1Spot the weightsFind what each value counts for: sales, days, volume, number of animals or an importance ratio.
2Turn values into totalsMultiply value × weight. For a mixture, this is the amount of salt, spice or profit.
3Add or removeAdd new items to the totals, or subtract removed ones, and update the total weight too.
4Divide and checkDivide the total by the total weight. The answer must lie between the smallest and largest values.
Concentration: the amount of salt, spice or sugar per 100 of the mixture, as a percentage.
Water or nothing added: counts as a part with concentration 0%, so it dilutes the mixture.
Adding one value: new average = (old total + new value) over (old count + 1).
Profit and averages
Q1
Chapter 10 Exercise 10.3 Question 1 Solution: A stationery shop owner made ₹8000 selling books, of which 30% is profit, and ₹1000 selling book covers, of which 50% is profit. What is the percentage of profit on the total sales?
Answer: ≈ 32.22%
Step 1 · Find the profit from each item
Item
Sales (₹)
Profit %
Profit (₹)
Books
8000
30%
2400
Book covers
1000
50%
500
Profit on books = 30100 × 8000 = ₹2400. Profit on covers = 50100 × 1000 = ₹500.
The answer lies between 30% and 50%, and it is close to 30% because books make up most of the sales.
The profit is about 32.22% of the total sales.
Q2
Chapter 10 Exercise 10.3 Question 2 Solution: A white stork's average daily distance over 20 days is 44.5 km. On the 21st day it flew 55 km. What is the average daily distance over these 21 days? Make a guess before you calculate.
Answer: 45 km
Step 1 · Make a guess
One new day is added to 20 days. The new day (55 km) is above the old average, so the average goes up. But one day among 21 can pull it up only a little. A good guess is a little above 44.5 km, maybe about 45 km.
Diagram
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Q2 · one new day (weight 1) barely moves an average built from 20 days
Step 2 · Calculate
The first 20 days have weight 20 and the 21st day has weight 1.
The average daily distance over 21 days is 45 km, very close to our guess.
Mixtures
Q3
Chapter 10 Exercise 10.3 Question 3 Solution: 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? Options: (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: (v)
Step 1 · Find how much salt and sugar there is
Salt is only in the first solution, and sugar is only in the second.
After mixing, the total volume is 600 + 300 = 900 mL.
Step 2 · Find each concentration
ƒsalt: 30900 × 100 ≈ 3.33%
ƒsugar: 24900 × 100 ≈ 2.67%
The same salt result comes from a weighted mean, because the sugar solution has 0% salt: 5 × 600 + 0 × 300900 ≈ 3.33%.
Step 3 · Match with the options
Both concentrations are lower than in the solution they came from, because mixing dilutes them. So (i) is not possible. Only option (v) has salt 3.33% and sugar 2.67%.
Option (v): salt 3.33% and sugar 2.67%.
Q4
Chapter 10 Exercise 10.3 Question 4 Solution: The spice concentration in 10 litres of pani is 8%. How much regular water should be mixed in so that the spice level becomes 34 of the original?
Answer: 103 L ≈ 3.33 L
Step 1 · Find the target
ƒ34 of 8% = 6%
The spice itself stays the same: 8100 × 10 = 0.8 L. Only the total volume grows.
Step 2 · Let x litres of water be added
Water has 0% spice. The mixture is 10 + x litres, with 0.8 L of spice.
ƒ0.810 + x = 6100
ƒ80 = 6(10 + x), so 80 = 60 + 6x, so 6x = 20, so x = 206 = 103
Step 3 · Check
The new volume is 10 + 103 = 403 L ≈ 13.33 L. Then 0.8 L of spice in 13.33 L is 6%, as required.
About 103 litres (3.33 L) of regular water should be added.
Ratings and weights
Q5
Chapter 10 Exercise 10.3 Question 5 Solution: Strength, flexibility and agility are combined in the ratio 4 : 5 : 6. Rashi scored 60, 65 and 70. Keerthi scored 55, 65 and 75. (i) Whose total is more, without calculating? (ii) Find their final marks.
Answer: Keerthi leads
Part (i) · Compare without calculating
Strength (w = 4)
Flexibility (w = 5)
Agility (w = 6)
Rashi
60
65
70
Keerthi
55
65
75
Flexibility is equal. Rashi is ahead by 5 in strength, which has weight 4, so she gains 5 × 4 = 20. Keerthi is ahead by 5 in agility, which has weight 6, so she gains 5 × 6 = 30. Keerthi's gain is bigger, so her total is more.
The difference is 1015 ≈ 0.67, which matches the weighted gains 30 − 20 = 10 divided by 15.
(i) Keerthi has the larger total. (ii) Rashi: 65.67, Keerthi: 66.33.
Q6
Chapter 10 Exercise 10.3 Question 6 Solution: Ten customers rated a restaurant from 1 to 5 stars. The metrics are combined with weights food : ambience : service = 6 : 5 : 4. Find the average rating.
Answer: ≈ 3.57
Step 1 · Write the table
5★
4★
3★
2★
1★
Food
5
3
2
0
0
Ambience
0
4
5
1
0
Service
1
2
2
4
1
Step 2 · Average rating of each metric
Each metric has 10 ratings, so we multiply each star value by its count, add, and divide by 10.
The result lies between 2.8 and 4.3, and it leans towards food because food has the biggest weight.
The average rating is about 3.57 out of 5.
Adding and removing values
Q7
Chapter 10 Exercise 10.3 Question 7 Solution: A facility has 60 langurs. Male langurs average 16.5 kg, female langurs average 13.8 kg and all langurs average 14.925 kg. Answer the parts below.
Answer: 25 males, 35 females
Guess first · More males or more females?
The overall average 14.925 is between 13.8 and 16.5. It is 14.925 − 13.8 = 1.125 away from the female average but 16.5 − 14.925 = 1.575 away from the male average. It is closer to the female average, so there are more females.
Diagram
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Q7 · the overall average sits closer to the group with more langurs
Part (i) · Which expression describes the situation?(a)
Let x be the number of males and y the number of females. The weighted mean uses the numbers as weights:
ƒ16.5x + 13.8yx + y = 14.925
This is option (a). Options (b), (c) and (d) divide by 60, by 2, or by 16.5 + 13.8. None of these is the total number x + y. So only (a) is correct. Options (b) to (d) also show a minus sign, which is impossible for an average of weights.
Part (ii) · How many male langurs?
Males x, so females 60 − x. The total weight is 14.925 × 60 = 895.5 kg.
ƒ16.5x + 13.8(60 − x) = 895.5
ƒ16.5x + 828 − 13.8x = 895.5, so 2.7x = 67.5, so x = 67.52.7 = 25
There are 25 males and 35 females, which agrees with our guess that females are more.
Part (iii) · A 15.2 kg female is admitted
The total weight of 35 females is 13.8 × 35 = 483 kg. Now there are 36 females.
ƒ483 + 15.236 = 498.236 ≈ 13.84 kg
Part (iv) · Two males (16.9 kg and 16.1 kg) are released
The total male weight is 16.5 × 25 = 412.5 kg. The two released weigh 16.9 + 16.1 = 33 kg. Now there are 23 males.
ƒ412.5 − 3323 = 379.523 = 16.5 kg
The average does not change, because the two released langurs average 332 = 16.5 kg, the same as the group.
Part (v) · One male loses 1 kg
Continuing from part (iv): the total male weight becomes 379.5 − 1 = 378.5 kg for 23 males.
ƒ378.523 ≈ 16.46 kg
If we start from the original 25 males instead, we get 411.525 = 16.46 kg, the same to two decimals.
(i) (a); (ii) 25 male langurs; (iii) 13.84 kg; (iv) 16.5 kg; (v) 16.46 kg.
Salinity
Q8
Chapter 10 Exercise 10.3 Question 8 Solution: Dorjee has 1 litre of Dead Sea water (salinity ≈ 34%). Other sources: groundwater ≈ 0.01% and purified drinking water ≈ 0.001%.
Answer: 11.334%, 3777 L, not possible
Part (i) · 1 L Dead Sea + 2 L purified drinking water
ƒ34 × 1 + 0.001 × 21 + 2 = 34.0023 = 11.334%
Part (ii) · Can we reach the salinity of groundwater (0.01%)?Possible
A mixture of two waters always has a salinity between the two. Here 0.001% < 0.01% < 34%, so yes, it is possible. Let x litres of purified water be added to 1 L of Dead Sea water.
ƒ34 + 0.001x1 + x = 0.01
ƒ34 + 0.001x = 0.01 + 0.01x, so 33.99 = 0.009x, so x = 33.990.009 ≈ 3776.67
About 3777 litres of purified water are needed. The answer is large because 34% is so much saltier than 0.01%.
Part (iii) · Can we reach 0.001% using Dead Sea water and groundwater?Not possible
Both ingredients are saltier than 0.001% (34% and 0.01%). Any mixture lies between 0.01% and 34%, so it can never be as low as 0.001%. It is not possible, no matter how much groundwater is added.
Diagram
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Q8 · a mixture always lies between the two things mixed
(i) 11.334%; (ii) possible, about 3777 L of purified water; (iii) not possible, because the mixture can never be less salty than groundwater.
Answers at a glance
Each answer is a total divided by a total weight, so the result always lies between the smallest and largest values.
Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 10, Exercise 10.3. The solutions, explanations and diagrams on this page are our own working.