NCERT Solutions for Class 8 Maths Chapter 4: Data Handling — Complete Guide

In an ungrouped frequency distribution, each distinct data value gets its own row, and we count how many times it appears (its frequency).

Example: The marks obtained by 20 students in a quiz are: $5, 8, 7, 5, 6, 8, 9, 7, 5, 8, 6, 7, 9, 5, 8, 7, 6, 8, 7, 9$.

Now we can instantly see that marks $7$ and $8$ were the most common (mode = $7$ and $8$).

When data has a wide range of values, we group them into class intervals (also called classes or bins).

Example: Heights (in cm) of 25 students: $138, 142, 151, 147, 155, 161, 143, 149, 158, 163, 145, 152, 141, 156, 148, 162, 144, 153, 159, 146, 157, 140, 150, 160, 154$.

Key Terms:
- Class interval (or class width): The range of each group. Here it is $5$.
- Lower class limit: The smallest value in the interval (e.g., $138$ in $138-143$).
- Upper class limit: The largest value in the interval (e.g., $143$ in $138-143$).
- Class mark (midpoint): $\dfrac{\text{Lower limit} + \text{Upper limit}}{2}$.

In the convention used in NCERT, the upper limit of one class equals the lower limit of the next (e.g., $138-143, 143-148$). A value like $143$ is included in the class $143-148$, not $138-143$.

Problem: The following data shows the monthly expenditure of a family. Draw a pie chart.

Solution:

Total $= 3000 + 2400 + 1200 + 1400 = 8000$

Central angles:
- Food $= \dfrac{3000}{8000} \times 360° = \dfrac{3}{8} \times 360° = 135°$
- Rent $= \dfrac{2400}{8000} \times 360° = \dfrac{3}{10} \times 360° = 108°$
- Education $= \dfrac{1200}{8000} \times 360° = \dfrac{3}{20} \times 360° = 54°$
- Others $= \dfrac{1400}{8000} \times 360° = \dfrac{7}{40} \times 360° = 63°$

Verification: $135° + 108° + 54° + 63° = 360°$ ✓

Draw a circle, use a protractor to mark sectors with these angles, and label each sector with the category name and amount.

Problem: A pie chart shows the favourite sports of $720$ students. The sector for Cricket has a central angle of $120°$, Football has $90°$, and Tennis has $60°$. How many students chose each sport? What angle does the remaining sector have, and how many students does it represent?

Solution:

Cricket: $\dfrac{120°}{360°} \times 720 = \dfrac{1}{3} \times 720 = 240$ students.

Football: $\dfrac{90°}{360°} \times 720 = \dfrac{1}{4} \times 720 = 180$ students.

Tennis: $\dfrac{60°}{360°} \times 720 = \dfrac{1}{6} \times 720 = 120$ students.

Remaining angle $= 360° - (120° + 90° + 60°) = 90°$.

Remaining students $= \dfrac{90°}{360°} \times 720 = 180$ students.

Verification: $240 + 180 + 120 + 180 = 720$ ✓

Problem: A survey of $500$ people found that $30\%$ prefer tea, $25\%$ prefer coffee, $20\%$ prefer juice, and the rest prefer milk. Draw a pie chart.

Solution:

Milk percentage $= 100\% - 30\% - 25\% - 20\% = 25\%$.

Central angles:
- Tea $= \dfrac{30}{100} \times 360° = 108°$
- Coffee $= \dfrac{25}{100} \times 360° = 90°$
- Juice $= \dfrac{20}{100} \times 360° = 72°$
- Milk $= \dfrac{25}{100} \times 360° = 90°$

Verification: $108° + 90° + 72° + 90° = 360°$ ✓

Number of people: Tea $= 150$, Coffee $= 125$, Juice $= 100$, Milk $= 125$. Total $= 500$ ✓.

Problem: The following table shows the marks obtained by $40$ students. Draw a histogram.

Solution:

Step 1: Mark class intervals ($0-10, 10-20, \ldots, 40-50$) on the x-axis.
Step 2: Mark frequencies ($0$ to $14$, with suitable scale) on the y-axis.
Step 3: Draw bars:
- $0-10$: height $= 4$
- $10-20$: height $= 8$
- $20-30$: height $= 12$
- $30-40$: height $= 10$
- $40-50$: height $= 6$

All bars touch each other. The tallest bar corresponds to the class $20-30$ (most students scored in this range).

Total verification: $4 + 8 + 12 + 10 + 6 = 40$ ✓

Problem: A histogram shows the weights (in kg) of students in a class. The bars have the following heights: $40-45$: $3$, $45-50$: $7$, $50-55$: $12$, $55-60$: $8$, $60-65$: $5$. Find: (a) The total number of students. (b) How many students weigh $50$ kg or more? (c) Which class interval has the highest frequency?

Solution:

(a) Total $= 3 + 7 + 12 + 8 + 5 = 35$ students.

(b) Students weighing $50$ kg or more: Classes $50-55$, $55-60$, $60-65$.
$12 + 8 + 5 = 25$ students.

(c) The class interval $50-55$ has the highest frequency ($12$), so it is the modal class.

Problem: Draw a histogram for the following data:

Solution:

The class intervals are unequal: $5, 5, 10, 20, 30$. The minimum class width is $5$.

For histograms with unequal class intervals, we adjust the frequency to frequency density:

• $10-15$: $\dfrac{6}{5} \times 5 = 6$
  - $15-20$: $\dfrac{8}{5} \times 5 = 8$
  - $20-30$: $\dfrac{12}{10} \times 5 = 6$
  - $30-50$: $\dfrac{10}{20} \times 5 = 2.5$
  - $50-80$: $\dfrac{9}{30} \times 5 = 1.5$

Draw bars with these adjusted heights. The widths of the bars will be proportional to the class intervals.

Note: This type of problem is less common in Class 8 CBSE exams but good to know for competitive exams.

Problem: Priya spends her day as follows: School $= 6$ hours, Homework $= 3$ hours, Play $= 2$ hours, Sleep $= 8$ hours, Others $= 5$ hours. Draw a pie chart.

Solution:

Total $= 6 + 3 + 2 + 8 + 5 = 24$ hours.

Central angles:
- School $= \dfrac{6}{24} \times 360° = 90°$
- Homework $= \dfrac{3}{24} \times 360° = 45°$
- Play $= \dfrac{2}{24} \times 360° = 30°$
- Sleep $= \dfrac{8}{24} \times 360° = 120°$
- Others $= \dfrac{5}{24} \times 360° = 75°$

Verification: $90° + 45° + 30° + 120° + 75° = 360°$ ✓

Problem: In a pie chart showing the colours of cars in a parking lot, the red sector has an angle of $80°$, blue has $100°$, white has $120°$, and black takes the rest. If there are $180$ cars in total, how many are black?

Solution:

Angle for black $= 360° - (80° + 100° + 120°) = 360° - 300° = 60°$.

Number of black cars $= \dfrac{60°}{360°} \times 180 = \dfrac{1}{6} \times 180 = 30$.

Answer: There are $30$ black cars.

Problem: The marks of $30$ students are: $23, 45, 67, 38, 52, 15, 71, 44, 59, 33, 27, 48, 62, 41, 55, 18, 73, 36, 51, 29, 64, 42, 57, 21, 49, 68, 35, 53, 46, 61$. Organise into a grouped frequency table with class intervals of $10$.

Solution:

The modal class is $40-50$ (highest frequency of $7$).

Problem: A histogram has bars with the following details: $0-10$ (height $5$), $10-20$ (height $8$), $20-30$ (height $15$), $30-40$ (height $12$), $40-50$ (height $10$). (a) Write the frequency table. (b) How many data points are there? (c) What fraction of data falls in the range $20-40$?

Solution:

(a) Frequency table:

(b) Total $= 5 + 8 + 15 + 12 + 10 = 50$.

(c) Data in $20-40 = 15 + 12 = 27$. Fraction $= \dfrac{27}{50}$.

A fair coin has two equally likely outcomes: Heads (H) and Tails (T).

$P(H) = \dfrac{1}{2}$ and $P(T) = \dfrac{1}{2}$.

For two coins tossed together, the sample space is: $\{HH, HT, TH, TT\}$ — $4$ outcomes.

For three coins: $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$ — $8$ outcomes.

In general, $n$ coins give $2^n$ outcomes.

A fair die has six equally likely outcomes: $\{1, 2, 3, 4, 5, 6\}$.

Some useful probabilities:
- $P(\text{even number}) = \dfrac{3}{6} = \dfrac{1}{2}$ (outcomes: $2, 4, 6$)
- $P(\text{prime number}) = \dfrac{3}{6} = \dfrac{1}{2}$ (outcomes: $2, 3, 5$)
- $P(\text{greater than 4}) = \dfrac{2}{6} = \dfrac{1}{3}$ (outcomes: $5, 6$)
- $P(\text{less than or equal to 3}) = \dfrac{3}{6} = \dfrac{1}{2}$ (outcomes: $1, 2, 3$)

For two dice thrown together, the total outcomes $= 6 \times 6 = 36$.

A standard deck has $52$ cards: $4$ suits (Hearts, Diamonds, Clubs, Spades) with $13$ cards each (A, 2-10, J, Q, K). Hearts and Diamonds are red; Clubs and Spades are black.

• $P(\text{red card}) = \dfrac{26}{52} = \dfrac{1}{2}$
  - $P(\text{king}) = \dfrac{4}{52} = \dfrac{1}{13}$
  - $P(\text{heart}) = \dfrac{13}{52} = \dfrac{1}{4}$
  - $P(\text{face card}) = \dfrac{12}{52} = \dfrac{3}{13}$ (J, Q, K of each suit)
  - $P(\text{ace of spades}) = \dfrac{1}{52}$

Problem: A die is thrown once. What is the probability of getting: (a) a number greater than $4$, (b) a number less than or equal to $2$, (c) the number $7$?

Solution:

Sample space $= \{1, 2, 3, 4, 5, 6\}$. Total outcomes $= 6$.

(a) Greater than $4$: $\{5, 6\}$. $P = \dfrac{2}{6} = \dfrac{1}{3}$.

(b) Less than or equal to $2$: $\{1, 2\}$. $P = \dfrac{2}{6} = \dfrac{1}{3}$.

(c) The number $7$: $\{\}$ (empty set — $7$ is not on a die). $P = \dfrac{0}{6} = 0$.

The event is impossible.

Problem: A bag contains $5$ red balls, $3$ blue balls, and $2$ green balls. A ball is drawn at random. Find the probability of: (a) a red ball, (b) not a green ball, (c) a blue or green ball.

Solution:

Total balls $= 5 + 3 + 2 = 10$.

(a) $P(\text{red}) = \dfrac{5}{10} = \dfrac{1}{2}$.

(b) $P(\text{not green}) = 1 - P(\text{green}) = 1 - \dfrac{2}{10} = \dfrac{8}{10} = \dfrac{4}{5}$.

(c) Blue or green balls $= 3 + 2 = 5$. $P(\text{blue or green}) = \dfrac{5}{10} = \dfrac{1}{2}$.

Problem: A coin is tossed $200$ times. Heads comes up $118$ times. Find the experimental probability of: (a) getting heads, (b) getting tails.

Solution:

(a) $P(\text{heads}) = \dfrac{118}{200} = \dfrac{59}{100} = 0.59$.

(b) Tails $= 200 - 118 = 82$. $P(\text{tails}) = \dfrac{82}{200} = \dfrac{41}{100} = 0.41$.

Note: The theoretical probability of each is $0.5$. The experimental values are close but not exactly $0.5$ — this is normal. As the number of trials increases, the experimental probability gets closer to the theoretical probability.

Problem: Two coins are tossed simultaneously. Find the probability of: (a) both heads, (b) at least one tail, (c) exactly one head.

Solution:

Sample space $= \{HH, HT, TH, TT\}$. Total outcomes $= 4$.

(a) Both heads: $\{HH\}$. $P = \dfrac{1}{4}$.

(b) At least one tail: $\{HT, TH, TT\}$. $P = \dfrac{3}{4}$.
Alternatively: $P(\text{at least one tail}) = 1 - P(\text{no tail}) = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.

(c) Exactly one head: $\{HT, TH\}$. $P = \dfrac{2}{4} = \dfrac{1}{2}$.

Problem: A spinner has $8$ equal sections numbered $1$ to $8$. Find the probability of: (a) getting an odd number, (b) getting a multiple of $3$, (c) getting a number greater than $6$.

Solution:

Total outcomes $= 8$. All equally likely.

(a) Odd numbers: $\{1, 3, 5, 7\}$. $P = \dfrac{4}{8} = \dfrac{1}{2}$.

(b) Multiples of $3$: $\{3, 6\}$. $P = \dfrac{2}{8} = \dfrac{1}{4}$.

(c) Greater than $6$: $\{7, 8\}$. $P = \dfrac{2}{8} = \dfrac{1}{4}$.

Problem: A card is drawn at random from a well-shuffled deck of $52$ cards. Find the probability of getting: (a) a red queen, (b) a face card, (c) not a king.

Solution:

(a) Red queens: Queen of Hearts and Queen of Diamonds $= 2$ cards.
$P(\text{red queen}) = \dfrac{2}{52} = \dfrac{1}{26}$.

(b) Face cards: J, Q, K of each suit $= 4 \times 3 = 12$ cards.
$P(\text{face card}) = \dfrac{12}{52} = \dfrac{3}{13}$.

(c) Kings $= 4$. Not a king $= 52 - 4 = 48$.
$P(\text{not a king}) = \dfrac{48}{52} = \dfrac{12}{13}$.
Alternatively: $1 - P(\text{king}) = 1 - \dfrac{4}{52} = 1 - \dfrac{1}{13} = \dfrac{12}{13}$.

Problem: A letter is chosen at random from the word MATHEMATICS. Find the probability that it is: (a) M, (b) a vowel, (c) a consonant.

Solution:

MATHEMATICS has $11$ letters: M, A, T, H, E, M, A, T, I, C, S.

(a) M appears $2$ times. $P(M) = \dfrac{2}{11}$.

(b) Vowels in MATHEMATICS: A, E, A, I $= 4$ vowels. $P(\text{vowel}) = \dfrac{4}{11}$.

(c) Consonants $= 11 - 4 = 7$. $P(\text{consonant}) = \dfrac{7}{11}$.

Verification: $P(\text{vowel}) + P(\text{consonant}) = \dfrac{4}{11} + \dfrac{7}{11} = 1$ ✓

Problem: Two dice are thrown together. Find the probability that the sum is: (a) $7$, (b) greater than $10$.

Solution:

Total outcomes $= 6 \times 6 = 36$.

(a) Pairs with sum $7$: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6$ outcomes.
$P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$.

(b) Pairs with sum $> 10$ (i.e., sum $= 11$ or $12$):
Sum $11$: $(5,6), (6,5) = 2$ outcomes.
Sum $12$: $(6,6) = 1$ outcome.
Total $= 3$ outcomes.
$P(\text{sum} > 10) = \dfrac{3}{36} = \dfrac{1}{12}$.

Problem: A school budget of Rs $2,00,000$ is allocated as: Teachers' salary $40\%$, Infrastructure $25\%$, Sports $15\%$, Library $10\%$, Others $10\%$. Find the central angles and the amount for each.

Solution:

Central angles: Salary $= 0.4 \times 360° = 144°$, Infrastructure $= 0.25 \times 360° = 90°$, Sports $= 0.15 \times 360° = 54°$, Library $= 0.1 \times 360° = 36°$, Others $= 0.1 \times 360° = 36°$.

Amounts: Salary $= 80{,}000$, Infrastructure $= 50{,}000$, Sports $= 30{,}000$, Library $= 20{,}000$, Others $= 20{,}000$.

Verification: Angles: $144 + 90 + 54 + 36 + 36 = 360°$ ✓. Amounts: $80000 + 50000 + 30000 + 20000 + 20000 = 200000$ ✓.

Problem: A box contains $4$ red, $6$ white, and $5$ blue marbles. One marble is drawn at random. Find the probability that it is: (a) white, (b) not red, (c) red or blue.

Solution:

Total marbles $= 4 + 6 + 5 = 15$.

(a) $P(\text{white}) = \dfrac{6}{15} = \dfrac{2}{5}$.

(b) $P(\text{not red}) = 1 - \dfrac{4}{15} = \dfrac{11}{15}$.

(c) Red or blue $= 4 + 5 = 9$. $P(\text{red or blue}) = \dfrac{9}{15} = \dfrac{3}{5}$.