Question
The following table shows the marks obtained by 30 students in a mathematics test. Find the mode of the data.
| Marks | Number of Students |
|---|---|
| 10 – 25 | 2 |
| 25 – 40 | 3 |
| 40 – 55 | 7 |
| 55 – 70 | 6 |
| 70 – 85 | 6 |
| 85 – 100 | 6 |
Solution — Step by Step
The modal class is the class interval with the highest frequency. Scanning the table, the 40–55 interval has frequency 7 — the largest.
So our modal class is 40–55.
From the modal class and its neighbours, we pull four values:
l= lower boundary of modal class = 40f₁= frequency of modal class = 7f₀= frequency of class before modal class = 3 (the 25–40 class)f₂= frequency of class after modal class = 6 (the 55–70 class)h= class width = 15
Why do we need the neighbours? The formula adjusts for how “dominant” the modal class is relative to the ones beside it. A modal class that barely edges out its neighbours sits closer to the boundary; one that towers over them pulls the mode toward its centre.
Substituting:
Numerator: 7 - 3 = 4
Denominator: 14 - 3 - 6 = 5
The mode of the marks is 52 marks.
Why This Works
In ungrouped data, mode is just the value that appears most. With grouped data, we don’t know the exact values inside a class — we only know the class frequencies. The formula estimates where the mode lies within the modal class by looking at the “pull” from neighbouring classes.
The denominator 2f₁ - f₀ - f₂ measures how much the modal class rises above both neighbours combined. The numerator f₁ - f₀ measures the rise from the left neighbour specifically. Their ratio tells us what fraction of the class width to add to l.
If f₀ and f₂ were both equal to f₁, the denominator would be zero — meaning the modal class doesn’t stand out at all and the formula breaks down. This edge case never appears in board exam problems, but it’s worth understanding why the formula has that structure.
Alternative Method — Spotting the Answer Quickly
In a time-crunch situation (like a board exam), we can estimate before calculating.
The modal class is 40–55, so the mode must lie between 40 and 55. The class before (frequency 3) is much smaller than the class after (frequency 6). This means the modal class is being “pulled right” — toward the higher boundary.
Adding 12 to 40 gives 52, which sits closer to 55 than to 40 — consistent with our expectation. This sanity check takes 10 seconds and catches calculation errors before they cost you marks.
In CBSE Class 10 boards, mode questions always use equal class widths. If you ever see unequal widths in practice material, the question is either wrong or beyond your syllabus. Don’t waste time trying to adapt the formula.
Common Mistake
The most common error is mixing up f₀ and f₂ — students swap the frequencies of the preceding and succeeding classes. This gives a wrong numerator and a wrong final answer, but the calculation looks correct so you won’t catch it by checking your arithmetic alone.
Always label clearly: f₀ is the class immediately before the modal class, f₂ is the class immediately after. Write these down explicitly before substituting. In this question, f₀ = 3 (25–40 class) and f₂ = 6 (55–70 class) — not the other way around.