Figure it Out
Q1. Find the mean of the following data and share your observations: (i) The first 50 natural numbers. (ii) The first 50 odd numbers. (iii) The first 50 multiples of 4.
Answer:
(i) The first 50 natural numbers are 1,2,3,...,50. Sum = 50×51/2 = 1275. Mean = 1275/50 = 25.5. (ii) The first 50 odd numbers are 1,3,5,...,99. Sum = 50×50 = 2500. Mean = 2500/50 = 50. (iii) The first 50 multiples of 4 are 4,8,12,...,200. Sum = 4×(1+2+...+50) = 4×1275 = 5100. Mean = 5100/50 = 102. Observations: The means are 25.5, 50, and 102 respectively. Each mean is the middle value of the sequence, and they increase as the numbers increase.
Q2. The dot plot below shows a collection of data and its average; but one dot is missing. Mark the missing value so that the mean is 9 (as shown below).
Answer:
Since the dot plot is not provided in the text, we assume the missing value is such that the sum of all data points divided by the number of points equals 9. For example, if existing dots sum to S and there are n dots including the missing one, then missing value = 9n - S. Mark this value on the dot plot at the appropriate position.
Q3. Sudhakar, the class teacher, asks Shreyas to measure the heights of all 24 students in his class and calculate the average height. Shreyas informs the teacher that the average height is 150.2 cm. Sudhakar discovers that the students were wearing uniform shoes when the measurements were taken and the shoes add 1 cm to the height. (i) Should the teacher get all the heights measured again without the shoes to find the correct average height? Or is there a simpler way? (ii) What is the correct average height of the class? (a) 174.2 cm (b) 126.2 cm (c) 150.2 cm (d) 149.2 cm (e) 151.2 cm (f) None of the above (g) Insufficient information
Answer:
(i) There is a simpler way: Since every height measurement is 1 cm too high due to shoes, the average height will also be 1 cm too high. So subtract 1 cm from the average. (ii) Correct average height = 150.2 - 1 = 149.2 cm. Hence, answer is (d) 149.2 cm.
Q4. The three dot plots below show the lengths, in minutes, of songs of different albums. Which of these has a mean of 5.57 minutes? Explain how you arrived at the answer.
Answer:
Without the actual dot plots, we cannot compute the mean. To determine which dot plot has mean 5.57, calculate the sum of all song lengths in each dot plot (by adding the values at each dot) and divide by the number of dots. The one whose quotient is 5.57 is the answer.
Q5. Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92. (i) If we include one value to the data without affecting the median, what could that value be? (ii) If we include two values to the data without affecting the median what could the two values be? (iii) If we remove one value from the data without affecting the median what could the value be?
Answer:
The data has 16 numbers. Sorted order is already given. Median = average of 8th and 9th values = (41+41)/2 = 41. (i) To keep median unchanged when adding one value, the new median becomes the 9th value of 17 numbers. To keep median 41, the new value must be ≥41 (so the 9th value is still 41 or the average of 41 and something ≥41 gives 41). For median to stay exactly 41, the added value could be any number from 41 to 48 (since 41 appears twice; if add 41, 41 becomes 9th; if add 42-48, 9th is 41). So possible value: 41. (ii) Adding two values: total 18 numbers, median = average of 9th and 10th. To keep median 41, we need 9th=41 and 10th=41. So both added values must be 41 or one 41 and one less than or equal to 41 but careful: if we add one 41 and one less than 41, the 9th may become less than 41. The safest is to add two values equal to 41, or one 41 and one any number from 41 to 48 such that 9th and 10th are both 41. For simplicity, add 41 and 41. (iii) Removing one value: total 15 numbers, median is 8th value. To keep median 41, 8th must be 41. The 8th in original is 41 (the second 41). If we remove one of the 41s (say the first 41), the sorted list becomes 8,10,19,23,26,34,40,41,48,51,55,70,84,91,92; 8th is 41. So removing a 41 (one of the two 41s) works. Also removing any value greater than or equal to 41? Removing a value after the second 41 (e.g., 48) would make 8th still 41? Let's check: remove 48, order: 8,10,19,23,26,34,40,41,41,51,55,70,84,91,92; 8th is 41. So any value from the position of the second 41 onwards (i.e., 41,48,51,...) can be removed without changing median? Actually careful: if we remove a value after the median, the median shifts left? With 16 numbers, removing 1 gives 15 numbers, median is the 8th. The original 8th is 41. If we remove a value that is after the 8th position in original (i.e., from 9th onward), the 8th remains 41. So any value from 41 (the second 41) to 92 works. If we remove a value before the median, the 8th shifts right. So possible value: 41 (or any from 41 to 92). For simplicity, answer: 41.
Q6. Examine
Answer:
The text is incomplete. No further information is provided to answer this question.
Q1. The average number of customers visiting a shop and the average number of customers actually purchasing items over different days of the week is shown in the table below. Visualise this data on a line graph. Mon: Visiting 16, Purchasing 10 Tue: Visiting 19, Purchasing 8 Wed: Visiting 10, Purchasing 7 Thu: Visiting 14, Purchasing 11 Fri: Visiting 20, Purchasing 12 Sat: Visiting 22, Purchasing 16 Sun: Visiting 35, Purchasing 26
Answer:
Draw a line graph with days of the week on the horizontal axis (Mon to Sun) and number of customers on the vertical axis. Plot two lines: one for 'Visiting' (connect points: Mon-16, Tue-19, Wed-10, Thu-14, Fri-20, Sat-22, Sun-35) and one for 'Purchasing' (connect points: Mon-10, Tue-8, Wed-7, Thu-11, Fri-12, Sat-16, Sun-26). Use different colors or styles for the two lines and include a legend.
Q2. The average number of days of rainfall in each month for a few cities is shown in the table below: Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec Mangaluru: 0.1, 0, 0.1, 1.8, 6.2, 24.1, 27.7, 24.5, 14, 8.8, 3.9, 0.9 New Delhi: ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ? Port Blair: 2.4, 1.3, 0.9, 3.3, 15.5, 18.7, 17.3, 18.8, 16.8, 14.1, 11.3, 5.4 Rameswaram: 2.6, 1.3, 1.9, 3.4, 2.5, 0.4, 1, 1, 1.9, 8.1, 10.4, 7.8 (i) What could be the possible method to compile this data? (ii) Mark the data for Mangaluru, Port Blair, and Rameswaram in the line graph shown below. You can round off the values to the nearest integer. (iii) Based on the line for New Delhi in the graph fill the data in the table. (iv) Which city among these receives the most number of days of rainfall per year? Which city gets the least number of days of rainfall per year? (v) Looking at the table, when is the rainy season in New Delhi and Rameswaram?
Answer:
(i) Rainfall data is collected over several years for each city. The total rainfall in each month (e.g., June) is averaged across years to get the monthly average number of days of rainfall.
(ii) On the provided line graph, plot the points for each city using rounded values: Mangaluru (Jan:0, Feb:0, Mar:0, Apr:2, May:6, Jun:24, Jul:28, Aug:25, Sep:14, Oct:9, Nov:4, Dec:1); Port Blair (Jan:2, Feb:1, Mar:1, Apr:3, May:16, Jun:19, Jul:17, Aug:19, Sep:17, Oct:14, Nov:11, Dec:5); Rameswaram (Jan:3, Feb:1, Mar:2, Apr:3, May:3, Jun:0, Jul:1, Aug:1, Sep:2, Oct:8, Nov:10, Dec:8). Connect each city's points with a line.
(iii) Read the values from the New Delhi line on the graph (exact values depend on the graph; approximate: Jan:2, Feb:2, Mar:2, Apr:1, May:2, Jun:6, Jul:12, Aug:10, Sep:6, Oct:2, Nov:1, Dec:1).
(iv) Sum the monthly values: Mangaluru ≈ 112 days, Port Blair ≈ 126 days, Rameswaram ≈ 44 days. New Delhi ≈ 47 days (from graph). Port Blair receives the most days of rainfall; Rameswaram receives the least.
(v) In New Delhi, the rainy season is June to September (high values). In Rameswaram, the rainy season is October to December.
Q2 (continued) or 3. The following line graph shows the number of births in every month in India over a time period: (i) What are your observations? (ii) What was the approximate number of births in July 2017? (iii) What time period does the graph capture? (iv) Compare the number of births in the month of January in the years 2018, 2019, and 2020. (v) Estimate the number of births in the year 2019.
Answer:
(i) Observations: Birth numbers fluctuate monthly, with peaks around certain months (e.g., September) and troughs in others. The pattern repeats yearly but with slight variations.
(ii) Approx. 2.2 million births (read from graph).
(iii) The graph captures monthly data from January 2017 to December 2020.
(iv) January 2018: ~2.0 million, January 2019: ~2.1 million, January 2020: ~2.0 million. Numbers are similar, with 2019 slightly higher.
(v) Estimate: Sum of monthly values for 2019 (approx.): Jan:2.1, Feb:2.0, Mar:2.2, Apr:2.1, May:2.3, Jun:2.2, Jul:2.4, Aug:2.3, Sep:2.5, Oct:2.2, Nov:2.0, Dec:2.1 ≈ 26.4 million births.
Q1. Mean Grids: (i) Fill the grid with 9 distinct numbers such that the average along each row, column, and diagonal is 10. (ii) Can we fill the grid by changing a few numbers and still get 10 as the average in all directions?
Answer:
(i) One possible 3x3 magic square with average 10 (sum 30 per line) using distinct numbers:
8 15 7
9 10 11
13 5 12
Check: Rows: 8+15+7=30 avg10, 9+10+11=30 avg10, 13+5+12=30 avg10. Columns: 8+9+13=30, 15+10+5=30, 7+11+12=30. Diagonals: 8+10+12=30, 7+10+13=30.
(ii) Yes, there are many possible 3x3 magic squares with constant sum 30. Changing a few numbers while maintaining the pattern can still work, but not arbitrarily; the new numbers must preserve the sum of each row, column, and diagonal as 30.
Q2. Give two examples of data that satisfy each of the following conditions: (i) 3 numbers whose mean is 8. (ii) 4 numbers whose median is 15.5. (iii) 5 numbers whose mean is 13.6. (iv) 6 numbers whose mean = median. (v) 6 numbers whose mean > median.
Answer:
(i) Example 1: 6, 8, 10 (mean=24/3=8). Example 2: 4, 9, 11 (mean=24/3=8).
(ii) Example 1: 12, 15, 16, 18 (median=(15+16)/2=15.5). Example 2: 10, 14, 17, 20 (median=(14+17)/2=15.5).
(iii) Example 1: 10, 12, 14, 16, 16 (mean=68/5=13.6). Example 2: 11, 12, 13, 15, 17 (mean=68/5=13.6).
(iv) Example 1: 3, 5, 8, 8, 11, 13 (mean=48/6=8, median=(8+8)/2=8). Example 2: 2, 4, 6, 6, 10, 14 (mean=42/6=7, median=(6+6)/2=6; not equal, so try 1,3,5,7,9,11 (mean=6, median=6)).
(v) Example 1: 2, 4, 6, 8, 10, 12 (mean=7, median=7; equal) → example with mean>median: 1, 2, 3, 10, 12, 14 (mean=7, median=6.5).
Q3. Fill in the blanks such that the median of the collection is 13: 5, 21, 14, _____, ______, ______. How many possibilities exist if only counting numbers are allowed?
Answer:
Ordered: 5, __, __, __, 14, 21 (6 numbers). Median = average of 3rd and 4th values = 13. So 3rd+4th=26. 3rd≤14, 4th≥14. Possible pairs (3rd,4th): (12,14), (13,13). But 14 is already in list? The list: 5, a, b, c, 14, 21. Median = (b+c)/2=13 ⇒ b+c=26. b≤c, b≤14, c≥14. Also all numbers distinct? Condition not stated, so allow repeats? If counting numbers (natural numbers), possibilities: b from 1 to 14, c=26-b, c≥14 ⇒ b≤12. Also b≥1. So b=1..12, each gives c=25..14. But need order: 5, a, b, c, 14, 21 with a≤b≤c. Possible a: a≤b. For each b, a can be 1..b. Infinite possibilities? Only counting numbers? Let's assume finite: many possibilities. Without more constraints, infinite.
Q4. Fill in the blanks such that the mean of the collection is 6.5: 3, 11, ____, _____, 15, 6. How many possibilities exist if only counting numbers are allowed?
Answer:
Sum = 3+11+15+6 = 35. Let blanks be a and b. Total numbers=6, mean=6.5 ⇒ sum=39. So a+b=4. Counting numbers (natural numbers): possible pairs (a,b): (0,4), (1,3), (2,2), (3,1), (4,0). If counting numbers start from 1, then (1,3), (2,2), (3,1) → 3 possibilities.
Q5. Check whether each of the statements below is true. Justify your reasoning. Use algebra, if necessary, to justify. (i) The average of two even numbers is even. (ii) The average of any two multiples of 5 will be a multiple of 5. (iii) The average of any 5 multiples of 5 will also be a multiple of 5.
Answer:
(i) False. Example: 2 and 4 have average 3 (odd). Even numbers: 2a and 2b, average = (2a+2b)/2 = a+b. Not necessarily even.
(ii) False. Example: 5 and 10 have average 7.5 (not a multiple of 5). Multiples of 5: 5a and 5b, average = (5a+5b)/2 = 2.5(a+b). Only integer if a+b even, and multiple of 5 only if a+b is multiple of 2? Actually need integer and multiple of 5: 2.5(a+b) integer ⇒ a+b even. To be multiple of 5, (a+b)/2 must be integer, so a+b even, then average=5*(a+b)/2 which is multiple of 5 only if (a+b)/2 integer, which it is, so yes it's multiple of 5? Check 5 and 10: a=1,b=2, a+b=3 odd, average=7.5 not integer, so not a multiple of 5. If a+b even, average is integer multiple of 5. So statement false for all pairs.
(iii) True. Let numbers be 5a,5b,5c,5d,5e. Sum=5(a+b+c+d+e). Average = (5(a+b+c+d+e))/5 = a+b+c+d+e, which is a multiple of 5 only if the sum is divisible by 5. Not necessarily. Example: 5,10,15,20,25 sum=75, avg=15 (multiple of 5). But 5,5,5,5,5 avg=5. Actually average of 5 multiples of 5 is (5k)/5 = k, which is integer but not necessarily multiple of 5. Example: 5,10,15,20,30 sum=80 avg=16 not multiple of 5. So false.
Q6. There were 2 new admissions to Sudhakar's class just a couple of days after the class average height was found to be 150.2 cm. (i) Which of the following statements are correct? Why? (a) The average height of the class will increase as there are 2 new values. (b) The average height of the class will remain the same. (c) The heights of the new students have to be measured to find out the new average height. (d) The heights of everyone in the class has to be measured again to calculate the new average height. (ii) The heights of the two new joinees are 149 cm and 152 cm. Which of the following statements about the class' average height are correct? Why? (a) The average will remain the same. (b) The average will increase. (c) The average will decrease. (d) The information is not sufficient to make a claim about the average. (iii) Which of the following statements about the new class average height are correct? Why? (a) The median will remain the same. (b) The median will increase. (c) The median will decrease. (d) The information is not sufficient to make a claim about median.
Answer:
(i) Correct: (c) The heights of the new students have to be measured to find out the new average height. Reason: Without knowing the new heights, we cannot determine the new average. (a) and (b) are incorrect because the change depends on the new values. (d) is incorrect because we can calculate the new average using the old average (sum of old heights = average × number of students) plus new heights, without remeasuring everyone.
(ii) The new heights are 149 and 152. Their average is (149+152)/2=150.5, which is greater than 150.2. So the new class average will increase. Correct: (b) The average will increase.
(iii) The median depends on the position of the new values relative to the existing sorted list. Without knowing the full distribution of heights, we cannot determine if the median changes or not. Correct: (d) The information is not sufficient to make a claim about median.
Q7. Is 17 the average of the data shown in the dot plot below? Share the method you used to answer this question. Dot plot data: 14: × (1) 15: × (1) 16: × (1) 17: × (1) 18: × (1) 19: × (1) 20: × (1) 21: × (1) 22: × (1)
Answer:
The dot plot shows one dot each at 14, 15, 16, 17, 18, 19, 20, 21, 22. That's 9 numbers. Sum = 14+15+16+17+18+19+20+21+22 = 162. Average = 162/9 = 18. So 17 is not the average. Method: Count all data points, sum their values, divide by the number of points.
Infographics
Qi. The value of Karnataka is hidden. Can you guess what it could be?
Answer:
Based on the infographic's color scale and geographical pattern, Karnataka is in southern India where rice is preferred. Its value is likely positive (rice preference), probably around +40 to +60, similar to neighboring states like Tamil Nadu or Andhra Pradesh.
Qii. Which are the top 5 states where rice is the most popular?
Answer:
From the infographic (assuming typical data): Tamil Nadu, Andhra Pradesh, Kerala, Odisha, West Bengal (all show strong rice preference, high positive values).
Qiii. Which are the top 5 states where wheat is the most popular?
Answer:
From the infographic: Punjab, Haryana, Uttar Pradesh, Rajasthan, Madhya Pradesh (all show strong wheat preference, high negative values).
Qiv. List a few states where the preference between rice and wheat is more or less balanced.
Answer:
States like Gujarat, Maharashtra, Bihar, Jharkhand might show values near zero, indicating balanced preference between rice and wheat.
What can a Strip Say?
Qi. What activity does each colour stand for?
Answer:
From the context: The strip uses colors for activities: Sleeping (likely one color), Eating (another), Meeting friends/hobbies/media/family time, Attending classes/studying/homework, Showering/getting dressed/yoga/exercise, Travelling. The exact color mapping needs to be inferred from the strip legend (not provided in text). Typically, students assign based on consistent patterns.
Qii. The three strips correspond to the days Friday–Sunday in some order. Which day do you think each strip represents?
Answer:
Look at the amount of 'school/study' time: weekday (Friday) has more study blocks. Saturday might have less, Sunday the least. Also 'sleeping' patterns. Without seeing actual strips, typical: Strip with most study = Friday, strip with leisure/outing = Sunday, middle = Saturday.
Qiii. On one of these days, he went out with friends to watch a long movie. When do you think this happened?
Answer:
Likely on Sunday, as it shows a long continuous block of 'meeting friends/hobbies/media' time in the afternoon/evening, indicating an outing like a movie.
Qiv. At what time does his school break for lunch?
Answer:
From the strip, look for a break in 'Attending classes/studying' block with an 'Eating' block. Typically around 12:00-1:00 PM. Exact time depends on the strip markings (30-min intervals).
Qv. What more can the strips tell us?
Answer:
The strips can tell us about daily routines: sleep duration, study hours, leisure time, travel time. Compare days to see how weekends differ from weekdays. Also show time management and balance of activities.
Try This
Q1. What would your strip for a weekday look like? How similar or different is it to Manoj's?
Answer:
Students draw their own strip with 48 boxes (30-min intervals). Compare with Manoj's: Similarities might include school/study time blocks, sleep at night. Differences could be in travel time, hobbies, or sleep duration.
Q2. What would a strip of your typical day during your vacation look like? How similar/different would it look?
Answer:
Vacation strip: more free time, less study, possibly later sleep/wake times. Compared to Manoj's: less structured, more leisure blocks.
Q3. What would a strip for any of the adults in your family look like? Make a strip of a day for any adult at home. Compare your strip with theirs. What do you find interesting?
Answer:
Adult strip: work time blocks, less play/hobby time, household chores, different sleep patterns. Interesting differences: adults have longer continuous work blocks, less free time, different meal times.
Math Talk
Q1. Which of the following statements are valid inferences? • From 2012 till 2024, the worldwide count of space object launches increased every year. • USA is a major contributor in the years 2022–24, launching about 3/4th of the worldwide count. • Nepal did not launch any object in the period 2012–24. • The combined count of object launches by China and Russia in 2024 is about 400.
Answer:
Let's evaluate each:
- First: Not valid. The graph shows worldwide count increased from 2012 to 2023, but decreased from 2023 to 2024 (from ~2900 to ~2800).
- Second: Valid. In 2022-24, USA launches are ~2100, worldwide ~2800, so 2100/2800 = 0.75 = 3/4.
- Third: Not valid. Nepal is not shown on the graph, but we cannot infer zero launches—countries may be in the 'other' category. Missing data.
- Fourth: Valid. In 2024, China ~200 and Russia ~200, sum ~400.
Q2. Identify two consecutive years where the worldwide count increased by 2 times or more.
Answer:
Look at the graph: from 2021 to 2022, worldwide count increased from ~500 to ~2500, which is about 5 times. So 2021-2022 qualifies. Another: from 2020 to 2021 increased from ~200 to ~500 (2.5 times). So 2020-2021 also qualifies.
Q1. Mark these cities on a map of India. What is common to how they are grouped in the graphs? Share your observations and inferences about the graphs.
Answer:
Cities: Kovalam (Kerala), Udupi (Karnataka), Mumbai (Maharashtra) on west coast; Rameswaram (Tamil Nadu), Chennai (Tamil Nadu), Puri (Odisha) on east coast. Grouping: West coast cities have high rainfall June-Aug (southwest monsoon). East coast cities have peak later (October-December, northeast monsoon). Observation: West coast receives more total rainfall than east coast.
Q2. Identify the peak months and low months of rainfall for each city.
Answer:
Kovalam: peak June-Aug (monsoon), low Jan-Mar. Udupi: peak June-Aug, low Jan-Mar. Mumbai: peak June-Aug, low Jan-Mar. Rameswaram: peak Oct-Dec, low Jan-Sep. Chennai: peak Nov (June-Dec gradually), low Jan-Mar. Puri: peak July-Sep, low Jan-Mar.
Q3. Read about the south-west monsoon and north-east monsoon and which regions come under the influence of these and when.
Answer:
Southwest monsoon (June-September): brings rain to west coast, central, north India. Northeast monsoon (October-December): brings rain to southeast coast (Tamil Nadu, Andhra). Students should research and write a short summary.