NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 10 How Quantities Combine Understanding Data Exercise 10.3

Exercise Set 10.3

Question 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 on books = 30% of ₹8000 = ₹2400.
Profit on book covers = 50% of ₹1000 = ₹500.
Total sales = ₹9000; total profit = ₹2900.
Profit % = \( \frac{2900}{9000} \times 100\% \approx 32.22\% \).
The profit on total sales is about 32.22%. (It is close to 30% because most sales were books.)

Question 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: 55 km is above 44.5 km, so the average will rise, but only slightly because it is just one day in 21. Something near 45 km.
Distance in the first 20 days = 44.5 × 20 = 890 km.
New average = \( \frac{890 + 55}{20 + 1} = \frac{945}{21} = 45 \).
The average over 21 days is 45 km per day.

Question 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?
(i) Salt: 5%, Sugar: 8%
(ii) Salt: 13%, Sugar: 3%
(iii) Salt: 6%, Sugar: 6%
(iv) Salt: 5.55%, Sugar: 8.88%
(v) Salt: 3.33%, Sugar: 2.67%
(vi) Salt: 4.1%, Sugar: 7.08%
Answer:
Mixture volume = 600 + 300 = 900 mL.
Salt comes only from the first solution: 600 × 5% = 30 units; the second solution has 0% salt.
Sugar comes only from the second solution: 300 × 8% = 24 units; the first has 0% sugar.
Salt = \( \frac{30}{900} \times 100\% \approx 3.33\% \); Sugar = \( \frac{24}{900} \times 100\% \approx 2.67\% \).
Option (v): Salt 3.33%, Sugar 2.67%. Each substance gets diluted because the other solution contains none of it.

Teacher's Note:
1. Treat 'no salt' as a 0% concentration in the weighted mean.
2. Both results must be lower than the original 5% and 8%.

Question 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 \( \left(\frac{3}{4}\right)^{\text{th}} \) of the original concentration?
Answer:
Target concentration = \( \frac{3}{4} \times 8\% = 6\% \).
Spice in 10 L of pani = 10 × 0.08 = 0.8 L; plain water adds no spice.
Let w litres of water be added: \( \frac{0.8}{10 + w} = 0.06 \).
\( 0.8 = 0.6 + 0.06w \Rightarrow 0.06w = 0.2 \Rightarrow w = \frac{10}{3} \).
Check: \( \frac{0.8}{10 + \frac{10}{3}} = \frac{0.8}{\frac{40}{3}} = 0.06 \) ✓
Add \( \frac{10}{3} \) L ≈ 3.33 litres of regular water.

Teacher's Note:
1. The amount of spice stays fixed; only the volume grows.
2. Set up spice ÷ new volume = target and solve.

Question 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:
(i) Compare aspect by aspect: Keerthi is 5 marks behind in strength (weight 4), level in flexibility, and 5 marks ahead in agility (weight 6).
Her gain sits on the heavier weight (6 > 4), so it outweighs her loss. Keerthi's total is more.
(ii) Rashi = \( \frac{60 \times 4 + 65 \times 5 + 70 \times 6}{15} = \frac{240 + 325 + 420}{15} = \frac{985}{15} \approx 65.67 \).
Keerthi = \( \frac{55 \times 4 + 65 \times 5 + 75 \times 6}{15} = \frac{220 + 325 + 450}{15} = \frac{995}{15} \approx 66.33 \).
Rashi ≈ 65.67, Keerthi ≈ 66.33.

Teacher's Note:
1. Equal gains and losses are decided by which carries the larger weight.
2. Difference in weighted totals = 5 × 6 − 5 × 4 = 10, matching 995 − 985.

Question 6. A restaurant collected ratings from 10 customers on a scale of 1 to 5. The number of customers giving each rating is:
Food: 5★ 5, 4★ 3, 3★ 2, 2★ 0, 1★ 0
Ambience: 5★ 0, 4★ 4, 3★ 5, 2★ 1, 1★ 0
Service: 5★ 1, 4★ 2, 3★ 2, 2★ 4, 1★ 1
What is the average rating if the metrics are combined with the weights food : ambience : service = 6 : 5 : 4?
Answer:
Step 1: Average for each metric (the number of customers is the weight of each star value).
Food = \( \frac{5 \times 5 + 4 \times 3 + 3 \times 2}{10} = \frac{43}{10} = 4.3 \).
Ambience = \( \frac{4 \times 4 + 3 \times 5 + 2 \times 1}{10} = \frac{33}{10} = 3.3 \).
Service = \( \frac{5 \times 1 + 4 \times 2 + 3 \times 2 + 2 \times 4 + 1 \times 1}{10} = \frac{28}{10} = 2.8 \).
Step 2: Combine with weights 6 : 5 : 4: \( \frac{4.3 \times 6 + 3.3 \times 5 + 2.8 \times 4}{15} = \frac{25.8 + 16.5 + 11.2}{15} = \frac{53.5}{15} \approx 3.57 \).
The restaurant's average rating is about 3.57 out of 5.

Teacher's Note:
1. Two layers of weighting: customer counts, then metric weights.
2. Check that each row of counts adds up to 10.

Question 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?
Average weight: male langurs 16.5 kg, female langurs 13.8 kg, all langurs 14.925 kg. Total number of langurs: 60.
(i) Which of the following expression(s) describes the given scenario?
(a) \( \frac{16.5x + 13.8y}{x + y} = 14.925 \)   (b) \( \frac{16.5x + 13.8y}{60} = 14.925 \)   (c) \( \frac{16.5x + 13.8y}{2} = 14.925 \)   (d) \( \frac{16.5x + 13.8y}{16.5 + 13.8} = 14.925 \)
(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:
Without calculating: equal numbers would put the overall average midway, at (16.5 + 13.8) ÷ 2 = 15.15 kg. The actual 14.925 kg is below that, nearer the female average, so there are more female langurs.
Let x = number of males and y = number of females, with x + y = 60.
(i) Overall average = total weight ÷ total number = \( \frac{16.5x + 13.8y}{x + y} \), so (a) is correct; since x + y = 60, (b) says the same thing and is also correct. (c) and (d) divide by the wrong quantity and are incorrect.
(ii) \( 16.5x + 13.8(60 - x) = 14.925 \times 60 \Rightarrow 2.7x + 828 = 895.5 \Rightarrow 2.7x = 67.5 \Rightarrow x = 25 \). 25 male langurs (and 35 females).
(iii) Females' total = 35 × 13.8 = 483 kg. New average = \( \frac{483 + 15.2}{36} = \frac{498.2}{36} \approx 13.84 \) kg.
(iv) Males' total = 25 × 16.5 = 412.5 kg. New average = \( \frac{412.5 - 16.9 - 16.1}{23} = \frac{379.5}{23} = 16.5 \) kg, unchanged, because the two released langurs average exactly 16.5 kg.
(v) The 23 males now total 379.5 − 1 = 378.5 kg, so the average = \( \frac{378.5}{23} \approx 16.46 \) kg.

Teacher's Note:
1. Compare the overall mean with the midpoint to decide which group is larger.
2. Removing values equal to the mean leaves the mean unchanged.
3. A loss of 1 kg lowers the mean by 1 ÷ (number in the group).

Question 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.
Salinity: Dead Sea ≈ 34%; other seas and oceans ≈ 3.5%; ground water ≈ 0.01%; purified drinking water ≈ 0.001%.
(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:
(i) Salinity = \( \frac{1 \times 34 + 2 \times 0.001}{1 + 2}\% = \frac{34.002}{3}\% \) ≈ 11.33%.
(ii) Yes. A mixture's salinity always lies between the salinities of the two liquids, and 0.01% lies between 0.001% and 34%.
Let w litres of purified water be added: \( \frac{34 + 0.001w}{1 + w} = 0.01 \Rightarrow 34 + 0.001w = 0.01 + 0.01w \Rightarrow 0.009w = 33.99 \).
\( w = \frac{33.99}{0.009} \approx 3776.67 \), so about 3777 litres of purified water are needed.
(iii) No. Purified water (0.001%) is less salty than both the Dead Sea water and groundwater (0.01%). A weighted mean can never fall below the smaller of the two values, so the mixture always stays above 0.01%.
Algebra agrees: \( \frac{34 + 0.01w}{1 + w} = 0.001 \Rightarrow 0.009w = -33.999 \), which gives a negative w. No quantity of groundwater works.

Teacher's Note:
1. A mixture always lands between its ingredients.
2. A negative answer signals an impossible target.
3. Dead Sea water is so salty that it needs thousands of litres to dilute.