MathsCBSE Class 10 · Standard · Board exam 2024, set 30/1/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2024: question 37

Set in the CBSE Class 10 board exam of 2024, paper 30/1/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedStatistics

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below shows the data of one such game where 48 balls were used before Tara said 'BINGO'.

Numbers announced	Number of times
0-15	8
15-30	9
30-45	10
45-60	12
60-75	9

Based on the above information, answer the following:

(i)
Write the median class.
1 mark
(ii)
When first ball was picked up, what was the probability of calling out an even number?
1 mark
(iii)
(a) Find median of the given data.
2 marks
OR
(iii)
(b) Find mode of the given data.
2 marks

Figure in the paper: Photograph-style image of BINGO balls labelled B, I, N, G, O over a background of numbered bingo cards. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Make a cumulative frequency column and find N/2.

Hint 2

For the probability, count the even numbers among 1 to 75 (all 75 balls are available at the first draw).

Hint 3

Use Median = l + ((N/2 − cf)/f)·h, and Mode = l + ((f₁ − f₀)/(2f₁ − f₀ − f₂))·h with the modal class 45–60.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Cumulative frequencies:
Class 0–15 15–30 30–45 45–60 60–75
f 8 9 10 12 9
cf 8 17 27 39 48
$N = 48$
(i) $\frac{N}{2} = 24$ . The first cf greater than 24 is 27, in class 30–45.
So the median class is 30–45.
1 mark
(ii) Balls are numbered 1 to 75, so there are 75 equally likely outcomes.
Even numbers are $2, 4, \dots, 74$ , which is 37 numbers.
$P (even) = \frac{37}{75}$
1 mark
(iii)(a) For the median class 30–45: $l = 30$ , $c f = 17$ , $f = 10$ , $h = 15$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 30 + \frac{24 - 17}{10} \times 15$
1 mark
(iii)(a) $= 30 + 10.5 = 40.5$
1 mark
(iii)(b) The highest frequency is 12, so the modal class is 45–60.
Here $l = 45$ , $f_{1} = 12$ , $f_{0} = 10$ , $f_{2} = 9$ , $h = 15$ .
½ mark
(iii)(b) Mode $= l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h = 45 + \frac{12 - 10}{24 - 10 - 9} \times 15$
1 mark
(iii)(b) $= 45 + \frac{2}{5} \times 15 = 45 + 6 = 51$
½ mark

Class	0–15	15–30	30–45	45–60	60–75
f	8	9	10	12	9
cf	8	17	27	39	48

Answer: (i) 30–45; (ii) 37/75; (iii)(a) 40.5; (iii)(b) 51

Where the marks are

(i) Uses $N /2 = 24$ to identify the median class 30–45 (1 mark)
(ii) Counts 37 even numbers out of 75 and gets probability $37/75$ (1 mark)
(iii)(a) Substitutes correctly in the median formula (1 mark)
(iii)(a) Gets median $= 40.5$ (1 mark)
(iii)(b) Identifies the modal class 45–60 (½ mark)
(iii)(b) Substitutes correctly in the mode formula (1 mark)
(iii)(b) Gets mode $= 51$ (½ mark)

Revise the chapter

Chapter: Statistics (notes, formulas, where marks are lost and every past question)
Topic: Median of Grouped Data

Practise this chapterBoard-style questions from Statistics, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 30/1/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2024: question 37

Hints

Step-by-step solution

Revise the chapter

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