MathsCBSE Class 10 · Statistics

Mode of Grouped Data for Class 10 Statistics

Updated 14 October 2026 · Fact-checked

The mode of grouped data is the most common value in a frequency table. You find the modal class, the class with the highest frequency, then use a formula with its lower limit, its width and the frequencies of the classes on either side. Boards ask you to find the mode, or a missing frequency when the mode is given.

The mode sits inside the tallest (modal) bar, where the two crossing lines meet, and it shifts towards the side whose neighbouring bar is shorter.

Illustration drawn for QuestionBuddy.

Understand Mode of Grouped Data

The mode is the value that occurs most often. In ungrouped data you just spot the most repeated number. In grouped data you cannot see the individual values. You only know how many observations fall in each class, so you need a different method.

First find the modal class, the class with the highest frequency. The mode must lie inside it, because that is where the data is most crowded. The syllabus avoids bimodal data, so there will always be one clear highest frequency.

Next you decide where inside the modal class the mode sits. Look at the two neighbours. If the class before has a bigger frequency than the class after, the crowd leans to the left, so the mode moves towards the lower limit. If the class after is bigger, the mode moves right. The formula $Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ does exactly this. The denominator is $(f_{1} - f_{0}) + (f_{1} - f_{2})$ , the sum of the two drops from the peak. The class width $h$ is split in the ratio of these drops. When $f_{0} = f_{2}$ , the mode lands exactly at the class mark.

Here $l$ is the lower limit of the modal class, $h$ its width, $f_{1}$ its frequency, $f_{0}$ the frequency of the class just before and $f_{2}$ the frequency of the class just after. All five come from the ordinary frequency column, never from cumulative frequencies.

The formula needs continuous classes. If classes are written like 11–20, 21–30, close the gap first: subtract $0.5$ from each lower limit and add $0.5$ to each upper limit. Then $l$ becomes $20.5$ , not $21$ . If the data is given as a cumulative table, convert it back to frequencies before you look for the modal class.

The mode connects to the rest of the chapter through the empirical relation $3 Median = Mode + 2 Mean$ , and it often appears in the same question as the mean or median. Use the mode when a question asks for the most common or most popular value, such as the most common shoe size or waiting time.

Mode of Grouped Data: formulas to remember

Mode of grouped data: $Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$; l = lower limit of modal class, h = class width, f1 = frequency of modal class, f0 = class before, f2 = class after.
Modal class: $Modal class = class with the highest f_{i}$; Use ordinary frequencies only. The mode always lies inside this class.
Denominator as two drops: $2 f_{1} - f_{0} - f_{2} = (f_{1} - f_{0}) + (f_{1} - f_{2})$; Shows why the class width is split in the ratio of the two drops from the peak.
Inclusive to continuous classes: $l_{new} = l - \frac{gap}{2}$; For classes like 11–20, 21–30 the gap is 1, so subtract 0.5 from every lower limit.
Missing frequency from a given mode: $(Mode - l) (2 f_{1} - f_{0} - f_{2}) = h (f_{1} - f_{0})$; Cross-multiplied form; substitute the known values and solve the linear equation.

How to solve Mode of Grouped Data questions

This method works for any board question on the mode of grouped data, including ones with a missing frequency.

1Check the classes. If there are gaps (like 1–10, 11–20), make them continuous by subtracting and adding half the gap. If a cumulative table is given, convert it to frequencies.
2Find the modal class: the class with the highest frequency. If the mode is given instead, the modal class is the class that contains the mode.
3Write down clearly: $l$ , $h$ , $f_{1}$ , $f_{0}$ and $f_{2}$ . Writing this values line clearly helps you earn step marks.
4Write the formula $Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ before substituting.
5Substitute and simplify the fraction first, then multiply by $h$ and add $l$ .
6If an unknown frequency appears, cross-multiply to get a linear equation and solve it.
7Check: the mode must lie inside the modal class, and it should lean towards the bigger neighbour. Write the answer with units.

Mode of Grouped Data in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Mode of grouped data

Suppose your school hostel warden surveys 60 students on how much mobile data (in GB) each uses in a week, to pick one common weekly data pack for the hostel. The results are: 0–2 GB: 5 students, 2–4 GB: 14, 4–6 GB: 20, 6–8 GB: 18, 8–10 GB: 3. Packs come in 5 GB, 6 GB and 8 GB. The warden's rule: choose the smallest pack that is at least the modal use, but if more than one-third of the students would still need more data than that pack, move up to the next pack.

Using the mode and the table, decide which pack the warden should choose, and find how many students would still need extra data with that pack.

Try it first, then show the solutionSolution

Highest frequency is 20, so modal class is 4–6.
$l = 4$ , $h = 2$ , $f_{1} = 20$ , $f_{0} = 14$ , $f_{2} = 18$ .
$Mode = 4 + \frac{20 - 14}{40 - 14 - 18} \times 2$
$= 4 + \frac{6}{8} \times 2 = 4 + 1.5 = 5.5$ GB.
The smallest pack that is at least 5.5 GB is the 6 GB pack.
Students needing more than 6 GB (classes 6–8 and 8–10): $18 + 3 = 21$ .
One-third of 60 is 20, and $21 > 20$ , so the rule says move up to the 8 GB pack.
With the 8 GB pack, only the 8–10 class needs extra: 3 students.

Answer: Modal use is 5.5 GB, so the 6 GB pack is the first choice; but 21 students (more than 20) exceed it, so the warden picks the 8 GB pack, leaving only 3 students needing extra data.

Memory hook: Most students crowd at one data level: the mode picks the pack, the count checks it.

Tricks to remember Mode of Grouped Data

Short memory aids. Say each one aloud once, then use it on a question the same day.

One-Zero-Two: before, peak, after
Read the subscripts left to right across the table: $f_{0}$ is before, $f_{1}$ is the peak, $f_{2}$ is after. The formula is (peak minus before) over (two peaks minus both neighbours).
Use it when: Stops you swapping $f_{0}$ and $f_{2}$ when substituting.
Lean towards the bigger neighbour
If the class after the peak is bigger than the class before, the mode is past the middle of the modal class; if smaller, before the middle.
Use it when: A quick check on your final answer without redoing the arithmetic.
Mode looks at frequency, median looks at cf
Mode: highest $f$ . Median: first cf reaching $\frac{n}{2}$ . Never mix the columns.
Use it when: Prevents picking the modal class from a cumulative column.

Mode of Grouped Data: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Mode of a simple distribution · 2 marks · Very short answerBoard level

The distance (in km) cycled daily by 45 students to reach school is given below. Find the modal distance.
Distance (km): 10–15, 15–20, 20–25, 25–30, 30–35
Number of students: 4, 9, 15, 11, 6

Try it first, then show the solutionSolution

Highest frequency is 15, so modal class is 20–25. (½ mark)
$l = 20$ , $h = 5$ , $f_{1} = 15$ , $f_{0} = 9$ , $f_{2} = 11$ . (½ mark)
$Mode = 20 + \frac{15 - 9}{30 - 9 - 11} \times 5 = 20 + \frac{6}{10} \times 5$ (½ mark)
$= 20 + 3 = 23$ km. (½ mark)

Answer: Modal distance = 23 km.

Example 2: Mode with inclusive classes · 3 marks · Short answerBoard level

The weights (in kg) of 40 parcels at a courier counter are given below. Find the modal weight.
Weight (kg): 11–20, 21–30, 31–40, 41–50, 51–60
Number of parcels: 3, 7, 13, 9, 8

Try it first, then show the solutionSolution

Classes have a gap of 1, so subtract 0.5 from lower limits and add 0.5 to upper limits: 10.5–20.5, 20.5–30.5, 30.5–40.5, 40.5–50.5, 50.5–60.5. (1 mark)
Highest frequency 13, so modal class is 30.5–40.5; $l = 30.5$ , $h = 10$ , $f_{1} = 13$ , $f_{0} = 7$ , $f_{2} = 9$ . (1 mark)
$Mode = 30.5 + \frac{13 - 7}{26 - 7 - 9} \times 10 = 30.5 + \frac{6}{10} \times 10$ (½ mark)
$= 30.5 + 6 = 36.5$ kg. (½ mark)

Answer: Modal weight = 36.5 kg.

Example 3: Missing frequency of the modal class · 3 marks · Short answerBoard level

The mode of the following distribution of marks is 52. Find the missing frequency $x$ , given that 40–60 is the modal class.
Marks: 0–20, 20–40, 40–60, 60–80, 80–100
Number of students: 8, 10, $x$ , 14, 6

Try it first, then show the solutionSolution

Modal class 40–60: $l = 40$ , $h = 20$ , $f_{1} = x$ , $f_{0} = 10$ , $f_{2} = 14$ . (½ mark)
$52 = 40 + \frac{x - 10}{2 x - 10 - 14} \times 20$ (½ mark)
$12 (2 x - 24) = 20 (x - 10)$ (½ mark)
$24 x - 288 = 20 x - 200$ (½ mark)
$4 x = 88$ , so $x = 22$ . (½ mark)
Check: $40 + \frac{12}{44 - 24} \times 20 = 40 + 12 = 52$ ; and 22 is the highest frequency, so 40–60 is indeed modal. (½ mark)

Answer: $x = 22$

Example 4: Two missing frequencies from mode and total · 5 marks · Long answerHarder board level

The marks of 44 students in a test are given below. The mode is 24.
Marks: 0–10, 10–20, 20–30, 30–40, 40–50
Number of students: 6, $p$ , 15, $q$ , 8
(i) Find $p$ and $q$ .
(ii) Explain why 20–30 is the modal class with your values, and state on which side of the class mark the mode lies.

Try it first, then show the solutionSolution

Total: $6 + p + 15 + q + 8 = 44$ , so $p + q = 15$ . (1 mark)
The mode 24 lies in 20–30, so it is the modal class: $l = 20$ , $h = 10$ , $f_{1} = 15$ , $f_{0} = p$ , $f_{2} = q$ . (½ mark)
$24 = 20 + \frac{15 - p}{30 - p - q} \times 10$ (½ mark)
$4 (30 - p - q) = 10 (15 - p)$ , so $120 - 4 p - 4 q = 150 - 10 p$ (½ mark)
$6 p - 4 q = 30$ , i.e. $3 p - 2 q = 15$ . (½ mark)
Substitute $q = 15 - p$ : $3 p - 30 + 2 p = 15$ , so $5 p = 45$ , $p = 9$ , $q = 6$ . (1 mark)
Frequencies are 6, 9, 15, 6, 8; 15 is the highest, so 20–30 is modal. Mode 24 is below the class mark 25 because the class before (9) is bigger than the class after (6). (1 mark)

Answer: $p = 9$ , $q = 6$ ; the mode 24 lies left of the class mark 25.

Example 5: Case study: waiting time at an auto stand · 4 marks · Case-basedHOTS / competency

Suppose a student volunteer records how long 80 commuters wait for a shared auto at a busy stand one morning. Two entries on the sheet got smudged, but the volunteer remembers that the modal waiting time was 11 minutes.
Waiting time (min): 0–4, 4–8, 8–12, 12–16, 16–20
Number of commuters: 6, 16, 25, $p$ , $q$
(i) Which class is the modal class? Give a reason. (1 mark)
(ii) Show that $p + q = 33$ . (1 mark)
(iii) Find $p$ and $q$ . The stand's target is that fewer than one-fifth of the commuters wait 16 minutes or more. Is the target met? (2 marks)

Try it first, then show the solutionSolution

(i) The mode 11 lies in 8–12, so the modal class is 8–12. (1 mark)
(ii) $6 + 16 + 25 + p + q = 80$ , so $47 + p + q = 80$ , giving $p + q = 33$ . (1 mark)
(iii) $l = 8$ , $h = 4$ , $f_{1} = 25$ , $f_{0} = 16$ , $f_{2} = p$ : $11 = 8 + \frac{25 - 16}{50 - 16 - p} \times 4$ (½ mark)
$3 (34 - p) = 36$ , so $34 - p = 12$ and $p = 22$ . (½ mark)
$q = 33 - 22 = 11$ . Check: 25 is still the highest frequency, so 8–12 is modal. (½ mark)
One-fifth of 80 is 16. Commuters waiting 16 minutes or more $= q = 11$ , and $11 < 16$ , so the target is met. (½ mark)

Answer: (i) 8–12, as it contains the mode 11 (ii) $p + q = 33$ (iii) $p = 22$ , $q = 11$ ; 11 commuters is fewer than 16, so the target is met.

Example 6: First class as the modal class · 2 marks · Very short answerBoard level

Suppose a bank counter records the waiting time of 40 customers. Find the modal waiting time.
Time (min): 0–10, 10–20, 20–30, 30–40
Number of customers: 16, 12, 8, 4

Try it first, then show the solutionSolution

Highest frequency is 16, so the modal class is 0–10, the first class. There is no class before it, so $f_{0} = 0$ . (½ mark)
$l = 0$ , $h = 10$ , $f_{1} = 16$ , $f_{0} = 0$ , $f_{2} = 12$ . (½ mark)
$Mode = 0 + \frac{16 - 0}{32 - 0 - 12} \times 10 = \frac{16}{20} \times 10$ (½ mark)
$= 8$ min, which lies inside 0–10. (½ mark)

Answer: Modal waiting time = 8 minutes.

Example 7: Mode from a 'less than' cumulative table · 3 marks · Short answerBoard level

The marks of 45 students in a class test are given as a 'less than' cumulative frequency table. Find the modal marks.
Marks: less than 10, less than 20, less than 30, less than 40, less than 50
Number of students: 5, 13, 28, 40, 45

Try it first, then show the solutionSolution

Convert to frequencies by subtracting consecutive cf: 0–10: 5, 10–20: $13 - 5 = 8$ , 20–30: $28 - 13 = 15$ , 30–40: $40 - 28 = 12$ , 40–50: $45 - 40 = 5$ . (1 mark)
Highest frequency 15, so modal class is 20–30: $l = 20$ , $h = 10$ , $f_{1} = 15$ , $f_{0} = 8$ , $f_{2} = 12$ . (1 mark)
$Mode = 20 + \frac{15 - 8}{30 - 8 - 12} \times 10 = 20 + \frac{7}{10} \times 10$ (½ mark)
$= 27$ marks. It is right of the class mark 25, as expected since 12 > 8. (½ mark)

Answer: Modal marks = 27.

Example 8: Missing frequency, then all three averages · 5 marks · Long answerHOTS / competency

Suppose a school records the marks (out of 100) of its Class 10 students in a practice test. The mean mark is 51.
Marks: 0–20, 20–40, 40–60, 60–80, 80–100
Number of students: 5, 15, $f$ , 12, 8
(i) Find $f$ .
(ii) Find the median.
(iii) Use the empirical relation to estimate the mode, then find the mode from the formula and compare the two.

Try it first, then show the solutionSolution

(i) Class marks 10, 30, 50, 70, 90. $\sum f_{i} = 40 + f$ and $\sum f_{i} x_{i} = 50 + 450 + 50 f + 840 + 720 = 2060 + 50 f$ . (½ mark)
$\frac{2060 + 50 f}{40 + f} = 51$ gives $2060 + 50 f = 2040 + 51 f$ , so $f = 20$ . Total $n = 60$ . (1 mark)
(ii) cf: 5, 20, 40, 52, 60. $\frac{n}{2} = 30$ , so the median class is 40–60 with $l = 40$ , $c f = 20$ , $f = 20$ , $h = 20$ . (½ mark)
$Median = 40 + \frac{30 - 20}{20} \times 20 = 50$ . (1 mark)
(iii) Empirical: $Mode = 3 (50) - 2 (51) = 48$ . (1 mark)
Formula: modal class 40–60, $f_{1} = 20$ , $f_{0} = 15$ , $f_{2} = 12$ : $Mode = 40 + \frac{20 - 15}{40 - 15 - 12} \times 20 = 40 + \frac{100}{13} \approx 47.7$ . The two agree closely, since the relation is only an approximation. (1 mark)

Answer: (i) $f = 20$ (ii) Median = 50 (iii) Empirical mode = 48; formula mode $\approx 47.7$ , very close.

Example 9: Full 5-mark question: mean, median and mode from one table · 5 marks · Long answerHarder board level

The daily travel time (in minutes) of 50 teachers is given below. Find the mean, median and mode.
Time (min): 0–10, 10–20, 20–30, 30–40, 40–50
Number of teachers: 5, 10, 20, 5, 10

Try it first, then show the solutionSolution

Table: $x_{i}$ = 5, 15, 25, 35, 45; $f_{i} x_{i}$ = 25, 150, 500, 175, 450; cf = 5, 15, 35, 40, 50. $\sum f_{i} = 50$ , $\sum f_{i} x_{i} = 1300$ . (1 mark)
$Mean = \frac{\sum f _{i} x _{i}}{\sum f _{i}} = \frac{1300}{50} = 26$ min. (1 mark)
$\frac{n}{2} = 25$ ; the first cf reaching 25 is 35, so median class is 20–30: $l = 20$ , $c f = 15$ , $f = 20$ , $h = 10$ . (½ mark)
$Median = 20 + \frac{25 - 15}{20} \times 10 = 20 + 5 = 25$ min. (1 mark)
Highest frequency 20, so modal class is 20–30: $l = 20$ , $f_{1} = 20$ , $f_{0} = 10$ , $f_{2} = 5$ , $h = 10$ . (½ mark)
$Mode = 20 + \frac{20 - 10}{40 - 10 - 5} \times 10 = 20 + \frac{10}{25} \times 10 = 24$ min. (1 mark)

Answer: Mean = 26 min, median = 25 min, mode = 24 min. Check: $3 (25) = 75$ and $24 + 2 (26) = 76$ , close, as the empirical relation expects.

Example 10: When the last class is modal · 2 marks · Very short answerBoard level

Find the mode of the following data.
Classes: 0–10, 10–20, 20–30, 30–40
Frequencies: 4, 8, 12, 20

Try it first, then show the solutionSolution

Highest frequency is 20, so the modal class is 30–40, the last class. There is no class after it, so $f_{2} = 0$ . (½ mark)
$l = 30$ , $h = 10$ , $f_{1} = 20$ , $f_{0} = 12$ , $f_{2} = 0$ . (½ mark)
$Mode = 30 + \frac{20 - 12}{40 - 12 - 0} \times 10 = 30 + \frac{80}{28} \approx 30 + 2.86 = 32.86$ . (1 mark)

Answer: Mode $\approx 32.86$ , which lies inside the last class 30–40.

Where students lose marks in Mode of Grouped Data

The slips examiners see most often, and what to write instead.

Mistake: Choosing the modal class from cumulative frequencies
Students build a cf column for the median and then read the mode from it too.
Wrong working
cf 8, 20, 40, 58, 60, so modal class 40–50.
Right working
Frequencies 8, 12, 20, 18, 2, so modal class 20–30, mode 28.
Do this: For the mode, look only at the frequency column.
Usually costs: Most of the marks for the question.
Mistake: Using the inclusive lower limit
The gap between classes like 30–40 written as 31–40 is ignored.
Wrong working
Classes 31–40: $l = 31$ , mode $= 31 + 6 = 37$ kg.
Right working
Continuous class 30.5–40.5: $l = 30.5$ , mode $= 36.5$ kg.
Do this: Make classes continuous before writing $l$ .
Usually costs: 1 mark for wrong $l$ and the final answer.
Mistake: Swapping $f_{0}$ and $f_{2}$
The subscripts are not linked to before and after in the student's mind.
Wrong working
Modal 20–25 with neighbours 9 and 11: $20 + \frac{15 - 11}{10} \times 5 = 22$ km.
Right working
$f_{0} = 9$ (before): $20 + \frac{15 - 9}{10} \times 5 = 23$ km.
Do this: Write $f_{0}$ , $f_{1}$ , $f_{2}$ in the order they appear in the table.
Usually costs: 1 to 1½ marks.
Mistake: Wrong denominator
Students write $f_{1} - f_{0} - f_{2}$ or $2 f_{1} - f_{0} + f_{2}$ .
Wrong working
$\frac{15 - 9}{15 - 9 - 11}$ gives a negative fraction.
Right working
$\frac{15 - 9}{2 ( 15 ) - 9 - 11} = \frac{6}{10}$ .
Do this: Remember the denominator is the sum of two drops: $(f_{1} - f_{0}) + (f_{1} - f_{2})$ .
Usually costs: Marks for substitution and answer.

Exam tips

Always write the line $l = \dots, h = \dots, f_{1} = \dots, f_{0} = \dots, f_{2} = \dots$ before substituting; it carries step marks even if arithmetic slips.
When the mode is given and a frequency is missing, state which class is modal and why (it contains the given mode).
In combined questions, one table with $f_{i}$ , $x_{i}$ , $f_{i} x_{i}$ and cf columns serves mean, median and mode together.
Check your answer lies inside the modal class and leans towards the bigger neighbour.
Keep the empirical relation $3 Median = Mode + 2 Mean$ ready, as it can link mean, median and mode.

Practice questions on Mode of Grouped Data

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 5 marks · Long answerHarder board level
The marks of 60 students are given below. The median is 25.5.
Marks: 0–10, 10–20, 20–30, 30–40, 40–50
Number of students: 5, $x$ , 20, $y$ , 10
(i) Find $x$ and $y$ .
(ii) Show that the mode is 24.
(iii) Find the mean, and check the empirical relation.
Hint
The total gives $x + y$ . The median class is 20–30, and its cf before it is $5 + x$ .
Answer
(i) $5 + x + 20 + y + 10 = 60$ , so $x + y = 25$ . Median class 20–30: $25.5 = 20 + \frac{30 - ( 5 + x )}{20} \times 10$ , so $5.5 = \frac{25 - x}{2}$ and $x = 14$ ; then $y = 11$ .
(ii) Modal class 20–30 (20 is the highest): $Mode = 20 + \frac{20 - 14}{40 - 14 - 11} \times 10 = 20 + \frac{6}{15} \times 10 = 24$ .
(iii) $\sum f_{i} x_{i} = 25 + 210 + 500 + 385 + 450 = 1570$ , so mean $= \frac{1570}{60} \approx 26.17$ . Check: $Mode + 2 Mean \approx 24 + 52.33 = 76.33$ and $3 Median = 76.5$ , very close.
Q2 · 3 marks · Short answerBoard level
The ages (in years) of 45 members of a yoga club are given. Find the modal age.
Age: 5–9, 10–14, 15–19, 20–24, 25–29
Members: 4, 10, 16, 12, 3
Hint
Make the classes continuous first; the modal class becomes 14.5–19.5.
Answer
$l = 14.5$ , $h = 5$ , $f_{1} = 16$ , $f_{0} = 10$ , $f_{2} = 12$ . $Mode = 14.5 + \frac{6}{10} \times 5 = 17.5$ years.
Q3 · 3 marks · Short answerHarder board level
The daily wages (in Rs) of 43 workers are given. The modal wage is Rs 152. Find the missing frequencies $x$ and $y$ .
Wage (Rs): 100–120, 120–140, 140–160, 160–180
Workers: 9, $x$ , 18, $y$
Hint
Use the total for one equation. Rs 152 lies in 140–160, so $f_{0} = x$ and $f_{2} = y$ in the mode formula.
Answer
Total: $9 + x + 18 + y = 43$ , so $x + y = 16$ . Mode: $152 = 140 + \frac{18 - x}{36 - x - y} \times 20$ ; here $36 - x - y = 20$ , so $12 = 18 - x$ , giving $x = 6$ and $y = 10$ . 18 is the highest frequency, so 140–160 is modal and Rs 152 checks out.
Q4 · 5 marks · Long answerHarder board level
The minutes of daily exercise of 50 residents of a colony are given below. Find the mean and the mode, and say which is larger and why the data suggests this.
Minutes: 0–10, 10–20, 20–30, 30–40, 40–50
Residents: 4, 12, 18, 14, 2
Hint
Use class marks 5, 15, 25, 35, 45 for the mean; modal class is 20–30.
Answer
$\sum f_{i} x_{i} = 20 + 180 + 450 + 490 + 90 = 1230$ , mean $= \frac{1230}{50} = 24.6$ min. Mode $= 20 + \frac{6}{36 - 12 - 14} \times 10 = 26$ min. Mode is larger: the few residents in 0–10 pull the mean down more than the 2 residents in 40–50 pull it up, while the mode leans towards the large class 30–40.
Q5 · 3 marks · Short answerHOTS / competency
For the data below a student found the mode as follows: "cf = 8, 20, 40, 58, 60. Highest is 60, so modal class is 40–50." Identify the error and find the correct mode.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 8, 12, 20, 18, 2
Hint
The last cf is always the total, so it can never decide the modal class.
Answer
Error: the modal class was chosen from cumulative frequencies. Use frequencies: highest is 20, so modal class is 20–30. $Mode = 20 + \frac{20 - 12}{40 - 12 - 18} \times 10 = 20 + 8 = 28$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a school canteen records how much 50 students spend (in Rs) during lunch break.
Spending (Rs): 0–20, 20–40, 40–60, 60–80, 80–100
Students: 6, 10, 16, 12, 6
(i) Name the modal class. (1 mark)
(ii) Find the modal spending. (1 mark)
(iii) The next day, 10 more students from a sports team buy lunch, and each spends between Rs 60 and Rs 80. Find the new modal class and the new mode, and explain why the mode moved. (2 marks)
Hint
Recount the frequencies after the team joins before choosing the modal class.
Answer
(i) 40–60 (highest frequency 16). (1)
(ii) $Mode = 40 + \frac{16 - 10}{32 - 10 - 12} \times 20 = 40 + 12 = 52$ rupees. (1)
(iii) The 60–80 class becomes $12 + 10 = 22$ , now the highest, so the modal class is 60–80 with $f_{1} = 22$ , $f_{0} = 16$ , $f_{2} = 6$ : $Mode = 60 + \frac{22 - 16}{44 - 16 - 6} \times 20 = 60 + \frac{120}{22} \approx 65.45$ rupees. The mode moved because the most crowded class changed. (2)
Q7 · 5 marks · Long answerHarder board level
The mean of the following distribution is 25.5. Find the missing frequency $f$ , and then find the mode.
Classes: 0–10, 10–20, 20–30, 30–40, 40–50
Frequencies: 4, 8, $f$ , 10, 4
Hint
Write $\sum f_{i}$ and $\sum f_{i} x_{i}$ in terms of $f$ using class marks 5, 15, 25, 35, 45, then use the mean.
Answer
$\sum f_{i} = 26 + f$ , $\sum f_{i} x_{i} = 20 + 120 + 25 f + 350 + 180 = 670 + 25 f$ . $25.5 (26 + f) = 670 + 25 f$ gives $663 + 25.5 f = 670 + 25 f$ , so $0.5 f = 7$ , $f = 14$ . Then 14 is the highest frequency, modal class 20–30: $Mode = 20 + \frac{14 - 8}{28 - 8 - 10} \times 10 = 20 + 6 = 26$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the modal class?Show answer
The class with the highest frequency.
2.State the mode formula for grouped data.Show answer
$Mode = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$
3.What is $f_{0}$ ?Show answer
Frequency of the class just before the modal class.
4.If $f_{0} = f_{2}$ , where is the mode?Show answer
At the class mark of the modal class.
5.Lower limit of class 21–30 for the mode formula?Show answer
$20.5$ , after making classes continuous.

Mode of Grouped Data: frequently asked questions

Can the mode lie outside the modal class?

No. If your answer is outside the modal class, you have used a wrong value or a wrong denominator. Recheck your substitution.

What if the first class is the modal class?

Then there is no class before it, so take the frequency before it as zero. The same applies to the last class with the frequency after it.

Will the board give data with two highest frequencies?

No. The syllabus says bimodal situations are to be avoided, so there will be one clear modal class.

Do I need to draw graphs to find the mode?

No. You find the mode by the formula only. Graphs such as ogives are not in the 2026-27 syllabus.