ICSE Class 10 Maths Mock Sample Question Paper 4 With Answers - Free PDF Download

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ICSE Class 10 Maths Mock Sample Paper 4 with Solutions

SECTION A

Attempt all questions from this section.

Question 1

(a) Without using trigonometry tables, prove that sin 37° cos 53° + cos 37° sin 53° = 1. [3]

Solution:

LHS = sin 37° cos 53° + cos 37° sin 53°

= sin 37° cos (90° – 37°) + cos 37° sin (90° – 37°)

= sin 37° sin 37° + cos 37° cos 37°

= sin²37° + cos²37°

= 1

= RHS

Hence proved.

(b) Find the mean proportional between (7 + √3) and (7 – √3). [3]

Solution:

Let x be the mean proportional between (7 + √3) and (7 – √3).

i.e. (7 + √3) : x = x : (7 – √3)

(x)(x) = (7 + √3)(7 – √3)

x² = (7)² – (√3)²

x² = 49 – 3

x² = 46

x = √46

(c) AB and CD are two chords of a circle intersecting at a point P outside the circle when produced, such that PA = 16 cm, PC = 10 cm, and PD = 8 cm. Find AB. [4]

Solution:

Given,

Two chords AB and CD meet at P when produced.

PA × PB = PC × PD

16 × PB = 10 × 8

PB = 80/16

PB = 5 cm

Now,

PA = AB + PB

16 = AB + 5

AB = 16 – 5

AB = 11 cm.

Question 2

(a) The common factor of 2x² + 5x + k and 2x² + 3x + l is (2x – 1). Find the values of k and l. [3]

Solution:

Let f(x) = 2x² + 5x + k and p(x) = 2x² + 3x + l

Given that, (2x – 1) is the factor of f(x) and g(x).

Thus, f(½) = 0 and p(½) = 0

Now,

f(½) = 0

2(½)² + 5(½) + k = 0

2(¼) + (5/2) + k = 0

(½) + (5/2) + k = 0

(6/2) + k = 0

3 + k = 0

k = -3

And

p(½) = 0

2(½)² + 3(½) + l = 0

2(¼) + (3/2) + l = 0

(½) + (3/2) + l = 0

(4/2) + l = 0

2 + l = 0

l = -2

Therefore, k = -3 and l = -2.

(b)

Solution:

Given,

(c) Solve 7 ≤ 4x + 2 ≤ 12, x ∈ R. Graph the solution set on the number line. [4]

Solution:

7 ≤ 4x + 2 ≤ 12, x ∈ R

7 – 2 ≤ 4x ≤ 12 – 2

5 ≤ 4x ≤ 10

(5/4) ≤ x ≤ (10/4)

Question 3

(a) The marks of 20 students in a test were as follows:

10, 15, 14, 11, 10, 8, 10, 6, 18, 19, 16, 14, 10, 3, 4, 20, 3, 10, 16, 10

Find:

(i) the mean

(ii) the median

(iii) the mode. [3]

Solution:

Given,

10, 15, 14, 11, 10, 8, 10, 6, 18, 19, 16, 14, 10, 3, 4, 20, 3, 10, 16, 10

Number of observations = 20

(i) Mean = Sum of observations/ Number of observations

= (10 + 15 + 14 + 11 + 10 + 8 + 10 + 6 + 18 + 19 + 16 + 14 + 10 + 3 + 4 + 20 + 3 + 10 + 16 + 10)/ 20

= 227/10

= 22.7

(ii) n = 20 (even)

Median = (½) [(n/2)th + (n/2 + 1)th observation]

= (½) [(20/2) th + (20/2 + 1)th observation]

= (½) [10th + 11th observation]

= (½) (19 + 16)

= 35/2

= 17.2

(iii) 10 has occurred the maximum number of times.

Therefore, mode = 10

(b) In the figure, part of a geometrical figure is given. Complete the figure so that the resulting figure is symmetrical about both the x-axis and the y-axis. [3]

Solution:

The complete figure so that the resulting figure is symmetrical about both the x-axis and the y-axis is:

(c) Show that the opposite angles of a cyclic quadrilateral are supplementary. [4]

Solution:

Let ABCD be a cyclic quadrilateral of a circle with centre O.

Join OB and OD.

Consider two opposite angles ∠BAD and ∠BCD.

We know that the angle subtended by the arc at the centre is twice the angle subtended by it at the remaining part of the circle.

∠BOD = 2∠BAD….(i)

Also,

reflex∠BOD = 2∠BCD….(ii)

Adding (i) and (ii),

2∠BAD + 2∠BCD = ∠BOD + reflex ∠BOD

2(∠BAD + ∠BCD) = 360°

∠BAD + ∠BCD = 360°/2

∠BAD + ∠BCD = 180°

Therefore, the opposite angles of a cyclic quadrilateral are supplementary.

Question 4

(a) Draw a circle of radius 2.5 cm. Draw two tangents to it inclined at an angle of 45° to each other. [3]

Solution:

Therefore, PA and PB the required tangents to the circle.

(b) Find the volume of a solid in the form of a right circular cylinder with hemispherical ends whose extreme length is 22 cm and diameter 3 cm. [3]

Solution:

Given,

Total length of the solid = 22 cm

Diameter of circular bases = 3 cm

Radius of cylinder = Radius of hemisphere = r = 3/2 = 1.5 cm

Height of cylinder(h) = Total length – Radius of 2 hemispheres

= 22 – 2(1.5)

= 19 cm

Volume of the solid = Volume of cylinder + 2 × Volume of hemisphere

= πr²h + 2 × (⅔)πr³

= (22/7) × 1.5 × 1.5 × 19 + (4/3) × (22/7) × 1.5 × 1.5 × 1.5

= 134.36 + 14.14

= 148.5 cm³

(c) Mr. Sagar’s savings bank account passbook entries are as follows:

Calculate the interest on the minimum balance on or after the 10th day of the month from April to September at 4 ½ % p.a. [4]

Solution:

Total = Rs. 28925

Rate of interest = R = 4 ½ % = 9/2% = 4.5%

Time = 6 months

I = PTR/100

= (28925 × 4.5 × 6)/ (12 × 100)

= 650.8125

Hence, the interest for six months is Rs. 650.8125.

Question 5

(a) Draw two intersecting lines AB and CD. Find the position of the point which is 2 cm away from AB and 1.8 cm away from CD. [3]

Solution:

AB and CD are two intersecting lines that intersect each other at the point O.

Construct a line EF which is parallel to AB which 2 cm away and GH which is parallel to CD (1.8 cm away) intersecting each other at point P.

Hence, P is the position of the required point.

(b) In how many years a sum of Rs. 6400 compounded quarterly at the rate of 5% p.a. will amount to Rs. 6561? [3]

Solution:

Given,

P = Rs. 6400

R = 5% p.a. = (5/4)% per quarter

Let n be the number of years.

A = Rs. 6561

A = P(1 + R/100)ⁿ

6561 = 6400 [1 + (5/400)]ⁿ

6561/ 6400 = [1 + (1/80)]ⁿ

6561/ 6400 = (81/80)ⁿ

(81/80)² = (81/80)ⁿ

⇒ n = 2

Hence, the required number of years is 2.

(c) Angle of elevation of a cloud from a point 20 m above the surface of a lake is 30°. The angle of depression of the reflection of the cloud in the lake from the same point is 60°. Calculate the height of the cloud above the lake. [4]

Solution:

Let A be the surface of the lake.

C be the cloud and D be its reflection.

B be the point of observation.

AB = EF = 20 m

CE = DE = h

AE = BF

In right triangle BFC,

tan 30° = CF/BF

1/√3 = (h – 20)/ BF

BF = √3(h – 20)….(i)

In right triangle BFD,

tan 60° = DF/BF

√3 = (h + 20)/ BF

BF = (h + 20)/ √3….(ii)

From (i) and (ii),

√3(h – 20) = (h + 20)/ √3

(√3)(√3)(h – 20) = h + 20

3(h – 20) – h = 20

3h – 60 – h = 20

2h = 20 + 60

2h = 80

h = 80/2

h = 40 m

Hence, the height of the cloud above the lake is 40 m.

Question 6

(a) In the figure, AX = 2BX and CX = 2XD. Prove that:

(i) ΔAXC and ΔBXD are similar

(ii) AC || DB [3]

Solution:

(i) Given,

AX = 2BX

CX = 2XD

∠AXC = ∠BXD (vertically opposite angles)

By SAS similarity criterion,

ΔAXC ~ ΔBXD

(ii) ∠XAC = ∠XDB (corresponding angles)

Therefore, AC || BD

(b) A manufacturer sold a dining table to a dealer for Rs. 8000. The dealer sold it to the shopkeeper at a profit of Rs. 2000. The shopkeeper sold it to the consumer at a profit of Rs. 3000. Find:

(i) the total VAT received by the government at 8%

(ii) the amount paid by the consumer inclusive of sales tax. [3]

Solution:

Given,

Selling price by manufacturer = Rs. 8000

Selling price by the dealer = Rs. 8000 + Rs. 2000 = Rs. 10000

Selling price by the shopkeeper = Rs. 10000 + Rs. 3000 = Rs. 13000

(i) Total VAT received by the government = (8/100) × Rs. 8000 + (8/100) × Rs. 10000 + (8/100) × Rs. 13000

= Rs. 640 + Rs. 800 + Rs. 1040

= Rs. 2480

(ii) The amount paid by the consumer inclusive of sales tax = Rs. 13000 + (8/100) × Rs. 13000

= Rs. 13000 + Rs. 1040

= Rs. 14040

(c) Two unbiased coins are tossed simultaneously. Find the probability of getting:

(i) two heads

(ii) one head

(iii) at least one head. [4]

Solution:

Given,

Two unbiased coins are tossed simultaneously.

Sample space = S = {HH, HT, TH, TT}

n(S) = 4

(i) Let A be the event of getting two heads.

A = {HH}

n(A) = 1

P(A) = n(A)/n(S) = ¼

(ii) Let B be the event of getting one head.

B = {HT, TH}

n(B) = 2

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

= 2/4

= ½

(iii) Let C be the event of getting at least one head.

C = {HT, TH, HH}

n(C) = 3

P(C) = n(C)/ n(S)

= ¾

Question 7

(a) Construct a ΔABC in which AB = AC = 3 cm and BC = 2 cm. Using a ruler and compasses only, draw the reflection A’BC of ΔABC in BC. Draw the lines of symmetry of the figure ABA’C. [3]

Solution:

Given,

AB = AC = 3 cm

BC = 2 cm

AA’ and BC are the two lines of symmetry.

(b)

Solution:

Given,

(c) Using the quadratic formula, solve: [(x – 1)/ (x – 2)] + [(x – 2)/ (x – 3)] = 4. [4]

Solution:

(x – 1)(x – 3) + (x – 2)(x – 2) = 4(x – 2)(x – 3)

x² – 3x – x + 3 + x² – 2x – 2x + 4 = 4(x² – 3x – 2x + 6)

2x² – 8x + 7 = 4x² – 20x + 24

4x² – 20x + 24 – 2x² + 8x – 7 = 0

2x² – 12x + 17 = 0

Comparing with the standard form ax² + bx + c = 0,

a = 2, b = -12, c = 17

Using quadratic formula,

x = [-b ± √(b² – 4ac)]/ 2a

x = [-(-12) ± √{(-12)² – 4(2)(17)}] / 2(2)

= [12 ± √(144 – 136)]/ 4

= [12 ± √8]/ 4

= [12 ± 2√2]/ 4

= 2(6 ± √2)/ 4

= (6 ± √2)/ 2

Question 8

(a) In the figure, P is a point on AB such that AP : PB = 4 : 3 and PQ || AC. Calculate the ratio of PQ : AC. [3]

Solution:

Given,

AP : PB = 4 : 3

PQ || AC

By BPT,

AP/PB = CQ/QB

CQ/QB = 4/3

⇒ BQ/BC = 3/7….(i)

Now,

∠PQB = ∠ACB (corresponding angles)

∠QPB = ∠CAB (corresponding angles)

By AA similarity,

ΔPBQ ~ ΔABC

⇒ PQ/AC = BQ/BC (by BPT)

⇒ PQ/AC = 3/7 [From (i)]

Therefore, PQ : AC = 3 : 7

(b) In what ratio does the point (-3, 7) divide the join of A(-5, 11) and B(4, -7)? [3]

Solution:

Let P(-3, 7) divide the line joining A(-5, 11) and B(4, -7) in the ratio λ : 1.

(-5, 11) = (x₁, y₁)

(4, -7) = (x₂, y₂)

M : n = λ : 1

Using the section formula,

P(x, y) = [(mx₂ + nx₁)/ (m + n), (my₂ + ny₁)/ (m + n)]

P(-3, 7) = [(4λ – 5)/ (λ + 1), (-7λ + 11)/ (λ + 1)]

Now,

-3 = (4λ – 5)/ (λ + 1)

-3(λ + 1) = 4λ – 5

4λ – 5 + 3λ + 3 = 0

7λ – 2 = 0

λ = 2/7

Hence, the required ratio is 2 : 7.

(c) Neha invests in 12% Rs. 25 shares of a company quoted at Rs. 36. Her income from this investment is Rs. 720. Calculate:

(i) the total amount of money invested by her in these shares

(ii) the number of shares bought by her

(iii) % return on her investment. [4]

Solution:

Let n be the number of shares.

Given,

Nominal value = Rs. 25

Market value = Rs. 36

Nominal value of n shares = Rs. 25n

Dividend = 12% of Rs. 25n

Income from the shares = Rs. 720

720 = (12/100) × 25n

n = (720 × 100)/ (25 × 12)

n = 240

Total investment = Market value × n

= Rs. 36 × 240

= Rs. 8640

Percentage of return on investment = (Income/ Total investment) × 100

= (720/ 8640) × 100

= 83.33%

Question 9

(a) Find the equation of a line that passes through (1, 3) and is parallel to the line y = -2x + 4. [3]

Solution:

The line parallel to y = -2x + 4 will have the same slope.

By comparing y = mx + c

Slope = m = -2

Given point is (1, 3)

Equation of the line passing through (x₁, y₁) and having slope m is:

y – y₁ = m(x – x₁)

y – 3 = -2(x – 1)

y – 3 = -2x + 2

y = -2x + 2 + 3

y = -2x + 5

(b) In the figure, O is the centre of the circle. If ∠PAO = 30° and ∠PBO = 40°, find:

(i) ∠APB

(ii) ∠AOB. [3]

Solution:

Given,

∠PAO = 30°

∠PBO = 40°

Join AB (chord) such that OA and OB are angle bisectors of ∠PAB and ∠PBA respectively.

Thus, ∠OAB = 30° and ∠OBA = 40°

In triangle AOB,

∠OAB + ∠OBA + ∠AOB = 180° (by the angle sum property of a triangle)

30° + 40° + ∠AOB = 180°

∠AOB = 180° – 30° – 40°

∠AOB = 110°

And

∠APB = (½) × ∠AOB

= (½) × 110°

= 55°

(c) The area of the quadrant OAB of a circle is 9 ⅝ cm². Calculate:

(i) OA

(ii) the perimeter of the quadrant. [4]

Solution:

(i) Let r be the radius of the quadrant of the circle.

i.e. OA = OB = r

Given,

Area of the quadrant OAB = 9 ⅝ cm² = 77/8 cm²

(πr2)/4 = 77/8

(22/7) × r² = (77 × 4)/ 8

r² = (77 × 7)/ (2 × 22)

r² = (7 × 7)/ (2 × 2)

r² = (7/2)²

r = 7/2 cm

Therefore, OA = 7/2 cm

(ii) Perimeter of quadrant = (¼) 2πr

= (¼) × 2 × (22/7) × (7/2)

= 11/2 cm

Question 10

(a) Find the value of m such that the lines 3xm + 3y = 5 and y = 1 – 2x are perpendicular to each other. [4]

Solution:

Given,

3xm + 3y = 5

3y = -3xm + 5

y = (-3mx/3) + (5/3)

y = -mx + (5/3)

Slope = m₁ = m

And

y = 1 – 2x

y = -2x + 1

Slope = m₂ = -2

Given that, the two lines are perpendicular to each other.

m₁m₂ = -1

(m)(-2) = -1

-2m = -1

m = ½

(b) Draw an Ogive for the following distribution and hence estimate the median.

Solution:

Less than cumulative frequency distribution table:

N = 50

N/2 = 50/2 = 25

Ogive:

Therefore, the median = 43

Question 11

(a) Show that the equation x² + 2px – 3 = 0 has real and distinct roots for all values of p. [3]

Solution:

Given,

x² + 2px – 3 = 0

Comparing with the standard form ax² + bx + c = 0,

a = 1, b = 2p, c = -3

Condition for real and distinct roots is:

b² – 4ac > 0

(2p)² – 4(1)(-3) > 0

4p² + 12 = 0

4(p² + 3) > 0

p² + 3 > 0

For any value of p, the above condition will be satisfied.

Thus, for any value of p, p² + 3 > 0 is always true.

Hence, the given equation has real and distinct roots for all the values of p.

(b) Prove that: [1/ (1 – sin θ)] + [1/ (1 + sin θ)] = 2 sec²θ. [3]

Solution:

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

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

= (1 + sin θ + 1 – sin θ) / (1² – sin²θ)

= 2/ (1 – sin²θ)

Using the identity sin²A + cos²A = 1,

= 2/cos²θ

= 2 sec²θ

= RHS

Hence proved.

(c) From the following frequency distribution, find mean, mode and median.

Solution:

Mean = ∑fx/ ∑f

= 381/25

= 15.24

N = ∑f = 25

N/2 = 25/2

Cumulative frequency greater than and nearest to N/2 is 14.

Median = 13

Highest frequency = 7

Hence, the mode = 13