By using Ganita Prakash Class 7 Solutions and Part 2 Chapter 5 Connecting the Dots Class 7 Question Answer, students can improve their problem-solving skills.
Class 7 Maths Ganita Prakash Part 2 Chapter 5 Solutions
Class 7 Maths Connecting the Dots Solutions
Class 7 Ganita Prakash Part 2 Chapter 5 Solutions Connecting the Dots
Figure it Out (Page : 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:
Number of attempts = 8
Number of bounces of the ball
= 6,2,9,5,4,6,3,5
Sum :
∑ x i = 6 + 2 + 9 + 5 + 4 + 6 + 3 + 5 = 40
To calculate Average Mean :
The average number of bounces the ball that Shreyas is able to make with his bat is 5.
Question 2.
Try the activity above on your own. Collect data for 7 or more attempts and find the average.
Solution:
The data for 7 attempts is 3, 5, 4, 7, 6, 4, 5
Sum (∑ x i) = 3 + 5 + 4 + 7 + 6 + 4 + 5 = 34
Number of attempts (N) = 7
Average number of bounces = 4.86 approx.
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:
Number of new flowers that bloomed daily over a week (7 days) are as: (2, 3, 1, 4, 2, 3, 2)
Sum (∑ x i) = 2 + 3 + 1 + 4 + 2 + 3 + 2 = 17
Number of days (N) = 7
Average number of flowers that bloomed per day = 2.43 approximately.
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:
Average running times
Both Nikhil and Sunil have the same average running time, so both ran equally fast.
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:
The total sum (∑ xi) of enrolments =
∑xi = 1555 + 1670 + 1750 + 2013 + 2040 + 2126 = 11154
Number of years = 6
Mean enrolment
Mean enrolment in the school during this period was 1859 students.
Figure it Out (Page: 112-113)
Question 1.
Find the median of onion prices in Yahapur and Wahapur.
Solution:
To find the median, the data for each town must be arranged in ascending order.
Yahapur prices (sorted) :
[(24,25,26,26,30,35,39,43,44,49,56,59)
Wahapur prices (sorted) :
[(16,17,18,19,23,30,35,38,39,42,52,60)
Yahapur median : 6th and 7th value are 35 and 39
Median = \(\frac{(35 + 39)}{2}\) = \(\frac{74}{2}\) = 37
Wahapur median : 6th and 7th value are 38 and 39
Median = \(\frac{ 38+39}{2}\) = \(\frac{ 77}{2}\) = 38.5
Median onion price in Yahapur is ₹37/kg Median onion price in Wahapur is ₹38.5/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:
The mean is calculated by summing all the data points and dividing by the total number of students surveyed (20).
Sum of data points :
∑xi = 0 + 1 + 0 + 4 + 8 + 0 + 0 + 2 + 1 + 1 + 5 + 3 + 4 + 0 + 0 + 10 + 25 + 2 + 2 + 4 = 72
Mean calculation :
Mean = Sum Number of students = 72/20 = 3.6.
The mean number of pets is 3.6.
Median
The median is the middle value of the data set when arranged in order. We first sort the 20 data points from least to greatest:
{0,0,0,0,0,0,1,1,1,2,2,2,3,4,4,4,5,8,10,25}
Since there are an even number of data points (20), the median is the average of the two middle values (the 10th and 11th values in the sorted list).
10th value: 2
11th value: 2
Median calculation :
Median = 2 + 2 / 2 = 2
The median number of pets is 2.
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:
The dot plot of height of trees.
Sum of heights of all trees = 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 = 1621
Number of trees = 29
= \(\frac{ 1621}{29}\) = 55.89 ft
Arranging heights 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.
Since there are odd number of values (29), the median is the middle value.
Therefore, Median = 56.
The heights of the date palm trees range from 43 to 67 feet, with most between 55 and 60 feet.
Thirteen trees are shorter than the mean height.
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) The mean or median cannot lie between 25 and 30. Because the mean is the average of all values, it is always between the minimum and maximum values, which are 3.09 litres and 20.5 litres, respectively. The median is the middle value, so it is also within the range of the data.
(b) No, the mean and median cannot be less than the minimum value or greater than the maximum value in a data set.
Question 5.
The weights of a few newborn babies are given in kgs. Fill the dot plot provided below. Analyse and compare this data.
Solution:
The weight of boys is between 2.6 kg and 4.1 kg. The weight of girls lies between 2.5 kg and 4 kg. The heaviest baby is a boy, and the lightest is a girl.
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:
Observations and Inferences:
(a) The girls’ heights are more widely spread, ranging from 126 cm to 158 cm, while the boys’ heights range from 130 cm to 148 cm.
(b) Both the shortest and the tallest students in the class are girls.
(c) The girls’ mean height is less than both the class average and the boys’ mean height, indicating that boys are taller than girls on average in this class.
(d) For boys, the mean is slightly less than the median (142.05 < 143), showing a small influence of lower values.
(e) For girls, the mean is slightly greater than the median (140.14 > 140), showing a small influence of higher values.
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.
Approximately how many times heavier is a sumo wrestler compared to a ballet dancer?
Solution:
Sumo wrestlers
Sum of weights = 295.2 + 250.7 + 234.1 + 221.0 + 200.9 = 1201.9 kg
Average weight = 1201.9 ÷ 5 ≈ 240.38 kg
Ballet dancers
Sum of weights = 40.3 + 37.6 + 38.8 + 45.5 + 44.1 + 48.2 = 254.5 kg
Average weights = 254.5 ÷ 6 ≈ 42.4 kg
Compare the average weights
A sumo wrestler is approximately 6 times heavier than a ballet dancer.
Figure it Out (Page: 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:
(a) The scale used is 1 unit = 16 km/hr.
(b) The peregrine falcon is the fastest animal shown in the given infographic. Overall, animals are fastest in air, slower on land, and slowest in water.
(c) The cheetah (103 km/hr) is about twice as fast as the flying fish (56 km/hr).
(d) Yes, the sailfish is about 4 times faster than the humpback whale because the speed of the sailfish is 109 km / hr, and that of the humpback whale is 26 km / hr (since 26 × 4 = 104, which is close to 109).
However, we cannot say that the sailfish is the fastest aquatic animal in the world; it is only the fastest among the animals shown in the infographic.
Question 2.
Preyashi asked her students ‘If you were to get a super power to become aquatic (water-borne), aerial (airborne), 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.
Solution:
First, organize the raw data by counting the frequency of each choice for both grades. The counts are as follows :
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?
Solution:
These days might belong to December and May.
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.)
(b) Notice how the graph is organised, what scale is used, and what patterns the data shows.
(c) How would you describe the character 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 – There were very few new registrations in Uttarakhand in 2023 and 2024, as the increase in the bar lengths is minimal”?
Solution:
(a)
(b) The scale used is 1 unit = 25000 vehicles.
(c) All states show growth in electric vehicle registration from 2022 to 2024, with larger increases in Gujarat, Delhi, and West Bengal, and smaller increases in Uttarakhand.
(d) Registrations in Assam in 2022 = 40,000 Registrations in Assam in 2023 = 60,000 Difference in registrations = 60,000- 40,000 = 20,000.
Therefore, Assam got 20,000 more registrations in 2023 compared to 2022.
(d) Registrations in West Bengal in 2022 = 11,000
Registrations in West Bengal in 2024 = 44,000
Since, \(\frac{ 44,000}{11,000}\) = 4
Therefore, the registrations increased four times in West Bengal from 2022 to 2024.
(e) Yes, the statement is correct.
Figure it Out (Page: 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.
Solution:
False
(b) The median number of pockets for the boys is more than that for the girls.
Solution:
True
(c) The mean number of pockets for the girls is more than that for the boys.
Solution:
False
(d) The maximum number of pockets for boys is greater than that for the girls.
Solution:
True
Question 2.
The following table shows the points scored by each player in four games:
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 number of points per game by A. Average A
= \(\frac{14+16+10+110}{4}\) = \(\frac{ 50}{4}\)
= 12.5 points per game
(b) To determine the best performer, we compare the average scores of all players.
Average A = 12.5 points/game
Average B = \(\frac{0+8+6+4}{4}\) = \(\frac{18}{4}\) = 4.5 points/ game
Average C = \(\frac{8+11+13}{3}\) = \(\frac{32}{3}\) = 10.67 points/ game
Comparing
(12.5 > 10.67 > 4.5); Player A has the highest average.
(c) The best performer based on the average points scored per game, is player A.
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:
Group 1
(a) Mean (Average) score :
Sum of scores :
85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 +
75 = 730
(b) Median score
First, arrange the scores in ascending order 39, 48, 75, 76, 81, 85, 85, 90, 95
Since there are an even number of scores (10), the median is the average of the two middle values (the 5th and 6th scores) :
Median1 = \(\frac{76+81}{2}\) = \(\frac{ 157}{2}\) = 78.5
Group 2
Mean :
Sum of scores :
68 + 59 + 73 + 86 + 47 + 79 + 90 + 93 + 86
= 681
= \(\frac{ 681}{9}\) = 75.67 (approx.)
Median score :
First, arrange the scores in ascending order :
47, 59, 68, 73, 79, 86, 86, 90, 93
Since there is an odd number of scores (9), the median is the single middle value (the 5th score) :
Median2 = 79
Comparison :
| Metric | Group 1 | Group 2 |
| Mean | 73 | 75.67(approx) |
| Median | 78.5 | 79 |
Comparison :
- Group 2 has a higher mean and median, so overall Group 2 performed better.
- Group 1 has a lower mean due to very low scores (39, 48), which affected its average.
Question 4.
Consider this data collected from a survey of a colony.
Solution:
Observations:
Cricket is the most popular sport for both watching and participating.
More people watch sports than participate in them across all categories.
Athletics has the lowest numbers for both watching and participating.
Basketball and swimming have similar participation numbers, but more people watch swimming than Basketball.
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:
17 is an odd number that cannot be divided into two perfectly equal groups.
Arranging the heights of students in increasing order.
101,102,106,109,110,110,112,115,115, 115,115,115,117,120,120,123,125
Mean = \(\left(\frac{17+1}{2}\right)^{\mathrm{th}}\) = 9th term
= 115 cm
Arranging to the conditions given in the question, the height used to divide the students into two groups be 112 < height < 115
If we take 114 cm as the required height, then
Group 1 (Less than 114 cm) – 101, 102, 106, 109, 110, 112
Total = 7 Students
Group 2 (More than 114 cm) – 115, 115, 115, 115, 115, 117, 120, 120, 123, 125
Total = 10 students
Question 6.
Describe the mean and median of heights of your class. You can visualise the heights on a dot plot. (Suggested answer)
Solution:
The data set in order is :
145, 148, 148, 150, 150, 150, 152, 152, 152, 152, 155, 155, 155, 155, 155, 158, 158, 158, 160, 160
Mean = Sum of all 20 heights = 3074 cm
Mean height = \(\frac{3074}{20}\) = 153.7 cm
Average height of a student in the class is 153.7
Median = is the average of the 10th or 11th values in the ordered list
- 10th value = 152 cm
- 11th value = 155 cm
Median height = \(\frac{152+155}{2}\) = 153.5 cm
So median tells us that half the students in the class are shorter than 153.5 cm , and half are taller than 153.5 cm. Most students cluster around the 152 cm to 155 cm range.
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:
The correct answer is (d) The mean height of students in the other section cannot be determined.
Explanation: The fact that both 7th grade sections have the same number of students (15 boys and 15 girls) and are in the same grade does not guarantee that their average heights are identical.
The mean height is calculated based on the specific individual heights of each student in that particular section. Since no data or constraints were provided for the heights in the second section, its mean height could be less than, equal to, or greater than 154.2 cm. The information given is insufficient to determine the other section’s mean height with certainty.
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) Estimated values for the number of skyscrapers in
New York – 38
Tokyo – 160
London – 305
(b) (i) Valid
(ii) Valid
(iii) Invalid
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.
Solution:
The estimates are quite accurate, as the difference between estimated and measured values is very small for all objects.
Average difference = \(\frac{1+0.5+2+1.5+1}{5}\)
= \(\frac{ 6}{5}\) = 1.2
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:
= \(\frac{ 5470}{17}\) = 321.76 ≈ 321 Sec.
Arranging in ascending order:
220, 230, 240, 270, 280, 290, 310, 320, 320, 330, 340, 360, 370, 380, 400, 400, 410
With 17 values, the median is the 9th value. So median = 320 sec.
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 m-any 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.