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NCERT Solutions Ganita Prakash (Part 2) Chapter 5 –116Section 5.1 The Balancing Act — Figure it Out

Book page 113 Updated on2026-09-05

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

All three lists are evenly spaced, so in each case the mean is simply the average of the first and last terms.

(i) 1, 2, 3, …, 50
Sum = 50 × 51 ÷ 2 = 1275
Mean = 1275 ÷ 50 = 25.5  (= (1 + 50) ÷ 2)

(ii) 1, 3, 5, …, 99
Sum of the first 50 odd numbers = 502 = 2500
Mean = 2500 ÷ 50 = 50  (= (1 + 99) ÷ 2)

(iii) 4, 8, 12, …, 200
Sum = 4 × (1 + 2 + … + 50) = 4 × 1275 = 5100
Mean = 5100 ÷ 50 = 102  (= (4 + 200) ÷ 2)

Observations

  • For every one of these lists the mean is the midpoint of the smallest and largest value, because the dots are spread symmetrically about the middle.
  • The mean of the first n odd numbers is n itself — here 50.
  • (iii) is (i) with every value multiplied by 4, and sure enough its mean is 4 × 25.5 = 102 — the scaling rule from page 108.
  • In (i) and (iii) the mean is not even a member of its own list — 25.5 is not a natural number and 102 is not a multiple of 4. A mean need not be one of the data values.
Tip: for any evenly spaced list you can pair the terms from the outside in — 1 with 50, 2 with 49, and so on. Every pair has the same sum, so the average of the whole list is the average of any one pair.
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

The missing value is 16.

Reading the ten dots that are printed: one at 4, one at 7, two at 8, five at 9 and one at 11.

Sum of the printed dots = 4 + 7 + (8 × 2) + (9 × 5) + 11
= 4 + 7 + 16 + 45 + 11 = 83
With the missing dot there are 11 values, and the mean must be 9, so
Total needed = 11 × 9 = 99
Missing value = 99 − 83 = 16
Balance check: measure the distances from 9. Below: 5 (from 4), 2 (from 7), 1 + 1 (from the two 8s) → total 9. Above: 2 (from 11) → total 2. The left side is 7 heavier, so the missing dot must sit 7 above the mean: 9 + 7 = 16. ✓
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 no need to measure anyone again. Every single reading is exactly 1 cm too large, because every student wore the same shoes. Subtracting a fixed number from every value lowers the mean by that same number, so simply take 1 cm off the average.

Correct average = 150.2 − 1 = 149.2 cm

The long way, as a check:
Measured total = 150.2 × 24 = 3604.8 cm
Total of the 24 shoe-thicknesses = 24 × 1 = 24 cm
Correct total = 3604.8 − 24 = 3580.8 cm
Correct average = 3580.8 ÷ 24 = 149.2 cm ✓

(ii) Option (d), 149.2 cm.

Why it happens: this is the ‘subtract a constant’ rule of page 107 doing real work. The error is systematic — the same for everybody — so it shifts the whole dot plot 1 cm to the right without changing its shape, and the balance point shifts with it. Had the shoes been of different thicknesses, this shortcut would not be available and the heights really would need measuring again.
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

Album A.

Reading the dots off the three plots (the scale is marked in steps of 0.5, and dots also sit halfway between two marks):

A: 5, 5, 5.25, 5.5, 5.75, 6, 6.5 → 7 songs
Sum = 5 + 5 + 5.25 + 5.5 + 5.75 + 6 + 6.5 = 39
Mean = 39 ÷ 7 = 5.571… ≈ 5.57 minutes ✓

B: 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5 → 8 songs
Sum = 19.25, mean = 19.25 ÷ 8 = 2.41 minutes

C: 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5 → 8 songs
Sum = 31.25, mean = 31.25 ÷ 8 = 3.91 minutes
A quicker way to reach the same answer: the mean must lie inside the range of the data. Every dot in B sits between 0.5 and 5, and every dot in C between 3.5 and 4.5, so neither can have a mean as large as 5.57. Only A has dots spread around 5.5, so A is the only candidate — then one calculation confirms it.
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 (in the given list) 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 list is already in order and has 16 values, so the median is the average of the 8th and 9th.

8th value = 41, 9th value = 41
Median = (41 + 41) ÷ 2 = 41

Note that there are seven values below 41, two 41s, and seven values above 41. That symmetry is the key to all three parts.

(i) Any value at all. With 17 values the median becomes the 9th.

  • Add something less than 41 (say 20): now eight values are below, so the 9th is 41. ✓
  • Add something greater than 41 (say 60): seven below, then 41, 41 — the 9th is 41. ✓
  • Add 41 itself: three 41s at positions 8, 9, 10 — the 9th is 41. ✓

(ii) One value 41 or less together with one value 41 or more — for example 20 and 60, or 41 and 100. With 18 values the median is the average of the 9th and 10th, and only a value added on each side keeps 41 in both those places. Two values both below 41 would push a smaller number into 9th place; two values both above would push a larger number into 10th place.

(iii) Any value at all. With 15 values the median is the 8th.

  • Remove one from below (say 8): six values below, then 41, 41 — the 8th is 41. ✓
  • Remove one from above (say 92): seven below, then 41 — the 8th is 41. ✓
  • Remove one 41: seven below, then the remaining 41 in 8th place. ✓
Why this data set is so stubborn: the two middle values are equal, and there is a matching count of seven values on either side. That gives the median a cushion — a single value added or removed nudges the middle position by one place, and 41 is still sitting there.
Q6.
Examine the statements below and justify if the statement is always true, sometimes true, or never true. (i) Removing a value less than the median will decrease the median. (ii) Including a value less than the mean will decrease the mean. (iii) Including any 4 values will not affect the median. (iv) Including 4 values less than the median will increase the median.
Answer

(i) Never true. Removing a value from below the middle leaves fewer numbers on the low side, so the middle position moves towards the larger values. The median can rise or stay put, but it cannot fall.

1, 2, 3, 4, 5 → median 3. Remove 1 → 2, 3, 4, 5 → median 3.5 (rose)
1, 3, 3, 3, 5 → median 3. Remove 1 → 3, 3, 3, 5 → median 3 (unchanged)

(ii) Always true. From the formula on page 105, the shift in the mean is (x − a) ÷ (n + 1). If x < a this is negative, whatever the data.

4, 6, 8 → mean 6. Include 2 → 20 ÷ 4 = 5  (fell by 1 = (2 − 6) ÷ 4)

(iii) Sometimes true. It depends on where the four values land.

1, 2, 3 → median 2. Include 1, 1, 3, 3 → 1, 1, 1, 2, 3, 3, 3 → median 2 ✓ unchanged
1, 2, 3 → median 2. Include 0, 0, 0, 0 → 0, 0, 0, 0, 1, 2, 3 → median 0 ✗ changed

(iv) Never true. Four values added below the median make the low side heavier in count, so the middle position slides down the sorted list. The median can fall or stay the same, never rise.

1, 2, 3, 4, 5 → median 3. Include 0, 0, 0, 0 → nine values, median = 5th = 1 (fell)
2, 2, 2, 2, 2 → median 2. Include 1, 1, 1, 1 → nine values, median = 5th = 2 (unchanged)
The pattern behind all four: the mean responds to size, so a rule about “less than the mean” settles its direction completely — statement (ii) is the only ‘always’. The median responds to counts, so adding or removing on one side moves the middle position in a fixed direction, but it may land on an equal value and appear not to move at all. That is why the median statements come out as ‘never’ or ‘sometimes’, never ‘always’.
Q7.
The mean of the numbers 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the value of y.
Answer
There are 8 numbers, so the total must be
8 × 10.375 = 83

Sum of the seven known numbers
= 8 + 13 + 10 + 4 + 5 + 20 + 10 = 70

y = 83 − 70 = 13

Check: 8 + 13 + 10 + 4 + 5 + 20 + 13 + 10 = 83, and 83 ÷ 8 = 10.375. ✓

Tip: 10.375 = 10 + 3/8, which is a clue that the denominator is 8 — a decimal that ends in .375 comes from eighths.
Q8.
The mean of a set of data with 15 values is 134. Find the sum of the data.
Answer
Mean = sum ÷ number of values
So sum = mean × number of values
Sum = 134 × 15 = 2010
Why it works: the mean is the fair share each of the 15 values would get if the total were divided equally. Fifteen shares of 134 rebuild the whole total: 134 × 15 = 2010. Notice that this tells you nothing about the individual values — thousands of different data sets have 15 values adding to 2010.
Q9.
Consider the data: 12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p. Which of the following number(s) could be p if the median of this data is 29? (i) 10 (ii) 25 (iii) 40 (iv) 100 (v) 29 (vi) 47 (vii) 30
Answer

p can be 40, 100, 29, 47 or 30 — that is options (iii), (iv), (v), (vi) and (vii).

Sort the ten known values:
8, 8, 12, 18, 25, 29, 35, 39, 47, 73
With p there are 11 values, so the median is the 6th value.

Five of the known values (8, 8, 12, 18, 25) are below 29. So the 6th value is 29 only when p does not squeeze in ahead of it.

pSorted order around the middle6th valueWorks?
108, 8, 10, 12, 18, 25, 29, …25No
258, 8, 12, 18, 25, 25, 29, …25No
298, 8, 12, 18, 25, 29, 29, …29Yes
308, 8, 12, 18, 25, 29, 30, …29Yes
408, 8, 12, 18, 25, 29, 35, …29Yes
478, 8, 12, 18, 25, 29, 35, …29Yes
1008, 8, 12, 18, 25, 29, 35, …29Yes
The general rule: the median is 29 exactly when p ≥ 29. Any p below 29 gives the 6th place to 25; a p between 26 and 28 would itself take the 6th place. And once p is 29 or more it does not matter how much more — 30 and 100 give the same median, which is the median’s whole character.
Q10.
The number of times students rode their cycles in a week is shown in the dot plot below. Four students rode their cycles twice in that week. (i) Find the average number of times students rode their cycles. (ii) Find the median number of times students rode their cycles. (iii) Which of the following statements are valid? Why? (a) Everyone used their cycle at least once. (b) Almost everyone used their cycle a few times. (c) There are some students who cycled more than once on some days. (d) Exactly 5 students have used their cycles more than once on some days. (e) The following week, if all of them cycled 1 more time than they did the previous week, what would be the average and median of the next week’s data?
Answer

Counting the dots in each column of the plot:

Times cycled01234567810Total
No. of students314775463242

The column above 2 has four dots, which matches the hint in the question.

(i)

Total rides = (0×3) + (1×1) + (2×4) + (3×7) + (4×7) + (5×5) + (6×4) + (7×6) + (8×3) + (10×2)
= 0 + 1 + 8 + 21 + 28 + 25 + 24 + 42 + 24 + 20 = 193
Average = 193 ÷ 42 = 4.6 rides (to one decimal place)

(ii) With 42 students the median is the average of the 21st and 22nd values.

Cumulative counts: 0 → 3, 1 → 4, 2 → 8, 3 → 15, 4 → 22
Positions 16 to 22 all hold the value 4, so the 21st and 22nd are both 4
Median = 4 rides

(iii)

  • (a) Not valid. Three dots sit above 0, so three students did not ride at all.
  • (b) Valid. 38 of the 42 students rode at least twice, and most cluster between 3 and 7 rides — ‘a few times’ describes the bulk of the class well.
  • (c) Valid. A week has only 7 days, so the three students who rode 8 times and the two who rode 10 times must have ridden more than once on some day.
  • (d) Not valid. We can be sure about those 5 students, but nothing rules out a student who rode 6 times riding twice on one day and not at all on two others. The plot gives weekly totals, not day-by-day counts, so ‘exactly 5’ cannot be justified.
  • (e) If everyone rides once more, every value goes up by 1, so both measures go up by 1:
    New average = 4.6 + 1 = 5.6  (check: 235 ÷ 42 = 5.595)
    New median = 4 + 1 = 5
Why the median beats the mean here: the average of 4.6 is dragged up by the two students who rode 10 times. The median of 4 says something more useful about a typical student — and note that 4.6 is not a possible number of rides for anyone, while 4 is.
Q11.
A dart-throwing competition was organised in a school. The number of throws participants took to hit the bull’s eye (the centre circle) is given in the table below. Describe the data using its minimum, maximum, mean and median.
Answer
No. of trials12345678910
No. of students10014912151010
Cumulative111261527425262
Minimum = 1 trial (one lucky student hit it on the first throw)
Maximum = 10 trials

Number of students = 1 + 0 + 0 + 1 + 4 + 9 + 12 + 15 + 10 + 10 = 62
Total trials = (1×1) + (4×1) + (5×4) + (6×9) + (7×12) + (8×15) + (9×10) + (10×10)
= 1 + 4 + 20 + 54 + 84 + 120 + 90 + 100 = 473
Mean = 473 ÷ 62 = 7.63 trials

Median = average of the 31st and 32nd values.
The cumulative count reaches 27 at 7 trials and 42 at 8 trials, so positions 28 to 42 all hold 8.
Median = 8 trials

Describing the data: hitting the bull’s eye was hard. The trials ran from 1 to 10, but only 2 of the 62 students managed it in fewer than 5 throws, while 47 needed 7 or more. Both the mean (7.63) and the median (8) sit near the top of the range, and the most common result was 8 trials. The mean is slightly below the median because that single 1-trial student pulls the average down a little; the median is unmoved by that one lucky throw.

Tip: when the mean is a little below the median, the data has a tail stretching to the left — a few unusually small values. Here that tail is the pair of students who hit the target in 1 and 4 trials.
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