This section provides detailed solutions for ICSE Class 10 Maths Long Answer Questions, specifically the second set from the ICSE Maths Solved Competency Focused Questions textbook. These 5-mark questions are designed to challenge your understanding by integrating concepts from various topics like polynomials, arithmetic progressions, and commercial mathematics. Unlike straightforward problems, these require you to apply multiple steps and demonstrate a clear, logical thought process. Mastering these types of questions is crucial for scoring high marks in your board exams, as they test your ability to analyse complex scenarios and present a well-structured solution. Our step-by-step guides will help you break down each problem and understand the method required to solve it effectively.
If you’re stuck on a challenging multi-part problem or want to confirm the correct method for a 5-mark question, you’ve come to the right place. This page contains solutions for all 10 questions from the Long Answer Questions 2 chapter. We understand that marks are awarded for showing the correct steps, so our solutions follow the exact methodology and presentation style expected by the ICSE board. Here, you will find clear, expertly prepared solutions to help you verify your work and prepare thoroughly for your examinations.
Long Answer Questions 2 (5 Marks Each)
Question 110
On seeing the below display board outside Pearl Stationary Shop, Chetan enters the shop to buy the following items:
| Pen | Pencil | Rainbow cover notebook | |
|---|---|---|---|
| Price | ₹5 each | ₹7 each | ₹ 200 each |
| Discount | 5% on a dozen pens | 10% on 20 pencils | —— |
| Premium | – | – | ₹50 on each notebook |
| Items purchased | 1 dozen | 20 pencils | 5 |
| GST | 18% | 12% | 12% |
The shopkeeper handed over the bill to Chetan saying that he has given further discount of 2% on total bill. Chetan became so happy hearing about the discount that he did not check the bill until he reached home. He later found out that though shopkeeper has given 2% discount as promised, he had also mischarged uniform 18% GST on all the items.
(a) Calculate :
(i) total selling price of all the items as per the offers displayed on the board.
(ii) total amount to be paid by Chetan including GST with correct rates.
(iii) actual amount charged by the shopkeeper.
(b) Did the shopkeeper overcharge Chetan? Justify your answer
Here’s how we break down Chetan’s bill item by item.
For pen :
No. of pens purchased = 12
Price of 12 pens = ₹5 × 12 = ₹60
Discount (D) : 5% on dozen pens.
D = \dfrac{5}{100} \times 60 = 3.Price after discount = ₹ 60 – ₹ 3 = ₹ 57.
GST = 18% = \dfrac{18}{100} \times 57 = ₹ 10.26
Total price : ₹ 57 + ₹ 10.26 = ₹ 67.26
For pencil :
No. of pencils purchased = 20
Price of 20 pencils = ₹ 7 × 20 = ₹ 140
Discount (D) : 10% on 20 pencils.
D = \dfrac{10}{100} \times 140 = ₹ 14.Price after discount = ₹ 140 – ₹ 14 = ₹ 126.
GST = 12% = \dfrac{12}{100} \times 126 = ₹ 15.12
Total price : ₹ 126 + ₹ 15.12 = ₹ 141.12
For notebook :
No. of notebooks purchased = 5
Price of 5 notebooks = ₹ 200 × 5 = ₹ 1000
Premium (P) : ₹ 50 on each notebook
Total premium : ₹ 50 × 5 = ₹ 250.
Price after premium = ₹ 1000 + ₹ 250 = ₹ 1250.
GST = 12% = \dfrac{12}{100} \times 1250 = ₹ 150.
Total price : ₹ 1250 + ₹ 150 = ₹ 1400.
(i) Total selling price of all items = ₹ 57 + ₹ 126 + ₹ 1250 = ₹ 1433.
Hence, total selling price = ₹ 1433.
(ii) Total price paid (including GST) = ₹ 67.26 + ₹ 141.12 + ₹ 1400 = ₹ 1608.38
Hence, total price paid (including GST) = ₹ 1608.38
(iii) On applying incorrect GST rates :
On pen :
Price after discount = ₹ 57.
GST = 18% = \dfrac{18}{100} \times 57 = ₹ 10.26
Total price : ₹ 57 + ₹ 10.26 = ₹ 67.26
On pencil :
Price after discount = ₹ 126.
GST = 18% = \dfrac{18}{100} \times 126 = ₹ 22.68
Total price : ₹ 126 + ₹ 22.68 = ₹ 148.68
For notebook :
Price after premium = ₹ 1250.
GST = 18% = \dfrac{18}{100} \times 1250 = ₹ 225.
Total price : ₹ 1250 + ₹ 225 = ₹ 1475.
When the GST rate is applied as 18% uniformly, bill need to be paid by Chetan :
= ₹ 1475 + ₹ 148.68 + ₹ 67.26
= ₹ 1690.94
Discount given on total bill = 2%
\dfrac{2}{100} \times 1690.94 = 33.82
Bill = ₹ 1690.94 – ₹ 33.82 = ₹ 1657.12
Hence, bill = ₹ 1657.12
(b) The bill should be of ₹ 1608.38 but instead it was of ₹ 1657.12
⇒ ₹ 1657.12 – ₹ 1608.38 = ₹ 48.78
Hence, the shopkeeper overcharged an amount of ₹ 48.78
Question 111
Using remainder and factor theorem, show that (2x + 3) is a factor of the polynomial 2x^2 + 11x + 12. Hence, factorise it completely. What must be multiplied to the given polynomial so that x^2 + 3x – 4 is a factor of the resulting polynomial? Also, write the resulting polynomial.
To apply the factor theorem, first set the given linear factor equal to zero:
2x + 3 = 0
⇒ 2x = -3
⇒ x = -\dfrac{3}{2}
Now plug this value of x into the polynomial 2x² + 11x + 12:
\begin{aligned}\Rightarrow 2 \times \Big(-\dfrac{3}{2}\Big)^2 + 11 \times \Big(-\dfrac{3}{2}\Big) + 12 \\ \Rightarrow 2 \times \dfrac{9}{4} - \dfrac{33}{2} + 12 \\ \Rightarrow \dfrac{9}{2} - \dfrac{33}{2} + 12 \\ \Rightarrow \dfrac{9 - 33 + 24}{2} \\ \Rightarrow \dfrac{0}{2} \\ \Rightarrow 0.\end{aligned}Since the remainder comes out to be 0.
∴ 2x + 3 is indeed a factor of the polynomial 2x² + 11x + 12.
Next, factorise 2x² + 11x + 12 completely:
⇒ 2x² + 8x + 3x + 12
⇒ 2x(x + 4) + 3(x + 4)
⇒ (2x + 3)(x + 4).
Now factorise the divisor polynomial x² + 3x − 4:
⇒ x² + 4x − x − 4
⇒ x(x + 4) − 1(x + 4)
⇒ (x − 1)(x + 4).
∴ (x − 1) and (x + 4) are the factors of x² + 3x − 4.
∴ Multiplying 2x² + 11x + 12 by (x − 1) makes it divisible by x² + 3x − 4.
⇒ (2x² + 11x + 12)(x − 1)
⇒ 2x³ − 2x² + 11x² − 11x + 12x − 12
⇒ 2x³ + 9x² + x − 12.
Hence, the resulting polynomial = 2x³ + 9x² + x − 12.
Question 112
The sequence 2, 9, 16, ….. is given.
(a) Identify if the given sequence is an AP or a GP. Give reasons to support your answer.
(b) Find the 20th term of the sequence.
(c) Find the difference between the sum of its first 22 and 25 terms.
(d) Is the term 102 belong to this sequence?
(e) If ‘k’ is added to each of the above terms, will the new sequence be in A.P. or G.P.?
(a) Calculate the gap between successive terms : 16 − 9 = 9 − 2 = 7.
Since every consecutive pair has the same difference.
Hence, the given sequence is an A.P.
(b) Using the standard formula,
aₙ = a + (n − 1)d
a₂₀ = 2 + (20 − 1) × 7
= 2 + 19 × 7
= 2 + 133
= 135.
Hence, 20th term of the sequence = 135.
(c) Using the sum formula,
Sₙ = \dfrac{n}{2}[2a + (n - 1)d]
Plugging in the known values :
Sum upto 22 terms :
\begin{aligned}S_{22} = \dfrac{22}{2}[2 \times 2 + (22 - 1) \times 7] \\ = 11 \times [4 + 21 \times 7] \\ = 11 \times [4 + 147] \\ = 11 \times 151 \\ = 1661.\end{aligned}Sum upto 25 terms :
\begin{aligned}S_{25} = \dfrac{25}{2}[2 \times 2 + (25 - 1) \times 7] \\ = \dfrac{25}{2} \times [4 + 24 \times 7] \\ = \dfrac{25}{2} \times [4 + 168] \\ = \dfrac{25}{2} \times 172 \\ = 25 \times 86 \\ = 2150.\end{aligned}S₂₅ − S₂₂ = 2150 − 1661 = 489.
Hence, difference between the sum of its first 22 and 25 terms = 489.
(d) Suppose the n-th term equals 102.
⇒ aₙ = a + (n − 1)d
⇒ 102 = 2 + 7(n − 1)
⇒ 102 = 2 + 7n − 7
⇒ 102 = 7n − 5
⇒ 102 + 5 = 7n
⇒ 7n = 107
⇒ n = \dfrac{107}{7} = 15\dfrac{2}{7}.
Since n must be a whole number, not a fraction.
Hence, 102 is not the term of the sequence.
(e) If a constant k is added to every term.
New sequence : 2 + k, 9 + k, 16 + k, …
The common difference still works out to 7.
Hence, sequence is in A.P.
Question 113
Given the equations of two straight lines, L1 and L2 are x – y = 1 and x + y = 5 respectively. If L1 and L2 intersects at point Q (3, 2). Find :
(a) the equation of line L3 which is parallel to L1 and has y-intercept 3.
(b) the value of k, if the line L3 meets the line L2 at a point P (k, 4).
(c) the coordinate of R and the ratio PQ : QR, if line L2 meets x-axis at point R.
(a) Rewrite L1 in slope-intercept form:
L1 : x − y = 1
⇒ y = x − 1
Comparing with y = mx + c, we extract:
⇒ m = 1.
Recall that parallel lines share identical slopes.
∴ Slope of L3 = 1.
We are also told,
L3 has y-intercept = 3.
∴ y = mx + c
⇒ y = 1·x + 3
⇒ y = x + 3.
Hence, equation of line L3 : y = x + 3.
(b) Solve the pair L2 : x + y = 5 and L3 : y = x + 3
⇒ x + y = 5 ………(1)
⇒ y = x + 3 ………(2)
Insert the expression for y from (2) into (1):
⇒ x + (x + 3) = 5
⇒ 2x + 3 = 5
⇒ 2x = 5 − 3
⇒ 2x = 2
⇒ x = \dfrac{2}{2}
⇒ x = 1.
Put this x back into equation (2):
⇒ y = 1 + 3 = 4.
⇒ (x, y) = (1, 4)
∴ P(k, 4) = (1, 4)
Hence, value of k = 1.
(c) We need the spot where L2 cuts the x-axis, call it R.
On the x-axis, y-coordinate = 0.
L2 : x + y = 5
⇒ x + 0 = 5
⇒ x = 5
⇒ R = (x, y) = (5, 0).
Given P = (1, 4), Q = (3, 2) and R = (5, 0)
Assume PQ : QR = k : 1
Apply the section formula,
\begin{aligned}\Rightarrow (x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\ \Rightarrow (3, 2) = \Big(\dfrac{k \times 5 + 1 \times 1}{k + 1}, \dfrac{k \times 0 + 1 \times 4}{k + 1}\Big) \\ \Rightarrow (3, 2) = \Big(\dfrac{5k + 1}{k + 1}, \dfrac{4}{k + 1}\Big) \\ \Rightarrow \dfrac{5k + 1}{k + 1} = 3 \text{ and } 2 = \dfrac{4}{k + 1} \\ \Rightarrow 5k + 1 = 3(k + 1) \text{ and } 4 = 2(k + 1) \\ \Rightarrow 5k + 1 = 3k + 3 \text{ and } 4 = 2k + 2 \\ \Rightarrow 5k - 3k = 3 - 1 \text{ and } 2k = 4 - 2 \\ \Rightarrow 2k = 2 \text{ and } 2k = 2 \\ \Rightarrow k = \dfrac{2}{2} = 1.\end{aligned}∴ PQ : QR = 1 : 1.
Hence, R = (5, 0) and PQ : QR = 1 : 1.
Question 114
In the figure given below (not drawn to scale), AD ∥ GE ∥ BC, DE = 18 cm, EC = 3 cm, AD = 35 cm. Find :
(a) AF : FC
(b) length of EF
(c) area(trapezium ADEF) : area(Δ EFC)
(d) BC ∶ GF
(a) In △ ACD and △ FCE,
⇒ ∠ACD = ∠FCE (Common angles)
⇒ ∠ADC = ∠FEC (Corresponding angles are equal)
∴ △ ACD ~ △ FCE (By A.A. axiom)
Recall the property,
Corresponding sides of similar triangles stay in proportion.
\begin{aligned}\therefore \dfrac{FC}{AC} = \dfrac{EC}{CD} \\ \Rightarrow \dfrac{FC}{AC} = \dfrac{EC}{EC + ED} \\ \Rightarrow \dfrac{FC}{AC} = \dfrac{3}{3 + 18} \\ \Rightarrow \dfrac{FC}{AC} = \dfrac{3}{21} \\ \Rightarrow \dfrac{FC}{AC} = \dfrac{1}{7}.\end{aligned}Let FC = x and AC = 7x.
From the diagram,
AF = AC − FC = 7x − x = 6x.
\therefore \dfrac{AF}{FC} = \dfrac{6x}{x} = \dfrac{6}{1} = 6 : 1.
Hence, AF : FC = 6 : 1.
(b) Because △ ACD ~ △ FCE
\begin{aligned}\therefore \dfrac{EF}{AD} = \dfrac{EC}{CD} \\ \Rightarrow \dfrac{EF}{35} = \dfrac{3}{21} \\ \Rightarrow EF = \dfrac{3}{21} \times 35 \\ \Rightarrow EF = \dfrac{105}{21} = 5 \text{ cm}.\end{aligned}Hence, EF = 5 cm.
(c) Recall the theorem,
The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.
\begin{aligned}\therefore \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \Big(\dfrac{DC}{EC}\Big)^2 \\ \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \Big(\dfrac{21}{3}\Big)^2 \\ \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \dfrac{441}{9} \\ \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \dfrac{49}{1}.\end{aligned}Let area of △ ADC = 49a and area of △ FCE = a.
Area of trapezium ADEF = Area of △ ADC − Area of △ FCE = 49a − a = 48a.
\therefore \dfrac{\text{Area of trapezium ADEF}}{\text{Area of △ FCE}} = \dfrac{48a}{a} = \dfrac{48}{1}.Hence, area of trapezium ADEF : area of △ EFC = 48 : 1.
(d) In △ AGF and △ ABC,
⇒ ∠AGF = ∠ABC (Corresponding angles are equal)
⇒ ∠GAF = ∠BAC (Common angles)
∴ △ AGF ~ △ ABC (By A.A. axiom)
Again, using proportionality of corresponding sides,
\therefore \dfrac{AC}{AF} = \dfrac{BC}{GF}From part (a),
⇒ FC = x and AC = 7x
⇒ AF = AC − FC = 7x − x = 6x.
\Rightarrow \dfrac{BC}{GF} = \dfrac{7x}{6x} = \dfrac{7}{6}.Hence, BC : GF = 7 : 6.
Question 115
(a) Construct the locus of a moving point which moves such that it keeps a fixed distance of 4.5 cm from a fixed-point O.
(b) Draw line segment AB of 6 cm where A and B are two points on the locus (a).
(c) Construct the locus of all points equidistant from A and B. Name the points of intersection of the loci (a) and (c) as P and Q respectively.
(d) Join PA. Find the locus of all points equidistant from AP and AB.
(e) Mark the point of intersection of the locus (a) and (d) as R. Measure and write down the length of AR.
(Use a ruler and a compass for this question.)
Remember this key idea,
Locus of a point at a fixed distance from a fixed point is the circumference of the circle with fixed point as center and fixed distance as radius.
Steps of construction :
- Mark point O.
- With center O and radius = 4.5 cm draw a circle.
- Use a ruler to mark point A.
- From point A, measure 6 cm along a straight line and mark point B.
- Draw a straight line connecting points A and B.
We know that, locus of points equidistant from two points is the perpendicular bisector of the line joining the two points. - Draw perpendicular bisector of AB.
- Mark the points P and Q where the perpendicular bisector intersects circle.
We know that, locus of points equidistant from two lines is the angle bisector of the angle between the lines. - Join PA.
- Draw AX, the angle bisector of ∠A.
- Mark point R as the intersection point of AX on the circumference of the circle.
- Measure AR.
Hence, AR = 4.8 cm.
Question 116
Construct a regular hexagon ABCDEF of side 4.3 cm and construct its circumscribed circle. Also, construct tangents to the circumscribed circle at points B and C which meets each other at point P. Measure and record ∠BPC.
(Use a ruler and a compass for this question.)
Keep in mind each interior angle of a regular hexagon = 120°.
- Draw a line segment AB = 4.3 cm.
- At A and B draw rays making an angle of 120° each and cut off AF = BC = 4.3 cm.
- At F and C, draw rays making angle of 120° each and cut off EF = CD = 4.3 cm.
- Join ED. Hence, ABCDEF is the required hexagon.
- Draw the perpendicular bisector of AB and AF. Let these bisectors meet at the point O.
- With O as center and radius equal to OA or OB draw a circle which passes through the vertices of the hexagon. This is the required circumcircle of hexagon ABCDEF.
- Draw the radius OB and OC.
- At point B, construct a line perpendicular to OB. This line is the tangent at B.
- At point C, construct a line perpendicular to OC. This line is the tangent at C.
- The two tangents will intersect at point P.
- Measure ∠BPC.
Hence, ∠BPC = 120°.
Question 117
A mathematics teacher uses certain amount of terracotta clay to form different shaped solids. First, she turned it into a sphere of radius 7 cm and then she made a right circular cone with base radius 14 cm. Find the height of the cone so formed. If the same clay is turned to make a right circular cylinder of height \dfrac{7}{3} cm, then find the radius of the cylinder so formed. Also, compare the total surface areas of sphere and cylinder so formed.
Start with a sphere of radius (r) 7 cm.
Volume of sphere = \dfrac{4}{3}πr^3
\begin{aligned}= \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \\ = \dfrac{4}{3} \times 22 \times 7^2 \\ = \dfrac{4312}{3} \text{ cm}^3.\end{aligned}Next, the same clay is reshaped into a right circular cone with base radius (r₁) 14 cm. Let its height be h cm.
Because the clay volume stays unchanged,
∴ Volume of cone = Volume of sphere
\begin{aligned}\Rightarrow \dfrac{1}{3}πr_1^2h = \dfrac{4312}{3} \\ \Rightarrow πr_1^2h = 4312 \\ \Rightarrow \dfrac{22}{7} \times 14^2 \times h = 4312 \\ \Rightarrow 22 \times 2 \times 14 \times h = 4312 \\ \Rightarrow h = \dfrac{4312}{22 \times 2 \times 14} \\ \Rightarrow h = \dfrac{4312}{616} = 7 \text{ cm}.\end{aligned}We are also told,
The same clay is turned into a right circular cylinder of height (h₁) \dfrac{7}{3} cm. Let its radius be r₂.
Again, equating volumes,
∴ Volume of cylinder = Volume of sphere
\begin{aligned}\Rightarrow πr_2^2h_1 = \dfrac{4312}{3} \\ \Rightarrow \dfrac{22}{7} \times r_2^2 \times \dfrac{7}{3} = \dfrac{4312}{3} \\ \Rightarrow 22 \times r_2^2 = 4312 \\ \Rightarrow r_2^2 = \dfrac{4312}{22} \\ \Rightarrow r_2^2 = 196 \\ \Rightarrow r_2 = \sqrt{196} = 14 \text{ cm}.\end{aligned}Total surface area of sphere = 4πr²
Total surface area of cylinder = 2πr₂(r₂ + h₁)
\begin{aligned}\Rightarrow \dfrac{\text{TSA sphere}}{\text{TSA cylinder}} = \dfrac{4πr^2}{2πr_2(r_2 + h_1)} \\ = \dfrac{2r^2}{r_2(r_2 + h_1)} \\ = \dfrac{2 \times 7^2}{14(14 + \dfrac{7}{3})} \\ = \dfrac{98}{14 \times \dfrac{49}{3}} \\ = \dfrac{98 \times 3}{14 \times 49} \\ = \dfrac{3}{7}.\end{aligned}Hence, height of cone = 7 cm, radius of cylinder = 14 cm and Total surface area of sphere : Total surface area of cylinder = 3 : 7.
Question 118
A tree (TS) of height 30 m stands in front of a tall building (AB). Two friends Rohit and Neha are standing at R and N respectively, along the same straight line joining the tree and the building (as shown in the diagram). Rohit, standing at a distance of 150 m from the foot of the building, observes the angle of elevation of the top of the building as 30°. Neha from her position observes that the top of the building and the tree has the same elevation of 60°.
Find the:
(a) height of the building
(b) distance between
- Neha and the foot of the building
- Rohit and Neha
- Neha and the tree
- building and the tree.
(a) Referring to the diagram,
tan 30° = \dfrac{AB}{AR}
Putting in the known values :
\begin{aligned}\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{150} \\ \Rightarrow AB = \dfrac{150}{\sqrt{3}} = \dfrac{150}{1.732} \\ = 86.6 \text{ m}.\end{aligned}Hence, height of the building = 86.6 m.
(b)
- Referring to the diagram,
tan 60° = \dfrac{AB}{AN}
Putting in the known values :
\begin{aligned}\Rightarrow \sqrt{3} = \dfrac{\dfrac{150}{\sqrt{3}}}{AN} \\ \Rightarrow AN = \dfrac{150}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} \\ \Rightarrow AN = \dfrac{150}{3} = 50\text{ m}.\end{aligned}Hence, distance between Neha and foot of the building = 50 m.
- Referring to the diagram,
RN = AR − AN = 150 − 50 = 100 m.
Hence, distance between Rohit and Neha = 100 m.
- Referring to the diagram,
tan 60° = \dfrac{ST}{NT}
Putting in the known values :
\begin{aligned}\Rightarrow \sqrt{3} = \dfrac{30}{NT} \\ \Rightarrow NT = \dfrac{30}{\sqrt{3}} \\ \Rightarrow NT = \dfrac{30}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\ \Rightarrow NT = \dfrac{30\sqrt{3}}{3} \\ \Rightarrow NT = 10\sqrt{3} \\ \Rightarrow NT = 10 \times 1.732 = 17.32 \text{ m}.\end{aligned}Hence, distance between Neha and the tree = 17.32 m.
(iv) Referring to the diagram,
AT = AN − NT = 50 − 17.32 = 32.68 m
Hence, distance between building and tree = 32.68 m.
Question 119
A life insurance agent found the following data of age distribution of 100 policy holders, where f is an unknown frequency.
| Age in years | No. of policy holders |
|---|---|
| 15-20 | 7 |
| 20-25 | 12 |
| 25-30 | 15 |
| 30-35 | 22 |
| 35-40 | f |
| 40-45 | 14 |
| 45-50 | 8 |
| 50-55 | 4 |
(a) If the mean age of the policy holders is 35.65 years, find the unknown frequency f.
(b) Find the median class of the distribution.
(a) From the problem statement,
Total no. of policy holders = 100
∴ 7 + 12 + 15 + 22 + f + 14 + 8 + 4 = 100
⇒ 82 + f = 100
⇒ f = 100 − 82 = 18.
Hence, f = 18.
(b) Cumulative frequency distribution table :
| Age in years | No. of policy holders | Cumulative frequency |
|---|---|---|
| 15-20 | 7 | 7 |
| 20-25 | 12 | 19 |
| 25-30 | 15 | 34 |
| 30-35 | 22 | 56 |
| 35-40 | 18 | 74 |
| 40-45 | 14 | 88 |
| 45-50 | 8 | 96 |
| 50-55 | 4 | 100 |
Since, n = 100, which is even.
Median = \dfrac{n}{2}\text{ th term} = \dfrac{100}{2} = 50th term.
Steps of construction :
- Plot class interval on x-axis and cumulative frequency on y-axis.
- Mark points (20, 7), (25, 19), (30, 34), (35, 56), (40, 74), (45, 88), (50, 96) and (55, 100).
- Draw a free hand curve passing through the points marked and starting from the lower limit of first class and terminating at upper limit of the last class.
- From point A = 50 draw a line parallel to x-axis touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at C.
From graph,
C = 33.75, which lies between 30-35.
Hence, median class = 30-35.