Official Class 7 Mathematics Worksheets: Ganita Prakash Part 2 Chapter 05 Connecting the Dots
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Chapter-wise Practice Material: Ganita Prakash Part 2 Chapter 05 Connecting the Dots
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Long Answer Questions
Question 1. Data begins with the right kind of question.
(i) Explain what a statistical question is and what a statistical statement is, giving one example of each in your own words.
(ii) Say whether each of the following is a statistical question or not, with a reason: (a) What is my roll number? (b) How many hours do the Grade 7 students of our school sleep at night? (c) How tall is the Qutub Minar? (d) What were the daily maximum temperatures in Chennai in May?
(iii) A shopkeeper notes the price of 1 kg of tomatoes on 30 days. Write one statistical statement he could make from this data.
Answer:
(i) A statistical question can be answered only by collecting data, because its answer is expected to vary from one case to another. Example: “How many glasses of water do the students of our class drink in a day?”
A statistical statement is a claim or summary about something, expressed through numbers, proportions or predictions. Example: “On average, a student of our class drinks about 6 glasses of water a day.”
(ii) (a) Not statistical. A roll number is one fixed value; no data has to be collected.
(b) Statistical. Different students sleep for different times, so the sleeping hours must be collected and analysed.
(c) Not statistical. The Qutub Minar has one fixed height; a single measurement answers it.
(d) Statistical. The maximum temperature changes from day to day, so a whole month's readings must be collected.
(iii) For example: “The price of tomatoes in our market was about ₹30 per kg on average during the month, and it rose above ₹40 only in the last week.” Such a statement sums up many values using the mean, the maximum and a pattern over time.
In simple words: A statistical question needs varying data; a statistical statement sums that data up in numbers.
Teacher's Note:
1. Check whether the answer varies to decide if a question is statistical.
2. Statistical statements often use 'about' or 'on average'.
3. Any sensible statement based on the 30 prices is acceptable in (iii).
Question 2. The arithmetic mean is also called the fair share of a collection.
(i) Write the formula for the arithmetic mean in words.
(ii) Five children of Group I collected 6, 9, 4, 11 and 10 seashells. Six children of Group II collected 7, 5, 9, 6, 8 and 7 seashells. If each group shares its shells equally among its own members, find how many shells one child of each group gets.
(iii) Explain, using this example, why the mean is described as an “equalising” value. (Ancient Indian texts called it samamiti, meaning “mean measure”.)
Answer:
(i) Arithmetic mean = (sum of all the values in the data) ÷ (number of values in the data). It is also called the average.
(ii) Group I: total = 6 + 9 + 4 + 11 + 10 = 40 shells, shared by 5 children. Share = 40 ÷ 5 = 8 shells each.
Group II: total = 7 + 5 + 9 + 6 + 8 + 7 = 42 shells, shared by 6 children. Share = 42 ÷ 6 = 7 shells each.
Group II collected more shells in all, yet each member's share is smaller, because the group is larger.
(iii) Before sharing, the children of Group I had unequal numbers: 6, 9, 4, 11 and 10. After the fair share, every child has exactly 8. The mean is the number each one would have if the highs and lows were evened out without changing the total (8 × 5 = 40). That is why it is an equalising value, and why it was called samamiti, the equal measure.
In simple words: Sharing the total equally gives the mean: 8 shells per child in Group I and 7 in Group II.
Teacher's Note:
1. Mean × number of values gives back the total.
2. A bigger total does not always mean a bigger share.
3. Levelling the highs and lows is the idea behind the mean.
Question 3. The marks scored by nine students in a test are 32, 45, 28, 50, 39, 41, 36, 48 and 30.
(i) State the rule for finding the median when the number of values is odd, and when it is even.
(ii) Find the median of the given nine marks.
(iii) A tenth student scored 52. Find the new median and explain why it is different from the median in part (ii).
Answer:
(i) First sort the data from smallest to largest. If the number of values is odd, the median is the single middle value, with equally many values below and above it. If the number of values is even, there is no single middle value, so the median is the average of the two middle values.
(ii) Sorted: 28, 30, 32, 36, 39, 41, 45, 48, 50. There are 9 values, so the middle one is the 5th value. Median = 39 marks. Four students scored below 39 and four above it.
(iii) New sorted list: 28, 30, 32, 36, 39, 41, 45, 48, 50, 52. Now there are 10 values, so we average the 5th and 6th values: median = (39 + 41) ÷ 2 = 40 marks.
The median rises from 39 to 40 because the new mark 52 is added at the top of the list. This pushes the centre of the sorted data slightly up, and with an even count the median now lies between two values instead of being one of them.
In simple words: Nine marks give median 39; adding 52 makes ten values, and the median becomes (39 + 41) ÷ 2 = 40.
Teacher's Note:
1. Odd count: one middle value; even count: average of two.
2. Adding a value at one end shifts the middle a little towards that end.
| Activity | Grade 6 | Grade 7 |
|---|---|---|
| Music | 12 | 8 |
| Dance | 9 | 14 |
| Sports | 20 | 18 |
| Art | 7 | 10 |
(ii) Which activity was chosen by the largest number of students taking both grades together? Show your working.
(iii) Why is a double bar graph more helpful here than two separate bar graphs?
Answer:
(i) Steps to draw the double bar graph:
1. Draw the horizontal and vertical lines. Mark the four activities (Music, Dance, Sports, Art) at equal gaps along the horizontal line.
2. Show the number of students along the vertical line. The largest value is 20, so the scale 1 unit = 2 students is convenient; 10 units then cover the whole data.
3. For each activity draw two bars of equal width side by side, touching each other: the first for Grade 6 and the second for Grade 7. Keep equal gaps between the clusters.
4. Use two different colours or patterns and show a key, so that the reader knows which bar is which grade. Give the graph a title.
(ii) Adding the two grades: Music 12 + 8 = 20, Dance 9 + 14 = 23, Sports 20 + 18 = 38, Art 7 + 10 = 17. Sports was chosen by the largest number, 38 students.
(iii) With the two bars standing next to each other, we can compare the grades activity by activity in a single glance; for instance, more Grade 6 students chose Music, while more Grade 7 students chose Dance. With two separate graphs the eye has to travel from one picture to the other, and the comparison becomes harder, especially if the two graphs use different scales.
In simple words: Draw paired bars with a scale of 1 unit = 2 students; Sports is the favourite with 38 students in all.
Teacher's Note:
1. Choose a scale that makes the largest value fit.
2. A key is essential in a double bar graph.
3. Add the two bars of a cluster to get the combined total.
Question 5. Riya scored 32, 38, 45, 40, 37 and 42 marks in six weekly tests of 50 marks. Aman scored 25, 48, 30, 50, 30 and 45 marks in the same six tests.
(i) Find the mean marks of each student.
(ii) Find the median marks of each student.
(iii) Using the range as well, write who you would call the more consistent performer and why. Does the student with the higher mean also have the smaller range here?
Answer:
(i) Riya: total = 32 + 38 + 45 + 40 + 37 + 42 = 234, so mean = 234 ÷ 6 = 39 marks.
Aman: total = 25 + 48 + 30 + 50 + 30 + 45 = 228, so mean = 228 ÷ 6 = 38 marks.
(ii) Riya sorted: 32, 37, 38, 40, 42, 45, so median = (38 + 40) ÷ 2 = 39 marks.
Aman sorted: 25, 30, 30, 45, 48, 50, so median = (30 + 45) ÷ 2 = 37.5 marks.
(iii) Range of Riya = 45 − 32 = 13 marks; range of Aman = 50 − 25 = 25 marks.
Riya's marks stay in a narrow band and her mean and median are both 39. Aman's marks jump between 25 and 50. So Riya is the more consistent performer.
Yes: here the student with the higher mean, Riya, also has the smaller range. Aman, however, has the highest single score of 50, which shows that consistency and a high peak are different things.
In simple words: Riya has the higher mean (39) and the smaller range (13), so she is the steadier performer.
Teacher's Note:
1. Use the range to judge consistency.
2. Mean and median together tell more than either alone.
3. Six values: average the 3rd and 4th sorted values.
Question 6. The number of visitors to a school science exhibition on six days was 62, 70, 66, 68, 64 and 210. The sixth day was a public holiday.
(i) Find the mean and the median number of visitors for all six days.
(ii) Find the mean and the median after leaving out the holiday figure.
(iii) Compare the two sets of answers and explain which measure of central tendency describes an ordinary day at the exhibition better.
Answer:
(i) Total = 62 + 70 + 66 + 68 + 64 + 210 = 540 visitors, so mean = 540 ÷ 6 = 90 visitors.
Sorted: 62, 64, 66, 68, 70, 210, so median = (66 + 68) ÷ 2 = 67 visitors.
(ii) The five remaining values are 62, 64, 66, 68, 70, with total 330. Mean = 330 ÷ 5 = 66 visitors; median = middle value = 66 visitors.
(iii) Dropping the outlier 210 pulls the mean down sharply, from 90 to 66, a fall of 24 visitors. The median barely moves, from 67 to 66.
With all six days included, the mean of 90 is greater than five of the six values, so it does not describe an ordinary day; the single holiday crowd has dragged it up. The median of 67 sits among the usual counts.
Hence the median describes an ordinary day better. When data contains an outlier, the mean shifts towards it while the median stays near the centre.
In simple words: The holiday crowd inflates the mean to 90, but the median of 67 stays close to a normal day's count.
Teacher's Note:
1. Check whether the mean is bigger than most of the values.
2. Outliers move the mean far more than the median.
Question 7. The double bar graph shows the milk (in litres) collected by two dairy societies over five months.
(i) What scale is used along the vertical line, and in which month was the difference between the two societies the greatest?
(ii) Find the total and the mean monthly collection of each society.
(iii) Describe the pattern shown by each society over the five months, and write one conclusion a dairy officer could draw from the graph.
Answer:
(i) The vertical line is marked 0, 100, 200, … up to 700, so the scale is 1 unit = 100 litres.
Monthly differences: January 450 − 400 = 50 L; February 500 − 420 = 80 L; March 470 − 400 = 70 L (Society Y higher); April 550 − 500 = 50 L; May 600 − 530 = 70 L.
The greatest difference is in February, when Society X collected 80 litres more than Society Y.
(ii) Society X: 450 + 500 + 400 + 550 + 600 = 2500 litres, so mean = 2500 ÷ 5 = 500 litres per month.
Society Y: 400 + 420 + 470 + 500 + 530 = 2320 litres, so mean = 2320 ÷ 5 = 464 litres per month.
(iii) Society X collects more in most months and has the higher mean, but its collection is uneven: it rises in February, drops sharply to 400 litres in March, and then climbs to its peak of 600 litres in May.
Society Y rises steadily every month, from 400 litres to 530 litres, with no fall at all.
A dairy officer could conclude that Society X is the bigger supplier but should find out what went wrong in March, while Society Y, though smaller, is growing steadily.
In simple words: X collects more (mean 500 L) but unevenly; Y (mean 464 L) grows steadily every month.
Teacher's Note:
1. Read each bar carefully against the gridlines.
2. A steady rise and a higher average are different strengths.
3. The biggest gap need not be in the month with the biggest values.
| Month | Hill station | Beach town |
|---|---|---|
| January | 20 | 60 |
| March | 35 | 45 |
| May | 70 | 25 |
| July | 40 | 20 |
| September | 30 | 35 |
| November | 25 | 55 |
(i) Step 1 — Identify what is given: state what the rows and columns show, a suitable scale for a double bar graph, and the highest and lowest value for each place.
(ii) Step 2 — Infer: describe the pattern for each place and explain why the two patterns are opposite to each other.
(iii) Find the total for each place. Can you be sure that the beach town had more tourists in the whole year? Give a reason.
Answer:
(i) Step 1 — Identify. Each row is a month (January, March, May, July, September, November) and the two columns give the number of tourists, in thousands, at the hill station and at the beach town.
For a double bar graph, the months go along the horizontal line and the number of tourists along the vertical line. The largest value is 70 thousand, so the scale 1 unit = 10 thousand tourists is suitable.
Hill station: highest 70 thousand in May, lowest 20 thousand in January. Beach town: highest 60 thousand in January, lowest 20 thousand in July.
(ii) Step 2 — Infer. The hill station is busiest in the hot months, rising from 20 thousand in January to a peak of 70 thousand in May and falling again by November.
The beach town behaves the opposite way: it is busiest in the cool months of January and November and emptiest in May and July.
People go to the hills to escape the summer heat and to the coast in pleasant winter weather, so the two patterns are mirror images of each other.
(iii) Hill station: 20 + 35 + 70 + 40 + 30 + 25 = 220 thousand. Beach town: 60 + 45 + 25 + 20 + 35 + 55 = 240 thousand.
No, we cannot be sure about the whole year, because only six of the twelve months are given. The missing months, such as February or December, could change the totals completely.
In simple words: Hill station peaks in summer, beach town in winter; with only six months given, the yearly comparison is uncertain.
Teacher's Note:
1. First describe the data, then interpret it.
2. Do not draw conclusions about data you do not have.
3. Seasons explain the opposite patterns.
Question 9. The heights (in cm) of 17 students of a class are 118, 126, 131, 122, 140, 128, 135, 124, 130, 119, 133, 127, 129, 121, 136, 125 and 132. The sports teacher wants to split the class into two groups of equal size: one group of students shorter than a certain height and one group taller than it.
(i) Arrange the data and find the median height.
(ii) Explain how the median helps the teacher make the split, and state how many students fall on each side of it.
(iii) Find the mean height as well. What does the closeness of the mean and the median tell you about this data?
Answer:
(i) In increasing order: 118, 119, 121, 122, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 135, 136, 140. As 17 is odd, the median is the 9th value, so the median height is 128 cm.
(ii) The median has as many values below it as above it. If the teacher uses 128 cm as the dividing height, 8 students are shorter than 128 cm (118 to 127) and 8 students are taller (129 to 140), so the two groups are equal.
The one student who is exactly 128 cm tall is on the boundary; the teacher can keep that student as a helper or put the student in either group, since 17 students cannot be split into two equal halves.
(iii) Total = 2176 cm, so mean = 2176 ÷ 17 = 128 cm. The mean and the median are exactly equal here.
This tells us that the data is balanced and evenly spread about its centre; there is no outlier pulling the mean to one side.
In simple words: The median 128 cm splits the class into 8 shorter and 8 taller students, and the mean is also 128 cm.
Teacher's Note:
1. For 17 values the middle is the 9th.
2. Equal mean and median suggest no outliers.
3. The median is the natural dividing line for two equal halves.
Question 10. An advertisement for a bakery shows a bar graph of the number of cakes sold: January 500, February 510, March 520 and April 530. On the graph, the vertical line does not start at 0; it starts at 490 and goes up to 540 with one unit equal to 10 cakes. Below the graph is written, “Our sales are shooting up!”
(i) Find the actual increase in sales from January to April, and the mean monthly sales.
(ii) Explain why the April bar looks about four times the January bar on this graph, and why that impression is misleading.
(iii) Suggest how the graph should be redrawn so that it gives an honest picture, and state one situation from the chapter in which starting the vertical line above 0 is a fair thing to do.
Answer:
(i) Increase from January to April = 530 − 500 = 30 cakes, only 30 extra cakes on a base of 500.
Total = 500 + 510 + 520 + 530 = 2060, so mean monthly sales = 2060 ÷ 4 = 515 cakes.
(ii) Because the vertical line starts at 490, each bar shows only the part above 490. The visible bar heights are 500 − 490 = 10, 510 − 490 = 20, 520 − 490 = 30 and 530 − 490 = 40, that is 1, 2, 3 and 4 units.
So the April bar is four times as tall as the January bar, and the eye reads “sales have become four times”. The truth is a rise of just 30 cakes, about 6 per cent. The words “shooting up” are not supported by the numbers.
(iii) Redraw the graph with the vertical line starting at 0 and a scale such as 1 unit = 100 cakes. The four bars will then be almost the same height, honestly showing a small, steady growth.
Starting above 0 is fair only when it is clearly marked and the reader knows it is a zoomed-in view, as in the chapter's graph of average heights where the line begins at 145 cm to show small differences more clearly.
In simple words: A cut-off axis makes a rise of 30 cakes look like a fourfold jump; an honest graph starts at 0.
Teacher's Note:
1. Always check where the vertical axis begins.
2. Compare actual values, not bar lengths, on a cut axis.
3. A zoomed-in graph must be clearly labelled.
Case Study Based Questions
Ms. Kalpana gave her Grade 7 class a short mathematics test of 20 marks. Ten students wrote the test, and she entered their marks in a score sheet so that she could decide whether the class needed another practice session. She told the class, “Your total is 150 marks. Before I tell you anything else, look at the sheet and try to describe how the class performed with just one number.” The marks of the ten students are shown below.
| Student | Aarav | Bela | Chetan | Divya | Esha | Farhan | Gita | Harsh | Ira | Jai |
|---|---|---|---|---|---|---|---|---|---|---|
| Marks (out of 20) | 12 | 15 | 18 | 11 | 15 | 14 | 19 | 16 | 13 | 17 |
(ii) Find the mean marks of the class.
(iii) Find the median marks of the class.
(iv) How many students scored more than the mean?
Answer:
(i) Highest marks = 19 (Gita), lowest marks = 11 (Divya), so range = 19 − 11 = 8 marks.
(ii) Mean = 150 ÷ 10 = 15 marks.
(iii) Sorted marks: 11, 12, 13, 14, 15, 15, 16, 17, 18, 19. With 10 values, median = (15 + 15) ÷ 2 = 15 marks.
(iv) Marks greater than 15 are 16, 17, 18 and 19, so 4 students scored more than the mean.
Since the mean and the median are both 15, the class performance is balanced, with no extreme score pulling the mean away.
In simple words: The marks range over 8, and both the mean and the median are 15.
Teacher's Note:
1. The total is given, so the mean needs just one division.
2. Ten values: average the 5th and 6th sorted values.
A weather club in a village notes the rainfall every evening with a simple rain gauge. During the first week of July the club recorded the readings given below. On Wednesday the sky was clear and the gauge showed no rain at all, so the members wrote 0 mm for that day. At the end of the week the club found that the total rainfall for the seven days was 70 mm, and decided to put up a chart of the readings on the school notice board.
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
|---|---|---|---|---|---|---|---|
| Rainfall (in mm) | 8 | 12 | 0 | 15 | 20 | 6 | 9 |
(ii) Find the mean daily rainfall for the week.
(iii) Find the median daily rainfall.
(iv) One member says, “Wednesday had no rain, so we should drop it and divide by 6.” Is he right? Give a reason.
Answer:
(i) Heaviest on Friday (20 mm); least on Wednesday (0 mm).
(ii) Mean = 70 ÷ 7 = 10 mm per day.
(iii) Sorted readings: 0, 6, 8, 9, 12, 15, 20. The middle (4th) value gives median = 9 mm.
(iv) No, he is not right. A reading of 0 mm is a real measurement: the gauge was read on Wednesday and showed no rain. It is not a day with a missing reading. Dropping it would wrongly give 70 ÷ 6 ≈ 11.7 mm and make the week look rainier than it was.
In simple words: 0 mm is a real reading, so the mean is 70 ÷ 7 = 10 mm and the median is 9 mm.
Teacher's Note:
1. Zero is a value and must be counted.
2. Only a missing reading is left out.
Long ago in Europe, a “foot” meant the length of a real human foot, so it was different for different people. To fix one length, sixteen men stood in a line, toe to heel, and the total length of that line was divided into sixteen equal parts. In other words, people used the arithmetic mean without calling it by that name. To try this out, eight students of Grade 7 measured the length of their own right foot in centimetres and found that the eight lengths add up to 188 cm.
| Student | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Foot length (in cm) | 22 | 24 | 23 | 25 | 22 | 26 | 24 | 22 |
(ii) Find the mean foot length of the eight students.
(iii) Find the median foot length.
(iv) If the eight students stand in a line toe to heel, the line will be 188 cm long. What length would one “student foot” be if this line were divided into eight equal parts? Which measure does this answer match?
Answer:
(i) Longest = 26 cm, shortest = 22 cm (a difference of 4 cm).
(ii) Mean = 188 ÷ 8 = 23.5 cm.
(iii) Sorted: 22, 22, 22, 23, 24, 24, 25, 26. With 8 values, median = (23 + 24) ÷ 2 = 23.5 cm.
(iv) One “student foot” = 188 ÷ 8 = 23.5 cm. This is exactly the mean foot length, just as the line of 16 men divided into 16 parts gave the mean of their foot lengths.
In simple words: Dividing the total length into equal parts is exactly finding the mean, 23.5 cm.
Teacher's Note:
1. Equal division of a total is the mean.
2. Here the mean and the median happen to be equal.
A Grade 7 student, Nandini, decided to find out how much time she spends on a screen. For one week she noted the number of hours she spent watching television or using a phone each day, including the time spent on online classwork. Her teacher had asked the class to collect such data because he wanted the students to frame good questions about their own habits. At the end of the week Nandini’s seven readings added up to 21 hours.
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
|---|---|---|---|---|---|---|---|
| Screen time (in hours) | 2 | 3 | 1 | 2 | 4 | 5 | 4 |
(ii) Find the median screen time per day.
(iii) On how many days was Nandini’s screen time more than the mean?
(iv) Write one statistical question that Nandini’s class could answer by collecting such data from all the students.
Answer:
(i) Mean = 21 ÷ 7 = 3 hours per day.
(ii) Sorted: 1, 2, 2, 3, 4, 4, 5. The middle (4th) value gives median = 3 hours.
(iii) Screen times above 3 hours are 4, 5 and 4 hours, on Friday, Saturday and Sunday, so on 3 days.
(iv) For example: “How many hours a day do the students of our class spend on a screen?” The answer varies from student to student, so it can be answered only by collecting and analysing data from everyone.
In simple words: Mean and median are both 3 hours, and 3 days were above the mean.
Teacher's Note:
1. Use the given total to find the mean quickly.
2. A good statistical question expects varying answers.
The agriculture officer of a block compared the wheat harvested by two villages over five years. Both villages have fields of nearly the same area, but Village B started using a new seed and better irrigation from 2022 onwards. The officer prepared the table below and planned to draw a double bar graph from it, using the scale 1 unit = 5 tonnes along the vertical line. He noticed at once that the totals of the two villages for the five years were exactly the same.
| Year | Village A (tonnes) | Village B (tonnes) |
|---|---|---|
| 2020 | 18 | 12 |
| 2021 | 22 | 16 |
| 2022 | 15 | 20 |
| 2023 | 25 | 24 |
| 2024 | 20 | 28 |
(ii) With the scale 1 unit = 5 tonnes, how many units tall will the bars for 25 tonnes and for 12 tonnes be?
(iii) Which village shows a steady improvement, and how does the table show it?
(iv) Since both villages have the same mean, can the officer say their performance over the five years was exactly alike? Give a reason.
Answer:
(i) Village A: total = 18 + 22 + 15 + 25 + 20 = 100 tonnes, mean = 100 ÷ 5 = 20 tonnes.
Village B: total = 12 + 16 + 20 + 24 + 28 = 100 tonnes, mean = 100 ÷ 5 = 20 tonnes.
(ii) Bar for 25 tonnes = 25 ÷ 5 = 5 units; bar for 12 tonnes = 12 ÷ 5 = 2.4 units.
(iii) Village B: its yields 12, 16, 20, 24, 28 rise by 4 tonnes every year without a single fall, which matches the new seed and irrigation.
(iv) No. Equal means only say that the five-year totals are equal. Village A goes up and down (18, 22, 15, 25, 20), with its worst year in 2022, while Village B improves every year and is well ahead of A in 2024. A single representative value hides this pattern, so the year-wise data must be looked at too.
In simple words: Both villages average 20 tonnes, but only Village B rises steadily every year.
Teacher's Note:
1. Same mean does not mean same pattern.
2. Bar height in units = value ÷ scale.
Kavya played in an inter-district cricket series of seven matches. She was unwell on the day of Match 4 and did not take the field at all, so the scorer put a dash against that match. In Match 2 she went in to bat but was out without scoring, so the scorer wrote 0 against it. At the end of the series the coach wanted a single number to describe Kavya’s batting, and the team manager wanted to know the middle value of her scores as well.
| Match | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Runs | 45 | 0 | 78 | — | 32 | 55 | 60 |
(ii) Find her mean runs per match.
(iii) Find the median of her scores.
(iv) Explain why the 0 of Match 2 is counted but the dash of Match 4 is not.
Answer:
(i) Kavya batted in 6 matches (all except Match 4). Total = 45 + 0 + 78 + 32 + 55 + 60 = 270 runs.
(ii) Mean = 270 ÷ 6 = 45 runs per match.
(iii) Sorted: 0, 32, 45, 55, 60, 78. With 6 values, median = (45 + 55) ÷ 2 = 50 runs.
(iv) The 0 of Match 2 is a genuine score: she batted and made no run, so that match counts. The dash of Match 4 is not a score at all; she never batted, so there is no value to include. Counting it would mean dividing by 7, giving 270 ÷ 7 ≈ 38.6, which unfairly lowers her average.
In simple words: Six innings give a total of 270, a mean of 45 and a median of 50; the dash is not a score.
Teacher's Note:
1. 0 is a value; a dash is no value.
2. Six values: average the 3rd and 4th sorted values.
Aslam noticed that the newspaper delivered at his home is not of the same size every day. For one week he counted the number of pages each morning. He found that the Saturday paper was unusually thick because of a special supplement on jobs, while the Sunday paper was the thinnest of the week. The seven counts added up to 126 pages. Aslam then wanted to decide which single number would describe a usual day’s newspaper best.
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
|---|---|---|---|---|---|---|---|
| Number of pages | 14 | 16 | 18 | 16 | 20 | 30 | 12 |
(ii) Find the mean number of pages per day.
(iii) Find the median number of pages per day.
(iv) Here the mean is greater than the median. Which value is responsible for this, and at which end of the data does it lie?
Answer:
(i) Range = 30 − 12 = 18 pages.
(ii) Mean = 126 ÷ 7 = 18 pages per day.
(iii) Sorted: 12, 14, 16, 16, 18, 20, 30. The middle (4th) value gives median = 16 pages.
(iv) The value responsible is 30, Saturday's paper with the job supplement, and it lies at the higher end of the data. It adds a lot to the total and pulls the mean up to 18, while the median stays at 16. So the median of 16 pages describes a usual day better.
In simple words: Saturday's 30 pages is a high outlier that lifts the mean to 18 above the median of 16.
Teacher's Note:
1. A high value makes mean > median.
2. The median suits a 'usual day' better here.
A weather station in Jodhpur records the temperature at fixed hours every day. The readings taken at five of these hours on two days, picked from two different months of the same year, are given below. The station wants to show the two days side by side in a double bar graph, using the scale 1 unit = 5 °C along the vertical line, so that a reader can compare the two days hour by hour.
| Time | 6 am | 9 am | 12 noon | 3 pm | 6 pm |
|---|---|---|---|---|---|
| Day 1 (in °C) | 18 | 24 | 31 | 35 | 29 |
| Day 2 (in °C) | 10 | 15 | 22 | 25 | 19 |
(ii) By how many degrees was Day 1 hotter than Day 2 at 3 pm?
(iii) Find the mean temperature of Day 1 and of Day 2 for these five readings.
(iv) With the given scale, how many units tall will the bar for the 3 pm reading of Day 1 be? Which of the two days is more likely to be from December?
Answer:
(i) Both days were hottest at 3 pm: Day 1 at 35 °C and Day 2 at 25 °C.
(ii) 35 − 25 = 10 °C hotter.
(iii) Day 1: total = 18 + 24 + 31 + 35 + 29 = 137, so mean = 137 ÷ 5 = 27.4 °C.
Day 2: total = 10 + 15 + 22 + 25 + 19 = 91, so mean = 91 ÷ 5 = 18.2 °C.
(iv) Bar for 35 °C = 35 ÷ 5 = 7 units. Day 2 is more likely to be from December, because every one of its readings is lower, and its early-morning 10 °C suits a cold desert winter morning.
In simple words: Both days peak at 3 pm; Day 1 averages 27.4 °C, Day 2 averages 18.2 °C and is the likely December day.
Teacher's Note:
1. Compare the two days hour by hour.
2. Units = value ÷ scale.
Question 19. A tree-planting drive
Four schools of a town joined a tree-planting drive run by the forest department. Every year the department publishes a clustered bar graph showing how many saplings each school planted. The graph for 2022, 2023 and 2024 is given below. School R plants the same number each year because it plants exactly as many saplings as its small nursery can raise. A local newspaper looked at the graph and wrote: “School R is the most dependable school in the drive, because its three bars are exactly equal.”
(i) Which schools increased their planting every year, and which school decreased its planting every year?
(ii) Find the total number of saplings planted by all four schools in 2024.
(iii) Find the mean number of saplings planted per year by School S.
(iv) Is the newspaper’s conclusion about School R justified by this data? Give two reasons.
Answer:
(i) School P (120, 150, 180) and School S (160, 200, 240) increased their planting every year. School Q (200, 160, 140) decreased every year. School R stayed at 90 in all three years.
(ii) Total in 2024 = 180 + 140 + 90 + 240 = 650 saplings.
(iii) School S planted 160 + 200 + 240 = 600 saplings in three years, so mean = 600 ÷ 3 = 200 saplings per year.
(iv) No, the newspaper's conclusion is not justified.
Reason 1: Equal bars only show that School R repeated the same number every year; they say nothing about how well it did. In fact School R planted the fewest saplings of all four schools in every year (90 each year), and its three-year total of 270 is the lowest, against 600 for School S.
Reason 2: “Dependable” is a judgement about effort and care, which a graph of numbers alone cannot settle; the graph does not even tell us how many saplings survived. Also, R's numbers are fixed by the size of its nursery, not by its effort.
In simple words: P and S rose every year and Q fell; R's equal bars show sameness, not dependability.
Teacher's Note:
1. Read each school's three bars from left to right.
2. Equal values are not the same as good performance.
3. The scale here is 1 unit = 50 saplings.
A health club surveyed the 40 students of one section of Grade 7 in a school and asked each of them how many hours they had slept the previous night. The replies, rounded to the nearest whole hour, are shown in the table below; the numbers of students add up to 40. Using this table, a local newspaper printed the headline: “Children in our city sleep only 7.4 hours a night — far below the 9 hours doctors advise.”
| Hours of sleep | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|
| Number of students | 2 | 6 | 12 | 14 | 6 |
(ii) Find the total number of hours of sleep of all 40 students, and hence the mean hours of sleep.
(iii) Find the median hours of sleep. (Remember that there are 40 values.)
(iv) Is the newspaper headline justified by this survey? Give two reasons, and suggest one change that would make the claim stronger.
Answer:
(i) Students who slept less than 7 hours = 2 + 6 = 8 students.
(ii) Total hours = 5 × 2 + 6 × 6 + 7 × 12 + 8 × 14 + 9 × 6 = 10 + 36 + 84 + 112 + 54 = 296 hours. Mean = 296 ÷ 40 = 7.4 hours.
(iii) With 40 values, the median is the average of the 20th and 21st sorted values. The first 2 values are 5, the next 6 reach the 8th value, and the twelve 7s reach the 20th value. So the 20th value is 7 and the 21st is 8, giving median = (7 + 8) ÷ 2 = 7.5 hours.
(iv) No, the headline is not justified.
Reason 1: The survey covers only 40 students of one section of one school, far too small a group to speak for all the children of a city.
Reason 2: It records just one night's sleep, which may have been unusual. Also, 20 of the 40 students slept 8 hours or more, so “only 7.4 hours” hides half the story.
To make the claim stronger, the club should survey many sections in several schools from different parts of the city, and record each student's sleep over a week rather than one night.
In simple words: The mean is 7.4 hours and the median 7.5 hours, but one section on one night cannot speak for a whole city.
Teacher's Note:
1. Multiply each value by its number of students to get the total.
2. For 40 values, average the 20th and 21st.
3. Small samples cannot support big claims.
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