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?
- Opening question 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 + 13.8yx + y = 14.925 (b) 16.5x + 13.8y60 = – 14.925
(c) 16.5x + 13.8y2 = – 14.925 (d) 16.5x + 13.8y16.5 + 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?
Step-by-step solution
| Male langurs | Female langurs | All langurs | |
|---|---|---|---|
| Average weight | 16.5 kg | 13.8 kg | 14.925 kg |
| Number of langurs | 60 |
Idea: The overall average is a weighted mean of the male and female averages, weighted by their numbers. It lies closer to the group with more members. Adding or removing animals changes the total weight and the count.
Opening question Without doing any computations, can you tell whether there are more male langurs or more female langurs? Or are they equal in number?
- If the numbers were equal, the overall average would be midway: 16.5 + 13.82 = 15.15 kg.½ mark
- The overall average 14.925 kg is below the midway value, closer to the female average 13.8 kg. The larger group pulls the average towards itself, so there are more female langurs.½ mark
(i) Which of the following expression(s) describes the given scenario?
(a) 16.5x + 13.8yx + y = 14.925 (b) 16.5x + 13.8y60 = – 14.925
(c) 16.5x + 13.8y2 = – 14.925 (d) 16.5x + 13.8y16.5 + 13.8 = – 14.925
- Let x = number of males, y = number of females. Total weight ÷ total number = overall average: 16.5x + 13.8yx + y = 14.925, which is (a). (b), (c) and (d) have a minus sign on the right, and a weighted mean of positive weights cannot be negative; (c) and (d) also divide by the wrong number. So, as printed, only (a) fits. The minus signs look like a printing slip: without the minus sign, (b) is also correct, because x + y = 60. (c) and (d) are wrong either way.½ mark
(ii) Find out how many male langurs are present.
- x + y = 60, so y = 60 − x. Total weight = 14.925 × 60 = 895.5 kg: 16.5x + 13.8(60 − x) = 895.5.1 mark
- 16.5x + 828 − 13.8x = 895.5 ⇒ 2.7x = 67.5 ⇒ x = 25. There are 25 male (and 35 female) langurs.½ mark
(iii) A female langur weighing 15.2 kg is admitted to the facility. What is the average weight of the female langurs after this?
- Total weight of the 35 females = 35 × 13.8 = 483 kg. After admission: 483 + 15.2 = 498.2 kg for 36 females.½ mark
- New average = 498.236 ≈ 13.84 kg.½ mark
(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?
- Total weight of 25 males = 25 × 16.5 = 412.5 kg. After release: 412.5 − 16.9 − 16.1 = 379.5 kg for 23 males, so the average = 379.523 = 16.5 kg. It does not change, because the two released langurs together weigh 33 kg = 2 × 16.5, exactly the average.½ mark
(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?
- There are now 23 males with total 379.5 kg. Losing 1 kg makes it 378.5 kg, so the average = 378.523 ≈ 16.46 kg (it falls by 123 kg). (If the 1 kg loss is applied to the original 25 males instead, the average is 16.5 − 125 = 16.46 kg too.)½ mark
Check: (ii): 16.5 × 25 + 13.8 × 3560 = 412.5 + 48360 = 895.560 = 14.925 ✓.
Answer to write in the exam
Opening question
Midway value = 16.5 + 13.82 = 15.15 kg
14.925 < 15.15, so overall average is closer to 13.8 kg (females)
∴ There are more female langurs
(i)
Average of all = total weighttotal number
16.5x + 13.8yx + y = 14.925
(b), (c), (d) equal a negative number, which is impossible for an average weight
∴ Expression (a) describes the scenario
(ii)
Let males = x, females = 60 − x
16.5x + 13.8(60 − x) = 14.925 × 60 = 895.5
2.7x + 828 = 895.5 ⇒ 2.7x = 67.5 ⇒ x = 25
∴ Number of male langurs = 25
(iii)
Total weight of 35 females = 35 × 13.8 = 483 kg
New total = 483 + 15.2 = 498.2 kg for 36 females
Average = 498.236 = 13.84
∴ New average weight of females ≈ 13.84 kg
(iv)
Total weight of 25 males = 25 × 16.5 = 412.5 kg
New total = 412.5 − 16.9 − 16.1 = 379.5 kg for 23 males
Average = 379.523 = 16.5
∴ Average weight of males = 16.5 kg
(v)
Total weight of 23 males = 379.5 − 1 = 378.5 kg
Average = 378.523 = 16.46
∴ Average weight of males ≈ 16.46 kg
Common mistakes that cost marks
- Saying there are more males because 16.5 is the bigger number. It is the closeness of 14.925 to 13.8 that shows the larger group.
- In (iii), dividing 498.2 by 35 instead of the new number of females, 36.
- In (v), subtracting 1 kg from the average instead of from the total, giving 15.5 kg.
How this can come in the exam
In a class, the boys’ average weight is 40 kg, the girls’ is 35 kg and the whole class’s is 38 kg. Which is true?
- There are more boys than girls
- There are more girls than boys
- There are equal numbers of boys and girls
- Nothing can be said
Show answer
(A) There are more boys than girls
38 is closer to 40 than to 35, so the boys’ group is larger. (Boys : girls = (38 − 35) : (40 − 38) = 3 : 2.)
A class of 30 students has boys averaging 50 marks and girls averaging 60 marks. The class average is 56. How many girls are there?
Show answer
Let boys = b: 50b + 60(30 − b) = 56 × 30 = 1680 (1 mark) ⇒ 1800 − 10b = 1680 ⇒ b = 12, so girls = 18 (1 mark).Try one yourself
A park has 50 deer. Males average 80 kg, females 60 kg, and all the deer together average 68 kg. How many males are there?
Show answer
80m + 60(50 − m) = 68 × 50 = 3400 ⇒ 20m = 400 ⇒ m = 20 males (30 females).
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