CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard

Question 35 Important

CBSE Class 10 Sample Paper for 2023 Boards - Maths Standard →

Class 10

Solutions of Sample Papers for Class 10 Boards

Question 35 - CBSE Class 10 Sample Paper for 2024 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at Dec. 13, 2024 by Teachoo

The median of the following data is 50. Find the values of ‘p’ and ‘q’, if the sum of all frequencies is 90. Also find the mode of the data.

Marks obtained	Number of students
20 - 30	p
30 - 40	15
40 - 50	25
50 - 60	20
60 - 70	q
70 - 80	8
80 - 90	10

This question is similar to Ex 13.3, 2 – Chapter 13 Class 10

Please check https://www.teachoo.com/1947/559/Ex-14.3--2---If-median-is-28.5--find-values-of-x-and-y./category/Ex-14.3/

Something went wrong!

The video couldn't load due to a technical hiccup.
But don't worry — our team is already on it, and we're working hard to get it back up ASAP.

Thanks for bearing with us!

Something went wrong!

The video couldn't load due to a technical hiccup.
But don't worry — our team is already on it, and we're working hard to get it back up ASAP.

Thanks for bearing with us!

Next: Question 36 (i) [Case based] → Share on WhatsApp

Transcript

Now, Median = l + (𝑵/𝟐 −𝒄𝒇)/𝒇 × h Since Median = 50 50 – 60 is Median Class Where N = ∑▒𝑓𝑖 l = h = cf = f = Putting values in formula Median = l + (𝑵/𝟐 −𝒄𝒇)/𝒇 × h 50 = 50 + (𝟒𝟓 − (𝒑 + 𝟒𝟎))/𝟐𝟎 × 10 50 = 50 + (5 −𝑝)/2 50 – 50 = (5 −𝑝)/2 0 = (5 −𝑝)/2 5−𝑝 = 0 p = 5 Also, ∑▒𝒇𝒊 = 78 + p + q Putting values 90 = 78 + 5 + q 90 = 83 + y 90 – 83 = q 7 = q q = 7 Hence p = 5, q = 7 Finding Mode Mode = l + (𝒇𝟏 −𝒇𝟎)/(𝟐𝒇𝟏 −𝒇𝟎 −𝒇𝟐) × h Modal class = Interval with highest frequency = 40 – 50 Here l = lower limit of modal class = 40 h = class-interval = 50 − 40 = 10 f1 = frequency of the modal class = 25 f0 = frequency of the class before modal class = 15 f2 = frequency of the class after modal class = 20 Putting these values in formula ∴ Mode = l + ((𝒇_𝟏 − 𝒇_𝟎)/(𝟐𝒇_𝟏 − 𝒇_𝟎 − 𝒇_𝟐 )) × h = 40 + {(𝟐𝟓 −𝟏𝟓)/(𝟐 ×𝟐𝟓 −𝟏𝟓 −𝟐𝟎)} × 10 = 40 + 10/(50−35) × 10 = 40 + 100/15 = 40 + 6.67 = 46.67

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo