CBSE Class 7 Mathematics Data Handling Worksheet Set 07

Question 1. Sachin scored the following number of runs in 10 innings 13, 55, 86, 50, 60, 68, 72, 32, 98, 99 find the mean of runs scored by him?
Answer: First, add up all the runs scored in the 10 innings:
\( \text{Sum of runs} = 13 + 55 + 86 + 50 + 60 + 68 + 72 + 32 + 98 + 99 = 633 \)
Now, divide this sum by the total number of innings (10) to find the mean:
\( \text{Mean} = \frac{633}{10} = 63.3 \)
In simple words: Add up all Sachin's runs to get 633. Divide this by 10 to get his average score of 63.3.

Exam Tip: Always double-check your addition of raw data. A single incorrect sum will throw off your entire mean calculation.

Question 2. Find the mean of first 12 whole numbers?
Answer: Whole numbers start from 0. The first 12 whole numbers are:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
First, find the sum of these 12 numbers:
\( \text{Sum} = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 66 \)
Next, divide the sum by 12 to calculate the mean:
\( \text{Mean} = \frac{66}{12} = 5.5 \)
In simple words: The first 12 whole numbers are 0 to 11. Their sum is 66, and dividing this by 12 gives an average of 5.5.

Exam Tip: Remember that whole numbers start at 0. If you start counting from 1, you are listing natural numbers instead.

Question 3. The temperature ( in degree Celsius) of Mumbai during a particular fortnight in the month of june is given below. 32, 44, 42, 43, 31, 39, 42, 37, 38, 41, 44, 40, 28, 35
(i) Find the range of the temperature in the above data.
(ii) On how many days was the temperature more than the mean temperature.
Answer: Let's sort the temperature data in ascending order first:
28, 31, 32, 35, 37, 38, 39, 40, 41, 42, 42, 43, 44, 44

(i) To find the range, subtract the lowest temperature from the highest temperature:
\( \text{Highest temperature} = 44^\circ\text{C} \)
\( \text{Lowest temperature} = 28^\circ\text{C} \)
\( \text{Range} = 44^\circ\text{C} - 28^\circ\text{C} = 16^\circ\text{C} \)

(ii) First, calculate the mean temperature for the 14 days (a fortnight):
\( \text{Sum of temperatures} = 32 + 44 + 42 + 43 + 31 + 39 + 42 + 37 + 38 + 41 + 44 + 40 + 28 + 35 = 526 \)
\( \text{Mean} = \frac{526}{14} \approx 37.57^\circ\text{C} \)
Now, count the number of days with a temperature greater than \( 37.57^\circ\text{C} \).
The temperatures above \( 37.57^\circ\text{C} \) are: 38, 39, 40, 41, 42, 42, 43, 44, 44.
This gives a count of 9 days.
In simple words: The temperature range is 16°C. On 9 of the days, it was warmer than the average of 37.57°C.

Exam Tip: Sorting your data first makes it much easier to find the highest, lowest, and middle values without missing any numbers.

Question 4. Find the mode and median of the following data
(i) 42, 15, 24, 21, 18, 15, 33, 29, 15, 18, 24, 26, 40, 41, 36, 15, 24 21
(ii) 8, 7, 6, 9, 8, 5, 7, 5, 3, 2, 1, 8, 6, 7, 2, 7, 8
Answer: Let's find the values for each part:

(i) First, arrange the 18 numbers in ascending order:
15, 15, 15, 15, 18, 18, 21, 21, 24, 24, 24, 26, 29, 33, 36, 40, 41, 42
- Mode: The number that appears most often is 15 (it appears 4 times).
- Median: Since there are 18 values (even), the median is the average of the 9th and 10th values:
\( \text{9th value} = 24 \), \( \text{10th value} = 24 \)
\( \text{Median} = \frac{24 + 24}{2} = 24 \)

(ii) First, arrange the 17 numbers in ascending order:
1, 2, 2, 3, 5, 5, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9
- Mode: Both 7 and 8 appear most often (4 times each). This data has two modes: 7 and 8.
- Median: Since there are 17 values (odd), the median is the middle term (9th value):
\( \text{Median} = 7 \)
In simple words: For part one, the mode is 15 and the median is 24. For part two, the modes are 7 and 8, and the median is 7.

Exam Tip: If two values tie for the highest frequency, both are modes. This is known as bimodal data.

Question 5. Find the mode of median 2, 6, 5, 3, 0, 4, 3, 2, 4, 5, 2, 4
Answer: Let's find both the mode and the median for this data.
First, arrange the 12 numbers in ascending order:
0, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6
- Mode: Both 2 and 4 appear most often (3 times each). Thus, the modes are 2 and 4.
- Median: Since there are 12 values, the median is the average of the 6th and 7th values:
\( \text{6th value} = 3 \), \( \text{7th value} = 4 \)
\( \text{Median} = \frac{3 + 4}{2} = 3.5 \)
In simple words: The most frequent numbers are 2 and 4. The middle of the sorted list is 3.5.

Exam Tip: When the total count of numbers is even, always add the two middle values and divide by 2 to get the median.

Question 6. Find the mean and median of the multiples of 4 between 10 and 40
Answer: The multiples of 4 strictly between 10 and 40 are:
12, 16, 20, 24, 28, 32, 36
There are 7 numbers in total.
- Mean: Add the numbers and divide by 7:
\( \text{Sum} = 12 + 16 + 20 + 24 + 28 + 32 + 36 = 168 \)
\( \text{Mean} = \frac{168}{7} = 24 \)
- Median: Since the 7 numbers are already in order, the median is the 4th (middle) term:
\( \text{Median} = 24 \)
In simple words: The average of these multiples of 4 is 24, and the middle number in the list is also 24.

Exam Tip: For any set of numbers with a constant gap (like multiples), the mean and median will always be equal to the exact middle number.

Question 7. Draw bar graph to represent the given information.
Answer: First, let's look at the given data:

To draw the bar graph:
1. Draw two axes - a horizontal axis (X-axis) for the months and a vertical axis (Y-axis) for the number of bags.
2. Choose a scale for the Y-axis. Since the values are multiples of 500, a scale of 1 unit = 500 bags is suitable. Label the Y-axis from 0 up to 3000.
3. Draw vertical bars of equal width for each month. The heights should match the values: July (1000), August (1500), September (1500), October (2000), November (2500), and December (1500).
In simple words: Use a scale where each block represents 500 bags. Draw vertical bars for each month up to the correct value.

Exam Tip: Always write down your chosen scale (e.g., 1 unit = 500 bags) in the top-right corner of your graph to score full marks.

Question 8. The performance of a student in SA1 and SA2 is given. Draw a double bar graph choosing appropriate scale and answer the following
(i) In which subject has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone low in any subject?
Answer: Let's look at the performance table:

To draw the double bar graph:
Choose a scale on the Y-axis where 1 unit = 10 marks. For each subject, draw two adjacent bars - one for SA1 and one for SA2 - using different colors or shading.

Let's analyze the performance changes (SA2 - SA1):
- English: \( 70 - 85 = -15 \) (dropped)
- Hindi: \( 65 - 60 = +5 \) (improved)
- Maths: \( 80 - 95 = -15 \) (dropped)
- Science: \( 85 - 60 = +25 \) (improved)
- Social: \( 90 - 70 = +20 \) (improved)

(i) The child improved the most in Science (an increase of 25 marks).

(ii) The child improved the least in Hindi (an increase of only 5 marks).

(iii) Yes, the performance went down in English (by 15 marks) and Maths (by 15 marks).
In simple words: The child showed the biggest jump in Science and the smallest in Hindi. Scores dropped in English and Maths.

Exam Tip: Use a clear key or legend to distinguish between the SA1 and SA2 bars on your double bar graph.

Question 9. Draw a double bar graph
Answer: First, let's examine the provided data:

To construct this double bar graph:
1. Draw the horizontal axis (X-axis) for the four schools (School A, B, C, D) and the vertical axis (Y-axis) for the number of children.
2. Choose a scale of 1 unit = 10 children on the Y-axis. Label the Y-axis from 0 to 80.
3. For each school, draw a pair of adjacent vertical bars. Color or shade one bar for "Walking" and the other for "Cycling".
- For School A, draw walking bar up to 60 and cycling bar up to 65.
- For School B, draw walking bar up to 55 and cycling bar up to 40.
- For School C, draw walking bar up to 70 and cycling bar up to 60.
- For School D, draw walking bar up to 45 and cycling bar up to 75.
In simple words: Draw pairs of bars for each school using a scale of 10. Shading will help separate walkers from cyclists.

Exam Tip: Ensure the widths of all bars and the gaps between the school pairs are kept perfectly equal for a neat presentation.

Question 10. Salaries of Affan & Ayaan are Rs.10,000 & Rs.5,000 per month. There expenditures are given below show the data on a double bar graph?
Answer: Let's review the monthly expenditure data:

To draw the double bar graph:
1. Label the horizontal axis (X-axis) with the expenditure items: Food, Clothing, Education, Rent, and Entertainment. Label the vertical axis (Y-axis) with the expense amount in Rs.
2. Choose an appropriate scale. A scale of 1 unit = Rs. 500 is suitable. Label the Y-axis from 0 to 3500.
3. Draw two adjacent bars for each item - one for Affan and one for Ayaan. Use different colors or patterns to identify each person's bar.
In simple words: Draw side-by-side bars for Affan and Ayaan on each expense item using a scale of Rs. 500.

Exam Tip: Always make sure to write down the axis titles clearly. Label the X-axis as "Expenditure Items" and the Y-axis as "Expenditure (Rs.)".

Question 11. When a die is thrown what is the probability of
(i) Getting a prime number
(ii) Getting a non prime number
(iii) Getting a number greater than 3.
Answer: A standard six-sided die has 6 possible outcomes: 1, 2, 3, 4, 5, 6.

(i) The prime numbers on a die are 2, 3, and 5. This gives 3 favorable outcomes.
\( \text{Probability} = \frac{3}{6} = \frac{1}{2} \)

(ii) The non-prime numbers on a die are 1, 4, and 6 (note that 1 is neither prime nor composite, but it is not a prime). This gives 3 favorable outcomes.
\( \text{Probability} = \frac{3}{6} = \frac{1}{2} \)

(iii) The numbers greater than 3 on a die are 4, 5, and 6. This gives 3 favorable outcomes.
\( \text{Probability} = \frac{3}{6} = \frac{1}{2} \)
In simple words: For each of these three questions, the chance of it happening is exactly half, or \( \frac{1}{2} \).

Exam Tip: Remember to always simplify your probability fractions to their lowest terms to secure full marks.

Question 12. Numbers 1 to 30 are written on 30 separate slips kept in a box and mixed well. One slip is chosen from the box what is the probability of
(i) Getting a one digit number
(ii) Getting an even number
(iii) Getting the number to be multiple of 3
(iv) Getting number greater than 15
Answer: The total number of outcomes is 30.

(i) The one-digit numbers from 1 to 30 are 1, 2, 3, 4, 5, 6, 7, 8, and 9 (9 outcomes):
\( \text{Probability} = \frac{9}{30} = \frac{3}{10} \)

(ii) The even numbers from 1 to 30 are 2, 4, 6, ..., 30 (15 outcomes):
\( \text{Probability} = \frac{15}{30} = \frac{1}{2} \)

(iii) The multiples of 3 from 1 to 30 are 3, 6, 9, 12, 15, 18, 21, 24, 27, and 30 (10 outcomes):
\( \text{Probability} = \frac{10}{30} = \frac{1}{3} \)

(iv) The numbers greater than 15 are 16 to 30 (15 outcomes):
\( \text{Probability} = \frac{15}{30} = \frac{1}{2} \)
In simple words: Out of 30 slips, the probability of getting a single digit is \( \frac{3}{10} \), an even number is \( \frac{1}{2} \times \), a multiple of 3 is \( \frac{1}{3} \), and a number over 15 is \( \frac{1}{2} \).

Exam Tip: Probability is calculated as \( \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} \). Write down this formula at the start of your answer.