Practicing Ganita Prakash Class 7 Solutions and RBSE Class 7 Maths Part 2 Chapter 5 Connecting the Dots Solutions Question Answer helps develop logical thinking and accuracy.
Ganita Prakash Class 7 Part 2 Chapter 5 Solutions
Class 7 Ganita Prakash Part 2 Chapter 5 Solutions
RBSE Class 7 Maths Ganita Prakash Part 2 Chapter 5 Solutions
In-text Questions
Page 98
Question 1.
Which of the following are statistical questions?
(a) What is the price of a tennis ball in India?
(b) How old are the dogs that live on this street?
(c) What fraction of the students in your class like walking up a hill?
(d) Do you like reading?
(e) Approximately how many bricks are in this wall?
(f) Who was the best bowler in the match yesterday?
(g) What was the rainfall pattern in Barmer last year?
Solution:
(a) Not a statistical question because it expects a single fixed answer rather than data collection and analysis.
(b) Yes, a statistical question because different dogs have different ages and data needs to be collected.
(c) Yes, a statistical question because we need to collect data from students to find the fraction.
(d) Not a statistical question because it expects a simple yes/no answer from one person.
(e) Can be a statistical question if we estimate by measuring and analyzing patterns.
(f) Can be a statistical question if we use data like wickets taken, runs conceded, economy rate, etc.
(g) Yes, a statistical question because it requires collecting and analyzing rainfall data ova- months.
Pages 98-99
Question 1.
The runs scored by Shubman and Yashasvi in a cricket series are given in the table below. Who do you think performed better?
| Match 1 | Match 2 | Match 3 | Match 4 | |
| Shubman | 0 | 17 | 21 | 90 |
| Yashasvi | 67 | 55 | 18 | 35 |
Solution:
Shreyas says, “Both their perfor¬mances are similar since Yashasvi scored more in the first and second matches, whereas Shubman scored more in the third and fourth”. Vaishnavi says, “I think Shubman performed better because he scored the highest number of run in a match-90!”.
Shreyas says, “No! Yashasvi batted better since the total number of runs he made is 175, while Shubman made only 128”.
Vaishnavi says, “Oh! Also, Yashasvi’s batting is more consistent-the difference between his maximum score and minimum score is lower”.
Question 2.
The table below shows the runs scored by these two players in another series. Who do you think performed better in this series?
| Match 1 | Match 2 | Match 3 | Match 4 | Match 5 | |
| Shubman | 23 | 07 | 10 | 52 | 18 |
| Yashasvi | 26 | 53 | 02 | – | 15 |
Vaishnavi says, “Here, Shubman performed better since his total is 110 runs, while Yashasvi’s total is 96 runs”. What do you think of Vaishnavi’s statement?
Solution:
Shreyas says, “But Yashasvi made 96 runs in 4 matches and Shubman made 110 runs in 5 matches”.
So, how do we say who performed better? It is often not simple to compare two groups of numbers and clearly say that one is better than the other.
Question 3.
Can a single number act as a representative of a group of numbers?
Solution:
Yes, a single number can act as a representative of a group of numbers. Such a number is called a measure of central tendency.
Page 100
Question 1.
Shreyas and 4 of his friends have collected the following numbers of guavas: 3, 8,10,5, and 4. Parag and 5 of his friends have collected the following numbers of guavas: 5, 4,6,3, 4, and 8. Each group will share their guavas equally amongst themselves. In which group will each member get a bigger share of guavas?
Solution:
To find this out, we first find out how many guavas each group has collected. Then we divide this total by the number of people in the group to get each member’s share. Shreyas’s group has collected 3 + 8 + 10 + 5 + 4 = 30 guavas.
Each member of Shreyas’s group gets 30 ÷ 5 = 6 guavas.
Parag’s group has collected 5 + 4 + 6 + 3 + 4 + 8 = 30 guavas.
Each member of Parag’s group gets 30 ÷ 6 = 5 guavas.
So, the members of Shreyas’s group get 1 more guava each than the members of Parag’s group.
Question 2.
Vaishnavi tracks the number of Hibiscus flowers blooming in her garden each day. The data for the last few days’ is 2, 7, 9, 4, 3. What is the average number of Hibiscus flowers blooming per day in Vaishnavi’s garden?
Solution:
The average
= \(\begin{aligned}
& =\frac{\text { The total number of Hibiscus flowers bloomed }}{\text { Number of days }} \\
& =\frac{2+7+9+4+3}{5}=5 .
\end{aligned}\)
Page 104
Question 1.
Find the average price of onions at Yahapur and Wahapur.
Solution:
Average price in Yahapur
= 458 + 12 = 38.17 rupees per kg
Average price in Wahapur
= 450 – 12 = 37.50 rupees per kg
Page 105
Question 1.
The heights of the family members of Yaangba and Poovizhi are as follows : Yaangba’s family : 169 cm, 173 cm, 155 cm, 165 cm, 160 cm, 164 cm.
Poovizhi’s family : 170 cm, 173 cm, 165 cm, 118 cm, 175 cm.
Find the average height of each family. Can we say that Yaangba’s family is taller than Poovizhi’s family?
Solution:
Average height of Yaangba’s family
= (169 + 173 + 155 + 165 + 160 + 164) ÷ 6
= 986 ÷ 6 = 164.3 cm
Average height of Poovizhi’s family
= (170 + 173 + 165 + 118 + 175) ÷ 5
= 801 ÷ 5 = 160.2 cm
Although most members in Poovizhi’s family are taller, their family’s average height is less because one child is much younger and not as tall as the rest of the family.
Their average height, 160.2 cm, is less than the heights of 4 out of 5 members.
Here, the average doesn’t seem to represent the data very well.
Page 107
Question 1.
Find the mean and median in Poovizhi’s data without the outlier value 118. What change do you notice?
Solution:
Without 118: Heights are 170, 173, 165, 175
Mean = (170 + 173 + 165 + 175) ÷ 4
= 683 ÷ 4 = 170.75 cm
Median = (170 + 173) ÷ 2 = 171.5 cm
Now the mean and median are close to each other and better represent the data.
Page 115
Question 1.
What is the scale used in this graph?
Solution:
The scale used in this graph is Rs. 10 per unit on y – axis.
Question 2.
Is it now easier to compare month- wise prices in both places?
Solution:
Yes, it is easier, as visual display makes it easier to compare the data carefully and easily.
Page 122
Question 1.
The horizontal line lists the overs starting from 1, and the vertical line indicates the runs scored in each over. The graph shows the number of runs scored per over as a double bar graph- each bar corresponding to a team. Let us call
them the blue team (denoted by blue) and the red team (denoted by red). The scale used for the runs per over is 1 unit = 5 runs. The circles shown on top of the bars indicate that a wicket fell in that over.
Answer the following questions based on the graph :
1. Can we tell who batted first? Who won the match?
2. How many runs did the blue team score in over 12?
3. In which over did the red team score the least number of runs?
4. Is it easy to tell the target set by the team batting first?
Solution:
1. I think blue team batted first. By getting the total of the runs scored by both the teams in each over, we can find out who wins.
2. 15 runs
3. In over 4
4. Not easy, as it is not directly visualised from the graph.
Pages 125-126
Question 1.
Following are the dot plots of heights of boys (in blue) and girls (in orange) of Grades 6, 7 and 8 (in that order) of two different schools. What do you notice? Share your observations.
Solution:
School A :
- Grade 6 : Mean (boys) = 134:8 cm, Mean (girls) = 137.78 cm
- Grade 7 : Mean (boys) = 141.8 cm, Mean (girls) = 141.83 cm
- Grade 8 : Mean (boys) = 149.35 cm, Mean (girls) = 147.81 cm
School B:
- Grade 6 : Mean (boys) = 149.84 cm, Mean (girls) = 150.2 cm
- Grade 7 : Mean (boys) = 156.14 cm, 4Mean (girls) = 155.41 cm
- Grade 8 : Mean (boys) = 156.14 cm, Mean (girls) = 156.83 cm
Observations :
- In School A, girls are taller than boys in Grades 6 and 7, but boys become taller in Grade 8.
- In School B, girls are taller than boys in Grade 6 and Grade 8 but boys are taller than girls in Grade 7.
- School B students are significantly taller than School A students across all grades.
Class 7 Maths Connecting the Dots Solutions
Figure it Out (Pages 101)
Question 1.
Shreyas is playing with a bat and a ball — but not cricket. He counts the number of times he can bounce the ball on the bat before it falls to the ground. The data for 8 attempts is 6, 2, 9, 5, 4, 6, 3, 5. Calculate the average number of bounces of the ball that Shreyas is able to make with his bat.
Solution:
Total bounces = 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5 = 40
Number of attempts = 8
Average = 40 = 8
= 5 bounces
Question 2.
Try the activity above on your own. Collect data for 7 or more attempts and find the average.
Solution:
Do it yourself.
Question 3.
Identify a flowering plant in your neighbourhood. Track the number of flowers that bloom every day over a week during its flowering season. What is the average number of flowers that bloomed per day?
Solution:
Do it yourself.
Question 4.
Two friends are training to run a 100 m race. Their running times over the past week are given in seconds — Nikhil : 17, 18, 17, 16, 19, 17, 18; Sunil: 20, 18, 18, 17, 16, 16, 17. Who on average ran quicker?
Solution:
Nikhil’s total time = 17 + 18 + 17 + 16 + 19 + 17 + 18 = 122 seconds
Nikhil’s average = 122 ÷ 7 = 17.43 seconds
Sunil’s total time = 20 + 18 + 18 + 17 + 16 + 16 + 17 = 122 seconds
Sunil’s average = 122 ÷ 7 = 17.43 seconds
Both ran at the same average speed.
Question 5.
The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2040, 2126. Find the mean enrolment in the school during this period.
Solution:
Total enrolment = 1555 + 1670 + 1750 + 2013 + 2040 + 2126 = 11154
Mean enrolment = 11154 ÷ 6 = 1859 students
Figure it Out (Pages 112-113)
Question 1.
Find the median of onion prices in Yahapur and Wahapur.
Solution:
Yahapur prices : 25, 24, 26, 28, 30, 35, 39, 43, 49, 56, 59, 44
Data in ascending order : 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59
Median = (35 + 39) ÷ 2 = 37 rupees per kg
Wahapur prices : 19, 17, 23, 30, 38, 35, 42, 39, 53, 60, 52, 42
Data in ascending order : 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60
Median = (38 + 39) ÷ 2 = 38.5 rupees per kg
Question 2.
Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, —, 10, 25, 2, —, 2, 4. Find the mean and median. How would you describe this data?
Solution:
Values (excluding —): 0, 1, 0,4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4
Total = 0 + 1 + 0 + 4 + 8 + 0 + 0 + 2 + 1 + 1 + 5 + 3 + 4 + 0 + 0 + 10 + 25 + 2 + 2 + 4 = 72
Number of values = 20
Mean = 72 ÷ 20 = 3.6 animals
Sorted: 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25
Median = (2 + 2) ÷ 2 = 2 animals
Most students have 0 to 4 animals. The value 3.6 is an outlier. The mean is affected by this outlier, making the median a better representative.
Question 3.
Rintu takes care of a date-palm tree farm in Habra. The heights of the trees (in feet) in his farm are given as: 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Fill the dot plot, and mark the mean and median. How would you describe the heights of these palm trees? Can you think of quicker ways to find the mean? How many trees are shorter than the average height?
Solution:
Total = 50 + 45 + 43 + 52 + 61 + 63 + 46 + 55 + 60 + 55 + 59 + 56 + 56 + 49 + 54 + 65 + 66 + 51 + 44 + 58 + 60 + 54 + 52 + 57 + 61 + 62 + 60 + 60 + 67 = 1611 feet
Number of trees = 29
Mean = 1621 ÷ 29 = 55.89 feet
Data in ascending order : 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55, 56, 56, 57, 58, 59, 60, 60, 60, 60, 61, 61, 62, 63, 65, 66, 67 Median = 15th value = 56 feet
Heights are fairly evenly distributed between
43 and 67 feet, with most trees between 50 and 62 feet.
Trees shorter than average height (55.55): 43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55 = 13 trees
Mark the data by yourself on the plot.
Question 4.
The daily water usage from a tap was measured. The usage in liters for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4.
(a) Can the mean or median daily usage lie between 25 and 30? Justify your claim using the meaning of mean and median.
(b) Can the mean or median be lesser than the minimum value or greater than the maximum value in a data?
Solution:
(a) No, because all values are less than 21. The mean is the average of all values, so it must be less than the maximum value (20.5). The median is the middle value when sorted, so it also cannot be greater than the maximum value.
(b) No, the mean and median always lie within the range of the data (between minimum and maximum).
Question 5.
The weights of a few newborn babies are given in kgs. Fill the dot plot provided below. Analyse and compare this data.
| Boys | 3.5 | 4.1 | 2.6 | 3.2 | 3.4 | 3.8 |
| Girls | 4.0 | 3.1 | 3.4 | 3.7 | 2.5 | 3.4 |
Solution:
The dots show the boys and girls is shown the cross.
We can analyse the data, by finding their mean and median.
Boys : Mean = (3.5 +4.1 + 2.6 + 3.2 + 3.4 + 3.8) + 6 = 20.6 – 6 = 3.43 kg
Sorted : 2.6, 3.2, 3.4, 3.5, 3.8, 4.1; Median = (3.4 + 3.5) ÷ 2 = 3.45 kg
Girls : Mean = (4.0 + 3.1 + 3.4 + 3.7 + 2.5 + 3.4) ÷ 6 = 20.1 – 6 =3.35 kg
Sorted : 2.5, 3.1, 3.4, 3.4, 3.7, 4.0; Median = (3.4 + 3.4) ÷ 2 =3.4 kg
Both groups have similar mean and median, indicating newborn boys and girls have comparable weights.
Question 6.
The dot plots of heights of another section of Grade 5 students of the same school are shown below. Can you share your observations? What can we infer from the dot plots and the central tendency measures?
Solution:
Whole class :
Mean = 141.21 cm Median = 142.5 cm Boys :
Mean = 142.05 cm Median = 143 cm
Girls :
Mean = 140.14 cm Median = 140 cm
Observations :
In this section, boys are taller than girls on average (opposite of the first section). Heights are more evenly distributed compared to the first section.
Question 7.
The weights of some sumo wrestlers and ballet dancers are : Sumo wrestlers : 295.2 kg, 250.7 kg, 234.1 kg, 221.0 kg, 200.9 kg. Ballet dancers : 40.3 kg, 37.6 kg, 38.8 kg, 45.5 kg, 44.1 kg, 48.2 kg. Approximate^ how many times heavier is a sumo wrestler compared to a ballet dancer?
Solution:
Average weight of sumo wrestlers
\(\begin{aligned}
& =\frac{295.2+250.7+234.1+221.0+200.9}{5} \\
& =\frac{1201.9}{5}=240.38 \mathrm{~kg}
\end{aligned}\)
Average weight of ballet dancers
\(\begin{aligned}
& =\frac{40.3+37.6+38.8+45.5+44.1+48.2}{6} \\
& =\frac{254.5}{6}=42.42 \mathrm{~kg}
\end{aligned}\)
Times heavier = \(\frac{240.38}{42.42}\)
= Approximately 5.7 times
Figure it Out (Pages 122-125)
Question 1.
The following infographic shows the speeds of a few animals in air, on land, and in water. Can we call this graph a bar graph?
(a) What is the scale used in this graph?
(b) What did you find interesting in this infographic? What do you want to explore further?
(c) Identify a pair of creatures where one’s speed is about twice that of the other.
(d) Can we say that a sailfish is about 4 times faster than a humpback whale? Can we say that a sailfish is the fastest aquatic animal in the world?
Solution:
Yes, we can call this a bar graph.
(a) Scale used is 16 km/h per unit.
(b) Do it yourself
(c) Examples — (i) Peregrine falcon and spine tailed swift, (ii) Flying fish and humpback whale, (d) The sailfish travels at about 109 km/h, while the humpback whale swims at around 26 km/h, making the sailfish approximately 4.2 times faster. Based on the given graph, the sailfish appears to be the fastest aquatic animal shown; however, we cannot conclude that it is the fastest in the world without additional data.
Question 2.
Preyashi asked her students ‘If you were to get a super power to become aquatic (water-borne), aerial (air-borne), or spaceborne which one would you choose?’. The responses are shown below. Some chose none. Draw a double-bar graph comparing how both grades chose each option. Choose an appropriate scale.
| Grade 5 | w, a, a, a, w, n, s, a. n, w, a, a, a, a, a, w, w, s, a, a, n, w, a, a, n |
| Grade 9 | n, w, s, a, s, w, s. s, a, a, w, s, s, a, s, a, n, w, s, s, a, w, a, w, a |
Solution:
Count for Grade 5: Air (a) = 13, Water (w) = 6, Space (s) = 2, None (n) = 4
Count for Grade 9 : Air (a) = 8, Water (w) = 6, Space (s) = 9, None (n) = 2
Question 3.
The temperature variation over two days in different months in Jodhpur, Rajasthan, is given below. Draw a double-bar graph. Use the scale 1 unit = 4°C. Can you guess which two months these days might belong to?
| 12 am | 3 am | 6 am | 9 am | 12 pm | 3 pm | 6 pm | 9 pm | |
| Day 1 | 20°C | 18°C | 16°C | 20°C | 26°C | 34°C | 30°C | 24°C |
| Day 2 | 37°C | 34 °C | 30°C | 33°C | 37°C | 43°C | 42°C | 39°C |
Solution:
Day 1 appears to be from a cooler month (like January or February).
Day 2 appears to be from a hotter month (like May or June).
Question 4.
The following clustered-bar graph shows the number of electric vehicles registered in some states every year from 2022 to 2024.
(a) The data (rounded-off to thousands) for the states of Gujarat and Delhi are given in the table below. Mark the corresponding bars on the bar graph. (It is enough if you place the top of the bars between the two appropriate vertical guidelines.)
| 2022 | 2023 | 2024 | |
| Gujarat | 69000 | 89000 | 78000 |
| Delhi | 62000 | 74000 | 81000 |
(b) Notice how the graph is organised, what scale is used, and what patterns the data shows.
(c) How would you describe the change for various states between 2022 and 2024?
(d) Approximately how many more registrations did Assam get in 2023 compared to 2022?
(e) How many times more did the registrations in West Bengal increase from 2022 to 2024?
(f) Is this statement correct-4There were very few new registrations in Uttarakhand in 2023 and 2024, as the increase in the bar lengths is minimal’?
Solution:
(a) Do it yourself.
(b) Scale : 1 cm = 25,000 vehicles registered. It shows that use of electric vehicles is increasing every year.
(c) The trend in increasing in all the states.
(d) In 2022 it is 40,000 (Approx.)
In 2023 it is 60,000 (Approx.)
More registration = 60,000 – 40,000 = 20,000
(e) In 2022 = 11,000 (Approx.)
In 2024 = 44,000 (Approx.) becomes 4 times.
(f) Yes, in the complete bar graph we can observe that Uttarakhand has minimum number of registrations signifying its minimal bar length.
Figure it Out Pages 129-134
Question 1.
The dot plots below show the distribution of the number of pockets on clothing for a group of boys and for a group of girls
Based on the dot plots, which of the following statements are true?
(a) The data varies more for the boys than for the girls.
(b) The median number of pockets for the boys is more than that for the girls.
(c) The mean number of pockets for the girls is more than that for the boys.
(d) The maximum number of pockets for boys is greater than that for the girls.
Solution:
(a) No, the data varies more for girls than boys.
(b) Yes, the median number of pockets for the boys is more than that for the girls.
(c) Mean for boys is greater than for girls.
(d) Maximum number of pockets for both girls and boys are same.
Question 2.
The following table shows the points scored by each player in four games :
| Player | Game 1 | Game 2 | Game 3 | Game 4 |
| A | 14 | 16 | 10 | 10 |
| B | 0 | 8 | 6 | 4 |
| C | 8 | 11 | Did not play | 13 |
Now answer the following questions :
(a) Find the average number of points scored per game by A.
(b) To find the mean number of points scored per game by C, would you divide the total points by 3 or by 4? Why? What about B?
(c) Who is the best performer?
Solution:
(a) Average score of A
\(=\frac{14+16+10+10}{4}=\frac{50}{4}\)
= 12.5
(b) Player C
C played only 3 games (missed Game 3).
Total points by C = 8+ 11 + 13 = 32
To find C’s average, we divide by 3, not 4, because the mean must be calculated using the number of games actually played.
Mean for C = 32 ÷ 3 ≈ 10.67 points per game
Player B
B played all 4 games, even though one score is zero.
Total points by B = 0 + 8 + 6 + 4 = 18
Mean for B = 18 ÷ 4 = 4.5 points per game
(c) Best performer Comparing averages :
A = 12.5 points per game
C ≈ 10.67 points per game
B = 4.5 points per game
Player A is the best performer, having the highest average score per game.
Question 3.
The marks (out of 100) obtained by a group of students in a General Knowledge quiz are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Another group’s scores in the same quiz are 68, 59, 73, 86, 47, 79, 90, 93 and 86. Compare and describe both the groups performance using, mean and median.
Solution:
Given Data
Group 1 – Scores :
85, 76, 90, 85, 39, 48, 56, 95, 81, 75
Group 2 – Scores :
68, 59, 73, 86, 47, 79, 90, 93, 86
Find Mean Group 1
Sum = 85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75
= 730
Number of students = 10
Mean = 730 – 10 = 73
Group 2
Sum = 68 + 59 + 73 + 86 + 47 + 79 + 90 + 93 + 86
= 681
Number of students = 9
Mean = 681 ÷ 9 ≈ 75.67
Find Median
Group 1 – Arrange scores in ascending order:
39, 48, 56, 75, 76, 81, 85, 85, 90, 95
Since there are 10 observations (even number):
Median = average of 5th and 6th values
= (76 + 81) ÷ 2 = 78.5
Group 2 – Arrange scores in ascending order:
47, 59, 68, 73, 79, 86, 86, 90, 93
Since there are 9 observations (odd number):
Median = 5th value = 79
Both the mean and median of Group 2 are higher than those of Group 1. This shows that Group 2 performed better overall in the quiz.
Question 4.
Consider this data collected from a survey of a colony.
| Favourite Sports | Cricket | Basket Ball | Swimming | Hockey | Athletics |
| Watching | 1240 | 470 | 510 | 430 | 250 |
| Participating | 620 | 320 | 320 | 250 | 105 |
Choose an appropriate scale and draw a double-bar graph. Write down your observations. Solution:
Solution:
Question 5.
Consider a group of 17 students with the following heights (in cm) :
106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101.
The sports teacher wants to divide the class into two groups so that each group has an equal number of students: one group has students with height less than a particular height and the other group has students with heights greater than the particular height. Suggest a way to do this. Can you guess the age of these students based on the tabular data in the ‘Telling Tall Tales’ section?
Solution:
The 17 students could not be divided into two equal groups because the groups can be formed of 8 or 9, not equal. For doing so, median is the best measure.
Sorted heights (cm):
101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125
One group can have first 6 students and second group can have next 8 students.
So a reasonable guess is that these students are around 5-7 years old (most likely about 6 years).
Question 6.
Describe the mean and median of heights of your class. You can visualise the heights on a dot plot.
Solution:
Do it yourself.
Question 7.
There are two 7th grade sections at a school. Each section has 15 boys and 15 girls. In one section, the mean height of students is 154.2 cm.
From this information, what must be true about the mean height of students in the other section?
(a) The mean height of students in the other section is 154.2 cm.
(b) The mean height of students in the other section is less than 154.2 cm.
(c) The mean height of students in the other section is more than 154.2 cm.
(d) The mean height of students in the other section cannot be determined.
Solution:
(d) The mean height of students in the other section cannot be determined.
No information is given about :
- The heights of boys and girls individually, or
- Whether the students in the two sections are similar in height.
Because the actual heights of students in the second section may be :
- equal to those in the first section,
- generally shorter, or
- generally taller,
- the mean height of the second section could be equal to, less than, or greater than 154.2 cm.
So, from the given information, it is not possible to determine the mean height of the other section
Question 8.
Standing tall in the storm.
(a) Write estimated values for the number of skyscrapers in New York, Tokyo, and London.
(b) Are the following statements valid?
(i) Only 12 cities have more skyscrapers than Mumbai.
(ii) Only 7 cities have fewer skyscrapers than Mumbai.
(iii) The tallest building in the world is in Hong Kong.
Solution:
(a) New York : 309, Tokyo : 168, London : 25 (b) (i) True
(ii) True
(iii) Can’t be said.
Question 9.
Estimate and then measure the objects listed in the following table. Draw a double bar graph based on the data. How accurate were your estimates? Find the average difference between the estimated and measured values.
| Object | Estimate (in cm) | Measure (in cm) | Positive Difference |
| Length of a pen | |||
| Length of an eraser | |||
| Length of your palm | |||
| Length of your geometry box | |||
| Length of your math notebook |
Solution:
Do it yourself.
Question 10.
Aditi likes solving puzzles. She recently started attempting the ‘Easy’ level Sudoku puzzles. The time she took (in seconds) to solve these puzzles are— 410, 400, 370, 340, 360, 400, 320, 330, 310, 320, 290, 380, 280, 270, 230, 220,
240. The first nine values correspond to Week 1 and the rest to Week 2.
(a) Construct a dot plot below showing the data for both weeks.
(b) Describe the mean, median, and any observations you may have about the data.
Solution:
Given Data
Week 1 (first 9 values, in seconds) :
410, 400, 370, 340, 360, 400, 320, 330, 310
Week 2 (next 8 values, in seconds) :
320, 290, 380, 280, 270, 230, 220, 240
(a) Do it yourself.
(b) Mean, Median, and Observations
Week 1 – Calculations
Sorted data :
310, 320, 330, 340, 360, 370, 400, 400, 410
Mean
Total = 3240
Mean 3240 ÷ 9 = 360 s
Median
Middle value (5th of 9) = 360 s
Week 2 – Calculations
Sorted data :
220, 230, 240, 270, 280, 290, 320, 380 Mean
Total = 2230
Mean – 2230 ÷ 8 ≈ 278.8 s
Median
Average of 4th and 5th values = (270 + 280) ÷ 2 = 275 s
Aditi showed significant improvement from Week 1 to Week 2, as reflected by lower average and median solving times.
Question 11.
Individual Project: Pick at least one of the following :
(a) How Long is a Sentence? Pick any two textbooks from different subjects. Choose any page with a lot of text from each book.
(i) Use a dot plot to describe how many words the sentences have on each page.
(ii) Compare the data of both the pages using mean and median.
(b) What is in a Name? Write down the names of all of your classmates. The following are some interesting things you can do with this data!
(i) Find the mean and median name length (number of letters in a name).
(ii) Visualise the data and describe its variability and central tendency.
(iii) Which starting letters are more popular? Which are less popular?
(iv) What is the median starting letter? What does this say about the number of names starting with the letters A – M and N – Z?
(v) Plot a double-bar graph showing the number of boys’ names and girls’ names that:
- start and end with vowels,
- start with vowels and end with consonants,
- start with consonants and end with vowels,
- start and end with consonants.
Solution:
Do it yourself.
Question 12.
Individual project (long term): This requires collecting data over 2 weeks or more.
In and Out : Track how many times you step out of your house in a day. Do this for a month.
(i) Describe the variability and central tendency of this data. Make a dot plot.
(ii) Do you find anything interesting about this data? Share your observations.
(iii) You can ask any of your family members or friends to do this as well.
Solution:
Do it yourself.
Question 13.
Small-group project: Pick at least one of the following. Make groups of 8 to 10. Collect data individually as needed. Put together everyone’s data and do the appropriate analysis and visualisation.
(a) Our heights vs. our family’s heights: Collect the heights of your family members.
(i) Make a dot plot showing heights of just your family members. Describe its variability and central tendency.
(ii) Make a double-bar graph showing each student’s height next to their family’s mean height.
(iii) Look at everyone’s data and share your observations.
(b) Estimating time: Check the time and close your eyes. Open them when you think 1 minute has passed (no counting). Note down after how many seconds you opened your eyes. Collect this data for yourself and for your family members. Repeat this activity to estimate 3 minutes.
(i) Make two dot plots (for 1 minute and 3 minutes) showing estimates of just your family members.
(ii) Mark these on the respective dot plots. Describe its variability and central tendency.
(iii) Make a double bar graph showing each family’s mean 1 minute estimate and mean 3 minute estimate.
(iv) Look at everyone’s data and share your observations.
Solution:
Do it yourself.
It’s Puzzle Time
Connect the Dots
A number lock has a 3-digit code. Find the code using the hints below.
Solution:
The correct digits can be 1, 4 and 5.
5 is rightly placed at ones place.
So, possible combination can be : 415.
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