NCERT Solutions for Class 7 Maths Chapter 3: Data Handling — Complete Guide with Step-by-Step Solutions

Consider the salaries in a small company: Rs. $10000, 12000, 11000, 13000, 95000$.

• Mean $= \frac{10000 + 12000 + 11000 + 13000 + 95000}{5} = \frac{141000}{5} = 28200$.
  - Median $= 12000$ (the middle value after arranging).

The mean (Rs. $28200$) is misleading because one very high salary ($95000$) pulls it up. The median (Rs. $12000$) is a much better representation of the "typical" salary.

Rule of thumb: If your data has outliers, prefer the median. If data is fairly uniform, the mean works well. For finding the most popular or common item, use the mode.

Problem: Find the mean of $5, 9, 8, 7, 6$.

Solution:

Answer: The mean is $7$.

Problem: A batsman scored $40, 55, 63, 45, 32$ in $5$ matches. Find the mean score.

Solution:

Answer: The mean score is $47$.

Problem: The mean of $6$ numbers is $21$. If one number is excluded, the mean of the remaining $5$ numbers becomes $20$. Find the excluded number.

Solution:
Sum of $6$ numbers $= 21 \times 6 = 126$.

Sum of remaining $5$ numbers $= 20 \times 5 = 100$.

Excluded number $= 126 - 100 = 26$.

Answer: The excluded number is $26$.

Problem: The mean of $10$ observations is $15$. If the mean of the first $6$ observations is $12$, find the mean of the remaining $4$ observations.

Solution:
Sum of $10$ observations $= 15 \times 10 = 150$.

Sum of first $6$ observations $= 12 \times 6 = 72$.

Sum of remaining $4$ observations $= 150 - 72 = 78$.

Mean of remaining $4$ $= \frac{78}{4} = 19.5$.

Answer: $19.5$.

Problem: A student scored $72, 85, 90, 68$ in $4$ tests. What score must she get in the $5$th test to have an overall mean of $80$?

Solution:
Required total for mean $80$ across $5$ tests $= 80 \times 5 = 400$.

Sum of first $4$ scores $= 72 + 85 + 90 + 68 = 315$.

Required $5$th score $= 400 - 315 = 85$.

Answer: She needs to score $85$.

Problem: Find the mean of the first $5$ natural numbers.

Solution:

Shortcut: The mean of the first $n$ natural numbers is $\frac{n + 1}{2}$. So for $n = 5$: $\frac{5 + 1}{2} = 3$.

Answer: $3$.

Problem: The mean of $5$ numbers is $12$. If $3$ is added to each number, what is the new mean?

Solution:
Original sum $= 12 \times 5 = 60$.

New sum $= 60 + 3 \times 5 = 60 + 15 = 75$.

New mean $= \frac{75}{5} = 15 = 12 + 3$.

Key insight: When a constant $k$ is added to every observation, the mean increases by $k$.

Answer: $15$.

Problem: Find the mean of the following data:

Solution:
Total observations $= 3 + 5 + 4 + 2 = 14$.

Sum $= 5 \times 3 + 10 \times 5 + 15 \times 4 + 20 \times 2 = 15 + 50 + 60 + 40 = 165$.

Answer: $11.8$ (approximately).

Problem: The heights of $7$ students (in cm) are $145, 152, 148, 160, 155, 149, 153$. Find the range and mean.

Solution:
Range $= \text{Maximum} - \text{Minimum} = 160 - 145 = 15$ cm.

Mean $= \frac{145 + 152 + 148 + 160 + 155 + 149 + 153}{7} = \frac{1062}{7} = 151.7$ cm.

Answer: Range $= 15$ cm, Mean $= 151.7$ cm.

Problem: The temperatures (in °C) for a week were $28, 30, 27, 29, 31, 28, 30$. Find the mean temperature.

Solution:

Answer: The mean temperature is $29°C$.

Problem: Find the mode of: $2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4$.

Solution:
First, count the frequency of each value:
- $0 \to 1$ time
- $2 \to 3$ times
- $3 \to 3$ times
- $4 \to 3$ times
- $5 \to 2$ times
- $6 \to 1$ time

Values $2, 3, 4$ each appear $3$ times (the highest frequency).

Answer: There are three modes: $2, 3, 4$. (This data is trimodal.)

Problem: The runs scored by $11$ players are: $7, 16, 121, 51, 101, 81, 1, 16, 9, 11, 16$. Find the mode.

Solution:
$16$ appears $3$ times, more than any other value.

Answer: Mode $= 16$.

Problem: Find the median of $13, 16, 12, 14, 19, 12, 14, 13, 14$.

Solution:
Step 1: Arrange in ascending order: $12, 12, 13, 13, 14, 14, 14, 16, 19$.

Step 2: $n = 9$ (odd). Median $= \left(\frac{n+1}{2}\right)$th value $= 5$th value.

Step 3: The $5$th value is $14$.

Answer: Median $= 14$.

Problem: Find the median of $3, 1, 5, 6, 3, 4$.

Solution:
Step 1: Arrange: $1, 3, 3, 4, 5, 6$.

Step 2: $n = 6$ (even). Median $= \frac{\left(\frac{n}{2}\right)\text{th value} + \left(\frac{n}{2}+1\right)\text{th value}}{2}$.

Answer: Median $= 3.5$.

Problem: For the data $5, 7, 9, 7, 3, 7, 5, 9, 3$, find the mean, median, and mode.

Solution:
Mean:

Median: Arrange: $3, 3, 5, 5, 7, 7, 7, 9, 9$. $n = 9$, so median $= 5$th value $= 7$.

Mode: $7$ appears $3$ times (most frequent). Mode $= 7$.

Answer: Mean $\approx 6.1$, Median $= 7$, Mode $= 7$.

Problem: Find the mode of $5, 8, 12, 3, 7$.

Solution:
Each value appears exactly once. No value is repeated.

Answer: There is no mode for this data set.

Problem: The shoe sizes of $20$ students are given:

Find the mode.

Solution:
Size $7$ has the highest frequency ($8$ students).

Answer: Mode $= 7$.

Problem: Find the median of $4, 4, 4, 4, 4, 4, 4$.

Solution:
All values are $4$. The data is already sorted. $n = 7$ (odd).

Median $= 4$th value $= 4$.

Answer: Median $= 4$. (When all values are the same, mean $=$ median $=$ mode $=$ that value.)

Problem: In a class, the marks obtained (out of $10$) are: $2, 3, 5, 5, 5, 6, 6, 7, 8, 10$. Which measure of central tendency best represents this data?

Solution:
Mean $= \frac{2 + 3 + 5 + 5 + 5 + 6 + 6 + 7 + 8 + 10}{10} = \frac{57}{10} = 5.7$.

Median $= \frac{5\text{th} + 6\text{th}}{2} = \frac{5 + 6}{2} = 5.5$.

Mode $= 5$ (appears $3$ times).

All three are close ($5, 5.5, 5.7$), so any would work. But since there are no extreme outliers and the data is fairly symmetric, the mean is appropriate.

Answer: The mean ($5.7$) is a good representative.

Problem: The ages of $5$ children in a group are $8, 9, 8, 10, 25$. Identify the outlier and explain which measure is best.

Solution:
$25$ is an outlier (much larger than the rest).

Mean $= \frac{8 + 9 + 8 + 10 + 25}{5} = \frac{60}{5} = 12$.

Median $= 9$ (after arranging: $8, 8, 9, 10, 25$).

Mode $= 8$.

The mean ($12$) is misleadingly high because of the outlier $25$. The **median ($9$)** is the best representative.

Answer: The median is best because it is not affected by the outlier.

Problem: The number of students in different classes is: Class 6 $= 120$, Class 7 $= 100$, Class 8 $= 90$, Class 9 $= 110$, Class 10 $= 80$. Draw a bar graph.

Solution:
Steps to draw:
1. Draw two axes — horizontal (X-axis: classes) and vertical (Y-axis: number of students).
2. Choose a scale: $1$ unit $= 20$ students.
3. Draw bars of appropriate height for each class:
- Class 6: $120 \div 20 = 6$ units high
- Class 7: $100 \div 20 = 5$ units high
- Class 8: $90 \div 20 = 4.5$ units high
- Class 9: $110 \div 20 = 5.5$ units high
- Class 10: $80 \div 20 = 4$ units high
4. Ensure equal width and equal gaps between bars.
5. Label both axes and give a title.

Problem: A bar graph shows rainfall (in mm) for different months: Jan $= 20$, Feb $= 30$, Mar $= 50$, Apr $= 70$, May $= 40$. Answer: (a) Which month had the highest rainfall? (b) What is the difference between the highest and lowest?

Solution:
(a) The tallest bar corresponds to April ($70$ mm).

(b) Difference $= 70 - 20 = 50$ mm (between April and January).

Problem: Data values are $250, 180, 300, 420, 150$. What scale would you use for a bar graph?

Solution:
The range is $150$ to $420$. A scale of $1$ unit $= 50$ works well:
- Smallest bar: $150 \div 50 = 3$ units
- Tallest bar: $420 \div 50 = 8.4$ units

This keeps all bars between $3$ and $9$ units, which fits nicely on paper.

Tip: Choose a scale that makes the tallest bar about $8$-$10$ units high.

Problem: A bar graph shows the marks of $5$ students: Anil $= 65$, Bina $= 80$, Chetan $= 72$, Deepa $= 90$, Esha $= 85$. (a) Who scored the highest? (b) How many scored above $75$?

Solution:
(a) Deepa scored the highest ($90$).

(b) Students scoring above $75$: Bina ($80$), Deepa ($90$), Esha ($85$) $= 3$ students.

Problem: A school's results for two years are:

How would you represent this?

Solution:
Use a double bar graph with two bars for each class (one for 2024, one for 2025), using different colours or shading.

This allows easy visual comparison between the two years for each class.

Observation: Every class improved from 2024 to 2025. Class 8 showed the biggest improvement ($80\% \to 88\%$, an increase of $8$ percentage points).

Problem: List the essential features every bar graph must have.

Solution:
Every bar graph must have:
1. Title — describes what the graph represents
2. X-axis label — what the categories are
3. Y-axis label — what the values represent (with units)
4. Scale — clearly stated (e.g., $1$ unit $= 10$)
5. Equal bar widths — all bars should be the same width
6. Equal gaps — spacing between bars should be uniform
7. Bars starting from the base line — all bars start from $0$

Common exam mistake: Forgetting to label the axes or state the scale. This costs marks.

Problem: A bar graph uses a scale where $1$ unit $= 1000$ people. If a bar is $4.5$ units tall, how many people does it represent?

Solution:

Answer: $4500$ people.

Problem: A bar graph shows $5$ bars with heights $3, 5, 2, 7, 4$ units respectively (scale: $1$ unit $= 20$). Convert this to a data table.

Solution:

Problem: A coin is tossed. What is the probability of getting heads?

Solution:
Total outcomes $= 2$ (Heads, Tails).
Favourable outcomes (Heads) $= 1$.

Answer: $\frac{1}{2}$ (or $0.5$ or $50\%$).

Problem: A die is thrown. What is the probability of getting a number greater than $4$?

Solution:
Total outcomes $= 6$ ($1, 2, 3, 4, 5, 6$).
Favourable outcomes (greater than $4$) $= 2$ ($5, 6$).

Answer: $\frac{1}{3}$.

Problem: A bag has $5$ red balls and $3$ blue balls. What is the probability of drawing a blue ball?

Solution:
Total balls $= 5 + 3 = 8$.
Favourable (blue) $= 3$.

Answer: $\frac{3}{8}$.

Problem: A die is thrown. Find the probability of: (a) getting $7$, (b) getting a number less than $7$.

Solution:
(a) A die has numbers $1$-$6$ only. Getting $7$ is impossible.

(b) All numbers $1$-$6$ are less than $7$. This is a certain event.

Key facts:
- $P = 0$ means the event is impossible.
- $P = 1$ means the event is certain.
- $0 \leq P \leq 1$ for any event.

Problem: The probability of rain tomorrow is $\frac{2}{5}$. What is the probability of no rain?

Solution:

Key rule: $P(\text{event}) + P(\text{not event}) = 1$.

Answer: $\frac{3}{5}$.

Problem: From a well-shuffled pack of $52$ cards, one card is drawn. Find the probability of getting a king.

Solution:
Total cards $= 52$.
Kings in a pack $= 4$ (one of each suit).

Answer: $\frac{1}{13}$.

Problem: A die is rolled. Find the probability of getting an even number.

Solution:
Total outcomes $= 6$.
Even numbers: $2, 4, 6$ $\to$ favourable outcomes $= 3$.

Answer: $\frac{1}{2}$.

Problem: A spinner has $8$ equal sectors numbered $1$ to $8$. Find the probability of landing on: (a) a prime number, (b) a multiple of $3$.

Solution:
(a) Prime numbers from $1$-$8$: $2, 3, 5, 7$ $\to 4$ favourable outcomes.

(b) Multiples of $3$ from $1$-$8$: $3, 6$ $\to 2$ favourable outcomes.

Problem: Two coins are tossed simultaneously. Find the probability of getting at least one head.

Solution:
Total outcomes when two coins are tossed: $\{HH, HT, TH, TT\} = 4$.

Favourable (at least one head): $\{HH, HT, TH\} = 3$.

Alternative: $P(\text{at least one head}) = 1 - P(\text{no head}) = 1 - \frac{1}{4} = \frac{3}{4}$.

Answer: $\frac{3}{4}$.

Problem: A jar contains $6$ red, $4$ green, and $5$ yellow marbles. One marble is drawn at random. Find the probability of getting: (a) a green marble, (b) a non-yellow marble.

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

(a) $P(\text{green}) = \frac{4}{15}$.

(b) Non-yellow marbles $= 6 + 4 = 10$.

Answer: (a) $\frac{4}{15}$, (b) $\frac{2}{3}$.

Answer 1: $\frac{14 + 18 + 22 + 16 + 20}{5} = \frac{90}{5} = 18$.

Answer 2: Sum of $7$ numbers $= 11 \times 7 = 77$. Sum of $6$ given numbers $= 8 + 13 + 15 + 9 + 10 + 12 = 67$. The $7$th number $= 77 - 67 = 10$.

Answer 3: Arrange: $3, 3, 5, 5, 5, 5, 7, 8, 9$. Mode $= 5$ (appears $4$ times). $n = 9$, median $= 5$th value $= 5$.

Answer 4: Numbers divisible by $3$: $3, 6$ $\to 2$ out of $6$. $P = \frac{2}{6} = \frac{1}{3}$.

Answer 5: Total $= 10$, non-blue $= 3 + 2 = 5$. $P(\text{not blue}) = \frac{5}{10} = \frac{1}{2}$.

Answer 6: When each observation is multiplied by $k$, the mean is also multiplied by $k$. New mean $= 8 \times 2 = 16$.