# CBSE 10th Math Sample Paper 2019 | Q28A

#### Median and Frequency

This 2019 CBSE class 10 Maths 4 mark question is from Statistics. An important 4-mark question from statistics in CBSE class 10 sample question paper. Concept: Computing median when a frequency distribution table is given.

Question 28A: The median of the following data is 525. Find the values of x and y if the total frequency is 100.

Class Interval Frequency Class Interval Frequency
0 - 100 2 500 - 600 20
100 - 200 5 600 - 700 y
200 - 300 x 700 – 800 9
300 - 400 12 800 - 900 7
400 - 500 17 900 - 1000 4

## NCERT Solution to Class 10 Maths

#### With Videos

Class Interval Frequency Cumulative Frequency
0 - 100 2 2
100 - 200 5 7
200 - 300 x 7 + x
300 - 400 12 19 + x
400 - 500 17 36 + x
500 - 600 20 56 + x
600 - 700 y 56 + x + y
700 - 800 9 65 + x + y
800 - 900 7 72 + x + y
900 - 1000 4 76 + x + y

It is given that n = 100
So, 76 + x + y = 100
i.e., x + y = 24 ------ 1

The median is 525, which lies in the class 500 – 600
Median = l + [$$frac{\frac{n}{2} − cf}{f}] \times h$ So, l = 500, f = 20, cf = 36 + x, h = 100 Using the formula: Median = l + [$\frac{\frac{n}{2} − cf}{f}] \times h$ 525 = 500 + $\frac{50 − 36 − x}{20}$ × 100 525 – 500 =$14 – x) × 5
25 = 70 – 5x
5x = 70 – 25 = 45
x = 9
From (1), x + y = 24 => 9 + y = 24
y = 15

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