the table:
- Salinity of Dead Sea water ≈ 34%
- Salinity of groundwater ≈ 0.01%
- Salinity of purified drinking water ≈ 0.001%
(i) Mixing 1 litre of Dead Sea water with 2 litres of purified drinking water
Total volume = 1 + 2 = 3 litres
Amount of salt contributed:
Dead Sea water = 34% of 1 L
Purified water = 0.001% of 2 L
Therefore,
Salinity of mixture
= [34 × 1 + 0.001 × 2] / 3
= 34.002 / 3
= 11.334%
Therefore, the salinity of the mixture is approximately 11.334%.
(ii) Can Dead Sea water and purified drinking water be mixed to get the salinity of groundwater?
Yes.
The salinity of groundwater, 0.01%, lies between the salinities of Dead Sea water (34%) and purified drinking water (0.001%). Therefore, an appropriate mixture can have a salinity of 0.01%.
Let x litres of purified drinking water be mixed with 1 litre of Dead Sea water.
Required salinity = 0.01%
So,
(34 × 1 + 0.001x) / (1 + x) = 0.01
34 + 0.001x = 0.01 + 0.01x
33.99 = 0.009x
x = 33.99 / 0.009
x ≈ 3776.67 litres
Therefore, Dorjee would need to mix approximately 3776.67 litres of purified drinking water with 1 litre of Dead Sea water.
(iii) Can Dead Sea water and groundwater be mixed to get the salinity of purified drinking water?
No, it is not possible.
Dead Sea water has a salinity of 34%, and groundwater has a salinity of 0.01%. Any mixture of the two must have a salinity between 0.01% and 34%.
However, purified drinking water has a salinity of only 0.001%, which is lower than the salinity of both liquids being mixed.
Therefore, no quantity of groundwater can be mixed with 1 litre of Dead Sea water to obtain a salinity of 0.001%.
Related Links: Chapter 10 Exercise-wise Solutions
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.1
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.2
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.4
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 Exercise 10.5
- NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 10 End-of-Chapter Exercises