Tales by Dots and Lines Class 8 Extra Questions Maths Part 2 Chapter 5 - #NCSOLVE 📚

During revision, students quickly go through Class 8 Maths Extra Questions Part 2 Chapter 5 Tales by Dots and Lines Class 8 Extra Questions with Answers for clarity.

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

Question 1.
Find the mean of the following data values:
4, 2, 8, 7, 1, 10, 9, 11, 3, 15
Solution:
Total number of data values = 10
Sum of the values = 4 + 2 + 8 + 7 + 1 + 10 + 9 + 11 + 3 + 15 = 70
We have, mean = \(\frac{\text { Sum of all data values }}{\text { Number of data values }}\)
= \(\frac {70}{10}\)
= 7

Question 2.
The mean of a data set: x₁, x₂, x₃, x₄, x₅, …,x_n is a, if each data value is increased by 7. What will be the new mean of the data set?
Solution:
Since, mean = \(\frac{x_1+x_2+x_3+\ldots+x_n}{n}\) = a
When each data point is increased by 5,
The new mean

Hence, the new mean of the data set is a + 7.

Question 3.
What is the median of the following data set?
3, 5, 7, 12, 15, 18, 20
Solution:
Here, the number of data values = 7
And the data is already arranged in ascending order.
The middle number in the sorted data set is 12.
So, the median is 12.

Question 4.
Find the missing number in the following data set so that the mean is 15:
12, 16,…., 20
Solution:
Let the missing number be x.
Mean = 15
⇒ \(\frac{12+16+x+20}{4}\) = 15
⇒ 48 + x = 60
⇒ x = 12
So, the missing number is 12.

Question 5.
If the median of the data set is 10, find the missing numbers:
5, 8, 10,…..,18, 22
Solution:
Since there are 6 data values and the median is 10 (the average of the middle two numbers), the missing number must be 10.

Question 6.
A class of 25 students has an average score of 72. If a new student joins the class with a score of 80, what will the new average be?
Solution:
New total score = (72 × 25) + 80
= 1800 + 80
= 1880
New average = \(\frac {1880}{26}\) = 72.31

Question 7.
The following graph shows the temperature forecast and the actual temperature for each day of a week.

(a) On which days was the forecast temperature the same as the actual temperature?
(b) What was the maximum forecast temperature during the week?
(c) What was the minimum actual temperature during the week?
(d) On which day did the actual temperature differ the most from the forecast temperature?
Solution:
(a) The forecast temperature is the same as the actual temperature on Tuesday, Friday, and Sunday.
(b) The maximum forecast temperature was 35°C.
(c) The minimum actual temperature was 15°C.
(d) The actual temperature differs the most from the forecast temperature on Thursday.

Question 8.
Observe the given line graph and answer the following questions.

(a) What does the given line graph show?
(b) Compare the circulation of Hindi and English newspapers on Day 2.
(c) On which days does the circulation of the Hindi newspaper exceed that of the English newspaper?
(d) On which day does the English newspaper have more circulation than the Hindi newspaper?
(e) What trend is seen in the circulation of the Hindi newspaper from Day 1 to Day 3?
(f) How does the circulation of the English newspaper change between Day 3 and Day 5?
Solution:
(a) The line graph shows the circulation of Hindi and English newspapers on weekdays.
(b) On Day 2, about 600 copies of the Hindi newspaper were circulated, while about 400 copies of the English newspaper were circulated; therefore, about 200 more copies of the Hindi newspaper were circulated.
(c) The circulation of the Hindi newspaper was higher on Day 1, Day 2, Day 4, Day 5, and Day 6.
(d) The circulation of the English newspaper was higher on Day 3 only.
(e) It shows a sharp decline, falling from 750 to 200 copies.
(f) It increases from 300 copies on Day 3 to 450 copies on Day 5, showing a steady rise.

Question 9.
The marks obtained by four students in a test are 24, 32, 28, and 36. Find the mean. How will the mean change if a student scoring 40 marks is included? Represent all the values on the dot slot after adding the new value.
Solution:
Mean = \(\frac{24+28+32+36}{4}\)
= \(\frac {120}{4}\)
= 30
After including 40 marks
New mean = \(\frac{24+28+32+36+40}{5}\)
= \(\frac {160}{5}\)
= 32
The mean increases as the newly added value is greater than the original mean.

Question 10.
The age of six men in a group are 26, 28, 29, 30, 34, and 38. Find the median.
If the man aged 38 years is removed from the group, what is the new median of the group?
Solution:
Ages in years = 26, 28, 29, 30, 34, 38
Number of observations = 6 (even)
Median (even observations) = average of 3rd and 4th observations
= \(\frac{29+30}{2}\)
= \(\frac {59}{2}\)
= 29.5
After removing the man aged 38.
Number of observations = 5 (odd)
New median = Value of \(\frac{(n+1)^{t h}}{2}\) observation
= Value of 3rd observation
= 29
The median of the group is 29 years.

Question 11.
The daily wages (in ₹) of four workers in a small manufacturing unit are ₹ 650, ₹ 700, ₹ 680, and ₹ 720 respectively. The management has decided to double the wages of all the workers. Find the new mean of the daily wages of the workers.
Solution:
Original wages = ₹ 650, ₹ 700, ₹ 680, ₹ 720
Original mean = \(\frac{650+700+680+720}{4}\)
= \(\frac {2750}{4}\)
= ₹ 687.50
When each wage is doubled.
New wages = ₹ 1300, ₹ 1400, ₹ 1360, ₹ 1440
New mean = \(\frac{1300+1400+1360+1440}{4}\)
= \(\frac {5500}{4}\)
= ₹ 1375
New mean of daily wages of the workers = ₹ 1375

Question 12.
A cricket team’s average score in six matches is 84 runs. The scores are 72, 78, 82, 88, and 92. Find the score of sixth match.
Solution:
Total runs in 6 matches = 84 × 6 = 504
Sum of given runs = 72 + 78 + 82 + 88 + 92 = 412
Runs in 6th match = 504 – 412 = 92
The score of the 6th match is 92.

Question 13.
Priya tracks the number of flowers bloomed in her garden. She calculates an average of 18 flowers per plant across 12 plants. Later, she finds that one plant’s count was recorded 2 more than the actual. What is the correct average of flowers per plant?
Solution:
Original total flowers = 18 × 12 = 216
Correct total = 216 – 2 = 214
Correct average = \(\frac {214}{12}\) = 17.8
The average flowers per plant is 17.8.

Question 14.
The following table shows the marks scored by students of class 8. Represent all data on a dot plot.

Find the mean and median marks of class 8.
Solution:
Mean = \(\frac{\text { Sum of observations }(f \times x)}{\text { Total number of observations }(f)}\)

For Median

As total observation (32) is even.
So median = average of \(\frac{n^{\text {th }}}{2}\) and \(\left(\frac{n}{2}+1\right)^{t h}\) terms
= average of \(\left(\frac{32}{2}\right)^{t h}\) and \(\left(\frac{32}{2}+1\right)^{t h}\) terms
= average of 16th and 17th terms

The 16th and 17th terms lie in the cumulative frequency corresponding to 30.
So median = \(\frac{30+30}{2}\) = 30

Question 15.
The weights (in kg) of four students are 24, 30, 36, and 42. Find the mean. What will be the new mean if a student weighing 48 kg joins the group?
Solution:
Number of students = 4
Mean = \(\frac{24+30+36+42}{4}\)
= \(\frac {132}{4}\)
= 33
The mean is 33 kg.
Mean after joining of a student weighing 48 kg
Now total students = 5
New mean = \(\frac{132+48}{5}\)
= \(\frac {180}{5}\)
= 36
The new mean will be 36 kg.

Question 16.
The mean of the numbers 14, 6, 8, 12, 10, 15, and p is 12. Find the value of p.
Solution:
Total numbers = 7
Sum of nupibers = 12 × 7 = 84
Sum of given numbers = 14 + 6 + 8 + 12 + 10 + 15 + p = 65 + p
⇒ 84 = 65 + p
⇒ p = 84 – 65
⇒ p = 19

Question 17.
The average (mean) temperature in a city over 5 days was 28°C. The temperatures for 4 of the days were 25°C, 30°C, 26°C, and 32°C. What was the temperature on the 5th day?
Solution:
Mean temperature = 28°C
Total temperature for 5 days = 28 × 5 = 140°C
Sum of temperatures for 4 days = 25 + 30 + 26 + 32 = 113°C
Temperature on the 5th day = 140 – 113 = 27°C

Question 18.
Find the median of 5, 9, 14, 18, 27, 21, 30, 32, 35, 39, 42, 46, 60, 54, 60.
If we include one value in the data without affecting the median, what could that value be?
Solution:
Arranging the data
5, 9, 14, 18, 21, 27, 30, 32, 35, 39, 42, 46, 50, 54, 60
Number of observations = 15 (odd)
Median = Value of \(\left(\frac{15+1}{2}\right)^{t h}\) term
= value of 8th term
= 32
One value that can be added without affecting the median – 32 or any value between 30 and 35.

Question 19.
A basketball player scored the following points in 6 matches:
18, 22, 26, 30, 34, and 40.
(i) Calculate the mean number of points scored by the player.
(ii) If the points scored in each match are increased by 5 times, find the new set of scores.
(iii) Calculate the mean of the new scores.
(iv) State the relation between the original mean and the new mean.
Solution:
Given scores 18, 22, 26, 30, 34, 40
(i) Mean = \(\frac{18+22+26+30+34+40}{6}\)
= \(\frac {170}{6}\)
= 28.3
(ii) New scores after increasing each value 5 times
90, 110, 130, 150, 170, 200
(iii) Mean of new score = \(\frac{90+110+130+150+170+200}{6}\)
= \(\frac {850}{6}\)
= 141.6
(iv) The new mean is 5 times the original mean.

Question 20.
The table shows the number of vehicles passing through a toll plaza in one hour on different days. The data was recorded to study the traffic flow pattern over a period of several days.

Using the above data,
(i) Calculate the mean number of vehicles passing per hour.
(ii) Find the median number of vehicles recorded during the given days.
Solution:


The mean number of vehicles passing per hour is 194.

(ii) Total number of observations = 20 (even)
So median = average of \(\left(\frac{20}{2}\right)^{t h}\) term and \(\left(\frac{20}{2}+1\right)^{t h}\) term
= average of 10th term and 11th term
The 10th and 11th observations lie in the cumulative frequency corresponding to 200.
So, Median = \(\frac{200+200}{2}\) = 200
The median number of vehicles recorded is 200.

Question 21.
The line graph below represents the amount of water (in litres) in a tank of a society at different times of the day.

(i) At what time was the water level highest in the tank?
(ii) During which time interval did the water level decrease?
(iii) How much water was added to the tank between 6 a.m. and 10 a.m.?
(iv) What was the water level at noon?
(v) What does the steep rise in the graph between 2 p.m. and 4 p.m. indicate?
Solution:
(i) The water level was highest at 4 p.m., when it was 600 litres.
(ii) The water level decreased between 10 a.m. and 2 p.m.
(iii) Water at 10 a.m. = 500 litres
Water at 6 a.m. = 200 litres
Water added = 500 – 200 = 300 litres.
(iv) The water level at 12 noon was 450 litres.
(v) It indicates that a large amount of water was added to the tank during this time.

Question 22.
The line graph shows the population (in thousands) of City A, City B, and City C from the year 2015 to 2020.

(i) Which city had the highest population in 2020?
(ii) Which city showed the fastest growth during the given period?
(iii) In which year did city B’s population cross 450 thousand?
(iv) Compare the population growth of city C from 2015 to 2020.
(v) What does the overall trend of the graph indicate?
Solution:
(i) City A had the highest population in 2020 (580 thousand).
(ii) City A showed the fastest growth.
(iii) In the year 2018.
(iv) The population growth of city C increased steadily from 300 thousand to 420 thousand.
(v) The graph indicates that the population of all three cities increased over time.

Question 23.
The table below shows the annual rainfall (in cm) recorded in two cities, P and Q, over a period of five years.

Visualise this data on a line graph.
What does the overall trend of rainfall in city Q show?
Solution:
Take a year on the horizontal axis (x-axis)
Take rainfall on the vertical axis (y-axis)

The rainfall in city Q shows an overall increasing trend with a slight decrease in 2022.

The post Tales by Dots and Lines Class 8 Extra Questions Maths Part 2 Chapter 5 appeared first on Learn CBSE.

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