CBSE Class 7 Mathematics Data Handling Assignment Set 08

VII - Mathematics Assignment No-1 - Data Handling

Q1. A die was thrown 20 times and the following outcomes were recorded.
6, 1, 2, 4, 2, 3, 2, 5, 4, 1
4, 5, 4, 6, 2, 3, 5, 4, 6, 4.
Prepare a frequency table and answer the following questions
(a) How many times does 4 appear?
(b) What is the frequency of 6?
Answer:
Frequency Table:

(a) The number 4 appears 6 times.
(b) The frequency of 6 is 3.
In simple words: Count how many times each face of the die appears in the given list to make the frequency table.

Exam Tip: Always cross-check the sum of frequencies with the total number of observations given in the problem.

Q2. Prepare a frequency table for the following data
1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 4, 2, 2, 1, 2, 3, 2, 3, 2
1, 1, 2, 1, 2, 3, 2, 4, 3, 3, 2, 2, 1, 5, 1, 1, 4, 2, 3,
6, 4
Which number has repeated maximum number of times?
Answer:
Frequency Table:

The number 2 has repeated the maximum number of times (14 times).
In simple words: The number with the highest frequency is the one that appears most often.

Exam Tip: Sum of all frequencies must equal the total count of data values (40).

Q3. Find the range of the following data
400, 425, 401, 414, 424, 436,
450, 413, 434, 443, 444, 445,
413, 404, 415.
Answer:
Highest value = 450
Lowest value = 400
Range = Highest value - Lowest value
Range = \(450 - 400 = 50\) (Note: The answer key in the source document contains a printing typo listing 424.06; the correct mathematical range is 50).
In simple words: Range is the difference between the maximum and minimum values in a dataset.

Exam Tip: Always arrange numbers mentally or scratchpad in order to quickly spot the highest and lowest values.

Q4. Find the mean of first 7 prime numbers.
Answer:
First 7 prime numbers are 2, 3, 5, 7, 11, 13, and 17.
Sum = \(2 + 3 + 5 + 7 + 11 + 13 + 17 = 58\)
Number of observations = 7
Mean = \(\frac{\text{Sum of observations}}{\text{Number of observations}} = \frac{58}{7} = 8.28\) (approx.)
In simple words: Add the first seven prime numbers and divide the total by 7.

Exam Tip: Remember that 1 is neither prime nor composite; the first prime number is 2.

Q5. Find the mean of first 10 natural numbers.
Answer:
First 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Sum = \(\frac{n(n+1)}{2} = \frac{10 \times 11}{2} = 55\)
Mean = \(\frac{55}{10} = 5.5\)
In simple words: Find the sum of numbers from 1 to 10 and divide by 10.

Exam Tip: Use the formula \(\frac{n+1}{2}\) to quickly find the mean of the first \(n\) natural numbers.

Q6. The marks obtained by a student in 5 subjects in his final examination are 67, 92, 33, 15, 87. Find his percentage of marks, if the above marks are out of 100 each.
Answer:
Total marks obtained = \(67 + 92 + 33 + 15 + 87 = 294\)
Maximum possible marks = \(5 \times 100 = 500\)
Percentage = \(\frac{294}{500} \times 100 = \frac{294}{5} = 58.8\%\)
In simple words: Add all obtained marks, divide by total maximum marks, and multiply by 100.

Q7. Calculate the mean of the following data
(i) 37, 54, 67, 32, 55, 65
(ii) 40, 50, 90, 32, 56, 75
Answer:
(i) Sum = \(37 + 54 + 67 + 32 + 55 + 65 = 310\)
Number of observations = 6
Mean = \(\frac{310}{6} = 51.66\) (approx.)

(ii) Sum = \(40 + 50 + 90 + 32 + 56 + 75 = 343\)
Number of observations = 6
Mean = \(\frac{343}{6} = 57.16\) (approx.)
In simple words: Mean is calculated by dividing the sum of all values by the total count of values.

Exam Tip: Round decimals to two places if the division is non-terminating.

Q8. The rainfall recorded in a certain city are given below. Calculate the average rainfall.
\(2.54\text{ cm}, 1.75\text{ cm}, 3.05\text{ cm}, 5.00\text{ cm}, 3.72\text{ cm}\) and \(4.34\text{ cm}\).
Answer:
Sum of rainfall = \(2.54 + 1.75 + 3.05 + 5.00 + 3.72 + 4.34 = 20.40\text{ cm}\)
Number of observations = 6
Average rainfall = \(\frac{20.40}{6} = 3.4\text{ cm}\)
In simple words: Add all the rainfall measurements and divide by the number of measurements.

Exam Tip: Align decimal points carefully when adding numbers with decimals.

Q9. Find average of odd numbers between 20 and 30.
Answer:
Odd numbers between 20 and 30 are 21, 23, 25, 27, and 29.
Sum = \(21 + 23 + 25 + 27 + 29 = 125\)
Number of observations = 5
Average = \(\frac{125}{5} = 25\)
In simple words: List the odd numbers between 20 and 30, add them up, and divide by 5.

Exam Tip: For any arithmetic sequence with an odd number of terms, the middle term is always the average.

Q10. The heights of 10 girls (in cm) are given below.
147, 136, 127, 151, 150, 138, 147, 150, 145, 155.
Answer the following questions.
(a) What is the height of the tallest girl?
(b) What is the height of the shortest girl?
(c) Find the mean height.
Answer:
(a) Height of the tallest girl = 155 cm.
(b) Height of the shortest girl = 127 cm.
(c) Sum of heights = \(147 + 136 + 127 + 151 + 150 + 138 + 147 + 150 + 145 + 155 = 1446\text{ cm}\)
Number of girls = 10
Mean height = \(\frac{1446}{10} = 144.6\text{ cm}\)
In simple words: Find max and min for parts (a) and (b), and divide the total sum by 10 for the mean in part (c).

Exam Tip: Always include proper units (like cm) in your final answers for word problems.