This page solves all six exercise 13.2 class 10 maths ncert solutions questions from the Statistics chapter, one by one, with the full working shown. Every solution opens with the formula, applies it to the exact data printed in the question, and ends with a clear final answer you can check your own homework against.
Each question in Exercise 13.2 asks for the mode of grouped data — and questions 1, 3 and 4 also ask for the mean by the direct method, so the page recaps both formulas first, then works through Q1 to Q6 in order, flagging the mistake students make most often in each one.
What Exercise 13.2 tests
Exercise 13.2 has six questions, and every one of them tests the mode of grouped data — finding the modal class and substituting into the mode formula. Three of the six (Q1, Q3, Q4) go further and ask you to find the mean by the direct method as well, using class marks.
Q1 and Q4 add a third demand: compare and interpret the two measures, which means writing a sentence that explains what each value tells you, not just quoting both numbers (NCERT, p. 186).
Three skills are actually being examined here:
- locating the modal class — the class with the highest frequency, never the one in the middle of the table;
- substituting l, h, f1, f0, f2 into the mode formula in the right order;
- choosing the direct method for the mean when the class marks are tidy, and computing \(\sum f_i x_i\) without arithmetic slips.
Mode and mean formulas this exercise needs
For grouped data you cannot read the mode directly from the frequencies — you can only locate the class with the maximum frequency, called the modal class, and then compute a value inside it. The mode is given by (NCERT, p. 185):
\[ \text{Mode} = l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h \]
where the symbols mean:
- l = lower limit of the modal class
- h = size of the class interval (all classes equal in these questions)
- f1 = frequency of the modal class
- f0 = frequency of the class immediately before the modal class
- f2 = frequency of the class immediately after the modal class
The mean of grouped data assumes every frequency is centred at its class mark, the mid-point of the class:
\[ \text{Class mark} = \frac{\text{Upper limit} + \text{Lower limit}}{2}, \qquad \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
The textbook notes that the mean obtained by the direct, assumed-mean and step-deviation methods is always the same — the method is a choice about arithmetic, not about the answer (NCERT, p. 177).
Mini worked example (new numbers, not from the textbook): a small frequency distribution has classes 10–20, 20–30, 30–40 with frequencies 5, 12, 8. The highest frequency is 12, so the modal class is 20–30, giving l = 20, h = 10, f1 = 12, f0 = 5, f2 = 8. Substituting:
\[ \text{Mode} = 20 + \left( \frac{12 – 5}{2 \times 12 – 5 – 8} \right) \times 10 = 20 + \frac{7}{11} \times 10 = 20 + 6.36 = 26.36 \]
Notice the order: numerator first, then the denominator \(2f_1 – f_0 – f_2\) computed in full, then multiply by h. Students who divide by \(f_1\) alone, or who swap f0 and f2, get a very different number.
Question 1: The following table shows the ages of the patients admitted in a hospital during a year:
| Age (in years) | 5–15 | 15–25 | 25–35 | 35–45 | 45–55 | 55–65 |
|---|---|---|---|---|---|---|
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Concept: the mode answers “what age is the most common among admitted patients”, while the mean answers “what is the average age”. The highest frequency is 23, so the modal class is 35–45 — the fourth class, not a middle one by position.
Step 1 (mode): read off l = 35, h = 10, f1 = 23 (modal class), f0 = 21 (class 25–35 before it), f2 = 14 (class 45–55 after it).
\[ \text{Mode} = 35 + \left( \frac{23 – 21}{2 \times 23 – 21 – 14} \right) \times 10 = 35 + \frac{2}{11} \times 10 = 35 + 1.82 = 36.82 \]
Final answer (mode): the mode is 36.82 years — the most common age bracket of admitted patients is just under 37.
Step 2 (mean): form a direct-method table.
Class marks are 10, 20, 30, 40, 50, 60.
\[ \sum f_i = 6 + 11 + 21 + 23 + 14 + 5 = 80 \]
\[ \sum f_i x_i = 6(10) + 11(20) + 21(30) + 23(40) + 14(50) + 5(60) = 60 + 220 + 630 + 920 + 700 + 300 = 2830 \]
\[ \bar{x} = \frac{2830}{80} = 35.375 \approx 35.38 \]
Final answer (mean): the mean age is 35.38 years.
Compare and interpret: the mode (36.82 years) is slightly higher than the mean (35.38 years). This says the largest number of patients admitted fall in the 35–45 age band (peaking near 37), while the overall average age of all admitted patients is about 35.4 years — the most common age is a little above the average.
Student tip: the modal class is chosen by frequency, not by position — here 35–45 has the maximum 23 patients even though 25–35 sits to its left with 21. Also, round sensibly at the end: the board accepts 36.82 (or 36.8) and 35.38 (or 35.4). Never leave the interpret sentence out; it carries a mark of its own.
Question 2: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
| Lifetimes (in hours) | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 | 100–120 |
|---|---|---|---|---|---|---|
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Concept: “modal lifetime” does not mean the average lifetime — it means the lifetime range that occurs most often. The total 225 is background information; the mode formula never uses it.
Step 1: the maximum frequency is 61, so the modal class is 60–80.
Hence l = 60, h = 20, f1 = 61, f0 = 52 (class 40–60), f2 = 38 (class 80–100).
Step 2: substitute each value in its labelled place.
\[ \text{Mode} = 60 + \left( \frac{61 – 52}{2 \times 61 – 52 – 38} \right) \times 20 = 60 + \frac{9}{32} \times 20 = 60 + 5.625 = 65.625 \]
Final answer: the modal lifetime of the components is 65.625 hours.
Student tip: the most common error here is taking f0 = 35 or f2 = 29 because they look like “neighbours” on the page. Check the rows that sit immediately beside the modal class: f0 = 52 and f2 = 38, nothing else.
Question 3: The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
| Expenditure (in ₹) | Number of families |
|---|---|
| 1000–1500 | 24 |
| 1500–2000 | 40 |
| 2000–2500 | 33 |
| 2500–3000 | 28 |
| 3000–3500 | 30 |
| 3500–4000 | 22 |
| 4000–4500 | 16 |
| 4500–5000 | 7 |
Concept: the expenditure values are large, but they are still multiples of 250, so the direct method stays comfortable — you multiply neat class marks by frequencies without huge products.
Step 1 (mode): the highest frequency is 40, so the modal class is 1500–2000.
Read l = 1500, h = 500, f1 = 40, f0 = 24 (class 1000–1500), f2 = 33 (class 2000–2500).
\[ \text{Mode} = 1500 + \left( \frac{40 – 24}{2 \times 40 – 24 – 33} \right) \times 500 = 1500 + \frac{16}{23} \times 500 = 1500 + 347.83 = 1847.83 \]
Final answer (mode): the modal monthly expenditure is ₹1847.83.
Step 2 (mean): class marks are 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750, and \(\sum f_i = 200\).
\[ \sum f_i x_i = 24(1250) + 40(1750) + 33(2250) + 28(2750) + 30(3250) + 22(3750) + 16(4250) + 7(4750) \]
\[ = 30000 + 70000 + 74250 + 77000 + 97500 + 82500 + 68000 + 33250 = 532500 \]
\[ \bar{x} = \frac{532500}{200} = 2662.5 \]
Final answer (mean): the mean monthly expenditure is ₹2662.50.
Student tip: keep the mode and mean workings separate — a common slip is using a class mark (like 1750) inside the mode formula. The mode uses only l, h and the three neighbouring frequencies; the mean uses the class marks. Also double-check that f0 = 24 and f2 = 33, the classes on either side of 1500–2000.
Question 4: The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
| Number of students per teacher | Number of states / U.T. |
|---|---|
| 15–20 | 3 |
| 20–25 | 8 |
| 25–30 | 9 |
| 30–35 | 10 |
| 35–40 | 3 |
| 40–45 | 0 |
| 45–50 | 0 |
| 50–55 | 2 |
Concept: the highest frequency is 10, so the modal class is 30–35. The special trap in this question is the two zero-frequency classes (40–45 and 45–50) sitting right after the modal class — a zero frequency is still a number and enters the formula as 0.
Step 1 (mode): l = 30, h = 5, f1 = 10, f0 = 9 (class 25–30), f2 = 3 (class 35–40).
\[ \text{Mode} = 30 + \left( \frac{10 – 9}{2 \times 10 – 9 – 3} \right) \times 5 = 30 + \frac{1}{8} \times 5 = 30 + 0.625 = 30.625 \]
Final answer (mode): the mode is 30.625 students per teacher.
Step 2 (mean): class marks are 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5, 52.5 and \(\sum f_i = 3 + 8 + 9 + 10 + 3 + 0 + 0 + 2 = 35\).
\[ \sum f_i x_i = 3(17.5) + 8(22.5) + 9(27.5) + 10(32.5) + 3(37.5) + 0(42.5) + 0(47.5) + 2(52.5) \]
\[ = 52.5 + 180 + 247.5 + 325 + 112.5 + 0 + 0 + 105 = 1022.5 \]
\[ \bar{x} = \frac{1022.5}{35} = 29.21 \]
Final answer (mean): the mean is 29.21 students per teacher.
Interpret: most states and U.T.s have about 30.6 students per teacher, while on average there are about 29.2 students per teacher across all the states — the most common ratio sits just above the national average.
Student tip: the zero frequencies in 40–45 and 45–50 are easy to misread as “no class”. They are real classes with frequency 0, and if one sat in the f0 or f2 position you would still substitute 0. Here they appear after the modal class, so they do not change f2 = 3.
The mode always lands inside the modal class 30–35 — if your answer fell outside it, you have substituted wrongly.
Question 5: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
| Runs scored | Number of batsmen |
|---|---|
| 3000–4000 | 4 |
| 4000–5000 | 18 |
| 5000–6000 | 9 |
| 6000–7000 | 7 |
| 7000–8000 | 6 |
| 8000–9000 | 3 |
| 9000–10000 | 1 |
| 10000–11000 | 1 |
Find the mode of the data.
Concept: only the mode is asked, so no mean table is needed. The class intervals are equal here (1000 wide), so the mode formula applies directly with h = 1000.
Step 1: the highest frequency is 18, so the modal class is 4000–5000.
Read l = 4000, h = 1000, f1 = 18, f0 = 4 (class 3000–4000), f2 = 9 (class 5000–6000).
\[ \text{Mode} = 4000 + \left( \frac{18 – 4}{2 \times 18 – 4 – 9} \right) \times 1000 = 4000 + \frac{14}{23} \times 1000 = 4000 + 608.70 = 4608.70 \]
Final answer: the mode of the data is 4608.7 runs — the largest group of top batsmen has scored around 4600 runs.
Student tip: you do not need the total number of batsmen (which you can work out as 49) anywhere in the mode formula. If you find yourself using it, you have drifted into the mean — the mode ignores the total frequency entirely.
Question 6: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
| Number of cars | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 | 60–70 | 70–80 |
|---|---|---|---|---|---|---|---|---|
| Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Concept: “100 periods each of 3 minutes” is context — it tells you how the observations were collected. It never enters the mode formula. The mode only needs the class with the highest frequency and its two neighbours.
Step 1: the maximum frequency is 20, so the modal class is 40–50.
Read l = 40, h = 10, f1 = 20, f0 = 12 (class 30–40), f2 = 11 (class 50–60).
\[ \text{Mode} = 40 + \left( \frac{20 – 12}{2 \times 20 – 12 – 11} \right) \times 10 = 40 + \frac{8}{17} \times 10 = 40 + 4.71 = 44.71 \]
Final answer: the mode of the data is 44.7 cars — the most common number of cars seen in a 3-minute period is just under 45.
Student tip: this is the classic trap question. The number 100 is deliberately placed to tempt you into using it. The mode formula contains only l, h, f1, f0 and f2 — the total frequency never appears, so 100 is simply ignored.
Method recap: choosing a mean method and locating the modal class
The textbook’s Remark on choosing a mean method is worth memorising (NCERT, p. 180): the result is identical by all three methods, so the choice is purely about arithmetic convenience.
| Data situation | Method to choose | Why |
|---|---|---|
| \(x_i\) and \(f_i\) are small | Direct method, \(\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}\) | Products are quick to compute, no extra columns needed |
| \(x_i\) and \(f_i\) are numerically large | Assumed mean method, \(\bar{x} = a + \dfrac{\sum f_i d_i}{\sum f_i}\) | Subtracting a fixed \(a\) shrinks the class marks to small numbers |
| Large values with a common factor \(h\) in \(d_i\) | Step-deviation method, \(\bar{x} = a + h\left(\dfrac{\sum f_i u_i}{\sum f_i}\right)\) | Dividing by \(h\) gives the smallest numbers of all |
| Unequal class sizes with large \(x_i\) | Step-deviation with a suitable \(h\) | Still works because the method does not need equal classes |
For the mode, a three-step check keeps the reading of f0, f1, f2 honest:
- Find f1 first — scan the frequency column, pick the largest number; the class it belongs to is the modal class.
- Step one row back for f0 and one row forward for f2 — never skip to the start or end of the table.
- Check the mode lies inside the modal class — if your result falls outside \(l\) to \(l + h\), a substitution is wrong.
Common mistakes in the mode formula
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Choosing the modal class by position, not frequency | Modal class = class with the highest frequency | Scan every frequency; the largest one wins |
| Swapping f0 and f2 | f0 = class before, f2 = class after the modal class | Verify the rows immediately on either side |
| Dividing by f1 instead of the full denominator | Denominator is \(2f_1 – f_0 – f_2\) | Recompute the whole denominator, then divide |
| Using the total frequency \(n\) in the mode | \(n\) never appears in the mode formula | The formula uses only l, h, f1, f0, f2 |
| Ignoring a zero-frequency neighbour | A zero frequency is still substituted as 0 | Treat 0 as a real frequency in the calculation |
After Exercise 13.2, the chapter moves on to the median of grouped data; if you are revising the whole chapter, the worked solutions here give you the mode and mean habits you will reuse there.
You can also step back and refresh the Class 10 Maths notes hub, check the full Class 10 study material for other subjects, or browse all CBSE notes across classes. For syllabus context you may also look at the previous chapter, Surface Areas and Volumes, and the next one, Probability.
Frequently asked questions on Exercise 13.2
Why is the class with the highest frequency called the modal class, and why does the mode always lie inside it?
The mode is the value that occurs most often in the data. In grouped data we cannot see individual values, only classes and their frequencies — so the class with the maximum frequency contains the densest cluster of observations and is the most likely home of the most frequent value.
The formula \(l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right)h\) adds a correction to the lower limit \(l\) that is always between 0 and the class size \(h\), so the resulting mode always lies inside the modal class. If your answer falls outside the class, recheck your substitution.
How do I choose the assumed mean if I need the step-deviation method?
Take \(a\) as one of the class marks that lies near the centre of the table — the textbook itself chooses the middle class mark in its worked example (NCERT, p. 175).
Any class mark works, and the final mean is identical no matter which one you pick, but a central choice makes the deviations \(d_i = x_i – a\) small and symmetric, which keeps the arithmetic tidy.
For the step-deviation form you then divide each \(d_i\) by the class size \(h\) and use \(\bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right)\).
When a question asks me to compare the mode and the mean, what must I write to get the full mark?
To earn the interpretation mark you must do two things: state both values clearly, then write one sentence saying what each one means in the context of the data.
For Question 1, that means naming the mode (36.82 years) as the age at which the most patients were admitted, and the mean (35.38 years) as the average age of all admitted patients, then noting which is higher. A sentence that quotes the numbers without saying what they represent is only half the answer.
What happens when the class just before or just after the modal class has zero frequency, as in Question 4?
You treat the zero as a real frequency and substitute it as 0. In Question 4 the zero-frequency classes (40–45 and 45–50) come after the modal class 30–35, so they do not affect the mode at all — the succeeding class 35–40 still has f2 = 3.
If, in another problem, the class before the modal class had zero frequency, you would set f0 = 0 and the formula would still run normally, giving a larger correction because there is no build-up on the left side.
Reference: NCERT Class 10 Mathematics textbook, chapter Statistics. You can verify the exact Exercise 13.2 question text and the mode and mean formulas against the official Rationalised textbook PDF on the NCERT official website.
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