CBSE Class 8 Mathematics Tales by Dots and Lines Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. The daily number of glasses of water drunk by a student over five days is 4, 6, 7, 11, 12.
(i) Find the mean of this data.
(ii) Mark the mean on a number line in your answer book and show that the total distance of the values lying to its left equals the total distance of the values lying to its right.
(iii) Suppose someone claims that 9 is also a ‘centre’ of this data because it lies between the smallest and the largest value. Explain, using the distances on the two sides, why 9 cannot be the centre and why such a centre is unique.
Answer:
(i) Sum = 4 + 6 + 7 + 11 + 12 = 40 and there are 5 values, so mean = 40 ÷ 5 = 8 glasses.
(ii) Mark 8 on a number line and dots at 4, 6, 7, 11 and 12.
Distances on the left of 8: 8 − 4 = 4, 8 − 6 = 2, 8 − 7 = 1; total = 7.
Distances on the right of 8: 11 − 8 = 3, 12 − 8 = 4; total = 7.
The totals are equal, so the data balances exactly at 8; this is why the mean is called the centre of the data.
(iii) At 9, the left-hand distances are 5 + 3 + 2 = 10 and the right-hand distances are 2 + 3 = 5. They are not equal, so 9 is not a balance point.
If we move right from the mean, every left distance grows and every right distance shrinks, so the two sides can never balance again; moving left does the opposite. Hence the balance point is unique, and 8 is the only centre.
In simple words: Only at 8 do the distances on both sides add up to the same amount, so 8 is the one and only centre.

Teacher's Note:
1. Being between the smallest and largest value is not enough to be the centre.
2. Show the distances on both sides to test any claimed centre.

Question 2. The runs scored by a batter in seven innings are 12, 15, 9, 20, 17, 11, 25.
(i) Arrange the data in order and find the median.
(ii) The batter scores 22 in the eighth innings. Find the new median and state the rule you used for an even number of values.
(iii) Explain why including a value greater than the old median pushes the median up, and what would have happened if the eighth score had been 6 instead.
Answer:
(i) Sorted: 9, 11, 12, 15, 17, 20, 25. With 7 values the median is the 4th value: 15 runs.
(ii) Including 22: 9, 11, 12, 15, 17, 20, 22, 25 — eight values. For an even number of values the median is the average of the two middle values (4th and 5th).
Median = \( \frac{15 + 17}{2} \) = 16 runs.
(iii) 22 is greater than 15, so after including it there are more values above 15 than below it; 15 can no longer be in the middle and the median shifts towards larger values. It rises only to 16, since the median depends on position, not on how big the new score is.
If the eighth score were 6: 6, 9, 11, 12, 15, 17, 20, 25, and the median = \( \frac{12 + 15}{2} \) = 13.5 — the median would fall, because a value below the median was included.
In simple words: A score above the middle pushes the median up a little; a score below it pulls the median down.

Teacher's Note:
1. The size of the new value does not matter, only which side it falls on.
2. Recount the positions after adding a value.

Question 3. Line graphs are used throughout this chapter to show how a quantity changes over time.
(i) State what a line graph is and why it is suited to data collected over time.
(ii) The number of objects launched into space each year by three countries and by the world is to be shown for 13 years. Explain why a clustered column graph is a poor choice here and a line graph a better one.
(iii) In such a graph, one country’s values are drawn with circular markers and another’s with square markers. Give two reasons why different marker shapes are used in addition to different colours.
Answer:
(i) A line graph plots each value as a point and joins consecutive points with straight segments. The points are placed in time order along the horizontal axis, so the rise and fall of the line shows directly how the quantity changed, and the steepness shows how fast.
(ii) With 13 years and 4 counts per year, a clustered column graph needs 13 × 4 = 52 columns in 13 clusters. It is crowded, and the trend of any one country is hard to follow. Four lines keep each country’s story in one continuous path, so trends and crossings are seen at a glance.
(iii) Reason 1: if the graph is printed in black and white or photocopied, colours look alike, but circles and squares can still be told apart.
Reason 2: readers who find it difficult to distinguish colours can still follow each line correctly. So shapes make the graph accessible to everybody.
In simple words: Line graphs show change over time neatly, and marker shapes keep lines distinguishable even without colour.

Teacher's Note:
1. Choose the display that keeps the graph uncluttered.
2. Good graphs work for every reader, in colour or not.

Teacher's Note:
1. Frequency tables need value × frequency for the mean.
2. Running totals find the median without listing every value.

Question 5. Besides dot plots, bar graphs and line graphs, this chapter uses two other ways of showing information.
(i) Manoj records his day on a paper strip of 48 coloured boxes running from midnight to midnight. State the length of time one box stands for, and explain how the strip lets you see at a glance how long he slept.
(ii) An infographic colours a map of India by the difference between the per capita consumption of rice and wheat, on a scale from −100 to +100. Explain what a value of +100 for a state does and does not mean.
(iii) A health worker has the number of children vaccinated in one village in each of 12 months. Which display would you advise her to use, and why?
Answer:
(i) 24 hours ÷ 48 = 30 minutes per box. All boxes of one activity have the same colour, so a long unbroken block of the sleeping colour shows at a glance how long he slept; counting those boxes and halving the count gives the hours.
(ii) +100 means this state has the largest difference in favour of rice, so rice is strongly preferred there.
It does not mean the state eats no wheat, nor that it eats 100 units of rice; the scale only compares the two foods with each other.
(iii) A line graph, with months on the horizontal axis and the number of children vaccinated on the vertical axis. The data is collected over time, so the line shows the month-to-month rise and fall and makes any seasonal pattern or sudden drop easy to spot.
In simple words: Each box is half an hour, +100 means a strong preference for rice (not zero wheat), and monthly data suits a line graph.

Teacher's Note:
1. Read the key or scale of any display before interpreting it.
2. Data over time is best shown with a line graph.

Question 6. The mean of 12 numbers is 38.5.
(i) Find the sum of the 12 numbers.
(ii) Six of these numbers add up to 210. Find the mean of the remaining six numbers.
(iii) The number 44 is now removed from the original collection of 12. Find the mean of the remaining 11 numbers, and explain, without calculating again, why this new mean had to be smaller than 38.5.
Answer:
(i) Sum = mean × number of values = 38.5 × 12 = 462.
(ii) The other six add up to 462 − 210 = 252, so their mean = 252 ÷ 6 = 42.
(iii) The remaining 11 numbers add up to 462 − 44 = 418, so the new mean = 418 ÷ 11 = 38.
Why it had to fall: 44 is greater than the old mean 38.5. Removing a value above the balance point takes weight off the right side, so the balance point moves left.
In symbols, if v > a is removed from n values with mean a, the new mean is \( \frac{na - v}{n - 1} \), which is less than a. Indeed 38 < 38.5.
In simple words: Taking away a number bigger than the average lowers the average.

Teacher's Note:
1. Total = mean × count is the key step in every part.
2. Removing a large value lowers the mean; removing a small one raises it.

Question 7. The dot plot shows the number of times each of 20 students visited the school library in one week.
(i) Find the total number of library visits made by the class and hence the mean number of visits per student.
(ii) Find the median number of visits.
(iii) Next week every student visits the library exactly one more time than this week. Write the new mean and the new median, and justify both answers without redoing the whole calculation.
Answer:
(i) From the dot plot: 2 students visited 0 times, 4 once, 6 twice, 5 three times, 2 four times and 1 five times; 2 + 4 + 6 + 5 + 2 + 1 = 20 students.
Total visits = 0×2 + 1×4 + 2×6 + 3×5 + 4×2 + 5×1 = 0 + 4 + 12 + 15 + 8 + 5 = 44.
Mean = 44 ÷ 20 = 2.2 visits per student.
(ii) With 20 values the median is the average of the 10th and 11th values. Running totals: up to 0 → 2; up to 1 → 6; up to 2 → 12.
So positions 7 to 12 have the value 2, including the 10th and 11th. Median = 2 visits.
(iii) Every value increases by 1, so the whole dot plot shifts one step right.
New mean = 2.2 + 1 = 3.2 visits, because the total rises by 1 × 20 = 20 and 64 ÷ 20 = 3.2.
New median = 2 + 1 = 3 visits, because the order of the students does not change and the middle values are each one more.
In simple words: The average is 2.2 visits and the middle is 2 visits; one extra visit each raises both by 1.

Teacher's Note:
1. Count the dots in each column as frequencies.
2. Adding the same number to all values shifts both mean and median by that number.

Question 8. Let a collection of n values x₁, x₂, x₃, …, x_n have mean a.
(i) Show using algebra that if a fixed number k is added to every value, the mean of the new collection is a + k.
(ii) Show using algebra that if every value is multiplied by a fixed number k, the mean of the new collection is ka.
(iii) The mean monthly electricity bill of a colony is ₹1250. Next year each bill rises by a fixed charge of ₹80 and then the whole amount is doubled because of a change in tariff units. Use the two results above to find the new mean bill.
Answer:
(i) We have \( \frac{x_1 + x_2 + \cdots + x_n}{n} = a \). The new values are x₁ + k, …, x_n + k, so the new mean is
\( \frac{(x_1 + k) + \cdots + (x_n + k)}{n} = \frac{x_1 + \cdots + x_n}{n} + \frac{kn}{n} = a + k \).
Every value shifts by k, so the balance point shifts by k too.
(ii) The new values are kx₁, …, kx_n, so the new mean is
\( \frac{kx_1 + \cdots + kx_n}{n} = \frac{k(x_1 + \cdots + x_n)}{n} = ka \), taking k out by the distributive property.
(iii) Adding ₹80 to every bill: mean = 1250 + 80 = ₹1330 (part (i)).
Doubling every bill: mean = 2 × 1330 = ₹2660 (part (ii)). The new mean bill is ₹2660.
In simple words: Adding k to every value adds k to the mean; multiplying by k multiplies the mean by k.

Teacher's Note:
1. Apply the changes in the same order as given.
2. Doubling after adding is not the same as adding after doubling.

Teacher's Note:
1. Many answers are possible; always show the check.
2. A single large value raises the mean but barely moves the median.

Teacher's Note:
1. Compare changes, not values, to find the biggest rise.
2. A table cannot tell us about anything it does not show.

Case Study Based Questions

Teacher's Note:
1. Sort the marks before finding the median.
2. A value above the mean always raises the mean.

Teacher's Note:
1. Look at differences between consecutive days for the biggest increase.
2. Data over time suits a line graph.

Teacher's Note:
1. A zero reading still counts as a value.
2. One large value affects the mean more than the median.

Teacher's Note:
1. The mean gives the total, and the total gives the missing value.
2. Include the recovered value before finding the median.

Teacher's Note:
1. A trend is not a guarantee about the future.
2. Read the latest values before judging who is ahead.

Teacher's Note:
1. Always check that the frequencies add to the total.
2. 'More than 3' excludes those at exactly 3.

Teacher's Note:
1. The range is a quick measure of how much values vary.
2. Where two lines cross, the order of the values changes.

Teacher's Note:
1. Row 1 holds headings, so the marks start in row 2.
2. Column letter first, row number second.

Teacher's Note:
1. A constant error shifts the mean and median by the same amount.
2. A balanced pair keeps both the mean and the median.

Teacher's Note:
1. Conclusions must stay within what the data covers.
2. Highest minus lowest is a quick way to judge steadiness.