JEE Main 2024 · 9 April · morning · Single correct · +4 / −1
The frequency distribution of the age of students in a class of 40 students is given below.
If the mean deviation about the median is 1.25, then 4x+5y is equal to :
| Age | 15 | 16 | 17 | 18 | 19 | 20 |
|---|---|---|---|---|---|---|
| No of Students | 5 | 8 | 5 | 12 | x | y |
- (a) 43
- (b) 46
- (c) 44
- (d) 47
Answer and solution
Answer (c) 44
Solution
Maths sir · 4 steps
The median is 18 whatever x and y are, so the mean deviation gives a linear equation.
- Use the total. 5 + 8 + 5 + 12 + x + y = 40, so x + y = 10.
- Find the median. With 40 students, the median is the average of the 20th and 21st ages. The cumulative frequencies up to ages 15, 16, 17 and 18 are 5, 13, 18 and 30. So the 20th and 21st students are both 18 years old, and the median is 18 whatever x and y are.
Age f |Age - 18| f × |Age - 18| 15 5 3 15 16 8 2 16 17 5 1 5 18 12 0 0 19 x 1 x 20 y 2 2y (36 + x + 2y)/40 = 1.25 ⇒ x + 2y = 14
- Solve both. Subtract x + y = 10 from x + 2y = 14 to get y = 4. Then x = 6.
- So. 4x + 5y = 24 + 20 = 44, which is option (c).
The idea to keep. When unknown frequencies appear, first check whether they can shift the median. If they cannot, the mean deviation about the median turns into a linear equation in the unknowns.
Answer from the NTA final answer key. Solution by PrepWiser.