Step-by-step solutions for Class 9 Mathematics Ganita Manjari Part 1I Chapter 10 Exercise Set 10.3 (Questions 1–8). Covers weighted mean with custom weights, combining profit percentages, updating moving daily averages, concentrations of chemical and salt mixtures, dilution problems, weighted customer ratings, and animal weight data.
When different components have different significance or volume (such as sales amounts, customer rating categories, or fitness criteria), each component value is multiplied by its custom weight and divided by the total sum of weights.
When the target combined concentration or average is specified, set up an algebraic equation using the weighted mean formula and solve for the unknown quantity or group count.
When two solutions with concentrations c₁ and c₂ (c₁ < c₂) are mixed together, the resulting concentration c must satisfy c₁ ≤ c ≤ c₂. A concentration outside this range cannot be obtained by mixing.
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?
Since the two sales amounts are different, we use the weighted mean formula from Chapter 10.
Here, the weights are the respective sales amounts.
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.
The average daily distance for the first 20 days is 44.5 km, and the distance travelled on the 21st day is 55 km.
Guess: The new average should be slightly greater than 44.5 km because the stork travelled 55 km on the 21st day.
Using the weighted mean formula:
where a, b are the averages and p, q are the respective numbers of days.
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?
First solution = 600 mL, with 5% salt.
Second solution = 300 mL, with 8% sugar.
Total volume of the mixture:
We use the weighted mean formula to calculate the concentration of each substance separately.
The first solution contains 5% salt, while the second solution contributes no salt.
The first solution contributes no sugar, while the second solution contains 8% sugar.
Correct option: (v)
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?
The initial concentration of spice is 8%, and the required concentration is 34 of the original concentration.
Let x litres of regular water be added. Since regular water contains no spice, its spice concentration is 0%.
Using the weighted mean formula:
6 = 10 × 8 + x × 010 + x
Regular water to be added = 3 13 litres
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.
The ratio of weights for strength, flexibility and agility is 4:5:6.
Rashi scored 60, 65, 70, while Keerthi scored 55, 65, 75, respectively.
Both scored 65 marks in flexibility.
Rashi scored 5 marks more than Keerthi in strength, but Keerthi scored 5 marks more than Rashi in agility. Since agility has a greater weight (6) than strength (4), Keerthi's advantage in agility outweighs Rashi's advantage in strength.
Keerthi's final marks will be higher.
Using the weighted mean formula:
Rashi's final marks = 65 23 ≈ 65.67
Keerthi's final marks = 66 13 ≈ 66.33
A restaurant collected ratings from 10 customers on a scale of 1 to 5. The resulting data is shown in the table below.
| 5★ | 4★ | 3★ | 2★ | 1★ | |
|---|---|---|---|---|---|
| Food | 5 | 3 | 2 | 0 | 0 |
| Ambience | 0 | 4 | 5 | 1 | 0 |
| Service | 1 | 2 | 2 | 4 | 1 |
What is the average rating if the metrics are combined with the weights food : ambience : service = 6:5:4?
The ratings from 10 customers are given in the question's table. The weights for food, ambience and service are in the ratio 6:5:4.
The average rating for each metric is:
Using the weighted mean formula to combine the three average ratings:
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?
| Male langurs | Female langurs | All langurs | |
|---|---|---|---|
| Average weight | 16.5 kg | 13.8 kg | 14.925 kg |
| Number of langurs | 60 |
Before calculating: The combined average is closer to the average weight of female langurs than to that of male langurs. Therefore, there are more female langurs than male langurs.
Let the number of male langurs be x and the number of female langurs be y.
Using the weighted mean formula:
Correct option: (a)
Since the total number of langurs is 60:
x + y = 60 ⇒ y = 60 − x
Using the combined average:
Number of female langurs:
Male langurs = 25, Female langurs = 35
The total weight of the 35 female langurs is:
After admitting one more female langur:
The initial total weight of the 25 male langurs is:
The total weight of the two released langurs is:
The remaining total weight and number of male langurs are:
Taking this as a continuation of part (iv), 23 male langurs remain, with a total weight of 379.5 kg.
After one langur loses 1 kg, the new total weight becomes:
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.
| Water Source | Salinity |
|---|---|
| Dead Sea | ≈ 34% |
| Other seas and oceans | ≈ 3.5% |
| Ground water | ≈ 0.01% |
| Purified drinking water | ≈ 0.001% |
The table in the textbook gives these approximate salinities:
Yes, it is possible because the required salinity, 0.01%, lies between 34% and 0.001%.
Let y litres of purified drinking water be mixed with 1 litre of Dead Sea water.
Using the weighted mean formula:
0.01 = 1 × 34 + y × 0.0011 + y
No, it is not possible.
The salinity of Dead Sea water is 34%, while the salinity of groundwater is 0.01%. When two liquids are mixed, the resulting salinity must lie between their individual salinities.
But purified drinking water has a salinity of 0.001%, which is lower than both 34% and 0.01%.
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