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ICSE Class 10 Maths Mock Sample Paper 2 with Solutions

**SECTION A**

**Attempt all questions from this section.**

**Question 1**

**(a) **Find the remainder when 2x^{3} – 3x^{2} + 7x – 8 is divided by x – 1. [3]

**Solution:**

Let p(x) = 2x^{3} – 3x^{2} + 7x – 8

g(x) = x – 1

Remainder = -2

**(b)** Find the value of cosec (65° + θ) – sec (25° – θ). [3]

**Solution:**

cosec (65° + θ) – sec (25° – θ)

= cosec [90° – (25° – θ)] – sec (25° – θ)

= sec (25° – θ) – sec (25° – θ)

= 0

**(c)** A shopkeeper buys an article at a discount of 25% from the wholesaler. The printed price of the article is Rs. 5000 and the rate of sales tax is 10%. The shopkeeper sells it to the customer at a discount of 5% of the printed price and charges the sales tax at the same rate. Find:

(i) the amount paid by the customer

(ii) the VAT (Value Added Tax) paid by the shopkeeper. [4]

**Solution:**

Given,

The printed price of the article = Rs. 5000

Discount = 25%

The purchase price for shopkeeper = Rs. 5000 – (25% of Rs. 5000)

= Rs. 5000 – (25/100) × Rs. 5000

= Rs. 5000 – 1250

= Rs. 3750

Tax paid by the shopkeeper = 10% of Rs. 3750 = Rs. 375

Price paid by the shopkeeper for article = Rs. 3750 + Rs. 375 = Rs. 4125

(i) Discount given to the customer = 5% of Rs. 5000 = Rs. 250

Price of article after discount = Rs. 5000 – Rs. 250 = Rs. 4750

Tax paid by the customer = 10% of Rs. 4750 = Rs. 475

Price paid by the customer = Rs. 4750 + Rs. 475 = Rs. 5225

(ii) VAT paid by the shopkeeper = Tax charged when selling – Tax paid when buying

= Rs. 475 – Rs. 375

= Rs. 100

**Question 2**

**(a)** Manoj opened a recurring deposit account with Punjab National Bank and deposited Rs. 500 per month for 3 years. The bank paid him Rs. 20220 on maturity. Find the rate of interest paid by the bank. [3]

**Solution:**

Given,

Monthly installment = P = Rs. 500

n = 3 years = 36 months

Let r be the rate of interest.

Maturity value = (P × n) + P × [n(n + 1)/ 2 × 12] × (r/100)

20220 = (500 × 36) + 500 × [(36 × 37)/ (2 × 12)] × (r/ 100)

20220 – 18000 = 500 × (111/2) × (r/100)

2220 = 555r/ 2

r = (2220 × 2)/ 555

r = 8%

Hence, the rate of interest paid by the bank is 8%.

**(b)** The volume and curved surface of a cylinder are equal numerically. If the height is 3 ½ times the radius of the base, find the radius. [3]

**Solution:**

Let r be the radius and h be the height of a cylinder.

Given,

h = 3 ½ × 2 = 7r/2

The volume of cylinder = Curved surface area of the cylinder (given)

πr^{2}h = 2πrh

r^{2} = 2r

r = 2 units

Hence, the radius of the base of the cylinder is 2 units.

**(c) **A solid consisting of a right circular cone, standing on a hemisphere, is placed upright in a right circular cylinder full of water and touches the bottom. Find the volume of water left in the cylinder having given that the radius of the cylinder is 3 cm and its height is 6 cm, the radius of the hemisphere is 2 cm and the height of the cone is 4 cm. Give your answer to the nearest cubic centimeters. [4]

**Solution:**

Given,

Height of the cylinder = H = 6 cm

Radius of the cylinder = R = 3 cm

Height of the cone = h = 4 cm

Radius of the hemisphere = Radius of cone = r = 2 cm

Volume of water in the cylinder when it is full = πr²h

= π × 3 × 3 × 6

= 54π cm^{3}

Volume of water displaced = volume of cone + volume of hemisphere

= (1/3) πr²h + (2/3)πr³

= (1/3)πr²(h + 2r)

= (1/3) π × 2 × 2(4 + 2 × 2)

= (1/3) π × 4 × 8

= 32π/3 cm³

Volume of water left in the cylinder = 54π – (32π/3)

= (162π – 32π)/ 3

= 130π/3

= (130/3) × (22/7)

= 136.19

Hence, the volume of water left in the cylinder is 136 cm^{3}. (approx.)

**Question 3**

**(a)** Using factor theorem, show that (x – 3) is a factor of x^{3} – 7x^{2} + 15x – 9. Hence, factorize the given expression completely. [3]

**Solution:**

Let p(x) = x^{3} – 7x^{2} – 15x – 9

For checking (x – 3) is a factor of p(x) or not, substitute x = 3 in p(x),

p(3) = (3)^{3} – 7(3)^{2} + 15(3) – 9

= 27 – 63 + 45 – 9

= 72 – 72

= 0

Therefore, (x – 3) is a factor of the given polynomial.

By dividing p(x) by (x – 3), the quotient = x^{2} – 4x + 3

x^{3} – 7x^{2} + 15x – 9 = (x – 3)(x^{2} – 4x + 3)

= (x – 3)(x^{2} – 3x – x + 3)

= (x – 3) [x(x – 3) – 1(x – 3)]

= (x – 3)(x – 3)(x – 1)

= (x – 3)^{2}(x 1)

**(b)** Find the 2 x 2 matrix X which satisfies the equation:

[3]

**Solution:**

Given,

**(c) **Solve the inequation:

12 + 1 ⅚ x ≤ 5 + 3x, x ∈ R

Represent the solution on a number line. [4]

**Solution:**

Given,

12 + 1 ⅚ x ≤ 5 + 3x

12 + (11/6)x ≤ 5 + 3x

12 – 5 ≤ 3x – (11x/6)

7 ≤ (18x – 11x)/6

7 ≤ 7x/6

x ≥ (7 × 6)/ 7

x ≥ 6

Therefore, {x : x ∈ R and ≥ 6}

**Question 4**

**(a)** In the figure, AD is the diameter of the circle. If ∠BCD = 130°, calculate:

(i) ∠DAB

(ii) ∠ADB

[3]

**Solution:**

Given,

AD is the diameter of the circle.

∠BCD = 130°

(i) ABCD is a cyclic quadrilateral. (from the given figure)

We know that the sum of the opposite angles of a cyclic quadrilateral is supplementary.

∠DAB + ∠BCD = 180°

∠DAB + 130° = 180°

∠DAB = 180° – 130°

∠DAB = 50°

(ii) The angle in a semicircle is a right angle.

Thus, in triangle ABD, ∠ABD = 90°

∠ABD + ∠ADB + ∠DAB = 180° (by the angle sum property of a triangle)

90° + ∠ADB + 50° = 180°

∠ADB = 180° – 90° – 50°

∠ADB = 40°

**(b) **Part of a geometrical figure is given in the diagram alongside. Complete the figure so that both the x-axis and y-axis are lines of symmetry of the completed figure. [3]

**Solution:**

The completed figure that shows both the x-axis and y-axis are lines of symmetry is:

**(c) **Calculate the mean wage correct to the nearest rupee for the following data:

Category |
A |
B |
C |
D |
E |
F |
G |

Wages in Rs per day |
50 |
60 |
70 |
80 |
90 |
100 |
110 |

No. of workers |
2 |
4 |
8 |
12 |
10 |
6 |
8 |

**Solution:**

Category |
Wages in Rs per day (x) |
No. of workers (f) |
fx |

A |
50 |
2 |
100 |

B |
60 |
4 |
240 |

C |
70 |
8 |
560 |

D |
80 |
12 |
960 |

E |
90 |
10 |
900 |

F |
100 |
6 |
600 |

G |
110 |
8 |
880 |

∑f = 50 |
∑fx = 4240 |

Mean = ∑fx/ ∑f

= 4240/ 50

= 84.5

Hence, the approximate mean wage is Rs. 85.

**SECTION – B**

**Attempt any four questions from this section.**

**Question 5**

**(a) **Solve using the quadratic formula: 6x^{2} + (12 – 8a)x – 16a = 0 [3]

**Solution:**

Given,

6x^{2} + (12 – 8a)x – 16a = 0

Comparing with the standard form ax^{2} + bx + c = 0

a = 6, b = 12 – 8a, c = -16a

Quadratic formula:

x = [-b ± √(b^{2} – 4ac) ] / 2a

x = [-(12 – 8a) ± √{(12 – 8a)^{2} – 4(6)(-16a)} ] / 2(6)

= [(8a – 12) ± √(144 + 64a^{2} – 192a + 384a)] / 12

= [(8a – 12) ± √(144 + 64a^{2} + 192a)]/ 12

= [(8a – 12) ± √(12 + 8a)^{2} ]/ 12

= [(8a – 12) ± (12 + 8a)] /12

x = (8a – 12 + 12 + 8a)/ 12, x = (8a – 12 – 12 – 8a)/ 12

x = 16a/12, x = -24/12

x = 4a/3, x = -2

**(b) **Deepika opened a savings bank account in a bank. Her passbook entries are shown below:

Date |
Particulars |
Withdrawals |
Deposits Rs |
Balance Rs |

Jan 1 |
B/F |
_ |
_ |
6360 |

Jan 12 |
By cash |
_ |
750 |
7110 |

Feb 15 |
To self |
5000 |
_ |
2110 |

June 6 |
To cheque |
354 |
_ |
1756 |

July 18 |
By cheque |
_ |
543 |
2299 |

She closed the account on 29 July and received Rs. 2354.20 as balance. Calculate the rate of interest. [3]

**Solution:**

Month Qualifying amount for interest

Jan Rs. 7110

Feb Rs. 2110

Mar Rs. 2110

Apr Rs. 2110

May Rs. 2110

June Rs. 1756

Total Rs. 17306

Let R be the rate of interest.

Interest = PTR/100

= (17306 × 1 × R)/ 100 × 12

= 17306 R/ 1200

Given that, Deepika received Rs. 2354.20 as balance.

2354.20 = 2299 + (17306R/ 1200)

2354.20 – 2299 = 17306 R/1200

55.2 = 17306R/1200

R = (52.2 × 1200)/ 17306

R = 3.8%

**(c) **Use graph paper for this question.

(i) Plot the points A(3, 5) and B(-2, -4). Use 1 cm = 1 unit on both the axes.

(ii) A’ is the image of A when reflected in the x-axis. Write down the coordinates of A’ and plot it on the graph paper.

(iii) B’ is the image of B when reflected in the y-axis followed by a reflection in the origin. Write down the coordinates of B’ and plot it on the graph paper.

(iv) Write down the geometrical name of the figure AA’BB’. [4]

**Solution:**

Given,

A(3, 5) and B(-2, -4)

A’ is the image of A when reflected in the x-axis.

B’ is the image of B when reflected in the y-axis followed by a reflection in the origin.

(ii) Coordinates of A'(3, -5)

(iii) Coordinates of B'(-2, 4)

(iv) The geometrical name of the figure AA’BB’ is an isosceles trapezium since AA’ || BB’ and AB’ = A’B.

**Question 6**

**(a)** Construct a triangle ABC in which BC = 6 cm, AB = 9 cm, and ∠ABC = 60°. Construct the locus of all points inside triangle ABC, which are equidistant from B and C. [3]

**Solution:**

P is the locus of all points inside triangle ABC which is equidistant from B and C.

**(b)** The compound interest, calculated yearly on a certain sum of money for the second year is Rs. 880 and for the third year, it is Rs. 968. Calculate the rate of interest and the sum of money. [3]

**Solution:**

Given,

CI for the second year = Rs. 880

CI for the third year = Rs. 968

Now,

Simple interest (SI) on Rs. 880 for one year = Rs. 968 – Rs. 880 = Rs. 88

SI = PTR/100

R = (88 × 100)/ (880 × 1)

R = 10%

Let p be the actual sum.

Amount after 2 years – Amount after 1 year = CI for the second year

p(1 + 10/100)^{2} – p(1 + 10/100) = 880

p(110/100)^{2} – p(110/100) = 880

p[(11/10)^{2} – (11/10)] = 880

p[(121/100) – (11/10)] = 880

p[(121 – 110)/ 100] = 800

p(11/100) = 880

p = 8000

Hence, the rate of interest is 10% and the actual sum of money is Rs. 8000.

**(c) **In the figure, not drawn to scale, TF is a tower. The elevation of T from A is x°, where tan x = ⅖ and AF = 200 m. The elevation of T from B, where AB = 80 m, is y°. Calculate:

(i) the height of the tower TF.

(ii) the angle y, correct to the nearest degree. [4]

**Solution:**

Given,

TF is a tower.

The elevation of T from A is x°, where tan x = ⅖ and AF = 200 m.

The elevation of T from B, where AB = 80 m, is y°.

(i) In right triangle AFT,

tan x = TF/AF

⅖ = TF/200

TF = (200 × 2)/ 5

TF = 80 m

Therefore, the height of the tower TF is 80 m.

(ii) BF = AF – AB = 200 – 80 = 120 m

TF = 80 m

In right triangle BFT,

tan y = TF/BF

tan y = 80/120

tan y = ⅔

y = tan^{-1}(⅔)

y = 33.7

Therefore, the angle y = 34° (approx.)

**Question 7**

**(a)** A bag contains 8 red, 6 white, and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is:

(i) red or white

(ii) not black

(iii) neither white nor black. [3]

**Solution:**

Given,

A bag contains 8 red, 6 white, and 4 black balls.

Total number of outcomes = n(S) = 18

i.e. 8 + 6 + 4 = 18

(i) Let A be the event of getting a red or white ball.

Number of outcomes favourable to A = n(A) = 14

i.e. 8 + 6 = 14

P(A) = n(A)/n(S) = 14/18 = 7/9

(ii) Let B be the event of getting a ball that is not black.

Number of outcomes favourable to B = 14

i.e. Total number of balls – Black balls = 18 – 4 = 14

P(B) = n(B)/ n(S) = 14/18 = 7/9

(iii) Let C be the event of getting neither white nor black balls.

Number of outcomes favourable to C = n(C) = 8

i.e. only red balls = 8

P(C) = n(C)/n(S) = 8/18 = 4/9

**(b)** A can do a piece of work in x days and B can do it in (x + 16) days. If both working together can do it in 15 days, find x. [3]

**Solution:**

Given,

A can do a piece of work in x days and B can do it in (x + 16) days.

A and B can finish the work in 15 days.

i.e x(x + 16)/ (x + x + 16) = 15

(x^{2} + 16x)/ (2x + 16) = 15

x^{2} + 16x = 15(2x + 16)

x^{2} + 16x = 30x + 240

x^{2} + 16x – 30x – 240 = 0

x^{2} – 14x – 240 = 0

x^{2} – 24x + 10x – 240 = 0

x(x – 24) + 10(x – 24) = 0

(x + 10)(x – 24) = 0

x = -10, x = 24

The number of days cannot be negative.

Therefore, x = 24

**(c)** In the figure, BC is parallel to DE. Area of triangle ABC = 25 cm^{2}, area of trapezium BCED = 24 cm^{2}, DE = 14 cm. Calculate the length of BC. [4]

**Solution:**

Given,

Area of triangle ABC = 25 cm^{2}

Area of trapezium BCED = 24 cm^{2}

Area of triangle ADE = Area of triangle ABC + Area of trapezium

= 25 + 24

= 49 cm^{2}

In ΔABC and Δ ADE,

BC ∥ DE,

∠ABC = ∠ADE (corresponding angles)

∠ACB = ∠AED (corresponding angles)

∠BAC = ∠DAE (common)

By AAA similarity.

ΔABC ~ ΔADE

ar(ΔABC)/ ar(ΔADE) = BC^{2}/DE^{2}

25/49 = BC^{2}/ (14)^{2}

BC^{2} = (25 × 196)/49

BC^{2} = 4900/49

BC^{2} = 100

⇒ BC = 10 cm

**Question 8**

**(a)** Using a ruler and compasses only, construct an isosceles ΔABC having base = 4 cm, vertical angle = 45°, and median through vertex equal to 2.8 cm. Draw the incircle of the triangle. [3]

**Solution:**

**(b) **Using the properties of proportion, solve for x:

**Solution:**

Given,

[√(a + x) + √(a – x)] / [√(a + x) – √(a – x)] = bUsing componendo and dividendo rule,

[√(a + x) + √(a – x) + √(a + x) – √(a – x)] / [√(a + x) + √(a – x) – √(a + x) + √(a – x)] = (b + 1)/ (b – 1)2√(a + x) / 2√(a – x) = (b + 1)/ (b – 1)

√(a + x)/ √(a – x) = (b + 1)/ (b – 1)

Squaring on both sides,

(a + x)/ (a – x) = (b + 1)^{2}/ (b – 1)^{2}

Again, using componendo and dividendo rule,

(a + x + a – x)/ (a + x – a + x) = [(b + 1)^{2} + (b – 1)^{2}] / [(b + 1)^{2} – (b – 1)^{2}]

2a/2x = 2(b^{2} + 12)/ 2(2b)

a/x = (b^{2} + 1)/2b

⇒ x/a = 2b/ (b^{2} + 1)

⇒ x = 2ab/ (b^{2} + 1)

**(c)** A man invests Rs. 8800 on buying shares of the face value of Rs. 100 each at a premium of 10% in a company. If he earns Rs. 1200 at the end of the year as a dividend, find:

(i) the number of shares he has in the company

(ii) the dividend percentage per share. [4]

**Solution:**

Given,

Total investment = Rs. 8800

Nominal value = Rs. 100

Market value = 10% of Rs. 100 + Rs. 100

= Rs. 10 + Rs. 100

= Rs. 110

(i) Number of shared purchased = 8800/110 = 80

(ii) Let d be the dividend percentage.

Nominal value of 80 shares = 80 × Rs. 100 = Rs. 8000

According to the given,

d% of Rs. 8000 = Rs. 1200

(d/100) × 8000 = 1200

d × 80 = 1200

d = 1200/80

d = 15%

Hence, the dividend percentage per share is 15%.

**Question 9**

**(a)** Show that the points A(1, 0), B(5, 3), C(2, 7), and D(-2, 4) are the vertices of a square. [3]

**Solution:**

Given,

A(1, 0), B(5, 3), C(2, 7), and D(-2, 4)

Using distance formula,

AB = √[(5 – 1)^{2} + (3 – 0)^{2}]

= √(16 + 9)

= √25

AB = 5

BC = √[(2 – 5)^{2} + (7 – 3)^{2}]

= √(9 + 16)

= √25

BC = 5

CD = √[(-2 – 2)^{2} + (4 – 7)^{2}]

= √(16 + 9)

= √25

CD = 5

DA = √[(1 + 2)^{2} + (0 – 4)^{2}]

= √(9 + 16)

= √25

DA = 5

All the sides are equal.

AC and BD are the diagonals of quadrilateral ABCD.

Midpoint of AC = [(1 + 2)/2, (0 + 7)/2] = (3/2, 7/2)

Midpoint of BD = [(5 – 2)/2, (3 + 4)/2] = (3/2, 7/2)

Midpoint of AC = Midpoint of BD

Therefore, the given points are the vertices of a square.

**(b) **In the figure, O is the centre of the circle and ΔABC is equilateral. Find

(i) ∠BDC

(ii) ∠BEC. [3]

**Solution:**

Given,

ΔABC is an equilateral triangle.

∠ABC = ∠BCA = ∠CAB = 60°

(i) We know that the angles subtended by an arc on the circumference on the same segment are equal.

∠BDC = ∠CAB = 60°

(ii) ABEC is a cyclic quadrilateral.

We know that the sum of opposite angles of a cyclic quadrilateral is supplementary.

∠BAC + ∠BEC = 180°

60° + ∠BEC = 180°

∠BEC = 180° – 60°

∠BEC = 120°

**(c)** A bucket is raised from a well by means of a rope that is wound around a wheel of diameter 77 cm. Given that the bucket ascends in 1 min 28 sec with a uniform speed of 1. 1 m/sec, calculate the number of complete revolutions the wheel makes in raising the bucket. Take π = 22/7 [4]

**Solution:**

Given,

Diameter of wheel = 77 cm

Radius of the wheel = r = 77/2 cm = 77/200 m

Time = 1 min 28 sec = (60 + 28) sec = 88 sec

Speed = 1.1 m/s

Distance = Speed × Time

= 1.1 × 88

= 96.8 m

Number of revolutions = Distance/ Circumference of wheel

= 96.8/ 2πr

= 96.8/ [2 × (22/7) × (77/200)]

= 96.8/2.42

= 40

Therefore, the number of complete revolutions the wheel makes in raising the bucket is 40.

**Question 10**

**(a) **Prove the following identity:

^{2}θ) [4]

**Solution:**

LHS = [1/ (sin θ + cos θ)] + [1/ (sin θ – cos θ)]

= [(sin θ + cos θ + sin θ – cos θ)/ (sin θ + cos θ)(sin θ – cos θ)]

= 2 sin θ/ (sin^{2}θ – cos^{2}θ)

= 2 sin θ/ [(1 – cos^{2}θ) – cos^{2}θ]

= 2 sin θ/ (1 – cos^{2}θ – cos^{2}θ)

= 2 sin θ/ (1 – 2 cos^{2}θ)

= RHS

**(b) **The following table shows the distribution of the heights of a group of factory workers:

Height (in cm) |
150 – 155 |
155 – 160 |
160 – 165 |
165 – 170 |
170 – 175 |
175 – 180 |
180 – 185 |

No. of workers |
6 |
12 |
18 |
20 |
13 |
8 |
6 |

(i) Determine the cumulative frequencies.

(ii) Draw the cumulative frequency curve on a graph paper. [6]

**Solution:**

(i) Cumulative frequency table:

Height (in cm) |
No. of workers (f) |
Cumulative frequency (cf) |

150 – 155 |
6 |
6 |

155 – 160 |
12 |
18 |

160 – 165 |
18 |
36 |

165 – 170 |
20 |
56 |

170 – 175 |
13 |
69 |

175 – 180 |
8 |
77 |

180 – 185 |
6 |
83 |

(ii) Plot the points (155, 6), (160, 18), (165, 36), (170, 56), (175, 69), (180, 77) and (185, 83) on the graph and then join them with a free hand to obtain a cumulative frequency curve.

**Question 11**

**(a) **PAT is a tangent to the circumcircle of ΔABC such that PT || BC. Show that AB = AC. [3]

**Solution:**

Given,

PAT is a tangent to the circumcircle of ΔABC such that PT || BC.

∠PAC = ∠ABC….(i) (angles in the alternate segment)

∠PAC = ∠ACB….(ii) (alternate interior angles)

From (i) and (ii),

∠ABC = ∠ACB

We know that the sides opposite to equal angles are equal.

AB = AC

**(b) **A die is thrown once. What is the probability that the number obtained is:

(i) even?

(ii) other than 4? [3]

**Solution:**

Given,

A die is thrown once.

Sample space = S = {1, 2, 3, 4, 5, 6}

n(S) = 6

(i) Let A be the event of getting an even number.

A = {2, 4, 6}

n(A) = 3

P(A) = n(A)/n(S)

= 3/6

= 1/2

(ii) Let B be the event of getting a number other than 4.

B = {1, 2, 3, 5, 6}

n(B) = 5

P(B) = n(B)/ n(S)

= 5/6

**(c) **Find the equation of the line passing through (-2, -4) and perpendicular to the line 3x – y + 5 = 0. [4]

**Solution:**

Given,

3x – y + 5 = 0….(i)

y = 3x + 5

Gradient of the line = 3

The slope of the which is perpendicular to (i) is m = -1/3

The equation of line passing through (-2, -4) and perpendicular to (i) is:

y – (-4) = m[x – (-2)]

y + 4 = (-1/3) (x + 2)

3(y + 4) = -x – 2

3y + 12 + x + 2 = 0

x + 3y + 14 = 0