CBSE Class 7 Mathematics Data Handling Worksheet Set 04

Question 1. Find the complement of the angles :
a) \( 56^\circ \)
b) \( 47^\circ \)
c) \( 12^\circ \)
d) \( 81^\circ \)
Answer: To find the complement of any angle, subtract its value from \( 90^\circ \):
a) \( 90^\circ - 56^\circ = 34^\circ \)
b) \( 90^\circ - 47^\circ = 43^\circ \)
c) \( 90^\circ - 12^\circ = 78^\circ \)
d) \( 90^\circ - 81^\circ = 9^\circ \)
In simple words: Subtract the angle from 90 to find its complement.

Exam Tip: Complementary angles always add up to \( 90^\circ \). Double-check your subtraction to avoid simple calculation errors.

Question 2. Find the supplement of the angles :
a) \( 135^\circ \)
b) \( 87^\circ \)
c) \( 39^\circ \)
d) \( 112^\circ \)
Answer: To find the supplement of any angle, subtract its value from \( 180^\circ \):
a) \( 180^\circ - 135^\circ = 45^\circ \)
b) \( 180^\circ - 87^\circ = 93^\circ \)
c) \( 180^\circ - 39^\circ = 141^\circ \)
d) \( 180^\circ - 112^\circ = 68^\circ \)
In simple words: Subtract the angle from 180 to find its supplement.

Exam Tip: Supplementary angles always add up to \( 180^\circ \). Be careful not to confuse them with complementary angles.

Question 3. Find the supplement of the angles :
a)
b)

Answer: We use the angle relationships shown in the figures:

a)
- Angle \( x \) and \( 49^\circ \) are vertically opposite angles, so:
\( x = 49^\circ \)
- Angle \( y \) and \( 49^\circ \) lie on a straight line, forming a linear pair:
\( y + 49^\circ = 180^\circ \implies y = 180^\circ - 49^\circ = 131^\circ \)
- Angle \( z \) and \( y \) are vertically opposite angles, so:
\( z = y = 131^\circ \)

b)
- The angles \( 60^\circ \), \( x \), and \( 25^\circ \) lie on a straight horizontal line, meaning they add up to \( 180^\circ \):
\( 60^\circ + x + 25^\circ = 180^\circ \implies x = 180^\circ - 85^\circ = 95^\circ \)
- Angle \( z \) and \( 60^\circ \) are vertically opposite angles, so:
\( z = 60^\circ \)
- Angle \( y \) and \( z \) form a linear pair on the straight horizontal line:
\( y + z = 180^\circ \implies y + 60^\circ = 180^\circ \implies y = 120^\circ \)
In simple words: Use straight line and opposite angle rules to calculate the missing values.

Exam Tip: Clearly write down the geometric reason (e.g., "vertically opposite angles" or "linear pair") next to each step of your calculation.

Question 4. Find x, if l || m
a)
b)

Answer: We use the properties of parallel lines cut by a transversal:

a)
- The interior angle corresponding to \( 105^\circ \) on line \( m \) is \( 105^\circ \).
- Since \( x \) and this corresponding angle lie on a straight line, they form a linear pair:
\( x + 105^\circ = 180^\circ \implies x = 180^\circ - 105^\circ = 75^\circ \)

b)
- Since lines \( l \) and \( m \) are parallel, the angle \( x \) on line \( m \) is corresponding to the angle \( 110^\circ \) on line \( l \):
\( x = 110^\circ \)
In simple words: Use parallel line laws to find the matching angles.

Exam Tip: Remember that alternate interior, alternate exterior, and corresponding angles are equal when two parallel lines are intersected by a transversal.

Chapter 3 : Data Handling

Question 1. Following are the weights ( in kg ) of 8 students of a class
48.5, 50, 44.5, 49.5, 50.5, 45, 51, 43
a) Find the mean weight.
b) What will be the mean weight if a student, whose weight is 62kg, is also included?
Answer:
a) First, find the sum of the weights of the 8 students:
\( \text{Sum} = 48.5 + 50 + 44.5 + 49.5 + 50.5 + 45 + 51 + 43 = 382\text{ kg} \)
Now, divide this sum by the total number of students (8):
\( \text{Mean} = \frac{382}{8} = 47.75\text{ kg} \)

b) If a student weighing 62 kg is added, the new total weight becomes:
\( \text{New Sum} = 382 + 62 = 444\text{ kg} \)
The total number of students increases to 9.
Now, calculate the new mean weight:
\( \text{New Mean} = \frac{444}{9} = 49.33\text{ kg} \)
In simple words: The average weight of the 8 students is 47.75 kg. Adding a 62 kg student raises the average weight to 49.33 kg.

Exam Tip: When adding new data, always remember to increase the total number of items (the denominator) by the correct count.

Question 2. Find the arithmetic mean of the scores
8, 6, 10, 12, 1, 3, 4, 4. Find the range of the data.
Answer:
1. To find the arithmetic mean:
First, calculate the sum of all the scores:
\( \text{Sum} = 8 + 6 + 10 + 12 + 1 + 3 + 4 + 4 = 48 \)
Next, divide the sum by the total number of scores (8):
\( \text{Mean} = \frac{48}{8} = 6 \)

2. To find the range of the data:
Identify the highest and lowest values:
\( \text{Highest value} = 12 \)
\( \text{Lowest value} = 1 \)
Subtract the lowest value from the highest value:
\( \text{Range} = 12 - 1 = 11 \)
In simple words: The average score is 6. The gap between the highest and lowest scores is 11.

Exam Tip: Write down the formulas for "mean" and "range" clearly in your solution to score partial marks even if you make a calculation mistake.

Question 3. Find the mean of the 1st three composite numbers.
Answer: The first three composite numbers are 4, 6, and 8.
First, find the sum of these three numbers:
\( \text{Sum} = 4 + 6 + 8 = 18 \)
Now, divide the sum by the total count (3):
\( \text{Mean} = \frac{18}{3} = 6 \)
In simple words: The first three composite numbers are 4, 6, and 8. Their average value is 6.

Exam Tip: Remember that 1 is neither prime nor composite, and 2 and 3 are prime numbers. So 4 is the first composite number.

Question 4. 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
a) What is the height of the tallest girl?
b) What is the height of the shortest girl?
c) What is the range of the data?
d) Find the mean height ?
e) Find the number of girls whose heights are less than the mean height ?
Answer: Let's sort the heights in ascending order to make analysis easier:
128, 132, 135, 139, 143, 146, 148, 149, 150, 151

a) The height of the tallest girl is \( 151\text{ cm} \).

b) The height of the shortest girl is \( 128\text{ cm} \).

c) The range of the data is the difference between the tallest and shortest heights:
\( \text{Range} = 151\text{ cm} - 128\text{ cm} = 23\text{ cm} \)

d) To find the mean height:
First, calculate the sum of all heights:
\( \text{Sum} = 143 + 148 + 135 + 150 + 128 + 139 + 149 + 146 + 151 + 132 = 1416\text{ cm} \)
Divide the sum by the number of girls (10):
\( \text{Mean height} = \frac{1416}{10} = 141.6\text{ cm} \)

e) Compare each height with the mean of 141.6 cm.
Heights less than 141.6 cm are: 128 cm, 132 cm, 135 cm, and 139 cm.
This gives a count of 4 girls.
In simple words: The tallest girl is 151 cm, the shortest is 128 cm, and the range is 23 cm. The average height is 141.6 cm, and 4 girls are shorter than this average.

Exam Tip: Sorting the data is extremely helpful because it instantly gives you the values for parts (a), (b), and (e).

Question 5. Find the mean of the 1st ten natural numbers.
Answer: The first ten natural numbers are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10
First, find the sum of these ten numbers:
\( \text{Sum} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 \)
Now, divide the sum by 10 to calculate the mean:
\( \text{Mean} = \frac{55}{10} = 5.5 \)
In simple words: The average of the numbers from 1 to 10 is 5.5.

Exam Tip: Natural numbers start from 1, unlike whole numbers which start from 0. Be careful with this vocabulary.

Question 6. Two different states of India’s experts of garments in the years 2000 to 2005 are given in the following table

(i) Draw a double bar graph to represent the data (ii) What are the total earnings in the years 2002 and 2004 both the states separately?
Answer: Let's address both parts:

(i) Double Bar Graph Construction:
- Plot the years 2000 to 2005 on the horizontal axis (X-axis) and the earnings in crores on the vertical axis (Y-axis).
- Choose a scale of 1 unit = 2 crores. Label the Y-axis from 0 to 16.
- For each year, draw two adjacent bars - one for Kerala and one for Karnataka, using different colors or shading patterns.

(ii) Total Earnings Calculation:
Let's find the combined total earnings for both states in each of those years:
- For the year 2002:
\( \text{Kerala} = 8\text{ crores} \)
\( \text{Karnataka} = 9\text{ crores} \)
\( \text{Total Combined Earnings} = 8 + 9 = 17\text{ crores} \)

- For the year 2004:
\( \text{Kerala} = 12\text{ crores} \)
\( \text{Karnataka} = 8\text{ crores} \)
\( \text{Total Combined Earnings} = 12 + 8 = 20\text{ crores} \)

Alternatively, if the total earnings across both years are calculated for each state separately:
- Kerala: \( 8 + 12 = 20\text{ crores} \)
- Karnataka: \( 9 + 8 = 17\text{ crores} \)
In simple words: In 2002, both states combined made 17 crores. In 2004, they made 20 crores. Separately, across both years, Kerala made 20 crores and Karnataka made 17 crores.

Exam Tip: In a double bar graph, always include a clear legend showing which bar represents which state.

Question 7. Find the mode, mean and median of the scores
4, 5, 6, 7,7, 8, 9, 13, 12, 8, 8, 9, 8, 10, 11
Answer: Let's sort the 15 scores in ascending order first:
4, 5, 6, 7, 7, 8, 8, 8, 8, 9, 9, 10, 11, 12, 13

- Mode: The score that appears most frequently in the data set is 8 (it appears 4 times).

- Median: Since there are 15 scores (which is an odd number), the median is the middle term (the 8th term):
\( \text{Median} = 8 \)

- Mean: First, find the sum of all scores:
\( \text{Sum} = 4 + 5 + 6 + 7 + 7 + 8 + 8 + 8 + 8 + 9 + 9 + 10 + 11 + 12 + 13 = 117 \)
Now, divide the sum by 15:
\( \text{Mean} = \frac{117}{15} = 7.8 \)
In simple words: The most common score is 8, the middle score is 8, and the average score is 7.8.

Exam Tip: Sorting the data at the very beginning is the easiest way to find both the median and the mode accurately.

Question 8. Marks obtained by two girls of VII A in final term exam (out of 100) as follows :

1) Draw a double bar graph to represent the following data. 2) Who did better in the examination
Answer: Let's address both sub-questions:

1) Double Bar Graph Guidelines:
- Plot the subjects on the horizontal axis (X-axis) and the marks (out of 100) on the vertical axis (Y-axis).
- Choose a scale of 1 unit = 10 marks. Label the Y-axis from 0 to 100.
- For each subject, draw two adjacent bars - one for Sunita and one for Vandhana - using different colors.

2) Comparison of performance:
Let's calculate the total marks obtained by each student:
- Sunita's total marks:
\( \text{Total} = 75 + 80 + 92 + 84 + 62 = 393 \)
- Vandhana's total marks:
\( \text{Total} = 72 + 84 + 92 + 65 + 70 = 383 \)
Since \( 393 > 383 \), Sunita performed better in the examination.
In simple words: Draw side-by-side bars for both girls across each subject. Sunita did better because her total score (393) is higher than Vandhana's total score (383).

Exam Tip: Calculate the sum of all elements to compare overall student performance unless you are asked to compare a specific subject.

Question 9. The number of hours of television programme watched on Sunday in 40 houses were as follows

1) Organise the following numbers in a tabular form. 2) Estimate the mean, median and mode of this distribution. 3) What is the range of the data?
Answer: Let's address the questions step-by-step:

1) Frequency Table:

2) Mean, Median, and Mode:
- Mode: The value with the highest frequency is 9 (frequency is 14).
- Median: Since there are 40 values (even number), the median is the average of the 20th and 21st values when sorted:
The 20th value is 8 and the 21st value is 9.
\( \text{Median} = \frac{8 + 9}{2} = 8.5\text{ hours} \)
- Mean:
\( \text{Sum of all values} = (3 \times 1) + (4 \times 4) + (5 \times 4) + (6 \times 4) + (7 \times 3) + (8 \times 4) + (9 \times 14) + (10 \times 6) \)
\( \implies \text{Sum} = 3 + 16 + 20 + 24 + 21 + 32 + 126 + 60 = 302 \)
\( \text{Mean} = \frac{302}{40} = 7.55\text{ hours} \)

3) Range of the data:
\( \text{Highest value} = 10 \)
\( \text{Lowest value} = 3 \)
\( \text{Range} = 10 - 3 = 7\text{ hours} \)
In simple words: Organize the numbers into a table. The average is 7.55 hours, the middle value is 8.5 hours, the most common value is 9 hours, and the range is 7.

Exam Tip: Be very careful when calculating the median for even-numbered datasets. Remember to find the average of the two middle terms.

Question 10. A bag contains 3 red and 2 blue marbles. A marble is drawn at random. What is the probability of drawing a blue marble.
Answer: Let's find the probability step-by-step:
Number of red marbles = 3
Number of blue marbles = 2
Total number of marbles = \( 3 + 2 = 5 \)
The probability of drawing a blue marble is the number of blue marbles divided by the total number of marbles:
\( \text{Probability} = \frac{2}{5} \)
In simple words: There are 2 blue marbles out of 5 total marbles, so the chance of pulling a blue marble is \( \frac{2}{5} \).

Exam Tip: Always state the formula for probability first: \( \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \).

Question 11. A box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non defective bulb ?
Answer: Let's find the probability step-by-step:
Total number of electric bulbs = 600
Number of defective bulbs = 12
Number of non-defective bulbs = \( 600 - 12 = 588 \)
The probability of taking out a non-defective bulb is:
\( \text{Probability} = \frac{588}{600} \)
Divide both numerator and denominator by 12 to simplify:
\( \text{Probability} = \frac{49}{50} \)
In simple words: There are 588 good bulbs out of 600, so the chance of getting a working bulb is \( \frac{49}{50} \).

Exam Tip: Always simplify your final fraction to its simplest form to ensure you get full marks.

Question 12. What is the probability of getting
(i) an even number
(ii) a multiple of 3
(iii) a number 3 or 4
(iv) an odd number
(v) a number between 3 and 6
Answer: A standard six-sided die has 6 possible outcomes: 1, 2, 3, 4, 5, 6.

(i) An even number:
The even numbers are 2, 4, and 6 (3 outcomes).
\( \text{Probability} = \frac{3}{6} = \frac{1}{2} \)

(ii) A multiple of 3:
The multiples of 3 are 3 and 6 (2 outcomes).
\( \text{Probability} = \frac{2}{6} = \frac{1}{3} \)

(iii) A number 3 or 4:
The favorable numbers are 3 and 4 (2 outcomes).
\( \text{Probability} = \frac{2}{6} = \frac{1}{3} \)

(iv) An odd number:
The odd numbers are 1, 3, and 5 (3 outcomes).
\( \text{Probability} = \frac{3}{6} = \frac{1}{2} \)

(v) A number between 3 and 6:
The numbers strictly between 3 and 6 are 4 and 5 (2 outcomes).
\( \text{Probability} = \frac{2}{6} = \frac{1}{3} \)
In simple words: Use the standard outcomes of rolling a die to find each probability fraction.

Exam Tip: For "between" questions, do not include the boundary numbers unless the question specifies "inclusive".