Get the simplified Class 8 Maths Extra Questions Part 2 Chapter 5 Tales by Dots and Lines Class 8 Extra Questions and Answers with complete explanation.
Class 8 Tales by Dots and Lines Extra Questions
Class 8 Maths Chapter 5 Tales by Dots and Lines Extra Questions
Tales by Dots and Lines Extra Questions Class 8
Very Short Answer Type Questions
Question 1.
Calculate and mark the mean of the collection of data below.
Answer:
From the figure, the observations are 5, 8 and 10.
We know that
Mean = \(\frac{\text { Total observations }}{\text { Number of observations }}=\frac{5+8+10}{3}\)
= \(\frac{23}{3}\)
= 7.66
Question 2.
A cricketer’s scores in 4 innings are 64, 32, 68, 92. Find the mean score.
Answer:
Given, scores in 4 innings are 64, 32,68 and 92.
Sum of scores = 64 + 32 + 68 + 92 = 256
Mean of scores = \(\frac{\text { Total scores }}{\text { Number of innings }}=\frac{256}{4}\) = 64
Question 3.
In a data set, the sum of values is 420 and the mean is 21. Find the number of values.
8, 11, 19, 14, 13, 18, 26, x, 2x, 6
Answer:
Given, sum of values = 420 and mean = 21
We know that
Mean = \(\frac{\text { Sum of values }}{\text { Number of values }}\)
21 = \(\frac{420}{\text { Number of values }}\)
Number of values = \(\frac{420}{21}\) = 20
Question 4.
The height of 6 girls in a group are 142 cm, 150 cm, 146 cm, 154 cm, x cm and 148 cm, their average height is 147. Then, find the value of x.
Answer:
Given heights (in cm) of 6 girls in a group are 142,150, 146, 154, x and 148.
Sum of heights = 142 + 150 + 146 + 154 + x + 148
= 740 + x
Mean (Average) = \(\frac{1}{2}\)
⇒ 147 = \(\frac{1}{2}\)
⇒ 147 × 6 = 740 + x
⇒ 882 = 740 + x
⇒ x = 882 – 740 = 142
Question 5.
Find the median of the following data.
42, 46, 69, 62, 91, 72, 74
Answer:
Given data 42, 46, 69, 62, 91, 72, 74.
Ascending order of the given data 42, 46, 62, 69, 72, 91, 74.
∴ Median = Middle value = 69
Question 6.
The scores in Mathematics test (out of 25) of 15 students is as follows.
18, 21, 23, 24, 19, 18, 17, 16, 15, 25, 22, 20, 9, 4, 6
Find the median of this data.
Answer:
For median, we have to arrange the given data in ascending order.
4, 6, 9, 15, 16, 17, 18, 18, 19, 20, 21, 22, 23, 24, 25
∴ Median = Middle observation = 18
Question 7.
In Row BIO : G10 the values are 39, 37, 35, 38, 36, 40. Find the sum.
Answer:
Given, Row BIO : G10 contains values 39, 37, 35, 38, 36, 40.
We know the formula of sum = SUM(B10 : G10)
= SUM (39, 37, 35, 38, 36, 40) = 225
Hence, the sum of given data is 225.
Question 8.
A line graph shows temperature at 8 AM as 150 C and at 2 PM as 24 °C. What is the rise in temperature?
Answer:
Given,
Temperature at 8 AM = 15°C
Temperature at 2 PM = 24°C
Rise in temperature = Later temperature – Earlier temperature
Rise = 24°C- 15°C =9°C
Therefore, the rise in temperature is 9° C.
Short Answer Type Questions
Question 1.
The dot plot below shows a set of data values but one dot is missing. Mark the missing value so that the mean of the data becomes 10.
Answer:
Let the missing date is x.
So, the given observations are 6, 8, 9,10,10,11,12 and A, and number of observation = 8
We know that
Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
10 = \(\frac{6+8+9+10+10+11+12+x}{8}\)
10 = \(\frac{66+x}{8}\)
80 = 66 + x
x = 80 – 66 = 14
Question 2.
If the mean of 4 observations x, 2x, 3x and 4x is 10. Then, find the mean of the last three observations.
Answer:
Given, \(\frac{x+2 x+3 x+4 x}{4}\) = 10
⇒ \(\frac{10 x}{4}\) = 10
⇒ 10x = 40
⇒ x = 4
∴ Mean of the last three observations
= \(\frac{2 x+3 x+4 x}{3}\)
= \(\frac{9 x}{3}\) = 3x
= 3 × 4 = 12
Question 3.
Find the value of x, if the mean of the following data is 36.
8, 11, 19, 14, 13, 18, 26, x, 2x, 6
Answer:
Given data 8,11,19, 14, 13, 18, 26, x, 2x, 6.
Sum of given data
= 8 + 11 + 19 + 14 + 13 + 18 + 26 + x + 2x + 6 = 115 + 3x
∴ Mean = \(\frac{\text { Sum of observations }}{\text { Number of observations }}\)
⇒ 36 = \(\frac{115+3 x}{10}\)
⇒ 36 × 10 = 115 + 3x
⇒ 360 – 115 = 3A
⇒ 3x = 245
⇒ x = \(\frac{245}{3}\) = 81.66
Question 4.
The median of the observations 6, 7, x, (x + 2), 12,14 arranged in ascending order is 9. Find the value of x.
Answer:
Total number of observation, n = 6, which is even.
⇒ 2x + 2 = 18
⇒ 2x = 16
⇒ x = 8
Question 5.
A paper-plane contest was held in a school. The number of attempts taken by students to make their paper plane fly through a hoop is shown in the table below.
Describe the data using its minimum, maximum, mean and median.
Answer:
From the table,
The smallest value in the data = 1
and the largest value in the data = 7
Now, sum = (1 x 2) + (2 x 3) + (3 x 5) + (4 x 7) + (5 x 6) + (6×4)+(7×2)
= 2 + 6 + 15 +28 + 30 +24 + 14
= 119
Mean = \(\frac{\text { Sum }}{\text { Total frequency }}\)
= \(\frac{119}{29}\)
≈ 4.10
Question 6.
The average number of visitors attending a science exhibition and the average number of visitors buying souvenirs on different days of the week is given in the table below. Visualise this data on a line graph.
Answer:
Long Answer Type Questions
Question 1.
Find the median of the following data.
Answer:
Let us arrange the date in ascending order of xt and make a cumulative frequency table.
Here, n = 33 (odd)
Median = Value of \(\left(\frac{n+1}{2}\right)\)th observation
= Value of \(\left(\frac{33+1}{2}\right)\)th observation
= Value of 17th observation
Since, the corresponding value of 17th observation of cumulative frequency in x, is 28.
Hence, the median is 28.
Question 2.
Obtain the median for the following frequency distribution.
Answer:
Here, the given data is in ascending order of xi.
The cumulative frequency table for the given data is
| xi | fi | cf |
| 1 | 8 | 8 |
| 2 | 10 | 18 |
| 3 | 11 | 29 |
| 4 | 16 | 45 |
| 5 | 20 | 65 |
| 6 | 25 | 90 |
| 7 | 15 | 105 |
| 8 | 9 | 114 |
| 9 | 6 | 120 |
Here, n = 120 (even)
∴ Median
= \(\frac{1}{2}\)[ Value of {(\(\frac{n}{2}\))th + (\(\frac{n}{2}\) + 1)th observation
= \(\frac{1}{2}\)[ Value of {(\(\frac{120}{2}\))th + (\(\frac{120}{2}\) + 1)th observation
= \(\frac{1}{2}\) [Value of 60th observation + Value of 61th observation]
Both 60th and 61th observations lie in the cumulative frequency 65 and its corresponding value of x is 5.
∴ Median = \(\frac{1}{2}\) (5 + 5) = 5
Question 3.
Primary source of energy for irrigation pumps (Rural).
Referring to the graph below, which of the following statements are valid? Why?
(i) In 1985, most rural irrigation pumps used diesel as the primary energy source.
(ii) From 1985 to 2025, the use of solar energy increased while the use of diesel decreased.
(iii) In 2005, approximately 45% of rural irrigation pumps used solar energy.
(iv) In 2025, diesel-powered pumps were more common than solar pumps.
Answer:
(i) Valid
In 1985, diesel = 80%, while solar = 5%.
Diesel was clearly the majority source.
(ii) Valid
Solar rises from = 5% (1985) → 90% (2025)
Diesel drops from « 80% (1985) → 5% (2025)
Trend is consistent and clear on the graph.
(iii) Valid
In 2005, the graph shows solar ≈ 45%
(iv) Invalid
In 2025, solar = 90% and diesel = 5%.
So, solar is far more common than diesel, the opposite of what the statement claims.
Question 4.
The following line graph shows the monthly rainfall (in mm) in a city during the year 2021. On the basis of above graph, answer the following questions
(i) What are your observations about the rainfall pattern during the year?
Answer:
Observations about the rainfall pattern.
Rainfall is low during the beginning of the year (January-March).
It increases gradually from April onwards.
The highest rainfall occurs during the monsoon months (June-August).
After August, rainfall decreases steadily till December.
(ii) In which month was the rainfall maximum?
Answer:
The rainfall is maximum in July 2021, at approximately 210 mm.
(iii) What was the approximate rainfall in July 2021?
Answer:
The rainfall in July 2021 is about 210 mm.
(iv) Compare the rainfall in the months of March, June, and October.
Answer:
Comparison of rainfall in March, June and October Rainfall in March—about 55 mm
Rainfall in June—about 185 mm
Rainfall in October—about 95 mm
So, rainfall is highest in June and lowest in March.
(v) Estimate the total rainfall received during the year 2021.
Answer:
Adding the approximate rainfall of all months
= 42 + 38 +55 +72 + 110 + 185 + 210 + 195 + 160 +95 + 60 + 48
= 1270 mm
Hence, total estimated rainfall in 2021 ~ 1270 mm.
Skill Based Questions
Question 1.
A class has an average height of 150.2 cm, when measured with shoes that add 1 cm to everyone’s height.
(a) Without re-measuring, find the correct average height without shoes.
Answer:
149.2 cm
(b) Explain briefly, why you do not need individual heights again.
Answer:
Do yourself
Question 2.
A data set has numbers 10, 10, 11, 17 with mean 12.
Two new numbers are added so that the mean remains 12.
(a) Write an equation relating the two new numbers x and y.
Answer:
x + y = 24
(b) Give one possible pair(x, y).
Answer:
(8,16)
Question 3.
Family size data for a class.
(a) Find the mean family size.
Answer:
5.22 (approx)
(b) Find the median family size (without expanding all 36 values).
Answer:
5
Question 4.
A line graph shows monthly maximum temperatures of Punjab and Kerala. It is observed that Punjab ranges roughly from 19°C (Jan) to 38°C (June), while Kerala stays between about 29°C and 33 °C year-round.
(a) Which state shows greater variation in maximum temperature?
Answer:
Punjab
(b) Which month is likely most uncomfortable (hottest) in Punjab?
Answer:
June
(c) Suggest one reason, why a line graph is better than a clustered bar graph here.
Answer:
Do yourself
Question 5.
A shop’s weekly data.
(a) On which day is the conversion (buyers as a fraction of visitors) highest?
Answer:
Thursday
(b) On which day is footfall highest but conversion not the highest?
Answer:
Sunday
Case Study Based Question
Question 1.
MrsAnanya, a mathematics teacher of Grade 8, conducted a unit test for her class. After correcting the answer sheets, she listed the marks (out of 50) scored by 11 students.
12,18, 20, 24, 26, 28, 30, 32, 35, 38, 47
She calculated the mean of the marks and found it to be 28. However, some students felt that the average score did not truly represent the performance of the class because one student scored much higher than the rest. Mrs Ananya decided to analyse the data further using the median and by examining, how the mean changes, when values are added or removed.
On the basis of the above information, answer the following questions.
(a) Find the median of the given marks and compare it with the mean.
Answer:
Given, marks obtained = 12,18, 20, 24, 26, 28, 30, 32, 35, 38, 47
Number of students = 11
Mean = 28
Since, the number of observations is odd (11), the median is the middle value, when the data is arranged in ascending order.
The middle position = \(\frac{11+1}{2}=\frac{12}{2}\) = 6th value
The 6th value is 28.
Hence, median is 28, which is equal to the mean (28).
(b) Explain how the highest score affects the mean of the data.
Answer:
Given, highest score = 47
Since, the mean is calculated using all values in the data. Since, 47 is much higher than most other scores, it pulls the mean towards the higher side. This makes the mean larger than what most students actually scored.
Hence, the highest score increases the mean and affects how well the mean represents the majority of the data.
(c) If the highest score (47) is removed, state whether the mean will increase, decrease or remains the same. Justify your answer.
Answer:
The original mean was 28. The value being removed 47 is greater than the mean. When a value greater than the mean is removed from a data set, the sum of remaining values decreases proportionally more than the number of values decreases, thus lowering the average.
(d) Which measure-mean or median-better represents the overall performance of the class? Give reasons.
Answer:
Since, most students scored near the middle, the median shows the class performance more accurately.
Hence, the median is a better measure to represent the overall performance of the class.