CBSE Class 8 Mathematics Data Handling Worksheet Set 01

Very Short Answer type Questions

Question. One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a face card.
Answer : 3/13

Question. Out of 400 bulbs in a box, 15 bulbs are defective. One bulb is taken at random from the box. Find the probability that the bulb is not defective.
Answer : 77/80

Question. What is the total allocation of degrees in a pie chart?
Answer : 360°

Question. Find the probability of getting even number between 10 to 25.
Answer : 1/2

Question. Two dice are thrown simultaneously. Find the probability of getting an even number as the sum.
Answer : 1/2

Question. A die is thrown once. Find the probability of getting a prime number.
Answer : 1/2

Question. A class has 40 students, 25 boys and 15 girls. A student of the class is selected at random as the monitor. Find the probability of selecting a girl student as the monitor.
Answer : 3/8

Question. What is the maximum value of the probability for an event?
Answer : 1

Question. From 1 to 50 find the probability of getting a multiple of 6.
Answer : 4/25

Question. A die is thrown once. Find the probability of getting a number lying between 2 and 6.
Answer : 1/2

Question. Complete the following :
The representation of statistical data by means of circle is known as ___________.
Answer : Pie Chart

Question. A die is thrown once. What is the probability of getting a multiple of 3.
Answer : 1/3

Question. Complete the following –
In pie chart we transform the data in terms of __________.
Answer : Angles

Question. A bag contains 9 red, 7 white and 4 black balls. A ball is drawn at random. Find the probability that the ball drawn is not red.
Answer : 11/20

Question. Three coins are tossed together. What are the total number of possible outcomes.
Answer : 8

Question 1. What is the probability that a number selected from the numbers 1, 2, 3, .........., 25 is a prime number, when each of the given numbers is equally likely to selected.
Answer: Total number of possible outcomes = 25
Prime numbers between 1 and 25 are: 2, 3, 5, 7, 11, 13, 17, 19, 23
Number of favorable outcomes (prime numbers) = 9
Probability = \(\frac{9}{25}\)
In simple words: Count how many prime numbers exist from 1 to 25, then divide by the total number 25.

Exam Tip: Remember that 1 is neither prime nor composite. Start counting primes from 2.

Question 2. Tickets numbered from 1 to 20 are mixed up together and then a ticket is drown at random. What is the probability that the ticket has a number which is a multiple of 3 or 7.
Answer: Total number of tickets = 20
Multiples of 3 between 1 and 20 = 3, 6, 9, 12, 15, 18 (Total 6 numbers)
Multiples of 7 between 1 and 20 = 7, 14 (Total 2 numbers)
Since there are no common multiples of 3 and 7 between 1 and 20, the total favorable outcomes = 6 + 2 = 8
Probability = \(\frac{8}{20} = \frac{2}{5}\)
In simple words: Find all numbers up to 20 that divide by 3 or 7, count them, and divide by 20.

Exam Tip: Always check if any numbers are shared between the two sets (though here there are none) to avoid double-counting.

Question 3. 17 cards numbered 1, 2, 3 ...... 17 are put in a box and mixed thoroughly. One person draws a card from the box. Find the probability that the number on the card is (a) odd (b) a prime (c) divisible by 3 (d) divisible by 3 and 2 both
Answer: Total number of cards = 17
(a) Odd numbers from 1 to 17 are: 1, 3, 5, 7, 9, 11, 13, 15, 17 (Total 9 numbers)
Probability of getting an odd number = \(\frac{9}{17}\)

(b) Prime numbers from 1 to 17 are: 2, 3, 5, 7, 11, 13, 17 (Total 7 numbers)
Probability of getting a prime number = \(\frac{7}{17}\)

(c) Numbers divisible by 3 from 1 to 17 are: 3, 6, 9, 12, 15 (Total 5 numbers)
Probability = \(\frac{5}{17}\)

(d) Numbers divisible by both 3 and 2 (i.e., multiples of 6) from 1 to 17 are: 6, 12 (Total 2 numbers)
Probability = \(\frac{2}{17}\)
In simple words: Analyze each condition separately, count the matching cards from 1 to 17, and put that count over 17.

Exam Tip: Divisible by both 2 and 3 means the number must be a multiple of their least common multiple, which is 6.

Question 4. A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random. Find the probability that it is (a) Black (b) red (c) not green
Answer: Total number of balls = \(5 + 8 + 4 + 7 = 24\)
(a) Number of black balls = 7
Probability of getting a black ball = \(\frac{7}{24}\)

(b) Number of red balls = 5
Probability of getting a red ball = \(\frac{5}{24}\)

(c) Number of green balls = 4
Number of balls that are not green = Total balls - Green balls = \(24 - 4 = 20\)
Probability of getting a non-green ball = \(\frac{20}{24} = \frac{5}{6}\)
In simple words: First find the total number of balls. Then divide the number of target balls by the total.

Exam Tip: For 'not' events, you can either subtract the favorable count from total or sum all other color counts.

Question 5. A child has a block in the shape of a cube with one letter written on each face as shown below: A B C D E A. The cube is thrown once. What is the probability of getting (i) A (ii) D
Answer: Total faces on the cube = 6
(i) Letter A appears 2 times on the cube.
Probability of getting A = \(\frac{2}{6} = \frac{1}{3}\)

(ii) Letter D appears 1 time on the cube.
Probability of getting D = \(\frac{1}{6}\)
In simple words: Count how many times the specific letter appears on the 6 faces of the cube.

Exam Tip: A standard cube has 6 faces, so the denominator is always 6 even if letters repeat.

Question 6. A letter is chosen at random from the letters of the word ASSASSINATION. Find the probability that the letter chosen is a (i) vowel (ii) consonant
Answer: Total number of letters in ASSASSINATION = 13 letters
Let us list the letters: A, S, S, A, S, S, I, N, A, T, I, O, N
Vowels in the word: A, A, A, I, I, O (Total 6 vowels)
Consonants in the word: S, S, S, S, N, T, N (Total 7 consonants)
(i) Probability of choosing a vowel = \(\frac{6}{13}\)
(ii) Probability of choosing a consonant = \(\frac{7}{13}\)
In simple words: Count every single letter in the word to get the total 13, then count how many are vowels and consonants.

Exam Tip: Count repeated letters individually as they appear in the given word unless stated otherwise.

Question 7. The number of members in 20 families of a township are 6, 8, 4, 3, 5, 6, 7, 4, 3, 4, 5, 6, 4, 5, 4, 3, 3, 6, 4 and 3. Prepare a frequency distribution table for the data and answer the following questions: (i) What is the smallest family size ? How many families are of this size ? (ii) What is the largest family size ? How many are of this size ? (iii) What is the most common family size ?
Answer:
Frequency distribution table:

(i) Smallest family size = 3 members. Number of families of this size = 6.
(ii) Largest family size = 8 members. Number of families of this size = 1.
(iii) Most common family size (highest frequency) = 4 members (frequency 7).
In simple words: Count how many times each family size appears and organize it into a table to easily read the smallest, largest, and most frequent values.

Exam Tip: Always verify that the sum of all frequencies equals the total number of observations given in the problem.

Question 8. The heights of 10 girls were measured in cm and the results were as follows: 143, 148, 135, 150, 128, 139, 149, 146, 151, 132. (i) What is the height of the tallest girl? (ii) What is the height of the shortest girl? (iii) What is the range of the data? (iv) Find the mean height. (v) How many girls are there whose heights are less than the mean height?
Answer:
Arranging the heights in ascending order: 128, 132, 135, 139, 143, 146, 148, 149, 150, 151
(i) Height of the tallest girl = 151 cm
(ii) Height of the shortest girl = 128 cm
(iii) Range = Highest value - Lowest value = \(151 - 128 = 23\) cm
(iv) Mean height = \(\frac{\text{Sum of all heights}}{\text{Total number of girls}}\)
Sum = \(128 + 132 + 135 + 139 + 143 + 146 + 148 + 149 + 150 + 151 = 1421\) cm
Mean height = \(\frac{1421}{10} = 142.1\) cm
(v) Heights less than 142.1 cm are: 128, 132, 135, 139
Number of girls whose heights are less than the mean height = 4
In simple words: Sort the numbers first to easily find the min, max, range, and average. Then count how many values fall below the average.

Exam Tip: Range is always maximum minus minimum, never the other way around.

Question 9. The following data give the pocket expenses of 100 students of a school: Weekly pocket expenses (in rupees): 30, 35, 45, 50, 55, 60, 65. Number of students: 6, 10, 14, 22, 35, 9, 4. Prepare a grouped frequency distribution of class intervals of equal width, taking one of the class intervals as 30 - 40.
Answer:
Given class width = 10 (since 30 - 40 has a width of 10).
Grouped frequency distribution table with class intervals of width 10:

In simple words: Group the individual pocket money values into ranges of 10 (like 30 to 40, 40 to 50) and add up their student counts.

Exam Tip: In continuous grouped frequency tables, upper limit values (like 40) are included in the next higher class interval (40 - 50).

Question 10. The following distribution table shows the performance of 270 candidates appearing for Army Education Corps intelligence test. I.Q: 55 - 69, 69 - 83, 83 - 97, 97 - 111, 111 - 125. No. of candidates: 20, 50, 75, 75, 50. Draw a histogram for this distribution.
Answer:
To draw a histogram for this distribution:
1. Take I.Q scores along the horizontal X-axis on a suitable scale (e.g., 1 cm = 14 units, starting with a kink or break mark from 0 to 55 since data starts at 55).
2. Take the number of candidates along the vertical Y-axis on a suitable scale (e.g., 1 cm = 10 candidates).
3. Construct adjacent rectangular bars where the width of each bar corresponds to the class interval width (14) and the height corresponds to the respective frequency.
In simple words: Draw touching bars on a graph where the width represents the I.Q range and the height represents the number of candidates.

Exam Tip: Since the X-axis does not start at 0 (starts at 55), remember to put a kink symbol (zigzag mark) on the X-axis near the origin.