CBSE Class 10 Sample Paper for 2019 Boards
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CBSE Class 10 Sample Paper for 2019 Boards
Last updated at July 20, 2026 by Teachoo
Question 28 (OR 1 st question)
The median of the following data is 525. Find the values of x and y if the total frequency is 100.
|
Class Interval |
Frequency |
|
0 – 100 |
2 |
|
100 – 200 |
5 |
|
200 – 300 |
x |
|
300 – 400 |
12 |
|
400 – 500 |
17 |
|
500 – 600 |
20 |
|
600 – 700 |
y |
|
700 – 800 |
9 |
|
800 – 900 |
7 |
|
900 – 100 |
4 |
Transcript
Question 28 (OR 1st question) The median of the following data is 525. Find the values of x and y if the total frequency is 100. Given that total frequency is 100 So, Sum of frequency = 100 2 + 5 + x + 12 + 17 + 17 + 20 + y + 8 + 7 + 4 = 100 76 + x + y = 100 x + y = 100 – 76 x + y = 24 Now, to find values of x and y, We first find median Since Median is 525 500 – 600 is the median class Median = l + (𝑛/2 − 𝑐𝑓)/𝑓 × h Where l = lower limit of median class = 500 h = class-interval = 100 − 0 = 100 n = ∑▒𝑓𝑖 = 100 cf = cumulative frequency of the class before median class = 36 + x f = frequency of the median class = 20 Now, Median = l + (𝑛/2 −𝑐𝑓)/𝑓 × h 525 = 500 + (100/2 − (36 + 𝑥))/20 × 100 525 – 500 = (50 − (36 + 𝑥))/20 × 100 25 = (14 − 𝑥)/20 × 100 25 = (14 – x) × 5 25 = 70 – 5x 5x = 70 – 25 5x = 45 x = 45/9 x = 5 Now, from (1) x + y = 24 Putting x = 5 5 + y = 24 y = 24 – 5 y = 19 Thus, x = 5, y = 19
