CBSE Class 8 Mathematics Tales by Dots and Lines Short Answer Questions with Answers

Short Answer Questions

Question 1. Two numbers are 12 and 19.
(i) Find their mean.
(ii) Show that the mean lies exactly halfway between the two numbers by comparing the distance from each number to the mean.
Answer:
(i) Mean = \( \frac{12 + 19}{2} \) = \( \frac{31}{2} \) = 15.5.
(ii) Distance from 12 to 15.5 = 15.5 − 12 = 3.5; distance from 15.5 to 19 = 19 − 15.5 = 3.5.
The two distances are equal, so 15.5 is exactly halfway between 12 and 19. For two numbers the mean is always the midpoint, since their sum is shared equally.
In simple words: The mean of two numbers sits right in the middle of them.

Teacher's Note:
1. For two numbers, mean = midpoint.
2. Equal distances on both sides confirm the balance.

Question 2. The mean of the collection 6, 8, 9, 13 is 9. Verify that 9 is the balance point of this data by finding the total distance of the values lying below 9 and the total distance of the values lying above 9. What does this equality tell you about the mean?
Answer:
Values below 9: 9 − 6 = 3 and 9 − 8 = 1, so total distance on the left = 3 + 1 = 4. (9 itself is at distance 0.)
Value above 9: 13 − 9 = 4, so total distance on the right = 4.
The totals are equal, so the data balances at 9.
This shows that the mean is the balance point of the data, not necessarily the midpoint of the smallest and largest values (that would be 9.5).
In simple words: The distances on the two sides of 9 add up to the same amount, so 9 is where the data balances.

Teacher's Note:
1. A value equal to the mean adds nothing to either side.
2. Balance point and midpoint are different ideas.

Question 3. Find the median of 14, 9, 21, 5, 18, 11. Write down the rule you used for a collection that has an even number of values, and explain why the middle value cannot be picked directly here.
Answer:
Sorted data: 5, 9, 11, 14, 18, 21. There are 6 values, an even number, so there is no single middle value; the 3rd and 4th values share the middle.
Median = \( \frac{11 + 14}{2} \) = 12.5.
Rule: for an even number of values, the median is the average of the two middle values of the sorted data.
In simple words: With six numbers, two sit in the middle, so we take their average.

Teacher's Note:
1. Sort first, then count.
2. An even count always gives two middle values.

Question 4. Find the mean of:
(i) the first 20 natural numbers,
(ii) the first 20 odd numbers.
Write one observation comparing the two answers.
Answer:
(i) Sum of 1 to 20 = \( \frac{20 \times 21}{2} \) = 210, so mean = 210 ÷ 20 = 10.5.
(ii) Sum of 1, 3, …, 39 = 20 × 20 = 400, so mean = 400 ÷ 20 = 20.
Observation: both lists are equally spaced, so the mean equals the average of the first and last values: \( \frac{1 + 20}{2} \) = 10.5 and \( \frac{1 + 39}{2} \) = 20.
In simple words: For evenly spaced numbers, the mean is just the average of the first and the last number.

Teacher's Note:
1. The sum of the first n odd numbers is n².
2. Equally spaced data is symmetric about its mean.

Question 5. The mean of the five numbers 18, 24, 31, 27 and x is 26. Find x. Show the two steps you used and check your answer by adding all five numbers.
Answer:
Step 1: sum of all five numbers = 26 × 5 = 130.
Step 2: sum of the four known numbers = 18 + 24 + 31 + 27 = 100, so x = 130 − 100 = 30.
Check: 18 + 24 + 31 + 27 + 30 = 130 and 130 ÷ 5 = 26, the given mean.
In simple words: Find the total from the mean, then take away the numbers you know.

Teacher's Note:
1. Mean × count gives the total.
2. Always check by recomputing the mean.

Question 6. The mean marks of 25 students in a test is 42. The teacher later gives 6 grace marks to every student. Find the new mean without adding the marks again, and explain in one or two sentences why your short method works.
Answer:
New mean = 42 + 6 = 48 marks.
Adding 6 to each of 25 students adds 6 × 25 = 150 to the total, so the mean rises by 150 ÷ 25 = 6.
In other words, the whole data shifts 6 marks to the right, and so does its balance point.
In simple words: If everyone gets 6 extra marks, the average also goes up by 6.

Teacher's Note:
1. Adding a constant to every value adds it to the mean.
2. No need to know individual marks.

Question 7. The mean of 8 readings is 7.5. Each reading is then doubled. Find the mean of the new collection and justify your answer using the idea that the sum of all the values also doubles.
Answer:
Old total = 7.5 × 8 = 60. After doubling every reading, new total = 2 × 60 = 120.
The number of readings is still 8, so new mean = 120 ÷ 8 = 15, which is twice the old mean.
In simple words: Doubling every reading doubles the total, so the average doubles too.

Teacher's Note:
1. Multiplying every value by k multiplies the mean by k.
2. The count does not change.

Question 8. While reading a graph, the chapter suggests a two-step process. Name the two steps and write, in one line each, what you look for in each step.
Answer:
Step 1 — Identify what is given: how the graph is organised, what each axis and the scale show, what colours or marker shapes stand for, and what pattern the data makes.
Step 2 — Infer and interpret: analyse each observation, explain what it means in the real situation, and write clear summary statements.
In simple words: First understand what the graph shows, then work out what it means.

Teacher's Note:
1. Never jump to conclusions before reading the axes.
2. Summary statements should be backed by the graph.

Teacher's Note:
1. Mean = Σ(value × frequency) ÷ Σfrequency.
2. Check that the frequencies add up to 20.

Teacher's Note:
1. Running totals tell you which value sits in each position.
2. For 40 values, the middle positions are 20 and 21.

Question 11. The mean of 9 observations is 54. A tenth observation of value 54 is now included.
(i) Write the new mean.
(ii) Explain your answer using the balance-point idea of the mean.
Answer:
(i) The new mean is still 54.
(ii) The new value sits exactly at the balance point, so it adds no distance to either side and the balance is undisturbed.
Check: old total = 54 × 9 = 486; new total = 486 + 54 = 540; 540 ÷ 10 = 54.
In simple words: Adding a value equal to the average does not change the average.

Teacher's Note:
1. A value at the mean is at distance 0.
2. Including or removing it leaves the mean unchanged.

Question 12. The mean mass of 20 parcels in a courier van was found to be 4.5 kg. Later it was discovered that the mass of one parcel had been written down as 6 kg more than its actual mass. Find the correct mean mass of the parcels.
Answer:
Recorded total = 4.5 × 20 = 90 kg.
One parcel was written 6 kg too heavy, so correct total = 90 − 6 = 84 kg.
Correct mean = 84 ÷ 20 = 4.2 kg.
Check: the mean falls by 6 ÷ 20 = 0.3 kg, and 4.5 − 0.3 = 4.2.
In simple words: Fix the total by removing the extra 6 kg, then divide by 20 again.

Teacher's Note:
1. Correct the total, not the mean directly.
2. An error of 6 spread over 20 parcels changes the mean by 0.3.

Question 13. Consider the data 3, 7, 10, 12, 16, 19, 22, 25.
(i) Find its median.
(ii) One more value is to be included so that the median does not change. Find that value and explain why it is the only possible one.
Answer:
(i) There are 8 values, so the median is the average of the 4th and 5th values: \( \frac{12 + 16}{2} \) = 14.
(ii) The value must be 14. With 9 values the median is the 5th value.
Any value 12 or less makes the 5th value 12; any value 16 or more makes it 16.
Only a value between 12 and 16 becomes the 5th value itself, and to keep the median 14 it must be 14.
In simple words: Only by adding 14 itself does the middle stay at 14.

Teacher's Note:
1. After adding, the count becomes odd, so the median is one actual value.
2. Test values below, between and above the middle pair.

Question 14. Consider the data 12, 15, 18, 22, 23.
(i) Find the mean, then write a pair of new values that can be included without changing the mean. State the condition such a pair must satisfy.
(ii) Which single value, if removed, would leave the mean unchanged?
Answer:
(i) Mean = \( \frac{12 + 15 + 18 + 22 + 23}{5} \) = \( \frac{90}{5} \) = 18.
Including, for example, 10 and 26: new total = 90 + 36 = 126 and 126 ÷ 7 = 18, so the mean is unchanged.
Condition: the two new values must add up to 36 (that is, their own mean is 18).
(ii) Removing 18 leaves 72 ÷ 4 = 18, so the mean is unchanged.
In simple words: Values that balance each other around 18 can be added, and 18 itself can be removed, without changing the mean.

Teacher's Note:
1. A pair adding to twice the mean keeps the mean the same.
2. Removing a value equal to the mean keeps the mean too.

Question 15. In a spreadsheet, a student’s marks in six subjects are entered in row 4 from column B to column G.
(i) Write the formula that gives the total marks of this student.
(ii) Explain in one line what the formula =AVERAGE(C2:C21) would compute if column C holds the Science marks of 20 students.
Answer:
(i) =SUM(B4:G4), which adds every cell from B4 to G4.
(ii) It adds the 20 Science marks in cells C2 to C21 and divides by 20, giving the class average in Science.
In simple words: SUM adds a row of marks; AVERAGE finds the mean of a column of marks.

Teacher's Note:
1. A row range keeps the number fixed; a column range keeps the letter fixed.
2. C2 to C21 is 20 cells.

Question 16. The line graph shows the number of books issued from a school library in the first six months of a year.
(i) Between which two consecutive months was the rise in the number of books issued the sharpest?
(ii) How does the steepness of a line segment help you answer part (i)?
Answer:
(i) Month-to-month changes: +30, −10, +50, +70, −30. The sharpest rise is from April to May, from 190 to 260, an increase of 70 books.
(ii) The months are equally spaced, so a steeper segment means a bigger change in the same one-month gap. The April–May segment climbs more steeply than any other, so it shows the largest increase.
In simple words: The steepest upward piece of the line shows the biggest rise: April to May.

Teacher's Note:
1. Compare changes, not the heights of the points.
2. Steeper upward segment = bigger increase.

Question 17. State whether each statement below is always true, sometimes true or never true, and justify your answer briefly.
(i) Including one value greater than the mean increases the mean.
(ii) Including two new values changes the median.
Answer:
(i) Always true. If n values have mean a and a value v > a is included, the new mean is \( \frac{na + v}{n + 1} \). Since v > a, na + v > na + a = (n + 1)a, so the new mean is greater than a.
On a dot plot, the new dot adds extra distance on the right, so the balance point must move right.
(ii) Sometimes true. Adding 4 and 5 to 1, 2, 3 gives 1, 2, 3, 4, 5 and the median rises from 2 to 3; but adding 0 and 10 gives 0, 1, 2, 3, 10 and the median stays 2.
If one new value is below and the other above the median, the middle usually does not move.
In simple words: A larger value always pulls the mean up, but two new values may or may not move the median.

Teacher's Note:
1. One example is enough to show 'sometimes'.
2. 'Always' needs a general reason, such as algebra.

Question 18. The mean marks of a class of 30 students is 62. The 12 girls in the class have a mean of 68 marks. Find the mean marks of the boys. Show the sums you used at each step.
Answer:
Total marks of the class = 62 × 30 = 1860.
Total marks of the 12 girls = 68 × 12 = 816.
Number of boys = 30 − 12 = 18, and their total = 1860 − 816 = 1044.
Mean marks of the boys = 1044 ÷ 18 = 58.
Check: 816 + 58 × 18 = 816 + 1044 = 1860 = 62 × 30.
In simple words: Find the class total, take away the girls’ total, and share what is left among the boys.

Teacher's Note:
1. Work with totals, not with means directly.
2. You cannot just average 62 and 68.

Question 19. Last month the mean mass of a group of 10 people was 58.4 kg and the median mass was 60 kg. This month, two of them have each lost 1.5 kg and one of them has gained 3 kg; nobody else has changed.
(i) What is this month’s mean mass?
(ii) What can be said about this month’s median mass? Give a reason.
Answer:
(i) Total change = −1.5 − 1.5 + 3 = 0 kg. The total and the number of people are both unchanged, so the mean is still 58.4 kg.
(ii) Nothing definite can be said. We are not told which three people changed.
If the changes are away from the middle, the 5th and 6th masses stay the same and the median remains 60 kg; if a middle person changed, or the order changes, the median can move a little. The information is not sufficient.
In simple words: The total stays the same so the mean stays 58.4 kg, but the median depends on who changed.

Teacher's Note:
1. Mean depends only on the total; median depends on positions.
2. Say 'cannot be determined' when the data is not enough.

Question 20. Every value of a collection with mean a is multiplied by 3 and then 4 is subtracted from each result.
(i) Using algebra with values x₁, x₂, …, x_n, show that the new mean is 3a − 4.
(ii) Verify your result for the data 2, 5, 8, 9.
Answer:
(i) Let \( \frac{x_1 + x_2 + \cdots + x_n}{n} = a \). The new values are 3x₁ − 4, 3x₂ − 4, …, 3x_n − 4, so the new mean is
\( \frac{(3x_1 - 4) + \cdots + (3x_n - 4)}{n} = \frac{3(x_1 + \cdots + x_n) - 4n}{n} = 3a - 4 \).
(ii) For 2, 5, 8, 9: a = 24 ÷ 4 = 6. New values: 2, 11, 20, 23, with mean 56 ÷ 4 = 14.
And 3a − 4 = 18 − 4 = 14. Verified.
In simple words: Do to the mean exactly what you did to every value: times 3, then minus 4.

Teacher's Note:
1. Multiply first, then subtract, in the same order as for the values.
2. A numerical check confirms the algebra.