Last Updated on November 13, 2026 by sanjjeett
Practice the complete ICSE Class 10 Mathematics 2018 Question Paper with Solutions, including algebra, geometry, trigonometry, statistics, graphs, constructions and commercial mathematics. Use this solved paper to revise important concepts, practise board-style questions and strengthen your exam preparation.
Attempt all questions from Section A and any four questions from Section B.
All working, including rough work, must be clearly shown and must be done on the same sheet as the rest of the answer.
Omission of essential working will result in loss of marks.
The intended marks for questions or parts of question are given in brackets [ ].
Mathematical tables are provided.
ICSE Mathematics 2018 — Section A
Find the value of ‘ \(\boldsymbol{x}\) ‘ and ‘ \(\boldsymbol{y}\) ‘ if : [3]
or
By equating \(2 x+6=10\) or \(x=2\)
Sonia had a recurring deposit account in a bank and deposited ₹ 600 per month for \(\mathbf{2} \frac{\mathbf{1}}{\mathbf{2}}\) years. If the rate of interest was 10% p.a., find the maturity value of this account. [3]
Cards bearing numbers 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20 are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card which is :
(i) a prime number.
(ii) a number divisible by 4 .
(iii) a number that is a multiple of 6.
(iv) an odd number. [4]
(i)
Hence, probability of a getting a prime number card
(ii) Number divisible by \(4=\{4,8,12,16,20\}\)
No. of favourable cards \(=5\) Hence, probability of a getting card, where number divisible by 4 .
(iii) Number which are multiple of \(6=\) \{6, 12, 18\}
No. of favourable cards \(=3\) Hence, probability of getting card, which is multiple of 6.
(iv) Odd number = 0 (No. odd card)
No. of favourable cards \(=0\) Hence, probability of getting an odd number card
The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the
(i) radius of the cylinder
(ii) volume of cylinder. (use \(\pi=\frac{\mathbf{2 2}}{\mathbf{7}}\) ) [3]
Given
(i)
Hence, radius of cylinder \(=21 \mathrm{~cm}\).
(ii) volume of cylinder \(=\pi r^{2} h\)
If \((\boldsymbol{k}-\mathbf{3}),(\mathbf{2} \boldsymbol{k}+\mathbf{1})\) and \((\mathbf{4} \boldsymbol{k}+\mathbf{3})\) are three consecutive terms of an A.P., find the value of \(\boldsymbol{k}\). [3]
For three consecutive terms of an A.P., the common difference should be same, i.e,.
\(P Q R S\) is a cyclic quadrilateral. Given \(\angle Q P S=\) \(73^{\circ}, \angle P Q S=55^{\circ}\) and \(\angle P S R=82^{\circ}\), calculate :
(i) \(\angle Q R S\)
(ii) \(\angle R Q S\)
(iii) \(\angle P R Q\) [4]
(i) Since, sum of opposite pairs of angles in cyclic quadrilaterals is 180°.
Hence,
(ii)
(PQRS is a cyclic quadrilateral)
(iii) Join \(P\) to \(R\)
In \(\triangle P S Q\),
Now, \(\angle P R Q=\angle P S Q(\because\) Angle in same segment \()\)
If \((\boldsymbol{x}+\mathbf{2})\) and \((\boldsymbol{x}+\mathbf{3})\) are factors of \(\boldsymbol{x}^{\mathbf{3}}+\boldsymbol{a} \boldsymbol{x}+\boldsymbol{b}\), find the values of ‘ \(\boldsymbol{a}\) ‘ and ‘ \(\boldsymbol{b}\) ‘. [3]
\((x+2)\) is a factor of given polynomial then \(x=-2\) will satisfy the polynomial
\((x+3)\) is also a factor of given polynomial then \(x=-3\) will satisfy the polynomial
On solving (i) and (ii), we get
Prove that \(\sqrt{\boldsymbol{\operatorname { s e c }}^{\mathbf{2}} \theta+\boldsymbol{\operatorname { c o s e c }}^{\mathbf{2}} \theta}=\boldsymbol{\operatorname { t a n }} \theta+\boldsymbol{\operatorname { c o t }} \theta\) [3]
[Since, \(1+\tan ^{2} \theta=\sec ^{2} \theta\) and \(1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\) ]
Using a graph paper draw a histogram for the given distribution showing the number of runs scored by 50 batsmen. Estimate the mode of the data : [4]
| Runs scored | 3000-4000 | 4000-5000 | 5000-6000 | 6000-7000 | 7000-8000 | 8000-9000 | 9000-10000 |
|---|---|---|---|---|---|---|---|
| No. of batsmen | 4 | 18 | 9 | 6 | 7 | 2 | 4 |
Solve the following inequation, write down the solution set and represent it on the real number line : [3]
or
or
or
or
∴ The solution set is \(x \in[-4,4]\) and \(x \in Z\)
If the straight lines \(\mathbf{3} \boldsymbol{x}-\mathbf{5} \boldsymbol{y}=\mathbf{7}\) and \(\mathbf{4} \boldsymbol{x}+\boldsymbol{a} \boldsymbol{y}+\) \(\mathbf{9}=\mathbf{0}\) are perpendicular to one another, find value of \(\boldsymbol{a}\). [3]
Eqn. (i)
∴
Eqn. (ii)
∴
Both lines are perpendicular to each other. ∴
or
or
Solve \(\boldsymbol{x}^{\mathbf{2}}+\mathbf{7} \boldsymbol{x}=\mathbf{7}\) and give your answer correct to two decimal places. [4]
Compare the equation to
Roots are real, different and irrational number, or
or
Taking (+) sign
Taking (-) sign
40 Marks
ICSE Mathematics 2018 — Section B
The \(\mathbf{4}^{\text {th }}\) term of a G.P. is 16 and the \(\mathbf{7}^{\text {th }}\) term is 128. Find the first term and common ratio of the series. [3]
Given,
Let first term of G.P. be a and common ratio be \(r\). so
Dividing (ii) by (i), we get
From (i)
A man invests ₹ 22,500 in ₹ 50 shares available at 10% discount. If the dividend paid by the company is 12%, calculate :
(i) The number of shares purchased.
(ii) The annual dividend received.
(iii) The rate of return he gets on his investment. Give your answer correct to the nearest whole number. [3]
Given, Total investment \(=\text{₹} 22,500\), Face value \(=\text{₹} 50\)
( \(\because 10 \%\) discount) Dividend \(=12 \%\) (i)
(ii) Annual dividend =No. of shares × Nominal value of share × dividend rate
(iii) Return percentage
Use graph paper for this question (Take 2 cm \(=\mathbf{1}\) unit along both \(\boldsymbol{x}\) and \(\boldsymbol{y}\) axis). \(A B C D\) is a quadrilateral whose vertices are \(A(2,2), B(2,-2), C(0-1)\) and \(D(0,1)\)
(i) Reflect quadrilateral \(A B C D\) on the \(y\)-axis and name it as \(A^{\prime} B^{\prime} C D\).
(ii) Write down the coordinates of \(A^{\prime}\) and \(B^{\prime}\).
(iii) Name two points which are invariant under the above reflection.
(iv) Name the polygon \(A^{\prime} B^{\prime} C D\). [4]
(i)
(ii) Co-ordinates of \(A^{\prime} \rightarrow(-2,2)\)
Co-ordinates of \(B^{\prime} \rightarrow(-2,-2)\)
(iii) Two invariant points are \(C(0,-1)\) and \(D(0,1)\)
(iv) \(A^{\prime} B^{\prime} C D\) is a trapezium quadrilateral
Using properties of proportion, solve for \(\boldsymbol{x}\). Given that \(\boldsymbol{x}\) is positive : [3]
Applying componendo \& dividendo
Squaring both sides,
∵ x is positive, hence \(x=\frac{5}{8}\)
If \(A=\left[\begin{array}{cc}2 & 3 \\ 5 & 7\end{array}\right], B=\left[\begin{array}{cc}0 & 4 \\ -1 & 7\end{array}\right]\), and \(C=\left[\begin{array}{cc}1 & 0 \\ -1 & 4\end{array}\right]\), find \(\boldsymbol{A} \boldsymbol{C}+\boldsymbol{B}^{\mathbf{2}}-\mathbf{1 0} \boldsymbol{C}\). [3]
Prove that \((\mathbf{1}+\boldsymbol{\operatorname { c o t }} \theta-\boldsymbol{\operatorname { c o s e c }} \theta)(\mathbf{1}+\boldsymbol{\operatorname { t a n }} \theta+\) \(\boldsymbol{\operatorname { s e c }} \theta)=\mathbf{2}\) [4]
L.H.S. \(=(1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta)\)
Find the value of \(\boldsymbol{k}\) for which the following equation has equal roots. [3]
Compare the equation to
Since, roots of the equation are equal
One map drawn to a scale of 1 : 50,000, a rectangular plot of land \(\boldsymbol{A} \boldsymbol{B} \boldsymbol{C} \boldsymbol{D}\) has the following dimensions. \(\boldsymbol{A} \boldsymbol{B}=\mathbf{6 ~ c m} ; \boldsymbol{B} \boldsymbol{C}=\mathbf{8 ~ c m}\) and all angles are right angles. Find :
(i) the actual length of the diagonal distance \(\boldsymbol{A} \boldsymbol{C}\) of the plot in km.
(ii) the actual area of the plot in sq km. [3]
(i) Since,
By Pythagorean theorem
Since, it is given that
Diagonal \(A C=5 \mathrm{~km}\)
(ii)
\(\boldsymbol{A}(\mathbf{2}, \mathbf{5}), \boldsymbol{B}(-\mathbf{1}, \mathbf{2})\) and \(\boldsymbol{C}(\mathbf{5}, \mathbf{8})\) are the vertices of a triangle \(\boldsymbol{A} \boldsymbol{B} \boldsymbol{C}\), ‘ \(\boldsymbol{M}\) ‘ is a point on \(\boldsymbol{A} \boldsymbol{B}\) such that \(A M: M B=1: 2\). Find the co-ordinates of ‘ \(M\) ‘. Hence find the equation of the line passing through the points \(\boldsymbol{C}\) and \(\boldsymbol{M}\). [4]
Let \(M\left(x_{1}, y_{1}\right)\) be the point which divides the line segment \(A B\) in the ratio 1 : 2. Hence, Coordinate of \(M\) are
Coordinates of \(M\) are (1, 4). Now equation of line passing through \(C(5,8)\) and M(1, 4) is
₹ 7500 were divided equally among a certain number of children. Had there been 20 less children, each would have received ₹ 100 more. Find the original number of children. [3]
Let the number of children be \(x\) and ₹ \(y\) is given to each children then
and
The number of children can not be – ve.
∴ Number of children \(=50\)
If the mean of the following distribution is 24, find the value of ‘ \(a\) ‘. [3]
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Number of students | 7 | \(a\) | 8 | 10 | 5 |
| Marks | \(\operatorname{Mid} x_{i}\) | \(f_{i}\) | \(f_{i} x_{i}\) |
|---|---|---|---|
| 0-10 | 5 | 7 | 35 |
| 10-20 | 15 | \(a\) | \(15 a\) |
| 20-30 | 25 | 8 | 200 |
| 30-40 | 35 | 10 | 350 |
| 40-50 | 45 | 5 | 225 |
| \(30+a\) | \(810+15 a\) |
Using ruler and compass only, construct a \(\triangle A B C\) such that \(B C=5 \mathrm{~cm}\) and \(A B=6.5 \mathrm{~cm}\) and \(\angle A B C=\mathbf{1 2 0}^{\circ}\).
(i) Construct a circum-circle of \(\triangle A B C\).
(ii) Construct a cyclic quadrilateral \(A B C D\), such that \(D\) is equidistant from \(A B\) and \(B C\). [4]
(a) Steps of construction :
(i) Draw a line segment \(B C=5 \mathrm{~cm}\).
(ii) Construct \(\angle A B C=120^{\circ}\).
(iii) Cut \(B A=6.5 \mathrm{~cm}\)
(iv) Join \(A\) to \(C\).
(v) Construct perpendicular bisectors of \(A B\) and \(B C\), intersecting at \(O\). Join \(A O\).
(vi) Taking \(O\) as centre, and \(O A\) as radius draw a circle, passing through \(A, B\) and \(C\).
(b) (i) Draw the bisector of \(\angle A B C\) such that it touches the circle at point \(D\).
(ii) Join \(A\) to \(D\) and \(C\) to \(D\).
(iii) \(A B C D\) is required cyclic quadrilateral.
Priyanka has a recurring deposit account of ₹ 1,000 per month at 10% per annum. If she gets ₹ 5,550 as interest at the time of maturity, find the total time for which the account was held. [3]
\(P=\text{₹} 1,000, R=10 \%\), p.a., S.I. \(=\text{₹} 5,550\)
The value of time cannot be – ve. Hence
or \(n=3\) years
In \(\triangle P Q R, M N\) is parallel to \(Q R\) and
(i) Find \(\frac{M N}{Q R}\).
(ii) Prove that \(\triangle O M N\) and \(\triangle O R Q\) are similar.
(iii) Find, area of \(\triangle O M N\) : Area of \(\triangle O R Q\). [3]
(i) Given \(M N \| Q R\), then
or ∴
or
or
⇒
(ii) \(M N \| Q R\) (Given)
\(\angle O M N=\angle O R Q\) (Alternate angle) \(\angle O N M=\angle O Q R\) (Alternate angle) \(\angle M O N=\angle R O Q \quad(\) Vert. Opp. angles) \(\therefore \triangle O M N \sim \triangle O R Q\) (by AAA similarity)
(iii) ∵ \(\triangle O M N\) and \(\triangle O R Q\) are similar triangle.
We know that ratio of the areas of two similar triangle is equal to ratio of the squares of the corres-ponding sides.
Hence, \(\operatorname{ar}(\triangle O M N): \operatorname{ar}(\triangle O R Q)=4: 25\)
The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is 7 cm. The height of the cylinder and cone are each of 4 cm. Find volume of the solid. [4]
Common radius \((r)=7 \mathrm{~cm}\) Height of cylinder=Height of cone
Volume of solid =volume of hemisphere + volume of cylinder + volume of cone
Use Remainder theorem to factorize the following polynomial : [3]
Let
The factors of the constant terms are ± 1, ± 2, ± 5 and ± 10. We have
So, \((x+1)\) is a factor of \(f(x)\). We now divide \(f(x)=2 x^{3}+3 x^{2}-9 x-10\) by \((x+1)\) to get the other factor of \(f(x)\).
In the figure given below ‘ \(O\) ‘ is the centre of the circle. If \(Q R=O P\) and \(\angle O R P=20^{\circ}\). Find the value of ‘ \(\boldsymbol{x}\) ‘ giving reasons. [3]
Given
and
∵
∵
(Radius of the circle) ∴
(From (i)) In \(\triangle O Q R\),
∴
Now
(By angle sums property)
Now \(\angle O Q R+\angle O Q P=180^{\circ}\) (By linear pair)
In \(\triangle P O Q\),
Now
(By angle sum property)
Now
The angle of elevation from a point \(\boldsymbol{P}\) of the top of a tower \(Q R, 50 \mathrm{~m}\) high is \(60^{\circ}\) and that of the tower \(P T\) from a point \(Q\) is \(30^{\circ}\). Find the height of the tower \(\boldsymbol{P} \boldsymbol{T}\), correct to the nearest metre. [4]
Let the height of the tower \(P T\) be \(h\) and distance \(P Q=x \mathrm{~m}\) In \(\triangle P R Q\),
or
In right \(\triangle P T Q\),
The \(\mathbf{4}^{\text {th }}\) term of an A.P. is 22 and \(\mathbf{1 5}^{\text {th }}\) term is 66. Find the first term and the common difference. Hence find the sum of the series to 8 terms. [4]
Let first term be a and common difference be \(d\).
Subtracting (ii) from (i)
From (i)
Thus,
Use Graph paper for this question. A survey regarding height (in cm) of 60 boys belonging to Class 10 of a school was conducted. The following data was recorded :
| Height in cm | 135140 | 140145 | 145150 | 150155 | 155160 | 160165 | 165-170 |
|---|---|---|---|---|---|---|---|
| No. of boys | 4 | 8 | 20 | 14 | 7 | 6 | 1 |
Taking \(\mathbf{2 ~ c m}=\) height of \(\mathbf{1 0 ~ c m}\) along one axis and \(\mathbf{2} \mathbf{~ c m}=\mathbf{1 0}\) boys along the other axis draw an ogive of the above distribution. Use the graph to estimate the following :
(i) the median
(ii) lower quartile
(iii) if above 158 cm is considered as the tall boys of the class. Find the number of boys in the class who are tall. [6]
| Height (in cm) | Frequency \(\boldsymbol{(} \boldsymbol{f} \boldsymbol{)}\) | Cumulative Frequency (c. f.) |
|---|---|---|
| 135-140 | 4 | 4 |
| 140-145 | 8 | 12 |
| 145-150 | 20 | 32 |
Here, \(n=60\)
(i) Median = \(\frac{60}{2}\) th term = 30th term = 149.5
(ii) Lower quartile \(=\frac{60}{4}\) th term \(=15\) th term = 145.5
(iii) Through mark of 158 on \(x\)-axis, draw a vertical line which meets the graph at a point on the ogive. Then through that point, draw a horizontal line which meets the \(y\)-axis at the mark of 48. Therefore, number of boys in the class who are tall \(=60-48=12\).
