NCERT Solutions for Class 7th Maths Chapter 5 Figure it Out — Representative Values

Book page 112 – 113 Updated on2026-09-19

Q1.

Find the median of onion prices in Yahapur and Wahapur.

Answer

Yahapur: ₹37. Wahapur: ₹38.50.

Yahapur sorted: 24, 25, 26, 28, 30, 35, 39, 43, 44, 49, 56, 59
12 values, so average the 6th and the 7th
Median = (35 + 39) ÷ 2 = 74 ÷ 2 = ₹37

Wahapur sorted: 17, 19, 23, 30, 35, 38, 39, 42, 42, 52, 53, 60
Median = (38 + 39) ÷ 2 = 77 ÷ 2 = ₹38.50

Why it happens: both towns have 12 monthly prices — an even number — so there is no single middle value and we average the two middle ones.

Compare with the means: Yahapur mean ₹38.17 > median ₹37, while Wahapur mean ₹37.50 < median ₹38.50. So Yahapur's few very high months (56, 59) pull its mean above its median, while Wahapur's very cheap months (17, 19) pull its mean below. By the median, Wahapur is the dearer town; by the mean, Yahapur is. The two towns are genuinely close, and you should say so rather than pick a winner.

Q2.

Sanskruti asked her class how many domestic animals and pets each had at home. Some of the students were absent. The data values are 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, —, 10, 25, 2, — , 2, 4. Find the mean and median. How would you describe this data?

Answer

Mean = 3.6 animals; Median = 2 animals.

First, the two dashes are absent students. They gave no value at all, so they are not counted — unlike a 0, which is a real answer meaning "no pets". That leaves 20 values.

Values: 0, 1, 0, 4, 8, 0, 0, 2, 1, 1, 5, 3, 4, 0, 0, 10, 25, 2, 2, 4

Sum = (six 0s) + (three 1s) + (three 2s) + 3 + (three 4s) + 5 + 8 + 10 + 25
= 0 + 3 + 6 + 3 + 12 + 5 + 8 + 10 + 25
= 72
Mean = 72 ÷ 20 = 3.6 animals

Sorted: 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 8, 10, 25
20 values → average the 10th and 11th
Median = (2 + 2) ÷ 2 = 2 animals

Describing the data:

Answers run from 0 to 25, so the range is 25.
Most of the class has very few animals — 12 of the 20 students have 2 or fewer, and 6 have none at all.
25 is a clear outlier, and 10 is a second, smaller one. The next value down is 8.
Mean 3.6 > median 2, exactly the signature of a high outlier.
The mean of 3.6 is bigger than the value of 16 of the 20 students, so it is a poor description of a typical home. The median 2 describes this class far better.

Why it happens: the single answer of 25 contributes 25 of the total 72 — more than a third of everything — all by itself. Take it out and the mean of the remaining 19 values falls to 47 ÷ 19 ≈ 2.47, while the median stays at 2.

Tip: the student with 25 animals probably keeps livestock — cows, goats, hens. That is not a mistake in the data, just a genuinely different kind of home. Outliers are often real and interesting; do not delete them, report them.

Q3.

Rintu takes care of a date-palm tree farm in Habra. The heights of the trees (in feet) in his farm are given as: 50, 45, 43, 52, 61, 63, 46, 55, 60, 55, 59, 56, 56, 49, 54, 65, 66, 51, 44, 58, 60, 54, 52, 57, 61, 62, 60, 60, 67. Fill the dot plot, and mark the mean and median. How would you describe the heights of these palm trees? Can you think of quicker ways to find the mean? How many trees are shorter than the average height?

Page 112 — the empty dot plot printed with this question, with its scale running from 0 to 80 feet.

Answer

Mean ≈ 55.9 feet, Median = 56 feet, and 13 trees are shorter than the average. There are 29 trees.

The dot plot. The printed number line runs from 0 to 80 in steps of 5, but every tree lies between 43 and 67, so all the dots sit in the middle of the line.

Height (ft)	43	44	45	46	49	50	51	52	54	55	56	57	58	59	60	61	62	63	65	66	67
Dots	1	1	1	1	1	1	1	2	2	2	2	1	1	1	4	2	1	1	1	1	1

Heights of the 29 date-palm trees. The tallest stack is at 60 feet. The mean (solid) and median (dashed) fall almost on the same spot.

The mean, the ordinary way:

Sum = 50 + 45 + 43 + 52 + 61 + 63 + 46 + 55 + 60 + 55 + 59 + 56 + 56 + 49 + 54
+ 65 + 66 + 51 + 44 + 58 + 60 + 54 + 52 + 57 + 61 + 62 + 60 + 60 + 67
= 1621 feet, over 29 trees
Mean = 1621 ÷ 29 = 55.9 feet (to one decimal place)

A quicker way — the assumed mean. All the heights are near 55, so measure everything from 55 instead of from 0.

Write each height as 55 + something:
−5, −10, −12, −3, +6, +8, −9, 0, +5, 0, +4, +1, +1, −6, −1,
+10, +11, −4, −11, +3, +5, −1, −3, +2, +6, +7, +5, +5, +12

Sum of the pluses = 6+8+5+4+1+1+10+11+3+5+2+6+7+5+5+12 = 90
Sum of the minuses = 5+10+12+3+9+6+1+4+11+1+3 = 64
Total = 90 − 64 = +26
Mean = 55 + 26 ÷ 29 = 55 + 0.9 = 55.9 feet

The median. Sorted, the 15th of the 29 heights is the middle one.

43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55, 56, 56, 57, 58, 59, 60, 60, 60, 60, 61, 61, 62, 63, 65, 66, 67
Median = 15th value = 56 feet

How many trees are shorter than the average, 55.9 feet?

43, 44, 45, 46, 49, 50, 51, 52, 52, 54, 54, 55, 55 → 13 trees
(the 14th tree, at 56 feet, is already taller than 55.9)

Describing the heights: the trees stand between 43 and 67 feet, a range of 24 feet. They cluster strongly in the 50s and low 60s, with the commonest height being 60 feet (4 trees). There is no outlier, and the mean (55.9) and median (56) agree almost exactly — the sign of a well-balanced set of data. Roughly half the farm is under 56 feet and half over.

Why the assumed-mean trick works: if you subtract 55 from every value you subtract 55 × 29 from the total, so the mean also goes down by exactly 55. Adding 55 back at the end restores it. The numbers you actually add up are small, so the arithmetic is far easier.

Q4.

The daily water usage from a tap was measured. The usage in liters for the first few days are: 5.6, 8, 3.09, 12.9, 6.5, 12.1, 11.3, 20.5, 7.4. (a) Can the mean or median daily usage lie between 25 and 30? Justify your claim using the meaning of mean and median. (b) Can the mean or median be lesser than the minimum value or greater than the maximum value in a data?

Answer

(a) No. Neither the mean nor the median can lie between 25 and 30, because the largest reading in the whole data is only 20.5 litres.

Data sorted: 3.09, 5.6, 6.5, 7.4, 8, 11.3, 12.1, 12.9, 20.5
Minimum = 3.09 Maximum = 20.5

Sum = 5.6 + 8 + 3.09 + 12.9 + 6.5 + 12.1 + 11.3 + 20.5 + 7.4 = 87.39
Mean = 87.39 ÷ 9 = 9.71 litres
Median = 5th of 9 values = 8 litres

The justification.

Mean. The mean is a fair share of the total. If every one of the 9 days had used the maximum, 20.5 litres, the total would be 9 × 20.5 = 184.5 and the mean would be 20.5. Since no day used more than 20.5, the total cannot exceed 184.5 and the mean cannot exceed 20.5. And 20.5 is already below 25.
Median. The median is one of the values themselves (or the average of two of them). Since every value lies between 3.09 and 20.5, the median must too.

(b) No — never. The mean and the median always lie between the minimum and the maximum of the data (they may equal one of them, but cannot go outside).

Why it happens: for the mean, think of the fair share. Sharing the total equally cannot give everyone more than the biggest single value, nor less than the smallest — if it did, the shares would not add back up to the same total. For the median, the reason is even simpler: it is chosen from the middle of the sorted list, so there is always at least one value at or below it and at least one at or above it.

Check it yourself: here the mean 9.71 and the median 8 both sit comfortably inside 3.09 to 20.5. Note also that mean > median, which warns us of the high reading of 20.5 litres — perhaps a washing day.

Q5.

The weights of a few newborn babies are given in kgs. Fill the dot plot provided below. Analyse and compare this data.

Boys	3.5	4.1	2.6	3.2	3.4	3.8
Girls	4.0	3.1	3.4	3.7	2.5	3.4

Page 113 — the empty dot plot printed with this question, with its scale running from 0 to 4.5 kg.

Answer

Six boys and six girls. The printed line runs from 0 to 4.5 in steps of 0.5, so all twelve dots sit in its right-hand part.

Birth weights in kilograms. Both sets crowd between 3 and 4 kg; the boys' dots reach a little further to the right.

Boys: 2.6, 3.2, 3.4, 3.5, 3.8, 4.1
Sum = 20.6 kg, Mean = 20.6 ÷ 6 ≈ 3.43 kg
Median = (3.4 + 3.5) ÷ 2 = 3.45 kg

Girls: 2.5, 3.1, 3.4, 3.4, 3.7, 4.0
Sum = 20.1 kg, Mean = 20.1 ÷ 6 = 3.35 kg
Median = (3.4 + 3.4) ÷ 2 = 3.4 kg

	Boys	Girls
Lightest	2.6 kg	2.5 kg
Heaviest	4.1 kg	4.0 kg
Range	1.5 kg	1.5 kg
Mean	3.43 kg	3.35 kg
Median	3.45 kg	3.40 kg

Analysis. The two groups are remarkably alike. Both spread over exactly the same width, 1.5 kg. The boys' mean and median are each about 0.05 kg higher than the girls' — less than the weight of a small egg. In both groups the mean and median are almost equal, so neither set has an outlier. The commonest weight among the girls is 3.4 kg, which occurs twice.

Why it happens: with only 6 babies in each group, a difference of 0.05 kg is far too small to mean anything. Moving one baby's weight by 0.3 kg would flip the result. The strongest honest claim is that these boys and girls weighed about the same at birth — roughly 3.4 kg each, between 2.5 kg and 4.1 kg.

Tip: whenever two averages differ by less than the natural variation inside each group, say so. Claiming "boys are heavier" from a gap of 0.08 kg would be reading far more into the data than it can support.

Q6.

The dot plots of heights of another section of Grade 5 students of the same school are shown below. Can you share your observations? What can we infer from the dot plots and the central tendency measures? Compare the heights of the two sections. Share your observations.

Section 2 dot plots of height in cm, page 113. The solid line marks the mean and the dashed line the median.

Note on the printed figure: in the book the ‘Whole class’ dot plot on page 113 is the Section 1 plot from page 109 printed again — its dots do not match the mean 141.21 and median 142.5 given beside it. The boys’ and girls’ plots are right, and taken together they do give a median of 142.5 and a mean of about 141.2.

Answer

Observations about this second section

The girls' heights are far more spread out — their dots run from about 126 cm to about 158 cm. The boys are packed between about 130 cm and 148 cm.
So in this section both the tallest and the shortest student are girls.
The boys' dots pile up heavily between 141 and 146 cm — that thick stack is why the boys' mean (142.05) sits above the girls' (140.14) even though the girls own both extremes.
For every group, mean < median (141.21 < 142.5; 142.05 < 143; 140.14 < 140 is the one exception — here mean > median by a tiny 0.14 cm). So in the whole class and among the boys, a few short students pull the mean slightly below the middle value.
In this section, the boys are taller on average than the girls.

Comparing the two sections

	Section 1 (page 109)	Section 2	Difference
Whole class mean	144.5 cm	141.21 cm	Section 1 taller by ≈ 3.3 cm
Whole class median	145 cm	142.5 cm	Section 1 taller by 2.5 cm
Boys' mean	142.94 cm	142.05 cm	Almost the same
Girls' mean	146.9 cm	140.14 cm	Section 1 girls taller by ≈ 6.8 cm
Taller group	Girls	Boys	Reversed!
Widest spread	Boys	Girls	Reversed!

Why it matters: the boys of the two sections are practically the same height on average. Almost the whole 3.3 cm gap between the sections comes from the girls. And notice that the answer to "are boys or girls taller?" flips from one section to the other in the same school and the same grade.

The strongest thing you can say: in Section 1 the girls are taller on average, in Section 2 the boys are. Since two classes of the same grade in the same school give opposite answers, this data cannot be used to say anything about boys and girls in general. A class of about 30 children is simply too small a group to settle a question about everybody.

Q7.

The weights of some sumo wrestlers and ballet dancers are: Sumo wrestlers: 295.2 kg, 250.7 kg, 234.1 kg, 221.0 kg, 200.9 kg. Ballet dancers: 40.3 kg, 37.6 kg, 38.8 kg, 45.5 kg, 44.1 kg, 48.2 kg. Approximately how many times heavier is a sumo wrestler compared to a ballet dancer?

Answer

About 5.7 times — roughly 6 times heavier.

Sumo wrestlers (5 people):
295.2 + 250.7 + 234.1 + 221.0 + 200.9 = 1201.9 kg
Mean = 1201.9 ÷ 5 = 240.38 kg

Ballet dancers (6 people):
40.3 + 37.6 + 38.8 + 45.5 + 44.1 + 48.2 = 254.5 kg
Mean = 254.5 ÷ 6 ≈ 42.42 kg

How many times heavier = 240.38 ÷ 42.42 ≈ 5.67

Why we must use the means: the two groups have different sizes — 5 wrestlers and 6 dancers — so their totals (1201.9 kg and 254.5 kg) cannot be compared directly. Dividing each by its own count puts one wrestler against one dancer, which is what the question asks.

A quick check: the lightest wrestler (200.9 kg) is still more than 4 times the heaviest dancer (48.2 kg), and the heaviest wrestler (295.2 kg) is more than 7 times the lightest dancer (37.6 kg). So the true ratio for any pair lies between about 4 and 7 — and 5.7 sits sensibly in the middle. Notice too that neither group has an outlier, so the means are trustworthy here.

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