MathsICSE Class 10 · Histograms and Ogive

Histogram and Mode: Drawing the Graph and Reading the Mode (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

Draw adjacent bars on the class boundaries, with heights equal to the frequencies. In the tallest bar, join each top corner to the top corner of the neighbouring bar on the opposite side; the lines cross. Drop a perpendicular from the crossing point to the x-axis: that reading is the mode.

Cross the two lines from the tallest bar's top corners to its neighbours' corners; the perpendicular from the crossing point gives the mode, not the bar's middl

Illustration drawn for QuestionBuddy.

Understand Histogram and Mode

The CISCE 2027 syllabus asks for histograms and for finding the mode from the histogram. It is a regular 3- to 5-mark graph question (2024 Q5(i), 2025 Q7(i)).

Drawing the histogram (equal class widths).

Take class boundaries on the x-axis and frequency on the y-axis, using the scale given in the question.
Mark the boundaries at equal spacing; label both axes.
Draw adjacent rectangles (no gaps): each bar's width is the class width, its height the frequency.
If the x-axis does not start at 0, draw a kink (a small zig-zag) between the origin and the first boundary.

Discontinuous classes. For 11–15, 16–20, …, there is a gap of 1 between classes, so subtract 0.5 from each lower limit and add 0.5 to each upper limit: 10.5–15.5, 15.5–20.5, … Draw the bars on these boundaries.

The mode from the histogram. The mode is not simply the middle of the tallest bar; it is pulled towards the taller neighbour.

1. Find the tallest bar (the modal class).
2. Join the top-right corner of the tallest bar to the top-right corner of the bar before it.
3. Join the top-left corner of the tallest bar to the top-left corner of the bar after it.
4. From the point where the two lines cross, draw a perpendicular to the x-axis and read the mode.

Why it works, and a check. The two lines form similar triangles whose heights are $f_{1} - f_{0}$ and $f_{1} - f_{2}$ , where $f_{1}$ is the tallest frequency and $f_{0}$ , $f_{2}$ its neighbours. So the crossing point divides the class width $h$ in the ratio $(f_{1} - f_{0}) : (f_{1} - f_{2})$ , and

mode $= L + \frac{f _{1} - f _{0}}{( f _{1} - f _{0} ) + ( f _{1} - f _{2} )} \times h$ , with $L$ the lower boundary of the modal class.

Use this only to check your graph reading; the question expects the graph construction. If $f_{0} = f_{2}$ , the mode is the class mark.

Reading a printed histogram. Read each bar's height on the y-axis scale and write the classes in full (for example 60–70 cm), with their frequencies.

Histogram and Mode: formulas to remember

Class boundaries (discontinuous): $lower - 0.5, upper + 0.5$; When consecutive classes differ by 1.
Mode check (crossing lines): $mode = L + \frac{f _{1} - f _{0}}{( f _{1} - f _{0} ) + ( f _{1} - f _{2} )} \times h$; $f_{1}$ tallest, $f_{0}$ before, $f_{2}$ after.
Equal neighbours: $f_{0} = f_{2} \Rightarrow mode = class mark$; The lines cross over the mid-point.
Bar height: $height = frequency$; Equal class widths only.

How to solve Histogram and Mode questions

A neat graph earns most of the marks: scale, kink, boundaries, bars, crossing lines, perpendicular, value.

1Convert discontinuous classes to boundaries if needed.
2Draw and label the axes with the given scale; draw a kink if the x-axis does not start at 0.
3Mark the boundaries at equal spacing and draw adjacent bars with heights equal to the frequencies.
4In the tallest bar, join top-right to the previous bar's top-right, and top-left to the next bar's top-left.
5Drop a perpendicular from the crossing point to the x-axis and read the mode.
6Check with the formula; your graph reading should agree to within a small amount.

Histogram and Mode 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.

Typical workout time at a gym

Suppose a gym records how long 60 members work out: 20–30 minutes (5), 30–40 (12), 40–50 (20), 50–60 (15), 60–70 (8).

Estimate the modal workout time from a histogram, to 1 decimal place.

Try it first, then show the solutionSolution

Tallest bar: 40–50, with neighbours 12 and 15.
Mode $= 40 + \frac{20 - 12}{( 20 - 12 ) + ( 20 - 15 )} \times 10 = 40 + \frac{8}{13} \times 10 \approx 46.2$ minutes.

Answer: About 46.2 minutes

Memory hook: The mode leans towards the taller neighbour (15 > 12).

Most common delivery time

Imagine a delivery app plots the times of 60 orders: 10–15 minutes (6), 15–20 (18), 20–25 (24), 25–30 (10), 30–35 (2).

What delivery time does the histogram show as most common?

Try it first, then show the solutionSolution

Tallest bar 20–25; neighbours 18 and 10.
Mode $= 20 + \frac{6}{6 + 14} \times 5 = 21.5$ minutes.

Answer: 21.5 minutes

Memory hook: A tall bar before the tallest pulls the mode backwards.

Tricks to remember Histogram and Mode

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

Make an X in the tallest bar
Right corner to the left neighbour's right corner, left corner to the right neighbour's left corner: the lines cross like an X.
Use it when: Drawing the mode lines the right way round.
Boundaries, not limits
Bars stand on boundaries; for 11–15, the bar runs from 10.5 to 15.5.
Use it when: Discontinuous classes.
Zig-zag when not zero
If the x-axis starts at 30, put a kink near the origin.
Use it when: Avoiding the scale deduction.

Histogram and Mode: solved examples

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

Example 1: Where the first bar stands · 1 mark · MCQBoard level

For the classes 1–10, 11–20, 21–30, the first bar of a histogram stands on:
(a) 1 to 10
(b) 0.5 to 10.5
(c) 0 to 10
(d) 1.5 to 9.5

Try it first, then show the solutionSolution

The gap between 10 and 11 is 1; take half of it on each side: 0.5 to 10.5.

Answer: (b)

Example 2: Histogram and mode · 3 marks · Short answerBoard level

Draw a histogram and use it to estimate the mode. (Take 2 cm = 10 units on the x-axis and 2 cm = 5 on the y-axis.)

Class	10–20	20–30	30–40	40–50	50–60	60–70
Frequency	6	11	19	14	8	4

Try it first, then show the solutionSolution

x-axis from 10 to 70 with a kink near the origin; bars of heights 6, 11, 19, 14, 8, 4 standing side by side.
Tallest bar 30–40. Join $(40, 19)$ to $(30, 11)$ and $(30, 19)$ to $(40, 14)$ .
The perpendicular from the crossing point meets the x-axis at about 36.
Check: $30 + \frac{8}{8 + 5} \times 10 \approx 36.2$ .

Answer: Mode ≈ 36

Example 3: Reading a printed histogram · 3 marks · Short answerHarder board level

A histogram of the daily rainfall (mm) over some days has adjacent bars on 0–4, 4–8, 8–12, 12–16 and 16–20 with heights 3, 7, 12, 9 and 5.
(a) Write the frequency table.
(b) Find the number of days.
(c) On how many days was the rainfall at least 4 mm but less than 12 mm?
(d) State the modal class.

Try it first, then show the solutionSolution

(a) 0–4: 3, 4–8: 7, 8–12: 12, 12–16: 9, 16–20: 5.
(b) $3 + 7 + 12 + 9 + 5 = 36$ days.
(c) $7 + 12 = 19$ days.
(d) Tallest bar: 8–12 mm.

Answer: (b) 36 (c) 19 (d) 8–12 mm

Example 4: Discontinuous classes · 4 marks · Short answerHarder board level

Draw a histogram and find the mode, to 1 decimal place.

Score	11–15	16–20	21–25	26–30	31–35
Players	5	12	18	10	4

Try it first, then show the solutionSolution

Boundaries: 10.5, 15.5, 20.5, 25.5, 30.5, 35.5; draw a kink before 10.5.
Tallest bar 20.5–25.5 (frequency 18), neighbours 12 and 10.
Crossing lines in this bar; the perpendicular meets the x-axis at about 22.6.
Check: $20.5 + \frac{6}{6 + 8} \times 5 \approx 22.6$ .

Answer: Mode ≈ 22.6

Example 5: Mode leaning towards a neighbour · 3 marks · Short answerHOTS / competency

For the distribution below, the modal class is 120–130.

Class	100–110	110–120	120–130	130–140	140–150
Frequency	7	15	21	18	9

(a) Without drawing, say whether the mode is above or below 125, and why.
(b) Find the mode to 1 decimal place.

Try it first, then show the solutionSolution

(a) The bar after the modal bar (18) is taller than the one before (15), so the crossing point is pulled to the right: the mode is above 125.
(b) Mode $= 120 + \frac{21 - 15}{( 21 - 15 ) + ( 21 - 18 )} \times 10 = 120 + \frac{6}{9} \times 10 \approx 126.7$ .

Answer: (a) Above 125 (b) ≈ 126.7

Example 6: Case study: a missing frequency · 4 marks · Case-basedHOTS / competency

A histogram is drawn for the table below, and the mode read from it is 22.

Class	0–10	10–20	20–30	30–40	40–50
Frequency	5	$x$	20	12	6

(a) Given that 20–30 is the modal class, write an equation for $x$ using the crossing-lines ratio.
(b) Solve it.
(c) Explain, using the bars, why the mode is below the class mark 25.

Try it first, then show the solutionSolution

(a) $22 = 20 + \frac{20 - x}{( 20 - x ) + ( 20 - 12 )} \times 10$ , so $\frac{20 - x}{28 - x} = 0.2$ .
(b) $20 - x = 5.6 - 0.2 x$ , so $0.8 x = 14.4$ and $x = 18$ .
(c) The bar before (18) is taller than the bar after (12), so the crossing lines meet left of the centre of the 20–30 bar.

Answer: $x = 18$

Where students lose marks in Histogram and Mode

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

Mistake: No kink, wrong scale, uneven spacing
The graph must use the given scale with equally spaced boundaries.
Wrong working
x-axis starting at 30 with no zig-zag; class intervals written under bars
Right working
Kink near the origin; boundaries 30, 40, 50, … at equal spacing
Do this: Write the scale on the graph and check every spacing with a ruler.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q7(i).
Mistake: Wrong lines for the mode
The lines must cross: right corner to the previous bar, left corner to the next bar.
Wrong working
Joining top-left of the tallest bar to top-left of the previous bar
Right working
Top-right of the tallest to top-right of the previous; top-left of the tallest to top-left of the next
Do this: Check that the two lines form an X inside the tallest bar.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q7(i): the mode mark.
Mistake: Mode and modal class mixed up
The modal class is an interval; the mode is a single value.
Wrong working
Modal class: 64
Right working
Modal class: 60–70; mode ≈ 64
Do this: Read the question word: 'class' wants an interval.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q5(i).
Mistake: Misreading bar heights
Each small square has a value set by the scale.
Wrong working
Reading 2.5 for a bar of height 3
Right working
Work out the value of one small division first
Do this: Write 'one small square = …' beside the axis before reading.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q5(i).

Exam tips

Asked in 2025 Q7(i) (3 marks): histogram of marks 30–80 for 164 students, scale given; the expected mode was about 64 ± 1 from the modal class 60–70.
Asked in 2024 Q5(i) (5 marks): a printed histogram of plant heights; candidates had to write the frequency table, state the modal class, read the mode and count plants in one class.
The CISCE examiners suggest cross-checking the mode by estimation: it lies near the tallest bar (and, by the crossing lines, leans towards the taller neighbour).
Write 'Mode ≈ …' beside the perpendicular on the graph as well as in your answer.

Practice questions on Histogram and Mode

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

Q1 · 1 mark · MCQBoard level
The x-axis of a histogram starts at 40 instead of 0. On the graph you must show:
(a) a kink near the origin
(b) an arrow on the y-axis
(c) a gap before the first bar
(d) bars of width 40
Hint
The axis does not start at zero.
Answer
(a)
Q2 · 3 marks · Short answerBoard level
Draw a histogram and estimate the mode. (Take 2 cm = 10 marks and 2 cm = 5 students.)
Marks 20–30 30–40 40–50 50–60 60–70 70–80
Students 8 14 22 30 18 6
Hint
Tallest bar 50–60; neighbours 22 and 18.
Answer
Mode ≈ 54 (check: $50 + \frac{8}{8 + 12} \times 10 = 54$ ).
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The mode found from a histogram always lies within the modal class.
Reason (R): The two lines used to find it cross inside the tallest bar.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Where does the perpendicular land?
Answer
(c)
Q4 · 4 marks · Short answerHarder board level
Draw a histogram and find the mode, to 1 decimal place.
Class 1–10 11–20 21–30 31–40 41–50
Frequency 4 9 16 11 5
Hint
Bars on 0.5–10.5, 10.5–20.5, ….
Answer
Modal bar 20.5–30.5; mode $\approx 20.5 + \frac{7}{7 + 5} \times 10 \approx 26.3$ .
Q5 · 4 marks · Short answerHarder board level
A histogram of the heights (cm) of a class has adjacent bars on 145–150, 150–155, 155–160, 160–165 and 165–170 with heights 4, 10, 15, 8 and 3.
(a) How many students are there?
(b) What percentage are at least 160 cm tall?
(c) Estimate the mode to 1 decimal place.
Hint
At least 160 means the last two bars.
Answer
(a) 40 (b) $\frac{11}{40} \times 100 = 27.5%$ (c) $155 + \frac{5}{5 + 7} \times 5 \approx 157.1$ cm
Q6 · 2 marks · Short answerHarder board level
In a histogram, the tallest bar stands on 30–40 with height 14, and the bars on either side both have height 6. Show that the mode is 35.
Hint
Equal neighbours make the X symmetric.
Answer
$30 + \frac{8}{8 + 8} \times 10 = 35$ : the lines cross over the mid-point of the bar, the class mark.
Q7 · 3 marks · Short answerHOTS / competency
For the classes 0–20, 20–40, 40–60, 60–80, 80–100 the frequencies are 4, 10, 16, $y$ , 6, and the mode read from the histogram is 52. Find $y$ .
Hint
Modal class 40–60.
Answer
$52 = 40 + \frac{6}{6 + ( 16 - y )} \times 20$ , so $\frac{6}{22 - y} = 0.6$ , $22 - y = 10$ and $y = 12$ .

Marks	20–30	30–40	40–50	50–60	60–70	70–80
Students	8	14	22	30	18	6

Class	1–10	11–20	21–30	31–40	41–50
Frequency	4	9	16	11	5

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 goes on the x-axis of a histogram?Show answer
Class boundaries, equally spaced.
2.Boundaries of 16–20?Show answer
15.5–20.5
3.Which corners are joined to find the mode?Show answer
Tallest bar's top-right to previous bar's top-right; tallest bar's top-left to next bar's top-left.
4.When is the mode exactly the class mark?Show answer
When the two neighbouring bars are equal.
5.What is a kink for?Show answer
To show the axis does not start at 0.

Histogram and Mode: frequently asked questions

How do I find the mode from a histogram?

In the tallest bar, join its top-right corner to the top-right corner of the bar before it and its top-left corner to the top-left corner of the bar after it. Drop a perpendicular from where the lines cross; the x-axis reading is the mode.

How do I draw a histogram for classes like 11–20, 21–30?

Convert them to boundaries by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit (10.5–20.5, 20.5–30.5), then draw adjacent bars on these boundaries.

Why is there a gap or zig-zag near the origin of a histogram?

The zig-zag, called a kink, shows that the x-axis does not start at zero. Bars themselves must be adjacent with no gaps between them.