ICSE Class 9 Maths Sample Question Paper 8 with Answers

ICSE Class 9 Maths Sample Question Paper 8 with Answers

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) The mean of 100 observations was found to be 30. If two observations were wrongly taken as 32 and 12 instead of 23 and 11, find the correct mean.
Answer:
Here, n = 100, \(\bar{x}\)= 30
∴ Incorrect = Σx= \(\bar{x}\) n = 30 x 100 = 3000.
∴ Correct Σ x = 3000 – (32 + 12) + (23 + 11)
= 3000 – 44 + 34 = 2990
∴ Correct mean = \(\frac{2990}{100}=29.9\)

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(b) Determine the rate of interest for a sum that becomes \(\frac{216}{125}\) times of itself in 3 years, compounded annually.
Answer:
Let principal be ? P and rate of interest be r% p. a. So,
ICSE Class 9 Maths Question Paper 8 with Answers 6

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(c) Without using tables, find the value of :
ICSE Class 9 Maths Question Paper 8 with Answers 33
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 7

Question 2.
(a) If \(x=\frac{3+\sqrt{7}}{2}\) find the value of \(4 x^{2}+\frac{1}{x^{2}}\)
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 8
ICSE Class 9 Maths Question Paper 8 with Answers 9

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(b) In the given figure, ABCD is a rectangle with sides AB = 8 cm and AD = 5 cm. Compute : (i) area of parallelogram ABEF, (ii) area of ΔEFG.
ICSE Class 9 Maths Question Paper 8 with Answers 1
Answer:
Given : AB 8 cm, AD = 5 cm.
(i) Area of parallelogram ABEF = Area of rectangle ABCD
(∵ they are on same base and between same parallels)
= (8 x 5) cm2 = 40 cm2
Area of ΔEFG = \(\frac{1}{2}\) x Area of parallelogram ABEF
(∵ both are on same base and between same parallels)
= \(\frac{1}{2}\) x 40 cm2 20 cm2

(c) Without using tables, find the value of :
\(\frac{(b+c)^{2}}{b c}+\frac{(c+a)^{2}}{c a}+\frac{(a+b)^{2}}{a b}\)
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 10

ICSE Class 9 Maths Sample Question Paper 8 with Answers

Question 3.
(a) Solve for x : 2x +3 + 2x+1 = 320.
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 11

(b) In the given figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of lengths 24 cm and 18 cm respectively.
ICSE Class 9 Maths Question Paper 8 with Answers 2
Answer:
Given: radius = OA = OC = 15 cm, AB || CD.
Let AB = 24 cm, CD = 18 cm.
We know perpendicular drawn from centre to the chord, bisects the chord
∴ M and N are mid-point of sides AB and CD respectively.
AM= \(\frac{1}{2}\) AB= \(\frac{1}{2}\) x24=12cm.
ICSE Class 9 Maths Question Paper 8 with Answers 13

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(c) Factorize : 4 (2a – 3)2 – 3 (2a – 3) (a – 1) – 7 (a – 1)2.
Answer:
Given expression is, 4 (2a – 3)2 – 3 (2a – 3) (a – 1) – 7 (a – 1)2
Let 2a – 3 = x and a – 1 = y
The expression becomes
= 4a2 – 3xy – 7y2 = 4x2 – (7 – 4) xy – 7y2
= 4x2 – 7xy + 4xy – 7y2
= x (4x – 7y) + y (4x – 7y)
= (4x – 7y) (x + y)
Substituting values of x and y, we have
= {4 (2a – 3) – 7 (a – 1)} {2a – 3 + a – 1)
= (8a -12 -7a + 7) (3a – 4) = (a – 5) (3a – 4).

Question 4.
(a) Solve for x : log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3.
Answer:
log {x + 5) + log {x – 5)
=4 log 2 + 2 log 3 log (x + 5) + log (x – 5)
= log 24 + log 32 log (x + 5) + log (x – 5)
= log 16 + log 9 log [(x + 5) (x – 5)]
= log (16 x 9) log(x2 – 25) – log 144
⇒ x2 – 25 = 144
⇒ x2 = 144 + 25 = 169
⇒ x = √169 = 13

(b) Solve simultaneously : \(2 x+\frac{x-y}{6}=2 ; x-\frac{(2 x+y)}{3}=1\)
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 14
ICSE Class 9 Maths Question Paper 8 with Answers 15
ICSE Class 9 Maths Question Paper 8 with Answers 16

(c) If 8 cot 915, find the value of: \(\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}\)
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 17

ICSE Class 9 Maths Sample Question Paper 8 with Answers

Question 5.
(a) The cost of enclosing a rectangular garden with a fence all around at the rate of ₹ 15 per metre is ₹ 5400. If the length of the garden is 100 m, find the area of the garden.
Answer:
Total cost of fendng = ₹ 5400
Rate = ₹ 15 per metre
Perimeter = \(\frac{5400}{15}\) = 360 m
Length, l= 100 m
Let breadth be b m.
2 (l + b) = 360
b = 180 \(\frac{360}{2} \) 100 = 80 m
Area = i x b = 100 m x 80 m = 8000 m2

(b) If 4 sin2 x° – 3 = 0 and x° is an acute angle, find (i) sin x° (ii) x°.
Answer:
Given: 4 sin2 x° – 3 = O
(i) 4 sin2 x° =3
sin2 x° = \(\frac{3}{4}\)
sin x°= \(\frac{\sqrt{3}}{2}\)

(ii) Now sin x°= \(\frac{\sqrt{3}}{2}\)
⇒ sin x°= sin 60°
⇒ x° = 60°

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(c) Draw a frequency polygon from the following data :

Age (in years)25-3030-3535-4040-4545-50
No. of doctors4060503520

Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 18
Question 6.
(a) If \(\frac{(x-\sqrt{24})(\sqrt{75}+\sqrt{50})}{\sqrt{75}-\sqrt{50}}=1\) find the value of x.
Answer:

ICSE Class 9 Maths Question Paper 8 with Answers 19ICSE Class 9 Maths Question Paper 8 with Answers 20

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(b) From the given figure, find the values of a and b.
ICSE Class 9 Maths Question Paper 8 with Answers 3
Answer:
Given : AD||BC
∠DBC = ∠ADB = a° (Alternate angles)
Now, a + 28° = 75° (Exterior angle is equal to sum of interior opposite angles)
⇒ a = 75° – 28° = 47°.
Also, ∠ABC + ∠BAD = 180° (Co-interior angles)
⇒ a + b + 90° = 180°
⇒ 47° + b + 90° = 180°
⇒ b = 180° – 137° = 43°
a = 47°, b = 43°

(c) Show that the points A (2, – 2), B (8, 4), C (5, 7) and D (- 1, 1) are the vertices of a
rectangle. Also, find the area of the rectangle.
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 21
i.e., opposite sides are equal and diagonals are equal
ABCD is a rectangle.
Hence Proved.
Area of rectangle = AB x BC = 6√2 x 3√2 = 36 sq. units.

Question 7.
(a) By using suitable identity, evaluate : (9.8).
Answer:
(9.8)1 = (10 – 0.2)3 = 103 – 3 x 102 x 0.2 + 3 x 10 x (0.2)2 – (0.2)3
= 1000 – 60 + 1.2 – 0.008 = 941.192.

(b) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB
and parallel to BC intersects AC at D. Show that (i) D is mid-point of AC (ii) MD L AC
(iii) CM = MA = \(\frac{1}{2}\) AB.
Answer:
Given : M is mid-point of AB, ∠C = 90°, MD||BC.
Join MC.
(i) ∵ M is mid-point of AB and MD|| BC
ICSE Class 9 Maths Question Paper 8 with Answers 23
∴ By the converse of mid-point theorem,
D is mid-point of AC.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(ii) ∠BCD + ∠CDM = 1800 (Co-interior angles, MDIIBC)
⇒ 90° + ∠CDM = 180° (∠BCD = 90°)
∠CDM= 180°-90°=90°
MD ⊥ AC. Hence Proved.

(iii) In ΔAMD and ΔCMD,
AD = CD (D is mid-point of AC)
∠ADM = ∠CDM (Each being 90°)
MD = MD (Common side)
∴ ΔAMD ≅ ΔCMD. (SAS axiom)
∴ AM = CM (c.p.c.t.)
Also, AM = \(\frac{1}{2}\) AB ( M is mid-point of AB)
∴ CM = AM = \(\frac{1}{2}\) AB.  Hence Proved.

(c) Solve \(x+\frac{1}{x}=2 \frac{1}{2}\)
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 24

ICSE Class 9 Maths Sample Question Paper 8 with Answers

Question 8.
(a) If : a = b2x, b – c2y and c = a2z, show that 8xyz = 1.
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 25

(b) In the given figure, ABC is a right triangle at C. If D is the mid-point of BC, prove that AB2 = 4AD2 – 3AC2.
ICSE Class 9 Maths Question Paper 8 with Answers 4
Answer:
Given :∠C = 90°, D is mid-point of BC.
In ΔABC, In ΔACD, ⇒ AB2 = AC2 + BC2 AD2
⇒ AC2 + CD2 CD2
⇒ AD2 – AC2 (Pythagoras theorem) … (i) (Pythagoras theorem)
⇒ \(\left(\frac{1}{2} \mathrm{BC}\right)^{2}\) = AD2 – AC2 (∵ D is mid-point of BC)
⇒ \(\frac{1}{4}\) BC2 = AD2 – AC2 4
⇒ BC2 = 4AD2 – 4AC2 …(h)
Using (i) and (ii), we have
AB2 = AC2 + 4AD2 – 4AC2
⇒ AB2 = 4AD2 – 3AC2
Hence Proved.

(c) Find the value of x, if tan 3x = sin 45° cos 45° + sin 30°.
Answer:
Given: tan 3x = sin 45° cos 45° + sin 30°
⇒  tan 3x \(=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2}\)
⇒ tan 3x = \(=\frac{1}{2}+\frac{1}{2}\)
⇒ tan 3x = 1
⇒ tan 3x = tan 45°
⇒ 3x = 45° ,
⇒ 45°
x = \(\frac{45^{\circ}}{3}\) = 15°

ICSE Class 9 Maths Sample Question Paper 8 with Answers

Question 9.
(a) Factorize : 12 – (x + x1) (8 – x – x2).
Answer:
12 – (x + x2) (8 – x – x2) = 12 – (x + x2) {8 – (x + x2)}
Let  x + x2 = a
⇒ 12 – a (8 – a) = 12 – 8a + a2
⇒ a2 – 8a + 12
⇒ a2 – (6 + 2) a + 12
= a2 – 6a – 2m + 12
= a (a – 6) – 2 (a – 6)
⇒ (a-6) (a- 2)
Substituting a = x + x2, we get
⇒ (x + x2 – 6) (x + x2 – 2)
⇒ (x2 + x – 6) (x2 + x – 2)
⇒ (x2 + 3x – 2x – 6) (x2 + 2x – x – 2)
⇒ {x (x + 3) – 2 (x + 3)} {x (x + 2) – 1 (x + 2)}
⇒ (x + 3) (x – 2) (x + 2) (x – 1)
⇒ (x – 1) (x + 2) (x – 2) (x + 3)

(b) A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Answer:
Let the speed of the train be x km/h and that of the car be y km/h.
Total distance travelled = 370 km.
Case I : Distance travelled by train = 250 km.
Distance travelled by car = (370 – 250) km = 120 km
∴ Time taken by train \( =\frac{250}{x} \mathrm{~h}\)
ICSE Class 9 Maths Question Paper 8 with Answers 26
ICSE Class 9 Maths Question Paper 8 with Answers 27
ICSE Class 9 Maths Question Paper 8 with Answers 28
ICSE Class 9 Maths Question Paper 8 with Answers 29

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(c) Express as a single logarithm :
2 log10 5 – log10 2 + 3log10 4+1
Answer:
2 log10 5 – log10 2 + 3 log10 4 + 1 = log10 52 – log10 2 + log10 43 + log10 10
\(=\log _{10}\left(\frac{5^{2} \times 4^{3} \times 10}{2}\right)\)
= log10 8000 = log10 (20)3 = 3 log10 20

Question 10.
(a) The value of a car purchased 2 years ago depreciates by 10% every year. Its present value is ₹ 1,21,500. Find the cost price of the car. What will be its value after 2 years ?
Answer:
ICSE Class 9 Maths Question Paper 8 with Answers 30

(b) Construct a quadrilateral ABCD given that AB = 4.5 cm, ∠BAD = 60°, ∠ABC = 105°, AC = 6.5 cm and AD = 5 cm.
Answer:
Given : AB 4.5 cm, Z BAD = 60°, Z ABC 105°, AC = 6.5 cm and AD – 5 cm. Steps of
construction :
(1) Draw AB = 4.5 cm.
(2) At A, draw ∠BAX = 60°.
(3) At B, draw ∠ABY = 105°.
ICSE Class 9 Maths Question Paper 8 with Answers 31
(4) From A, cut BY at C such that AC = 6.5 cm.
(5) From A, cut AX at D such that AD = 5 cm.
(6) Join CD.
Hence, ABCD is the required quadrilateral.

(c) Factorize : x9 + y9.
Answer:
= x9 + y9 = (x3)3 + (y3)3 = (x3 + y3) {(x3)2 – x3y3 + (y3)2}
= (x + y) (x2 – xy + y2) (x6 – x3y3 + y6).

Question 11.
(a) In the given figure, ABCD is a parallelogram. Find the values of x, y and z.
ICSE Class 9 Maths Question Paper 8 with Answers 5
Answer:
Given : ABCD is a parallelogram.
AB = CD
⇒ 3x – 1 = 2x + 2
⇒ 3x – 2x =2 + 1
⇒ x = 3
Also, ∠D = ∠B = 102° ( ∵ Opposite angles are equal) Exterior
In ΔACD, y = 50° + 102° (∵ angle is equal to sum of interior opposite angles)
= 152°
and ∠A + ∠D = 180°
⇒ z + 50° + 102° = 180°
⇒ z = 180° – 152° = 28°
⇒ x = 3, y = 152° z = 28°

ICSE Class 9 Maths Sample Question Paper 8 with Answers

(b) Draw the graph of 3x + 2 = 0 and 2y – 1 = 0 on the same graph sheet. Do these lines intersect ? If yes, find the point of intersection.
Answer:
Given: 3x+2=0 ……..(i)
and 2y – 1 = 0 ……. (ii)
From (i), 3x = – 2
= \(x=\frac{-2}{3}\)
It is a straight line parallel to Y-axis at \(x=\frac{-2}{3}\)
From (ii), 2y = 1
⇒ y = \(\frac{1}{2}\)
It is a straight line parallel to Y-axis at y = \(\frac{1}{2}\)
ICSE Class 9 Maths Question Paper 8 with Answers 32

(c) Prove that (sin A + cos A)2 + (sin A – cos A)2 = 2.
Answer:
L. H.S. = (sin A + cos A)2 + (sin A – cos A)2
= sin2A + cos2 A + 2 sin A cos A + sin2A + cos2A – 2 sin A cos A = 2 (sin2A + cos2A)
= 2 x 1
= 2 = R.H.S.

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 7 with Answers

ICSE Class 9 Maths Sample Question Paper 7 with Answers

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) If, in a ∆ABC, AB = 3 cm, BC = 4 cm and ∠ABC = 90°, find the values of cos C, sin C and
tan C.
Answer:
Given : AB = 3 cm, BC = 4 cm, ∠ABC = 90°
By Pythagoras theorem,
AC2 = AB2 + BC2 = 32 + 42 = 25
ICSE Class 9 Maths Question Paper 7 with Answers 7

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(b) A man purchased an old scooter for ₹ 16,000. If the cost of the scooter after 2 years depreciates to ₹ 14,440, find the rate of depreciation.
Answer:
Present value (V0) = ₹ 16,000
Value after 2 year (V1) = ₹ 14,440
∴ n =2
Let r be the rate of depreciation.
ICSE Class 9 Maths Question Paper 7 with Answers 8

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(c) Prove that √2 + √5 is irrational.
Answer:
Let us assume that √2 + √5 is a rational number.
Then \(\sqrt{2}+\sqrt{5}=\frac{a}{b}\)
Where a and b co-prime positive integers.
\(\frac{a}{b}-\sqrt{2}=\sqrt{5}\)
ICSE Class 9 Maths Question Paper 7 with Answers 9

ICSE Class 9 Maths Sample Question Paper 7 with Answers

Question 2.
(a) If \(x=\frac{1}{x-2 \sqrt{3}}\) , find the values of (i) x – \(\frac{1}{x}\) (ii) x + \(\frac{1}{x}\).
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 10

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(b) In the given figure, ABC is an equilateral triangle. Find the measures of angles marked by x, y and z.
ICSE Class 9 Maths Question Paper 7 with Answers 1
Answer:
Given : ABC is an equilateral triangle.
∠ABC = ∠ACB = ∠B AC = 60°.
Now, ∠BAD + ∠ADB = ∠ABC (Ext. angle is equal to sum of int. opp. angles)
⇒ x + 40° = 60°
⇒ x = 60° – 40°
⇒ x = 20°.
Also, ∠CAE + ∠AEC = ∠ACB (Ext. angle is equal to sum of int. opp. angles)
⇒ y + 30° = 60°
⇒ y = 60° – 30°
⇒ y = 30°
and ∠ACE +∠ACB = 180° (Linear Pair)
⇒ z + 60° = 180°
⇒ z = 180° – 60°
⇒ z = 120°

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(c) Solve \(\frac{2}{3} x^{2}-\frac{1}{3} x-1=0\)
Answer:
\(\frac{2}{3} x^{2}-\frac{1}{3} x-1=0\)
\(3 \times \frac{2}{3} x^{2}-3 \times \frac{1}{3} x-3 \times 1=3 \times 0\)
⇒ 2x2 – x – 3 = 0
⇒ 2 x 2 – (3 – 2)x -3=0
⇒ 2x2 – 3x + 2x – 3 = 0
⇒ x (2x – 3) + 1 (2x – 3) = 0
⇒ (2x – 3) (x + 1) = 0
⇒ 2x-3=0 or x + 1= 0
⇒ x= \(\frac{3}{2}\) or x =-1
⇒ x= \(\frac{3}{2}\) or -1

Question 3.
(a) Factorize : a3 – b3 – a + b.
Answer:
a3 -b3 – a + b = (a-b) (a2 + ab + b2) – (a-b) = (a-b) (a2 + ab + b2 – 1).

(b) Draw a histogram to represent the following :

Class Interval40 – 4848-5656-6464-7272 – 80
Frequency1525353010

Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 11

ICSE Class 9 Maths Sample Question Paper 7 with Answers
(c) Prove that \(\sqrt{\frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}}}=\tan 30^{\circ}\)
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 12

Question 4.
(a) Simplify: \(\frac{5^{2(x+6)} \times(25)^{-7+2 x}}{(125)^{2 x}}\)
Answer:

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(b) In the figure, DE||BC. Prove that (i) Area of ΔACD = Area of ΔABE (ii) Area of ΔOBD = Area of ΔOCE.
ICSE Class 9 Maths Question Paper 7 with Answers 2
Answer:
Given DE || BC
Area of ΔBCD = Area of ΔBCE
(Triangles on same base and between same parallels have equal area) Now, Area of ΔACD + Area of ΔBCD = Area of ΔABE + Area of ΔBCE
⇒ Area of ΔACD = Area of ΔABE (∵ Area of ABCD = Area of ABCE).
Hence Proved.

(ii) Area of ABCD = Area of ABCE [From (i)]
⇒ Area of ABCD – Area of ΔOBC = Area of ΔBCE – Area of ΔOBC
(Subtracting area of ΔOBC from both side)
⇒ Area of ΔOBD = Area of ΔOCE.
Hence Proved

(c) If log10 x + \(\frac{1}{3}\) log10 y = 1, express y in terms of x.
Answer:
Given log10 x + \(\frac{1}{3}\) log10y = 1
log10 x + log10 y1/3 = log10 10
log10 (xy1/3) = log10 10

ICSE Class 9 Maths Question Paper 7 with Answers 14

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) The mean of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. Find the correct mean of the observations.
Answer:
We know,
Σx =\(\bar{x}\) x n
Incorrect ∑ x = 35 x 9 = 315
Correct ∑ x =315 – 18 + 81 = 378
Correct mean = \(\frac{378}{9}=42\)

(b) In the given figure, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find (i) radius of the circle (ii) Length of chord CD.

ICSE Class 9 Maths Question Paper 7 with Answers 3
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 15
⇒ 169 =144 ÷ CN2
⇒ CN2 = (169 – 144) = 25
⇒ CN= √25 =5
⇒ CD =2 CN (∵ N is mid-point of CD)
⇒ 2 x 5 = 10cm.

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(c) If \(x^{2}+\frac{1}{x^{2}}=83\) find the value of \(x^{3}-\frac{1}{x^{3}}\)
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 16

Question 6.
(a) A cumulative frequency distribution is given below. Convert this into a frequency distribution table.

MarksBelow 45Below 60Below 75Below 90Below 105Below 120
No. of Students08234885116

Answer:

MarksNo. of StudentsClass IntervalFrequency
Below 4500-450
Below 60845 – 608 (8-0)
Below 752360 – 7515 (23 – 8)
Below 904875 – 9025 (48 – 23)
Below 1058590 -10537 (85 – 48)
Below 120116105 – 12031 (116 – 85)

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(b) Half the perimeter of a garden, whose length is 4 more than its width, is 36 m. Find the dimensions of the garden.
Answer:
Let length and breadth of the garden be x m and y m respectively.
According to the question,
x = 4 + y …(i)
and x + y = 36 …(ii)
Substituting x = 4 + y in equation (ii), we get
4 + y + y = 36
2y = 36 – 4
y = \(\frac{32}{2}\) = 16
Substituting y= 16 in equation (i), we get
x = 4 + 16 = 20
∴ Length = 20 m and breadth = 16 m.

(c) If x and y are rational numbers and \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=x-y \sqrt{3}\)
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 17

Question 7.
(a) Factorize : (x2 + y2 – z2)2 – 4x2y2.
Answer:
(x2 + y2 – z2)[1] – 4x2y2 = (x2 + y1 – z2)2 – (2xy)2
= (x2 + y2 – z2 + 2xy) (x2 + y2 – z2 – 2xy)
= {(x2 + y2 + 2xy) – z2} {(x2 + y2 – 2xy) – z2}
= {(x + y)2 – (z)2}  – y)2 – (z)2}
= {x + y + z) {x + y – z) {x – y + z) {x – y – z).

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(b) Prove that in a right angled triangle, the median drawn to the hypotenuse is half the hypotenuse in length.
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 19
(c) Find the value of x if 3 cot2 (x – 5°) = 1.
Answer:
3 cot2 (x – 5°) =1
1 cot2 (x – 5°) = \(\frac{1}{3}\)
cot (x – 5°) = \(\frac{1}{\sqrt{3}}\)
cot (x – 5°) = cot 60°
x – 5°= 60°
x = 60° + 5°
x = 65°

Question 8.
(a) Solve: \(\frac{x+y}{x y}=2 ; \frac{x-y}{x y}=1\)
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 22
ICSE Class 9 Maths Question Paper 7 with Answers 21

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(b) Construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.
Answer:
Given: AB =5.1cm, BC = 7cm and ∠ABC = 75°
Steps of construction:
(1) Draw BC=7cm.
(2) At B, draw ∠ XBC = 75°
(3) From B, cut-off BA = 5.1 cm on BX.
(4) From C, draw an arc of radius 5.1 cm.
(5) From A, draw an arc of 7 cm to cut the arc from C at D.
(6) Join CD and AD.
Hence, ABCD is the required parallelogram.
ICSE Class 9 Maths Question Paper 7 with Answers 20

(c) Calculate the distance between A (7, 3) and B on the X-axis whose abscissa is 11.
Answer:
Given : A (7, 3)
∵ B lies on the X-axis whose abscissa is 11, the coordinates of B are (11, 0)
\(\mathrm{AB}=\sqrt{(11-7)^{2}+(0-3)^{2}}=\sqrt{4^{2}+(-3)^{2}}=\sqrt{16+9}=\sqrt{25}\)
= 5 Units.

ICSE Class 9 Maths Sample Question Paper 7 with Answers

Question 9.
(a) A sum of money ₹ 15,000 amounts to ₹ 16,537.50 in x years at the rate of 5% p.a. compounded annually. Find x.
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 26

(b) In the given figure, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.
ICSE Class 9 Maths Question Paper 7 with Answers 4
Answer:
Given: ∠PSR = 90°, PQ = 10 cm, QS = 6 cm, RQ = 9 cm.
∴ In Δ PQ2, PQ2 = PS2 + QS2 (Pythagoras theorem)
102 =PS2+62
PS2 = 100 – 36
PS = √64 = 8cm
In ΔPRS, PR2 = PS2 + RS2 (Pythagoras theorem)
PR2 = 8 + (9 + 6)2 = 64 + 225 = 289
PR=√289=17cm.

(c) In the given figure, ACB is a semicircle whose radius is 10.5 cm and C is a point on the semicircle at a distance of 7 cm from B. Find the area of the shaded region.
ICSE Class 9 Maths Question Paper 7 with Answers 5
Answer:
For semi-circle,
r = 10.5 cm
∴ Area =\(\text { Area }=\frac{1}{2} \pi r^{2}=\frac{1}{2} \times \frac{22}{7} \times(10.5)^{2}=173.25 \mathrm{~cm}^{2}\)
For triangle ABC,
AB2 = BC2 + AC2 (Pythagoras theorem, ∠C = 90°)
(2 x 10.5)2 = 72 ÷ AC2
AC2 =441 – 49 =392
AC = 19.8 cm.
Area = x BC x AC = x 7 x 19.8 = 69.3 cm2
The area of shaded region = (173.25 – 69.3) cm2
= 103.95 cm2

ICSE Class 9 Maths Sample Question Paper 7 with Answers

Question 10.
(a) If a2 + b2 + c2 – ab – be – ca = 0, prove that a = b = c.
Answer:
Given a2 + b2 + c2 – ab – be – ca =0
⇒ 2 (a2 + b2 + c2 – ab – be – ca) = 0
⇒  2a2 + 1b2 + 2c2 -2ab – 2bc – 2ca = 0
⇒ (a2 – 2ab + b2) + (b2 – 2be + c2) + (c2 – 2ca + a2) = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 =0
The above expression is possible only if
⇒ (a- b)2 = 0 Ab- c)2 = 0, (c – a)2 = 0
a-b =0, b – c = 0, c-a = 0
a = b,b = c, c = a
a = b = c.
Hence Prove.

(b) Solve graphically x + 3y = 6; 2x – 3y = 12 and hence find the value of a, if Ax + 3y = a
Answer:
x+3y=6 ………. (i)
2x – 3y = 12 ….. (ii)
from (i)  x = 6 – 3y

X630
y012

∴ (6, 0), (3, 1), (0, 2)
From (ii),
2x = 3y + 12
x = \(\frac{3 y+12}{2}\)

X630
y012

(6, 0), (3, – 2), (0, – 4)
These points are piotted in the graph.
ICSE Class 9 Maths Question Paper 7 with Answers 23

The two lines intersect at the point (6, 0).
∴ x = 6, y = 0
Now 4x + 3 y = a
⇒ 4 x 6 + 3 x 0 = a
24 + 0 =a
⇒ a = 24

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(c) Given, 1008 = 2p.3q.7r, find the values of p, q, r and hence evaluate 2p.3q.7-r÷192.
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 24

Question 11.
(a) If log \(\frac{x-y}{2}=\frac{1}{2} \)(log x + log y), prove that x2 + y2 = 6xy.
Answer:
ICSE Class 9 Maths Question Paper 7 with Answers 25
x2 + y2 – 2xy = 4xy
x2 + y2 – 4xy + 2 xy
x2 + y2 = 6 xy.
Hence Proved

(b) In a pentagon ABCDE, AB||ED and ∠B = 140°. Find ∠C and ∠D if ∠C: ∠D = 5:6.
ICSE Class 9 Maths Question Paper 7 with Answers 6
Answer:
Given : AB||ED, ZB = 140°, ∠C : ∠D = 5:6.
Let  ∠C =5x, ∠D = 6x.
Now,∠A+∠E= 180° (Co-interior angles, AB||ED)
Also, ∠A+ ∠B+ ∠C+ ∠D+ ∠E= (5-2) x 180°
(∠A + ∠E) + ∠B + ∠C + ∠D
= 3 x 180° 180° + 140° + 5x + 6x = 540°
11 x = 540° – 320°
\(x=\frac{220^{\circ}}{11}\)
∠C = 5x = 5 x 20° = 100°
∠D = 6x = 6 x 20° = 120°

ICSE Class 9 Maths Sample Question Paper 7 with Answers

(c) Factorize : 4 a3b – 44 a2b + 112
Answer:
4 a3b  – 44 a2b + 112 ab = 4 ab (a2 – 11a + 28)
= 4 ab {(a2  – (7 + 4) a + 28)}
= 4ab(a2 – 7a – 4a+28)
= 4ab {a (a – 7) – 4(a – 7))
= 4ab (a – 7) (a – 4).

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 6 with Answers

ICSE Class 9 Maths Sample Question Paper 6 with Answers

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) Prove that log (1 + 2 + 3) = log 1 + log 2 + log 3.
Answer:
log (1 + 2 + 3) = log 6 = log (1 x 2 x 3)
= log 1 + log 2 + log 3.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(b) In the given figure, CD is a diameter which meets the chord AB in E such that
AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 1
Answer:
Given : AE = BE = 4 cm, CE = 3 cm
Let r be the radius (OB = OC)
OE = OC – CE = r – 3.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 6
⇒ OB2 = OE2 + BE2 (Pythagoras theorem)
⇒ r2 = (r – 3)2 + 42 r2
⇒ r2 – 6r + 9 + 16
⇒ 6r = 25
\(r=\frac{25}{6}=4 \frac{1}{6} \mathrm{~cm}\)

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) If ₹ 6,400 is invested at 6 \(\frac{1}{4}\) % p.a. compound interest, find (i) the amount after 2 years (ii) the interest earned in 2 years.
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 7

Question 2.
(a) Evaluate tan x and cos y from the given figure.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 2
Answer:
In ΔACD, AC2 = AD2 + CD2
132 – 52 + CD2
⇒ CD2 = 169 – 25 = 144
⇒ CD = 12.
In A BCD, BC2 = CD2 + BD2
= 144 + 162 = 144 + 256 = 400
BC =20
ICSE Class 9 Maths Sample Question Paper 6 with Answers 8

(b) ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that the altitudes are equal.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 3
Answer:
Given: AC=AB
∴ In ΔBEC and ΔCFB,
∠C=∠B (∵ AB = AC)
∠BEC = ∠CFB (Each being a right angle)
BC = BC (Common side)
∴ ΔBFC  ≅ ΔCFB (AAS axiom)
∴ BE = CF (c.p.ct.)
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) The mean of 5 observations is 15. If the mean of first three observations is 14 and that of the last three is 17, find the third observation.
Answer:
Mean of 5 observations = 15
∴ Sum of 5 observations = 15 x 5 = 75
Mean of first 3 observations = 14
∴ Sum of first 3 observations = 14 x 3 = 42
Mean of last 3 observations = 17
∴ Sum of last 3 observations = 17 x 3 = 51
∴ The third observation = (42 + 51) – 75 = 18.

Question 3.
(a) Factorize : x4 + 4
Answer:
x4 + 4 = (x4 + 4x2 + 4) – 4x2 = {(x2)2 + 2 .x. 2 + (2)2} – (2x)2
= (x2 + 2)2 – (2x)2 = (x2 + 2 + 2x) (x2 + 2 – 2x)
= (x2 + 2x+ 2) (x2 – 2x+ 2)

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(b) Evaluate : \(\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \cdot \tan 60^{\circ}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 9

(c) Simplify:\((81)^{3 / 4}-3 \times(7)^{0}-\left(\frac{1}{27}\right)^{-2 / 3}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 10

Question 4.
(a) If x \(\frac{2}{x}\) = 5, find the value of \(x^{3}-\frac{8}{x^{3}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 11

(b) If the hypotenuse of a right angled triangle is 6 m more than twice the shortest side and third side is 2 m less than hypotenuse, find the sides of the triangle.
Answer:
Let the shortest side be x m.
Then, Hypotenuse =(2x+6)cm,thirdside=2x+6-2=(2x+4)m.
∴ (2x + 6)2 = (2x + 4)2 + x2 (Using Pythagoras theorem)
= (2x)2 + 2.2x.6 + 62 = (2x)2 + 2.2xA +42 + x2
= 4x2 + 24x + 36 = 4x2 + 16x + 16 + x2
= 24x – 16x = x2 + 16 – 36
= x2 – 8x – 20 = 0
= x2 – (10 – 2) x – 20 =0
= x2 – 10x + 2x – 20 =0
= x (x – 10) + 2 (x – 10) = 0
= (x – 10) (x + 2) = 0
= x – 10 =0 or x + 2 = 0
= x = 10 or x = – 2
∴ x = 10 (∵ x cannot be negative)
∴ 2x + 6 = 2 x 10 + 6 = 26
and 2x + 4 = 2 x 10 + 4 = 24
Therefore, the sides are 10 m, 26 m and 24 m.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) Simplify: \(\frac{3}{\sqrt{6}+\sqrt{3}}-\frac{4}{\sqrt{6}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 12
ICSE Class 9 Maths Sample Question Paper 6 with Answers 13

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Solve : log10 6 + log10 (4x + 5) = log10 (2x + 7) +1
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 14

(b) 3 men and 4 women can do a piece of work in 14 days while 4 men and 6 women can do it in 10 days. How long would it take 1 woman to finish the work ?
Answer:
Let 1 man take x days and 1 woman take y days to finish the work.
∴ In 1 day, 1 man does = \(\frac{1}{x} \) work and 1 woman does = \(\frac{1}{y}\) work.
So, 3 men and 4 women do=\(3 \times \frac{1}{x}+4 \times \frac{1}{y}=\frac{3}{x}+\frac{4}{y}\)
It is given that 3 men and 4 women finish the work in 14 days.
\(\frac{3}{x}+\frac{4}{y}=\frac{1}{14}\) ………….(i)
Also, 4 men and 6 women do the work in 10 days.
= \(\frac{4}{x}+\frac{6}{y}=\frac{1}{10}\) …………(ii)
Multiplying equation (i) by 4 and equation (ii) by 3, we get
ICSE Class 9 Maths Sample Question Paper 6 with Answers 16

∴ One woman finish the work in 140 days.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) There are two regular polygons with number of sides equal to (n – 1) and (n + 2). Their exterior angles differ by 6°. Find the value of n
Answer:
For first polygon,
ICSE Class 9 Maths Sample Question Paper 6 with Answers 17
\(\frac{3}{n^{2}+2 n-n-2}=\frac{1}{60}\)
n2  + n – 2 = 180
n2 + n- 182 = 0
n2  + (14 – 13) n – 182 = 0
n2  + 14n – 13n – 182 = 0
n(n + 14) -13 (n + 14) = 0
(n + 14) (n – 13) = 0
n + 14 = 0 or n – 13 =0
n = -14 = 0  or n = 13 (∵n cannot be negative)
∴ n = 13.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

Question 6.
(a) If \(a^{2}+\frac{1}{a^{2}}=7\), find the value of \(a^{2}-\frac{1}{a^{2}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 18

(b) Construct a trapezium ABCD in which AD || BC, Z B = 60°, AB = 5 cm, BC = 6.2 cm, and CD = 4.8 cm.
Answer:
Given : AD||BC, ZB = 60°, AB = 5 cm, BC = 6.2 cm, and CD = 4.8 cm.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 19
Steps of construction :
(1) Draw BC = 6.2 cm.
(2) At B, draw ∠CBX = 60° and cut off BA = 5 cm.
(3) At A, draw exterior ∠XAY = 60° such that AY||BC.
(4) From C, cut-off AY at D such that CD = 4.8 cm and join CD.
Hence, ABCD is the required trapezium.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) The inner dimensions of a closed wooden box are 2 m, 1.2 m and 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1m3 of wood costs ₹ 5400.
Answer:
The inner dimensions of the closed box are 2 m, 1.2 m, 0.75 m.
Inner volume = (2 x 1.2 x 0.75) m3 = 1.8 m3
Thickness of the box = 2.5 cm = 2.5/100m= 0.025 m
∴ Outer dimensions are (2 + 2 x 0.025) m, (1.2 + 2 x 0.025) m, (0.75 + 2 x 0.025) m
i.e. 2.05 m, 1.25, 0.8 m.
∴ Outer volume = (2.05 x 1.25 x 0.8) m3 = 2.05 m3
Volume of wood = (2.05 – 1.8) m3 0.25 m3
Cost of 1 m3 of wood = ₹ 5400
Cost of 0.25 m3 of wood = ₹5400 x 0.25
= ₹ 1350

Question 7.
(a) Solve : 4x2 + 15 =16x
Answer:
4x2+15 =16x
4x2 – 16x+15 =0
4x2 – (10-t-6)x+15 =0
4x2 10x – 6x+15 =0
= 2x(2x – 5)- 3(2x – 5) =0
(2x – 5)(2x – 3) =0
2x – 5 =0 or 2x – 3=0
2x =5 or 2x=3
ICSE Class 9 Maths Sample Question Paper 6 with Answers 20

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(b) Find graphically the vertices of the triangle whose sides have equations
2y – x = 8, 5y – x = 14 and y – 2x = 1.
Answer:
Given equations are,
2y – x =8 ……….(i)
5y – x =14 …(ii)
and y – 2x =1 …(iii)
From(i), x =2y – 8
ICSE Class 9 Maths Sample Question Paper 6 with Answers 21
∴ (- 6, 1), (- 4, 2), (- 2, 3)
From (ii), x=2 y-8
ICSE Class 9 Maths Sample Question Paper 6 with Answers 22
(- 4, 2), (1, 3), (6, 4)
From (iii) y=2 x+1
ICSE Class 9 Maths Sample Question Paper 6 with Answers 23
∴ (1, 3), (2, 5), (- 1, – 1)
These points are plotted on the graph.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 24
The three lines intersect at point (- 4, 2), (1, 3) and (2, 5) which are the required vertices of triangle formed by them

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) If 3tan2 θ-1=0′, find cos 2θ, given that θ is acute.
Answer:
Given:  3tan2 θ-1=0
tan2θ = 1/3
tan θ   =\(\frac{1}{\sqrt{3}}\)]
⇒ tanθ = tan 30°
θ = 30°
cos2θ = cos (2 x 30°) = cos 60° =\(\frac{1}{2}\)

Question 8.
(a) Solve for x : 3(2x + 1) – 2x+2 + 5 = 0.
Answer:
⇒ 3(2x + 1) – 2x + 2 + 5 =0
⇒ 3.2x + 3 – 2x. 22 + 5 =0
⇒ 3.2x – 4.2x + 8=0
⇒ -2x = – 8
⇒ 2x = 23
⇒ x =3

(b) Find the area of a triangle whose perimeter is 22 cm, one side is 9 cm and the difference of the other two sides is 3 cm.
Answer:
One side = 9 cm, perimeter = 22 cm.
Let other two sides be a cm and b cm and a > b.
According to the question,
a + b + 9 = 22
⇒ a + b = 13 ………..(i)
and a – b =3 (Given) ………(ii)
Adding equations (i) and (ii), we have
2 a = 16 ⇒ a = 8
Subtracting equation (ii) from equation (i), we have
2b = 10 ⇒ b = 5
The sides are a = 8 cm, b = 5 cm, c = 9 cm
ICSE Class 9 Maths Sample Question Paper 6 with Answers 25

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) Insert four irrational numbers between 2√3 and 3√2
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 26

Question 9.
(a) Form a cumulative frequency distribution table from the following data by exclusive method taking 4 as the magnitude of class intervals.
31, 23, 19, 29, 20, 16, 10, 13, 34, 38, 33, 28, 21, 15, 18, 36, 24, 18, 15, 12, 30, 27, 23, 20, 17, 14, 32, 26, 25, 18, 29, 24, 19, 16, 11, 22, 15, 17, 10, 25.
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 27

(b) Solve simultaneously : \(\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{4}{y}=-\frac{1}{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 28
ICSE Class 9 Maths Sample Question Paper 6 with Answers 33

(c) The diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that area of ΔOAD = area of ΔOBC. Prove that ΔBCD is a trapezium.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 4
Answer:
Given: Area of ΔOAD = area of ΔOBC.
Draw DM ⊥ AB,CN ⊥AB.
∴ DM || CN (∵Both DM, CN are perpendicular to AB)
Now,
Area of ΔOAD = Area of ΔOBC
Area of ΔOAD + Area of ΔOAB = Area of ΔOBC + Area of ΔOAB
ICSE Class 9 Maths Sample Question Paper 6 with Answers 34

ICSE Class 9 Maths Sample Question Paper 6 with Answers

Question 10.
(a) If the interest is compounded half yearly, calculate the amount when the principal is ₹ 7400, the rate of interest is 5% p. a. and the duration is one year.
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 35

(b) Find the coefficient of x2 and x in the product of (x – 2) (x – 3) (x – 4).
Answer:
Given :  (x -2) (x – 3) {x – 4)
Here,  a = – 2, b = – 3, c = -4
Coefficient of x2 = a + b + c = (- 2) + (- 3) + (- 4) = – 9
Coefficient of x = ab + be + ca = (- 2) (- 3) + (- 3) (- 4) + (- 4) (- 2)
= 6 + 12 + 8 = 26

(c) If the figure given, ABCD is a trapezium in which AB || DC. P is the mid-point of AD and PR || AB. Prove that PR = \(\frac{1}{2} (AB + CD)\).
ICSE Class 9 Maths Sample Question Paper 6 with Answers 5
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 36

Question 11.
(a) Factorize : a3 + 3a2b + 3ab2 + 2b3.
Answer:
a3 + 3a2b + 3 ab2 + 2b3 = (a3 + 3 a2b + 3ab2 + b3) + b3
= (a + b)3 + (b)3 = (a + b + b) {(a + b)2 – (a + b)b + b2}
= (a + 2b) (a2 + 2ab + b2 – ab – b2 + b2)
= (a + 2b) {a2 + ab + b2)

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(b) In the point A (2, – 4) is equidistant from the points P (3, 8) and Q (- 10, y), find the values of y.
Answer:
Given points are A (2, – 4), P (3, 8), Q (- 10, y)
AQ = AP
⇒ AQ2 = AP2
(- 10 – 2)2 + (y + 4)2
= (3 – 2)2 + (8 + 4)2 144 + (y + 4)2
= 1 + 144 (y + 4)2 = 1 y + 4 = ±1
y + 4= 1 or y + 4 = -1 y = – 3 or y = – 5
y = – 3 or – 5

(c) Simplify: \(\sqrt[a b]{\frac{x^{a}}{x^{b}}} \cdot b \sqrt[x]{\frac{x^{b}}{x^{c}}} \cdot \sqrt[c a]{\frac{x^{c}}{x^{a}}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 37

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 5 with Answers

ICSE Class 9 Maths Sample Question Paper 5 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) At what rate % p.a. will sum of ₹ 4000 yield ₹ 1324 as compound interest in 3 years ?
Answer:
(a) Given : P = ₹ 4,000, C.I. = ₹  1,324, n = 3 years.
Let r be the rate % p.a.
Now A = P + C.I. = ₹ (4,000 + 1,324) = ₹ 5,324.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 5
ICSE Class 9 Maths Sample Question Paper 5 with Answers 6

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(b) If x = 2 + √3 , prove that x2 – 4x + 1 = 0.
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 7

(c) How many times will the wheel of a car having radius 28 cm, rotate in a journey of 88 km
Answer:
Given : r = 28 cm and
Distance = 88 km = 88 x 1,000 x 100 cm = 88,00,000 cm
Now, Distance covered in 1 rotation = Circumference of wheelICSE Class 9 Maths Sample Question Paper 5 with Answers 8

ICSE Class 9 Maths Sample Question Paper 5 with Answers

Question 2.
(a) Factorize : \(x^{2}+\frac{1}{x^{2}}-11\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 9

(b) From the adjoining figure, find the value of x.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 1
Answer:
In ΔACD,
AC = CD  (Given)
⇒ ∠ADC = ∠CAD (Angles opposite to equal sides)
Now, ∠ADC + ∠CAD + ∠ACD = 180° (Sum of angles in a triangle is 180°)
2 ∠ADC + 56° = 180°    (∠ADC – ∠CAD)
⇒ 2∠ADC = 180° – 56°
⇒ ∠ADC \(\frac{124^{\circ}}{2}\)
∠ADC = \(\frac{124^{\circ}}{2}\) = 62°

In ΔABD, AD = BD (Given)
∠ABD = ∠ BAD (Angles opposite to equal sides)
Now, ∠ABD + ∠BAD = ∠ADC
(Exterior angle is equal to sum of interior opposite angles)
2∠ABD =62°
∠ABD\(\frac{62^{\circ}}{2}\)=31°
In ΔABC, ∠A + ∠B + ∠ C = 180° (Sum of angles in a triangle is 180°)
x°+31°+56°=180°
x° = 180°- 87°= 93°

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) Simplify : \(\frac{5 .(25)^{n+1}-25 .(5)^{2 n}}{5 \cdot(5)^{2 n+3}-(25)^{n+1}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 10

Question 3.
(a) If θ = 30°, verify that cos 3 θ = 4 cos3 θ -3 cos θ .
Answer:
L.H.S. = cos 3θ = cos (3 x 30°) = cos 90° = 0 ………..(1)
L.H.S. = 4cos3 θ -3cosθ= 4 cos3 30° – 3 cos 30°
ICSE Class 9 Maths Sample Question Paper 5 with Answers 11
From (i) and (ii)
L.H.S. = R.H.S.
Hence Proved.

(b) Solve by cross multiplication method :
x – 3y – 7 = 0; 3x – 3y = 15.
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 12
ICSE Class 9 Maths Sample Question Paper 5 with Answers 13

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) Prove that: \(\log \frac{11}{5}+\log \frac{14}{3}-\log \frac{22}{15}=\log 7\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 14
= log 11 – log 5 + log 14 – log 3 – log 22 + log 15
= log 11 – log 5 + log (2 x 7)- log 3 – log (2 x 11) + log (3 x 5)
= log 11 – log 5 + log 2 + log 7 – log 3 – log 2 – log 11 + log 3 + log 5
= log 7 = R.H.S
Hence Proved.

Question 4.
(a) If a2 – 3a – 1 = 0, find the value of a + \(a^{2}+\frac{1}{a^{2}}\)
Answer:
(a)
ICSE Class 9 Maths Sample Question Paper 5 with Answers 15

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(b) Construct a combined histogram and frequency polygon for the following data :
ICSE Class 9 Maths Sample Question Paper 5 with Answers 2
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 16
ICSE Class 9 Maths Sample Question Paper 5 with Answers 17

(c) Of two unequal chords of a circle, prove that longer chord is nearer to the centre of the circle.
Answer:
Given : AB > CD, OM⊥AB, ON ⊥ CD.
Join OA and OC.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 18

We know perpendicular drawn from the centre to the chord bisects the chord.
∴ AM= \(\frac{1}{2}\)AB
and CN = \(\frac{1}{2}\)CD
Now, AM > CN(∵ AB > CD)
In ΔOAM OA2 = AM2 + OM2 (Pythagoras theorem)
In ΔOCN OC2 = CN2 + ON2 (Pythagoras theorem)
AM2 + OM2 =CN2 +ON2(∵ OA = OC, Radii)
⇒ OM2 – ON2 = – (AM2 – CN2)
⇒ OM2 – ON2 <O (∵AM > CN)
⇒ OM2 < ON2
⇒ OM < ON
i.e., longer chord is nearer to the centre.
Hence Proved

ICSE Class 9 Maths Sample Question Paper 5 with Answers

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Factorize : \(\frac{y^{6}}{343}+\frac{343}{y^{6}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 19

(b) The diagonals AC and DB of a parallelogram intersect at O. If P is the mid-point of AD,
prove that (i) PO || AB (ii) PO = \(\frac{1}{2}\)CD.
Answer:
Given : In parallelogram ABCD, diagonals AC and BD intersect at O. P is the mid-point of AD.
(i) Since diagonals of a parallelogram bisect each other
∴ O is the mid-point of DB
Now, in ΔADC
P and O are mid-points of sides AD and BD respectively
By mid-point theorem,
ICSE Class 9 Maths Sample Question Paper 5 with Answers 20

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(ii) Also, by mid-point theorem
ICSE Class 9 Maths Sample Question Paper 5 with Answers 21

(c) If θ is acute and 3sin θ = 4cos θ, find the value of 4sin2 θ – 3cos2 θ + 2.
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 22
ICSE Class 9 Maths Sample Question Paper 5 with Answers 23

Question 6.
(a) If the points A (4,3) and B (x, 5) are on the circle with centre C (2, 3), find the value of x
Answer:
Given : A (4, 3), B (x, 5), C (2, 3).
∵ C (2, 3) is the centre,
∴ AC = BC (Redii)
AC2 = BC2
(4 – 2)2 + (3 – 3)2 =(x- 2) + (5 – 3)2
= 4 + 0 =(x – 2)2 + 4
(x-2)2 =0
x – 2 =0
x =2.

(b) ABCD is a trapezium with AB | | CD, and diagonals AC and BD meet at O.
Prove that area of ΔDAO = area of ΔOBC.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 3
Answer:
Given : AB||CD, diagonals AC and BD meet at O.
AB||DC
∴  Area of ΔABD = Area of ΔABC (Triangles on same base and between same parallels are equal in area)
∴ Area of ΔDAO + Area of ΔOAB = Area of ΔOBC + Area of Δ OAB (Addition area axiom)
⇒ Area of ΔDAO = Area of Δ OBC.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) Simplify: \(\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 24

Question 7.
(a) If x + y = 10 and x2 + y1 = 58, find the value of x3 + y3.
Answer:
Given :
x + y =10, x2 + y2 = 58.
(x + y)2 = x2 + y2 + 2xy
⇒ 102 = 58 + 2xy
2xy = 100 – 58
⇒ xy = \(\frac{42}{2} \) = 21
x3 + y3 = (x + y)3 – 3xy (x + y)
= 103 – 3 x 21 x 10
= 1000 – 630 = 370.

(b) The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Answer:
Let the larger supplementary angle be
Then, smaller supplementary angle = 180° – x According to the question,
x – (180° – x) = 18°
⇒ x – 180° + x =18°
⇒ 2x = 18° + 180°
⇒ x= \(\frac{198^{\circ}}{2}\) = 99°
∴ 180° –  x = 1800 99° = 81°
The supplementary angles are 990 and 81°.

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) The mean of 5 numbers is 20. If one number is excluded, mean of the remaining numbers becomes 23. Find the excluded number.
Answer:
Mean of 5 numbers = 20
∴Sum of 5 numbers = 20 x 5 = loo.
1f one number is excluded,
Then, Mean of 4 numbers = 23
Sum of 4 numbers = 23 x 4 = 92
The excluded number = 100 – 92 = 8.

Question 8.
(a) Solve for x : 9 x 3X = (27)2x-5
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 25

(b) In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is
4 : 3, find the sides.
Answer:
Given: Hypotenuse = 20 cm
and ratio of the other two sides = 4:3
Let the other two sides be 4x and 3x.
∴ By Pythagoras theorem,
ICSE Class 9 Maths Sample Question Paper 5 with Answers 26
4x – 4 x 4 = 16
3x = 3 x 4 = 12
∴ The required sides are 16 cm and 12 cm.

(c) Without using tables, find the value of :
ICSE Class 9 Maths Sample Question Paper 5 with Answers 35
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 27

ICSE Class 9 Maths Sample Question Paper 5 with Answers

Question 9.
(a) In what time will a sum of ₹ 8000 becomes ₹ 9261 at the rate of 10% p. a., if the interest is compounded semi-annually?
Answer:
Given : P = X 8,000, A = ? 9,261, r = 10% p.a.
Let n be the number of years.
∵ C.I. is compounded semi-annually,
ICSE Class 9 Maths Sample Question Paper 5 with Answers 28

(b) Construct a regular hexagon of side 2.2 cm.
Answer:
Each side = 2.2 cm.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 29
Steps of construction :
(1) Draw AB 2.2 cm
(2) At A and B, draw angle of 120°.
(3) From A and B, cut-off arcs of 2.2 cm each.
(4) At C, draw 120° and cut it off at D so that CD = 2.2 cm.
(5) At D, draw 120° and cut-off DE = 2.2 cm.
(6) Join EF.
Then, ABCDEF is the required hexagon.

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) Solve graphically : 2x – 3y + 2 = 4x + 1=3x-y + 2
Answer:
Given: 2x – 3y + 2 = 4x + 1=3x-y + 2
∴ 2x – 3y + 2 = 4x + 1 and 4x + 1= 3x-y + 2
⇒ 4x – 2x = – 3y + 2 – 1 and 4x-3x = -y + 2- 1
⇒  2x = 1 – 3y
ICSE Class 9 Maths Sample Question Paper 5 with Answers 30
ICSE Class 9 Maths Sample Question Paper 5 with Answers 31
The two lines intersect at the point (2, – 1).
x =2, y = -1

Question 10.
(a) Express (x2 -5x + 7) (x2 + 5x – 7) as a difference of two squares.
Answer:
(x2 – 5x + 7) (x2 + 5x – 7) = {x2 – (5x – 7)} {x2 + (5x – 7)}
= (x2)2 – (5x – 7)2

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(b) Simplify: \((64)^{2 / 3}-\left(\frac{1}{81}\right)^{-1 / 4}+8^{2 / 3} \cdot\left(\frac{1}{2}\right)^{-1} \cdot 3^{0}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 32

(c) In a pentagon ABCDE, BC | | ED and ∠B: ∠A: ∠E = 5:3:4. Find ∠B.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 4
Answer:
Given: ∠B: ∠A: ∠E =5:3:4
Let Now, ∠B = 5x, ∠A = 3x, ∠E = 4x.
∠C + ∠D = 180°(Co-interior angles; BC||ED)
Sum of angles in a figure with number of sides ‘n’ = (n – 2) x 180°
In pentagon, sum of angles = (5 – 2) x 180°
= 3 x 180° = 540°
∴ ∠ A + ∠B + ∠C + ∠D + ∠E = 540°
⇒ 3x + 5x + 180° + 4x = 540°
⇒ 12x = 540° – 180°
⇒ \(x=\frac{360^{\circ}}{12}\)
⇒ x = 30°
∴ ∠B = 5x = 5 x 30° = 150°

Question 11.
(a) If p + q = 10 and pq = 21, find 3 (p2 + q2).
Answer:
p + q = 10, pq = 21.
∴ p2 + q2 = (p + q)2 – 2pq = 102 – 2 x 21 = 100 – 42 = 58
3 (p2 + q2) = 3 x 58
= 174.

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(b) Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm
Answer:
Let length of each of equal sides be a and that of base be b.
b  = 6 cm(Given)
and Perimeter   = 16 cm
⇒ a + b  = 16
⇒ 2a   + 6  = 16
⇒ 2a = 16 – 6
ICSE Class 9 Maths Sample Question Paper 5 with Answers 33

(c) Prove that : \(\frac{1}{1+\tan ^{2} \theta}+\frac{1}{1+\cot ^{2} \theta}=1\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 34
= cos2 θ + sin2 θ=1
Hence Proved.

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 4 with Answers

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Expand : \(\left(\frac{2}{3} x-\frac{3}{2 x}-1\right)^{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 5

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(b) A person invests ₹ 10,000 for two years at a certain rate of interest, compounded annu­ally. At the end of one year, this sum amounts to ₹ 11,200. Calculate :
(i) The rate of interest p. a.
(ii) The amount at the end of second year.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 3

(c) Factorize : 64x6 – 729y
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 4

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Question 2.
(a) If a2 + b2 = 7ab, prove that 2 log (a + b) = log 9 + log a + log b.
Answer:
Given : a2 + b2 .= 7ab
⇒ a2 + b2 – 9ab – 2ab
⇒ a2 + b2 + 2ab = 9 ab
⇒ (a + b)2 = 9 ab
Taking log of both sides, we get
log (a + b)2 = log (9ab)
⇒ 2 log (a + b) = log 9 + log a + log b.

(b) Prove that: \(\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1=0\)
Answer:
To prove:
\(\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1=0\)
Consider L.H.S. = \(\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1\)
= tan2 θ- sec2 θ +1
= (1 + tan2 θ) – sec2 θ (∵ 1 + tan2 0 = sec2 0)
= sec2 θ – sec2 θ = θ= R.H.S.
Hence Proved

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) In the given figure, two chords AB and CD of a circle intersect at P. If AB = CD, prove that arc AD = arc CB.
Answer:
AB = CD
Minor arc AB = Minor arc CD
Minor arc AB – minor arc BD = Minor arc CD – Minor arc BD
⇒ arc AD = arc CB.
Hence Proved.

Question 3.
(a) In the given figure, ∠ BCD = ∠ADC and ∠BCA = ∠ADB.
Show that: (i)ΔACD ≅ ΔBDC (ii) BC = AD (iii) ∠A = ∠B.
ICSE Class 9 Maths Sample Question Paper 4 with Answers 1
Answer:
Given: ∠BCD = ∠ADC
and ∠BCA = ∠ADB
=> ∠BCA + ∠BCD = ∠ADB + ∠BCD
=> ∠BCA + ∠BCD = ∠ADB + ∠ADC (v ∠BCD = ∠ADC)
=> ∠ACD = ∠BDC
In ΔACD and ΔBDC
(i) ∠ADC = ∠BCD (Given)
CD = CD (Common side)
⇒ ∠ACD = ∠BDC (Proved above)
∴ ΔACD ≅ ΔBDC (ASA axiom)
(ii) ∴BC = AD (c.p.c.t.)
(iii) ∴ ∠A = ∠B. (c.p.c.t.)
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(b) \(a=\frac{2-\sqrt{5}}{2+\sqrt{5}} \text { and } b=\frac{2+\sqrt{5}}{2-\sqrt{5}}, \text { find } a^{2}-b^{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 6
ICSE Class 9 Maths Sample Question Paper 4 with Answers 7

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, – 1) are the vertices of a square ABCD.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 8

Question 4.
(a) The mean height of 10 girls in a class is 1.38 m and the mean height of 40 boys is 1.44 m. Find the mean height of 50 students of the class.
Answer:
Given : Mean height of 10 girls = 1.38 m
∴ Sum of heights of 10 girls = 1.38 x 10 = 13.8 m
and Mean height of 40 boys = 1.44 m
∴ Sum of heights of 40 boys = 1.44 x 40 = 57.6 m.
∴ Sum of heights of 50 students = 13.8 +57.6 = 71.4 m.
∴ Mean heights of 50 students \(\frac{71.4}{50}\)
=1.428m

(b) In the given figure, AABC is a right triangle with ∠C = 90° and D is mid-point of side BC. Prove that AB2 = 4AD2 – 3AC2.
ICSE Class 9 Maths Sample Question Paper 4 with Answers 2
Answer:
Given ∠C = 90°, D is the mid-point of BC.
∴ CD = BD
∴  In ΔABC,
AB2 = AC2 + BC(Pythagoras theorem)
= AC2 + (2CD)2 (∵ CD = BD = \(\frac{1}{2}\) BC)
AB2 = AC2 + 4 CD2  …(i)
In ΔACD,
AD2 = AC2 + CD2  (Pythagoras theorem)
⇒ CD2 = AD2 – AC2 …(ii)
From equations (i) and (ii), we get
AB2 = AC2 + 4 (AD2 – AC2)
= AC2 + 4AD2 – 4AC2
= 4AD2 – 3AC2
 Hence Proved.

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) The following observation have been arranged in ascending order.
3, 6, 7, 10, x, x + 4, 19, 20, 25, 28 . If the median of the data is 13, find the value of x.
Answer:
Given : Numbers in ascending order : 3, 6, 7, 10, x, x + 4, 19, 20, 25, 28.
Median = 13
Here, n = 20.
ICSE Class 9 Maths Sample Question Paper 4 with Answers 9

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Factorize : 5x2 + 17xy – 12y2.
Answer:
5x2 + 17xy – 12y2 = 5x2 + (20 – 3) xy – 12y2
= 5x2 + 20xy – 3xy – 12y2
= 5x(x + 4y) – 3y (x + 4y)
= (x + 4y) (5x – 3y).

(b) If twice the son’s age in years is added to the father’s age, the sum is 70. But if twice the  father’s age is added to the son’s age, the sum is 95. Find the ages of father and son.
Answer:
Let father’s age be x years and that of son’s be y years.
By 1st condition, x + 2y = 70 …(i)
By 2nd condition, 2x + y = 95 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y = 190
Subtracting equation (iii) from equation (i), we get
ICSE Class 9 Maths Sample Question Paper 4 with Answers 10
⇒ 40 +2y = 70
⇒ 2y = 70 – 40
⇒ \(y=\frac{30}{2}=15\)
∴ Father’s age is 40 years and son’s age is 15 years.

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.
Answer:
Given: E, F are mid-points of sides AC and AB, respectively.
ICSE Class 9 Maths Sample Question Paper 4 with Answers 11
We know, the line joining the mid-points of any two sides of a triangle is parallel to the third side
FE || BC
⇒ FQ || BP
Now, since F is mid-point of AB and FQ || BP
∴ By converse of mid-point theorem,
⇒ Q is the mid-point of AP.
AQ = QP. Hence Proved.

Question 6.
(a) If a + \(\frac{1}{a} = p,\) prove that \(a^{3}+\frac{1}{a^{3}}=p\left(p^{2}-3\right).\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 12

(b) The side of a square exceeds the side of another square by 3 cm and the sum of the areas of the two squares is 549 cm2. Find the perimeters of the squares.
Answer:
Let the side of one square be x cm
Then, side of other square = (x + 3) cm.
Area of two squares are x2 cm2 and (x + 3)2 cm2, respectively.
According to the question,
x2 + (x + 3)2 = 549 ⇒ x2 + x2 + 6x + 9 = 549
⇒ 2x2 + 6x – 540 = 0
⇒ x2 + 3x – 270 = 0
⇒ x2 + (18 – 15)x – 270 =0
⇒ x2 + 15x – 15x – 270 =0
⇒ x (x + 18) – 15 (x + 18) =0
⇒ (x + 18) (x – 15) =0
⇒ x = – 18 or 15
∴ x – 15      (∵ x cannot be negative)
∴ x+ 3 =15+ 3 = 18
∴ Perimeter of one square = 4 x 15 = 60 cm
and Perimeter of other square = 4 x 18 = 72 cm

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) Simplify :
\(\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\cos \left(90^{\circ}-\theta\right)}{\sec \left(90^{\circ}-\theta\right)}-3 \tan ^{2} 30^{\circ}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 13

Question 7.
(a) If log10 a = b, express 102b3 in terms of a.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 14

(b) ΔABC and ΔDBC are on the same base BC with A, D on opposite sides of BC. If area of ΔABC = area of ΔDBC, prove that BC bisects AD.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 15

(c) If cos θ + sec θ = 2, show that cos8 θ + sec8 θ = 2.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 16

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Question 8.
(a) If each interior angle is double the exterior angle, find the number of sides.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 17

(b) Solve by the substitution method :
5x + 4y – 4 = 0; x – 20 = 12y.
Answer:
Given: 5x+4y – 4 =0 ………..(i)
and x – 20 =12y ………… (2)
From (ii), x = 12y + 20 ……. (3)
Putting x = 12 + 20 in equation (i), we have
5(12y+20)+4y – 4 =0
⇒ 60y+100+4y – 4=0
⇒ 64y = – 96
⇒ \(y=-\frac{96}{64}=-\frac{3}{2}\)
From (iii), x = 12 x \(\left(\frac{-3}{2}\right)\) +20 = 2
x = 2,y = \(\frac{-3}{2}\)

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) If x = 2 + √3 , find the value of x – \(\frac{1}{x}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 18

Question 9.
(a) Evaluate : \(x^{2 / 3} \cdot y^{-1} \cdot z^{1 / 2}\) when x – 8, y = 4 and z = 25
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 19
(b) Construct a ΔABC is which base AB 5 cm, ∠A = 30° and AC – BC – 2.5 cm.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 20
Steps of construction :
(1) Draw base AB = 5 cm.
(2) Draw ∠BAX = 30°
(3) From AX, cut-off AD = 2.5 cm
(4) Join BD.
(5) Draw the perpendicular bisector of BD to cut AX at C.
(6) Join BC.
Thus, ABC in the required triangle.

(c) Simplify: \(\left(2 x-\frac{1}{2 x}\right)^{2}-\left(2 x+\frac{1}{2 x}\right)\left(2 x-\frac{1}{2 x}\right)\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 21

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Question 10.
(a) Simplify: \(\left(a^{m-n}\right)^{m+n} \cdot\left(a^{n-l}\right)^{n+l} \cdot\left(a^{l-m}\right)^{l+m}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 22

(b) If area of a semi-circular region is 1232 cm2, find its perimeter.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 23

(c) If x = 15°, evaluate : 8 sin cos 4x. sin 6x.
Answer:
Given: x=15°
∴ 8 sin 2xcos 4x.sin 6x = 8 sin (2 x 15°). cos (4 x 15°) . sin (6 x 15°)
= 8 sin 30° . cos 60° . sin 90°
\(=8 \times \frac{1}{2} \times \frac{1}{2} \times 1=2\)

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Question 11.
(a) A farmer increases his output of wheat in his farm every year by 8%. This year he pro­duced 2187 quintals of wheat. What was his yearly produce of wheat 2 years ago?
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 24

(b) Draw the graph of the equation 3x – y = 4.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 25
ICSE Class 9 Maths Sample Question Paper 4 with Answers 26

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) Evaluate : (99.9)2 – (0.1)1
Answer:
(99.9)2 – (0.1)2 = (99.9 + 0.1)
(99.9 – 0.1) = 100 x 99.8
= 9980.

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 3 with Answers

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Factorize : 8(a – 2b)2 – 2a + 4b – 1
Answer:
8 (a – 2b)2 -2a + 4b – 1 = 8 (a- 2b)2 -2(a – 2b) – 1
a – 2b = x
Then, the expression becomes
= 8x2 – 2x – 1
= 8x2 – 4x + 2x – 1
= 4x (2x – 1) + 1 (2x – 1)
= (2x – 1) (4x + 1)
= {2 (a- 2b) – 1} {4 (a – 2b) + 1} (Putting x-a-2b)
= (2a – 4b – 1) (4a- 8b + 1)

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(b) The mean of 6 observations is 17.5. If five of them are 14, 9, 23, 25 and 10, find the sixth observation.
Answer:
Mean of 6 observations is 17.5
5 observations are 14, 9, 23, 25, 10
Let 6th observation be x
ICSE Class 9 Maths Sample Question Paper 3 with Answers 5

(c) If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ-1.
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 6

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 2.
(a) A man borrowed ₹15,000 for 2 years. The rate of interest for the two successive years are 8% and 10% respectively. If he repays ₹6,200 at the end of first year, find the outstanding amount at the end of the second year.
Answer:
(a) For 1st year : P = ₹ 15,000, R = 8% p.a.
\(\mathrm{I}=\frac{15000 \times 8 \times 1}{100}\) = ₹ 16,200
A = P + I = 15,000 + 1,200 =₹16,200
Amount of money repaid = ₹ 6,200
For 2nd year : P = 16,200 – 6,200 = ₹10,000, R = 10% p.a.
\(\mathrm{I}=\frac{10,000 \times 10 \times 1}{100}\) = = ₹1,000
∴ A =P+I=10,000+1,000=11,000
∴ The amount outstanding at the end of 2nd year = 11, 000.

(b) Solve for x if log2 (x2 – 4) = 5.
Answer:
Given: log2(x2 – 4) = 5
x2 – 4 =25
x2 =32 + 4
X = ±√36= ±6.

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) In the following figure, AB is a diameter of a circle with centre O. If chord AC = chord AD,
prove that (i) arc BC = arc DB (ii) AB is bisector of ∠CAD.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 1

Answer:
(i) Given: chord AC = chord AD
⇒ arc AC = arc AD …(i)
Also, arc ACB = arc ADB (AB is a diameter) …(ii)
Subtracting (i) from (ii),
arc ACB – arc AC = arc ADB – arc AD
⇒ arc BC = arc DB. Hence Proved.

(ii) ∵ arc BC = arc BD (Proved above)
∴ ∠BAC = ∠DAB
⇒ AB is bisector of ∠CAD.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 3.
(a) Prove that: \(\frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}}=\frac{3}{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 7

(b) Given that 16 cot A = 12, find the value of \(\frac{\sin A+\cos A}{\sin A-\cos A}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 8

(c) If the altitudes from two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 9

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 4.
(a) If a + b + 2c = 0, prove that a3 + b3 + 8c3 – 6abc = 0.
Answer:
Given : a + b + 2c = 0
⇒ a + b = – 2c
Cubing both sides, we get
(a + b)3 = (- 2c)3
⇒ a3 + b3 + 3ab (a + b) = – 8c3
⇒ a3 + b3 + 3ab (- 2c) = – 8c3
⇒ a3 + b3 + 8c3 – 6abc = 0.
Hence Solved.

(b) Express \(0.1 \overline{34}\) in the form \(\frac{p}{q}\) ,p, q ∈ Z and q ≠ 0.
Answer:
Let x = \(0.1 \overline{34}\) = 0.1343434…
Multiplying both sides of (i) by 10, we get
10x = 1.343434 ………(i)
Multiplying both sides of (ii) by 100, we get
1000x = 134.3434 ………….(ii)
Subtracting (ii) from (iii), we get
1000x – 10x= 134.3434 … – 1.3434 ……………
ICSE Class 9 Maths Sample Question Paper 3 with Answers 10

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) If the sides are in the ratio 5 : 3 : 4, prove that it is a right angled triangle.
Answer:
Ratio of sides = 5:3:4
Let the length of sides be 5x, 3x, 4x.
Here,(3x)2 + (4x)2 = 9x2 + 16x2 = 25x2 = (5x)2
∴ By Pythagoras theorem, the triangle is right angled
Hence Proved.

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) On what sum of money will the difference between compound interest and simple inter­est for 2 years be equal to ₹25 if the rate of interest charged for both is 5% p.a. ?
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 12

(b) Show by distance formula that the points A (-1, -1), B (2, 3) and C (8,11) are collinear.
Answer:
Given points are A (-1, -1), B (2, 3), C (8, 11).
Now
ICSE Class 9 Maths Sample Question Paper 3 with Answers 11
AB + BC = 5 + 10 = 15
⇒ AB + BC = AC
The points are collinear.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) Factorize : a6 – 26a3 – 27.
Answer:
a6 – 26a3 – 27 = a6 – (27 – 1) a3-27 = a6 – 27a3 + a3 – 27
= a3 (a3 – 27) + 1 (a3 – 27) = {a3 – 27) (a3 + 1)
= (a3 – 33) (a3 + 13)
= (a – 3) (a2 + 3a + 9) (a + 1) (a2 – a + 1)

Question 6.
(a) Simplify: \(\frac{\left(x^{a+b}\right)^{2} \cdot\left(x^{b+c}\right)^{2} \cdot\left(x^{c+a}\right)^{2}}{\left(x^{a} \cdot x^{b} \cdot x^{c}\right)^{4}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 13

(b) In the given figure AABC, D is the mid-point of AB, E is the mid-point of AC. Calculate :
(i) DE, if BC = 8 cm.
(ii) ∠ADE, if ∠DBC = 125°.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 2
Answer:
Given : D is mid-point of AB, E is the mid-point of AC, BC = 8 cm, ∠DBC = 125°.
The line joining the mid-points of any two sides of a triangle is parallel to the third and is equal the half of it.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 14

(c) If a and b are rational numbers, find the values of a and b :
\(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 16
ICSE Class 9 Maths Sample Question Paper 3 with Answers 17

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 7.
(a) Draw a histogram from the following data :

Weight (in kg)40-4445-4950-5455-5960-6465-69
No. of students28121064

Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 18
ICSE Class 9 Maths Sample Question Paper 3 with Answers 19

(b) Solve:
83x – 67y = 383
67x – 83y = 367.
Answer:
83x – 67y = 383 ………….(i)
67a – 83y = 367 ………….(ii)
Adding (i) and (ii), we get
150x – 150y = 750
x – y = 5 ………. (iii)
Subtracting (ii) from (i), we get
16x + 16y = 16
x + y = 1 ……… (iv)
Adding (iii) and (iv), we get
2x = 6
X = 3
Putting x = 3 in (iv), we get
3 + y = 1
⇒ y = 1 – 3 = -2.
x = 3, y = – 2

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) In the following figure, area of parallelogram AFEC is 140 cm2. State, giving reason, the area of (i) parallelogram BFED (ii) ABFD.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 3
Answer:
Given : Area of parallelogram AFEC = 140 cm2.
(i)  Area of parallelogram BFED = Area of parallelogram AFEC
( ∵ They are on same base and between same parallels)
= 140 cm2

(ii) Area of Δ BFD = \(\frac{1}{2}\) x Area of parallelogram BFED
(∵ They are on same base and between same parallels)
\(\frac{1}{2}\) x 140 cm2 = 70 cm2.

Question 8.
(a) If log10 a = m and log10 b = n, express \(\frac{a^{3}}{b^{2}}\) in terms of m and n
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 20

(b) Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Hence, find the area of the region bounded by these lines and X-axis.
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 21
ICSE Class 9 Maths Sample Question Paper 3 with Answers 22
ICSE Class 9 Maths Sample Question Paper 3 with Answers 23

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) Factorize : \(8 x^{3}-\frac{1}{27 y^{3}}\)
Answre:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 24

Question 9.
(a) If \(\frac{x^{2}+1}{x}=2 \frac{1}{2}\) find the values of \((i) x-\frac{1}{x}
(ii) x^{3}-\frac{1}{x^{3}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 25

(b) In the following figure, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 4
Answer:
Given : OA = 3.5 cm, OD = 2 cm
Area of shaded region = Area of quadrant AOB – Area of ΔAOD.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 26
= 9.625 – 3.5 = 6.125 cm2.

(c) If a + b + c = 9 and ab + be + ca = 40, find the value of a2 + b2 + c2.
Answer:
Given: a + b + c = 9 ab + bc + ca = 40
We know, (a+b+c)2 =a2+ b2 + c2 +2(ab+bc+ca)
(9)2 =a2+ b2 + c2+ 2 x 40
a2+ b2 + c2 = 81 – 80 = 1

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 10.
(a) Solve: \((\sqrt{2})^{2 x+4}=8^{x-6}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 27

(b) Construct a rhombus whose diagonals are 5 cm and 6.8 cm.
Answer:
Given, diagonals are 5 cm and 6.8 cm.
Steps of construction :
(1) Draw AC = 5 cm.
(2) Draw perpendicular bisector PQ of AC which intersect AC at O.
(3) From POQ, cut-off OB = OD = \(\frac{6.8}{2}\) = 3.4 cm.
(4) Join the points A, B, C, D.
Then, ABCD is the required rhombus.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 28

(c) In a quadrilateral ABCD, AO and BO are the bisectors of ∠A and ∠B respectively.
Prove that ∠AOB = \(\frac{1}{2}\) (∠C + ∠D).
Answer:
Given, AO and BO are bisectors of ∠A and ∠B respectively.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 29
ICSE Class 9 Maths Sample Question Paper 3 with Answers 30

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 11.
(a) Find the value of log5√5 (125).
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 31

(b) The sum of a two-digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
Answer:
Let the digits in tens and units place be x and y respectively.
The number = 10x + y
The number obtained by reversing digits = 10y + x By 1st condition,
(10 + y) + (10y + x) = 165
⇒ 11x + 11y = 165
⇒ x + y =15 …(ii)
By 2nd condition, x-y =3 …(iii)
or y-x =3 …(iv)
Adding (ii) and (iii), we get 2x = 18
⇒ x =9
Putting x = 9 in (ii), we get y = 15 – 9 = 6
Again, adding (ii) and (iv), we get
2 y =18
⇒ y =9
Putting y = 9 in (ii), we get x = 15 – 9 = 6.
Substituting these values in (i), we get
The number = 10 x 9 + 6 or 10 x 6 + 9 = 96 or 69

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) If the area of an equilateral triangle is 81√3 cm2, find its perimeter.
Answer:
Given : Area of equilateral triangle = 81√3 cm2
Let the length side of each of equilateral triangle be a cm.ICSE Class 9 Maths Sample Question Paper 3 with Answers 32

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 2 with Answers

ICSE Class 9 Maths Sample Question Paper 2 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) If x – 3 – 2√2, find the value of x2 + -y.
Answer:
Given = ICSE Class 9 Maths Sample Question Paper 2 with Answers 7
ICSE Class 9 Maths Sample Question Paper 2 with Answers 8

(b) Factorize : 9x2 – 4 (y + 2x)2
Answer:
9x2 -4(y + 2x)2 = (3x)2 – {2 (y + 2x)}2
= (3x)2 – (2y + 4x)2
= (3x + 2y + 4x) (3x – 2y – 4x)
= (7x + 2y) (-x -2y)
= – (x + 2y) (7x + 2y).

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(c) The area enclosed between the concentric circles is 770 cm2. Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
Answer:
Given : The area enclosed between the circles = 770 cm2
Radius of outer circle (R) = 21 cm.
Let radius of inner circle be r.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 9

Question 2.
(a) A man invests ₹46,875 at 4% p.a. compound interest for 3 years. Calculate :
(i) The interest for the first year.
(ii) The amount at the end of the second year.
(iii) The interest for the third year.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 10

(b) If ax = by = cz and b2 = ac, Prove that \( y=\frac{2 x z}{x+z}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 11

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(c) In the following figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 1
Answer:
Given: AD = BC,
∠OAD = ∠OBC = 90°
In Δ OAD and ΔOBC,
AD =BC (Given)
ΔOAD = ΔOBC (Given)
Δ AOD =Δ BOC (Vertically opposite angles)
∴ ΔOAD ≅ ΔOBC (SAS axiom)
∴ OA = OB (c.p.c.t.)
Hence, CD bisects AB. Hence Proved.

Question 3.
(a) Solve the following equations by cross multiplication method :
3x – 7y = – 10, – 2x + y = 3.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 13

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(b) Find the value of :
2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°
Answer:
2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°
\(=2 \sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2}+2 \sqrt{3} \times \frac{1}{2} \times \sqrt{3}-1\)

(c) Construct a frequency polygon for the following frequency distribution using a graph sheet.

Marks40 – 5050-6060 – 7070-8080 – 9090 – 100
No. of Students5813975

Use 1 cm – 10 marks and 1 cm = 5 students.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 14
ICSE Class 9 Maths Sample Question Paper 2 with Answers 15

Question 4.
(a) Express as a single logarithm :
2 log 3 – \(\frac{1}{2}\) log 64 + log 16.
(b) If \(x+\frac{1}{x}=3\),evaluate \(x^{3}+\frac{1}{x^{3}}\)
(c) Prove that the line joining mid-points of two parallel chords of a circle passes through the centre of the circle.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 16

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(c) Given: AB || CD, M and N are mid-points of sides AB and CD respectively.
Construction: Join OM, ON and draw a straight line parallel to AB and CD.
Since, line segment joining the mid-point of the chord with centre of the circle is perpendicular to the chord
ICSE Class 9 Maths Sample Question Paper 2 with Answers 17
∴ OM ⊥ AB and ON ⊥ CD
⇒ ∠AMO = 90° and ∠NOE = 90°
Now, ∠MOE = 90° (Co-interior angles, OE || AB)
∠NOE = 90° (Co-interior angles, OE || CD)
∠MOE + ∠NOF =90° + 900 = 1800
So, MON is a straight line passing through the centre of the circle.
Hence Proved.

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Find a point on the Y-axis which is equidistant from the points A (6, 5) and B (- 4, 3).
Answer:
Given : A (6, 5), B (- 4, 3).
Let the point on the Y-axis be P (0, b).
According to the question,
AP = BP
⇒ AP2 = BP2
⇒ (6 – 0)2 + (5 – b)2 = (- 4 – 0)2 + (3 – b)2
⇒ 36 + 25 – 10b + V- = 16 + 9 – 6fo + b2
⇒ – 10b + 6b = 25 – 61
⇒  -4b =-36
⇒ b = 9
Required Point = (0,9)

(b) In the following figure, area of parallelogram ABCD is 29 cm2. Calculate the height of parallelogram ABEF, if AB = 5.8 cm.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 2
Answer:
Area of parallelogram ABCD = 29 cm2, AB = 5.8 cm.
Area of parallelogram ABEF = 29 cm2 (area of parallelograms on same base are equal)
⇒ AB x Height = 29 cm2
⇒ \(\text { Height }=\frac{29 \mathrm{~cm}^{2}}{\mathrm{AB}}=\frac{29 \mathrm{~cm}^{2}}{5.8 \mathrm{~cm}}=5 \mathrm{~cm}\)

(c) A sum of money doubles itself at compound interest in 15 years. In how many years will it become eight times ?
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 18
ICSE Class 9 Maths Sample Question Paper 2 with Answers 19

ICSE Class 9 Maths Sample Question Paper 2 with Answers

Question 6.
(a) Construct the quadrilateral ABCD, given that AB = 5 cm, BC = 2.5 cm, CD = 6 cm,
∠BAD = 90° and the diagonal AC = 5.5 cm.
Answer:
Given : AB = 5 cm, BC = 2.5 cm, CD = 6 cm, Z BAD 90°,
AC = 5.5 cm.
Steps of construction :
(1) Draw AB 5 cm.
(2) At A, draw ∠BAP = 90°.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 20

(3) From B and A, draw arcs of lengths 2.5 cm and 5.5 cm, respectively which intersect at C.
(4) From C, cut-off AP at D such that CD = 6 cm.
Thus, ABCD is the required quadrilateral.

(b) Factorize : (a + 1) (a + 2) (a + 3) (a + 4) – 3.
Answer:
(a + 1) (a + 2) (a + 3) (a + 4) – 3 = (a + 1) (a + 4) (a + 2) (a + 3) – 3
= (a2 + 5a + 4) (a2 + 5a + 6) – 3
= (p + 4) (p + 6) – 3  (Putting a2 + 5a = p)
= p2 + 6p + 4p + 24 – 3
= p2 + 10p + 21
= p2 + (7 + 3) p + 21
= p2 + 7p + 3p + 21
= p2 (p + 7) + 3 (p + 7) = (p + 7) (p + 3)
= (a2 + 5a + 7) (a2 + 5a + 3) (∵ p = a2 + 5a)

(c) In the following figure, D and E are mid-points of the sides AB and AC respectively. If BC = 6 cm and ∠B = 72°, compute (i) DE (ii) ∠ADE.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 3
Answer:
Given : BC= 5.6 cm, ∠B = 72° and D, E are mid-points of sides AB, AC, respectively, (i)
The line joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 21

ICSE Class 9 Maths Sample Question Paper 2 with Answers

Question 7.
(a) Evaluate without using tables :
\(\left(\frac{\cos 47^{\circ}}{\sin 43^{\circ}}\right)^{2}+\left(\frac{\sin 72^{\circ}}{\cos 18^{\circ}}\right)^{2}-2 \cos ^{2} 45^{\circ}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 22

(b) Solve:
\(\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 23
ICSE Class 9 Maths Sample Question Paper 2 with Answers 24

(c) In ΔABC, ∠ACB = 90°, AB = c unit, BC = a unit,
AC = b unit, C perpendicular to AB and CD = p unit.
Prove that \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
ICSE Class 9 Maths Sample Question Paper 2 with Answers 4
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 25

Question 8.
(a) If \(x^{2}+\frac{1}{x^{2}}=27\),find the values of :
(i) \(x+\frac{1}{x}\)
(ii) \(x-\frac{1}{x}\)
(iii) \(x^{2}-\frac{1}{x^{2}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 27

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(b) If 1 is added to the numerator of a fraction, it becomes \(\frac{1}{5}\). If 1 is subtracted from the denominator, it becomes \(\frac{1}{7}\). Find the fraction.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 28

(c) Find the mean and median of the numbers :
41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.
Answer:
41, 39, 52, 48, 54, 62, 46, 52, 40, %, 42, 40, 98, 60, 52.
∴ ∑ x =822, n=15
∴ \(\text { Mean }=\frac{\sum x}{n}=\frac{822}{15}=54.8\)
Rearranging in ascending order, we get
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
ICSE Class 9 Maths Sample Question Paper 2 with Answers 29

Question 9.
(a) The volume of a cuboidal block of silver is 10368 cm3. If its dimensions are in the ratio 3:2:1, find :
(i) Dimensions of the block.
(ii) Cost of gold polishing its entire surface at ₹0.50 per cm2.
Answer:
(a) (i) Ratio of dimensions = 3:2:1
Let its length, breadth and height be 3x cm, 2x cm and x cm respectively.
Volume of block = 3× x 2x × x = 10368
⇒ 6x3 = 10368
\(x^{3}=\frac{10368}{6}=1728 \Rightarrow x=12\)
Length (l) =3x = 3 × 12=36cm
Breadth (b) = 2x = 2 × 12=24cm
Height (h) =x = 12 cm

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(ii) Total surface area =2(lb+lh+bh)=2(36 x 24+36 x 12+24x 12)
= 2 (864 + 432 + 288)
= 2 x 1584 = 3168 cm2
∵ Rate of gold polishing = ₹ 0.50 = ₹\(\frac{1}{2}\)
Total cost of gold polishing of entire surface
ICSE Class 9 Maths Sample Question Paper 2 with Answers 30

(b) Factorize : 2 – y (7 – 5y).
Answer:
2 – y (7 – 5y) = 2-7y + 5y2
= 2 – (5 + 2) y + 5y2 = 2 – 5y – 2y + 5y2
= 1 (2 – 5y) – y (2 – 5y) = (2 – 5y) (1 – y)

(c) Solve graphically : x – 2y = 1; x + y – 4.
Answer:
x – 2y =1 ……….(i)
x + y = 4 ……………(ii)
from (i)
ICSE Class 9 Maths Sample Question Paper 2 with Answers 31
∴ The points are (1, 0), (3, 1), (5, 2)

From (ii)
ICSE Class 9 Maths Sample Question Paper 2 with Answers 32
The points are (4, 0), (3, 1), (2, 2)
These points are plotted on the graph.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 33
The two straight lines intersect at (3, 1)
∴ x=3,y=1

ICSE Class 9 Maths Sample Question Paper 2 with Answers

Question 10.
(a) From the adjoining figure, find the values of :
(i) cot2 x – cosec2 x
(ii) \(\tan ^{2} y-\frac{1}{\cos ^{2} y}\)
Answer:
(a) In AABD,
AB2 = AD2 + BD2 (By Pythagoras theorem)
= 42 + 32 = 16 + 9 = 25
AB = √25 = 5.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 38

(b) \(\text { If } \frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}, \text { prove that }: a^{a} . b^{b} \cdot c^{c}=1 \text { . }\)
Answer:
\(\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k(\text { say })\)
log a =k(b – c); log b = k(c-a); log c = k(a-b)
Now, a log a + b log c + c log c = ak (b – c) + bk (c – a) + ck (a – b)
⇒ log ab + log bb + log cc = kab – kac + kbc – kab + kac – kbc
⇒ log (aa. bb . cc) = 0
⇒ log (aa. bb . cc)= log 1
⇒ aa. bb. cc = 1.
Hence Proved.

(c) Prove that √2 is not a rational number.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 5
Answer:
Let us assume that√2 is a rational number.
If \(\sqrt{2}=\frac{p}{q}, p, q \in \mathrm{I}\) have no comman factor and C”q≠ 0.
\(2=\frac{p^{2}}{q^{2}} \Rightarrow p^{2}=2 q^{2} \Rightarrow p^{2}\) C”is an even integer
⇒ p is an even integer
⇒ p = 2m, where m∈I
⇒ p2 = 4m2 ⇒ 2y2 = 4m2 ⇒ q2 = 2m2
⇒ q2 is an even integer
⇒ y is an even integer.
Thus, p and q are both even integers and therefore, have a common factor 2 which contradicts that p and q have no common factor.
√2 is not a rational number.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 2 with Answers
Question 11.
(a) Simplify: \(\left(a+\frac{1}{a}\right)^{2}-\left(a-\frac{1}{a}\right)^{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 35

(b) In the given figure, ABCD is a trapezium. Find the values of x and y.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 36

(c) Simplify: \(\frac{(25)^{3 / 2} \times(243)^{3 / 5}}{(16)^{5 / 4} \times(8)^{4 / 3}}\)
ICSE Class 9 Maths Sample Question Paper 2 with Answers 6
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 37

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 1 with Answers

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Max Marks :80
[2 Hours]

General Instructions

  • Answers to this Paper must be written on the paper provided separately.
  • You will not be allowed to write during the first 15 minutes.
  • This time is to be spent in reading the question paper.
  • The time given at the head of this Paper is the time allowed for writing the answers.
  • Section A is compulsory. Attempt any four questions from Section B.
  • The intended marks for questions or parts of questions are given in brackets [ ].

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) Rationalize the denominator : \(\frac{14}{5 \sqrt{3}-\sqrt{5}}\) [3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 11

(b) Factorize the given expression completely : 6×2 + 7x – 5 [3]
Answer:
6 x2+ 7x-5 = 6x2 + (10 – 3)* – 5
– 6x2 + 10x- 3x – 5
= 2x(3x + 5) – 1(3x + 5)
= (3x + 5) (2x – 1).

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) In the given figure, AB = \(\frac{1}{2}\) BC, where BC = 14 cm. Find : [4]
(i) Area of quadrilateral AEFD
(ii) Area of ΔABC
(iii) Area of semicircle
Hence find the area of shaded region. Use 7π = \(\left(\text { Use } \pi=\frac{22}{7}\right)\)
ICSE Class 9 Maths Sample Question Paper 1 with Answers 1
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 12

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 2.
(a) Mr. Ravi borrows ₹ 16,000 for 2 years. The rate of interest for the two successive years are 10% and 12% respectively. If he repays ₹ 5,600 at the end of first year, find the amount outstanding at the end of the second year. [3]

(b) Simplify: \(\left(\frac{8}{27}\right)^{-\frac{1}{3}} \times\left(\frac{25}{4}\right)^{\frac{1}{2}} \times\left(\frac{4}{9}\right)^{0}+\left(\frac{125}{64}\right)^{\frac{1}{3}}\) [3]

(c) In the given figure, ABCD is a parallelogram. AB is produced to P, such that AB = BP and PQ is drawn parallel to BC to meet AC produced at Q.
Given AB = 8 cm, AD = 5 cm, AC = 10 cm,
(i) Prove that point C is mid-point of AQ.
(ii) Find the perimeter of quadrilateral BCQP. [4]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 2
Answer:
(a) Here, P = ₹ 16000
For first year: R = 10% , T = 1 year
∴ \(\text { Interest }=\frac{16000 \times 10 \times 1}{100}= 1600\)
Amount = ₹ (16000 + 1600) = ₹ 17600
∴ Amount repaid = ₹ 5600.

ICSE Class 9 Maths Sample Question Paper 1 with Answers

For Second Year :
P = (17600 – 5600) = 12000, R = 12% , T = 1 year
∴ Intrest = \(\frac{12000 \times 12 \times 1}{100}\) = ₹1440
∴ Amount =(12000+1440) = ₹13440

(b)
ICSE Class 9 Maths Sample Question Paper 1 with Answers 13
(c) Given : ABCD is parallelogram, AB = BP, AB = 8 cm, AD = 5 cm, AC = 10 cm.
(i) ∵ AB = BP
∴ B is mid-point of AP
Also, BC || PQ (Given)
AC = CQ (By mid-point theorem)
∴ C is mid-point of AQ. Hence Proved.

(ii) BP = AB = 8 cm (Given)
BC = AD = 5 cm (∵ ABCD is a parallelogram)
CQ = AC = 10 cm [From part, (i)]
PQ = 2 BC = 2×5 = 10 cm(By mid-point theorem)
∴ Perimeter of quadralateral BCQP = BP + PQ + CQ + BC

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 3.
(a) Solve following pairs of linear equations using cross-multiplication method : [3]
5x – 3y = 2
4x + 7y = – 3
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 14

(b) Without using tables, evaluate : [3]
\(4 \tan 60^{\circ} \sec 30^{\circ}+\frac{\sin 31^{\circ} \sec 59^{\circ}+\cot 59^{\circ} \cot 31^{\circ}}{8 \sin ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 15

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) Construct a frequency polygon for the following frequency distribution, using a graph sheet. [4]

Marks40-5050-6060-7070-8080-9090-100
No. of students7182637206

Use : 1 cm = 10 marks, 1 cm = 5 students
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 16
ICSE Class 9 Maths Sample Question Paper 1 with Answers 17

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 4.
(a) Evaluate : 3 log 2 – \(\frac{1}{3}log 27 + log 12 – log 4 + 3 log 5\). [3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 18

(b) If x –\(\frac{1}{x}\) =3, evaluate x3 – \(\frac{1}{x^{3}} \)[3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 19

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) In the given diagram, O is the centre of the circle and AB is parallel to CD. AB = 24 cm
and distance between the chords AB and CD is 17 cm. If the radius of the circle is 13 cm, find the length of the chord CD.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 3
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 20
ICSE Class 9 Maths Sample Question Paper 1 with Answers 21

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Section – B [40 Marks]
(Attempt any four questions from this Section)

Question 5.
(a) Find the coordinates of the points on Y-axis which are at a distance of 5√2 units from
the point (5, 8). [3]
Answer:
(a) Let the coordinates of the point on Y-axis be (0, y).
Distance = 5 √2
⇒\( \sqrt{(0-5)^{2}+(y-8)^{2}}=5 \sqrt{2} \)
Squaring both sides, we get
(0-5)2 + (y-8)2 = (5-√2)2
⇒ 25 + y2 – 2 y-8 + 64 = 50
⇒ y2 – 16y + 89 – 50 = 0
⇒ y2 – 16y + 39 = 0
⇒ y2 – (13 + 3)y + 39 = 0
⇒ y2-13y-3y+ 39 =0
⇒ y(y – 13) – 3(y – 13) = 0
⇒ (y – 13) (y – 3) = 0
⇒ y-13=0 or y-3 = 0
⇒ y = 13 or y = 3.
.’. The required point is (0,13) or (0, 3).

(b) In the given figure, BC is parallel to DE. Prove that area of ΔABE = Area of ΔACD. [3]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 4
Answer:
Given: BC || DE
∴ Area of ΔBCE = Area of ΔBCD
(Triangles, on same base and between the same parallels are equal in area)
⇒ Area of ΔBCE + Area of ΔABC = Area of ΔBCD + Area of ΔABC
(Adding area of ΔABC to both sides) .
⇒ Area of ΔABE = Area of ΔACD. Hence Proved.

(c) A stun of ₹ 12,500 is deposited for 1 \(\frac{1}{2}\) years, compounded half-yearly. It amounts to ₹ 13,000 at the end of first half year. Find : [4]
(i) The rate of interest
(ii) The final amount. Give your answer correct to the nearest rupee.
Answer:
P = ₹ 12,500, A = ₹ 13,000, T = – year.
∴ Interest for \(\frac{1}{2}\) year
= ₹ (13000 – 12500) = ₹ 500.
(i) Let R be the rate of interest.
∴ \(\mathrm{R}=\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{500 \times 100}{12500 \times \frac{1}{2}}=8\)
∴ The rate of interest = 8 % p.a.

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(ii) Now, n = 1 \(\frac{1}{2}\)years =\( \frac{3}{2}\)years.
C.I. is calculated half-yearly,
\(\mathrm{R}=\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{500 \times 100}{12500 \times \frac{1}{2}}=8\)

Question 6.
(a) Construct a parallelogram ABCD in which AB = 6.4 cm, AD = 5.2 cm and the
perpendicular distance between AB and DC is 4 cm. [3]
Answer:
(a) Given : AB = 6.4 cm, AD = 5.2 cm,
Perpendicular distance between AB and DC is 4 cm.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 23
Steps of construction :
(1) Draw a line segment XY and take any point P on it.
(2) At P, draw a perpendicular PZ and cut-off PD = 4 cm.
(3) From D, cut-off XY at A such that DA = 5.2 cm.
(4) From A, cut-off XY at B such that AB = 6.4 cm.
(5) From B and D, draw arcs of 5,2 cm and 6.4 cm radii respectively which intersect at C.
(6) Join AD, BC and CD to obtain the required parallelogram ABCD.

(b) Factorize : 4a2 – 9b2 – 16c2 + 24be [3]
Answer:
4a2 – 9b2 – 16c2 + 24 be =4a2– (9b2 – 14bc + 16c2)
= (2a)2 – {(3b)2 – 2-3b-4c + (4c)2}
= (2a)2 – (3b – 4c)2
= (2a + 3b – 4c) (2a – 3b + 4c).

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) In the given diagram, ABCD is a parallelogram, ΔAPD and ΔBQC are equilateral triangles.
Prove that: . [4]
(i) ∠PAB = ∠QCD
(ii) PB = QD

ICSE Class 9 Maths Sample Question Paper 1 with Answers 5
Answer:
Given : ABCD is parallelogram, ΔAPD and ΔBQC are equilateral triangles.
(i) ∠DAB = ZBCD (Opp. angles of a || gm are equal)
⇒ ∠DAB + ∠PAD = ∠BCD + ∠BCQ (∠PAD = ∠BCQ = 60°)
⇒ ∠PAB = ∠DCQ. Hence Proved.

(ii) In ΔPAB and ΔQCD,
AB DC (Opp. sides of aIgm are equal)
∠PAB = ∠QCD [From (i)
AP = CQ (∵AP=AD=BC=CQ)
∠PAB ≅ ΔQCD (SAS axiom)
PB = QD (c.p.c.t.)
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 7.
(a) Solve for x : sin2 x + cos2 30° = \(\frac{5}{4}\); where 0° ≤ x ≤ 90° [3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 24

(b) Evaluate for x :\(\left(\sqrt{\frac{5}{3}}\right)^{x-8}=\left(\frac{27}{125}\right)^{2 x-3}\) [3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 25

(c) In the given figure, triangle ABC is a right angle triangle with ∠B = 90° and D is mid­point of side BC. Prove that AC2 = AD2 + 3 CD2. [4]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 6
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 26
ICSE Class 9 Maths Sample Question Paper 1 with Answers 27

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 8.
(a) In the given figure, ∠ABC = 66°, ∠DAC = 38°. CE is perpendicular to AB and AD is perpendicular to BC. Prove that CP > AP. [3]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 7
Answer:
Given: ∠ABC = 66°, ∠DAC = 38°, CE ⊥AB, AD ⊥ BC.
In ∠ABD, ∠BAD + ∠ABD = ∠ADC (Exterior angle is equal to sum
of interior opposite angles)
∠BAD+66°=90°
∠BAD=90°- 66°=24°.
In ∠SACE, ∠ACE + ∠AEC + ∠CAE = 180° (Sum of angles in a triangle is 180°)
∠ACE + 90° + (24° + 38°) = 180°
∠ACE + 152° = 180°
∠ACE = 180° – 152° = 28°.
Now, ∠CAP > ∠ACP ( 38°> 28°)
CP > AP (In a triangle, greater angle has greater side opposite to it)
Hence Proved.

(b) Mr. Mohan has ₹ 256 in the form of ₹ 1 and ₹ 2 coins. If the number of ₹ 2 coins are three more than twice the number of ₹ 1 coins, find the total value of ₹ 2 coins. [3]
Answer:
Total amount = ₹ 256
Let the no. of ₹ 1 coins be x and that of ₹ 2 coins be y.
∴ Value of x coins = ₹ 1 × x = ₹  x
Value of y coins = ₹ 2 x y = ₹ 2y.
∴ x + 2y = 256
Also, y = 3 + 2x
Using equation (ii) in (i), we have
Also, y=3+2x
Using equation (ii) in (i), we have
⇒ x+2(3+2x)= 256
⇒ x+6+4x= 256
⇒ 5x =256 – 6
⇒ x=\(\frac{250}{5}\)=50.
Putting the value of x in equation (ii), we get
y =3+2x 50 =3+ 100 = 103.
∴ Total value of ₹ 2 coins = ₹ 2y
=₹ 2x 103
=₹ 206.

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) Find (i) mean and (ii) median for the following observations : [4]
10, 47, 3, 9, 17, 27, 4, 48, 12, 15
Answer:
Given observations are 10, 47, 3, 9, 17, 27, 4, 48, 12, 15.
Here, n 10
(i) Σx = 192
\(\text { Mean }=\frac{\Sigma x}{n}=\frac{192}{10}=19.2\)

(ii) Rearranging the observations in ascending order, we have
3, 4, 9, 10, 12, 15, 17, 27, 47, 48
ICSE Class 9 Maths Sample Question Paper 1 with Answers 28

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 9.
(a) Three cubes are kept adjacently, edge to edge. If the edge of each cube is 7 cm, find total surface area of the resulting cuboid. [3]
Answer:
Given : Length of each side of cube = 7 cm
For cuboid, 7cm
l= (7 + 7 + 7) cm = 21 cm
b = 7 cm, h = 7 cm.
We know, Total surface area = 2 (lb + bh + Ih)
= 2 (21 x 7 + 7 x 7 + 21 x 7)
= 2 (147+ 49 + 147) = 2 x 343 = 686 cm2
ICSE Class 9 Maths Sample Question Paper 1 with Answers 29

(b) In the given figure, arc AB = twice (arc BC) and ∠AOB = 80°. Find : [3]
(i) ∠BOC
(ii) ∠OAC
ICSE Class 9 Maths Sample Question Paper 1 with Answers 8
Answer:
(i) Given: Arc AB = 2 (arc BC),∠AOB =80°
∠AOB=2∠BOC
∠BOC = \(\frac{1}{2}\) ∠AOB
\(\frac{1}{2}\) × 80°
=40°

(ii) In ΔAOC
OA = OC (Radii)
⇒ ∠OCA = ∠OAC (Angles opposite to equal
sides are equal)
Now, ∠OAC + ∠AOC + ∠OCA = 180° (Angle sum property)
∠OAC + (∠AOB + ∠BOC) + ∠OAC = 180° (∠OAC= ∠OCA)
= 2∠OAC + (80° + 40°) = 180°
2∠OAC + 120° = 180°
2∠OAC = 180° – 120° = 60°
∴ ∠OAC =\(\frac{60^{\circ}}{2}\) 3o°

(c) Solve graphically the following system of linear equations (use graph sheet): [4]
x – 3y = 3
2x + 3y = 6
Also, find the area of the triangle formed by these two lines and the Y-axis.
Answer:
x – 3y = 3 …………….. (i)
2x + 3y = 6 ………. (ii)
from equation (i)
x = 3y + 3

X3.0-3
y0-1-2

The points are (3, 0), (0, – 1), (- 3, – 2).
From equation (ii),
⇒ 2x = 6 – 3y
⇒ \(x=\frac{6-3 y}{2}\)

ICSE Class 9 Maths Sample Question Paper 1 with Answers

X30-3
y024

The points are (3, 0), (0, 2), (- 3, 4).
These points are plotted on the graph.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 30
The two lines intersect at the point (3, 0).
∴ x=3,y=0
Triangle formed by the lines (i), (ii) and Y-axis is ABC.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 31

Question 10.
(a) Each interior angle of a regular polygon is 135°. Find : [3]
(i) The measure of each exterior angle.
(ii) Number of sides of the polygon.
(iii) Name the polygon.
Answer:
(a) Given: Each interior angle = 135°
(i) Exterior angle = 180° – 135° = 45°
ICSE Class 9 Maths Sample Question Paper 1 with Answers 32
(iii) The polygon is a regular octagon.

(b) If log 4 = 0.6020, find the value of log 80. [3]
Answer:
Given : log 4 = 0.6020
⇒ log 22 = 0.6020
⇒ 2 log 2 = 0.6020
⇒ log 2
\(=\frac{0.6020}{2}\)
Now, log 80 = log (8 x 10) = log 8 + log 10
= log 23 + log 10 = 3 log 2 + log 10
= 3 x 0.3010 + 1 = 1.9030

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) Evaluate x and y from the figure diagram. [4]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 9
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 33

Question 11.
(a) ΔABC is an isosceles triangle such that AB = AC. D is a point on side AB such that
BC = CD. Given ∠BAC = 28°. Find the value of ∠DCA. [3]
(b) Prove that opposite angles of a parallelogram are equal. [3]
(c) The cross-section of a 6 m long piece of metal is shown in the figure. Calculate : [4]
(i) The area of the cross-section
(ii) The volume of the piece of metal in cubic centimetres.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 10
Answer:
(a) Given : AB = AC, BC = CD, ∠BAC = 28°
ICSE Class 9 Maths Sample Question Paper 1 with Answers 34
Since, AB = AC
∠ABC = ∠ACB. (Equal sides have equal angles opposite to them)
∠ABC + ∠ACB + ∠BAC = 1800 (Sum of angleš in a triangle is 1800)
∠ABC + ∠ABC + 28° = 180°
2∠ABC =180°-28°
∠ABC= \(\frac{152^{\circ}}{2}\)=76°
∠BDC = ∠CBD = 76°
Now, ∠ACD + ∠CAD = ∠BDC (Exterior angle is equal to sum of interior opposite angles)
∠ACD + 28° = 76°
∠ACD = 76° – 28° = 480

(b) Given : A parallelogram ABCD.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 35

To prove:∠A = ∠C and ∠B = ∠D.
Proof: AB II DC, AD II BC ( ABCD is a parallelogram)
∠A + ∠D = 1800 (Co-interior angles) …(i)
and ∠D + ∠C = 180° (Co-interior angles) …(ii)
From (i) and (ii),∠A + ∠D =∠D + ∠C
∠A=∠C
Similarly,∠B = ∠D.Hence Proved.

(c) In triangle, length of equal sides (a) = 5 cm, base (b) = 8 cm.
In rectangle, Length (L) = 8 cm, Breadth (B) = 6.5 cm.
(i) The area of cross-section = Area of rectangle + Area of triangle
ICSE Class 9 Maths Sample Question Paper 1 with Answers 36

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(ii)  Length of metal = 6 m = 600 cm.
Volume = Area of cross-section x Length
= 64 cm2 x 600 cm
= 38400 cm3.

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Section -A (40 Marks)
Attempt all questions

Question 1
(a) What do you mean by reusability feature.
(b) Write the java expression for the roots of the quadratic equation.
(c) Name any two jump statments and their usages.
(d) (i) Name the mathematical function which is used to find the cosine of an angle given in radians.
(ii) Name a string functions which removes the blank spaces provided in the prefix and suffix of a string.
(e) What do you mean by dynamic initialization of an array? Give an example.
Answer:
(a) Reusability feature is related to inheritance in java where derived class can use the methods and common data from the base class.

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

(b) double r1 = (-b + Math.sqrt(b * b – 4 * a * c)) / (2 * a);
double r2 = (-b – Math.sqrt(b * b – 4 * a * c)) / (2 * a);

(c) (i) break;
(ii) continue
for(int i =1; i< 10; i++)
{
if(i*3 == 9)
break;
}
for(int i =1; i< 10; i++)
{
if(i%2 == 0)
continue;
System.out.println(i);
}

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

(d) Math.cos(b) trim( ).
Answer:
Math.cos(b)
trim( ).

(e) When the array is initialized during the run time of program its called dynamic initialization of array.
E.g.
int a[ ] = new int[10];

Question 2
(a) Consider the following code
class lol
{
public static int m=3,y=4; public int a =10,b = 15;
}
(i) Name the variables for which each of object of the class will have its own distinct copy.
(ii) Name the variables that are common to all the objects of the class.
Answer:
(i) m and y
(ii) a and b.

(b) Distinguish between constructor and method.
Answer:

ConstructorMethod
Automatically called during the creation of the object.Class Object needs to explicitly call the member functions.
It bears the same name as that of class.It doesn’t have same name.

(c) What are the values of a and b after the following function is executed if values passed are 30 and 50.
void pass(inta, int b)
{
a = a+b; b= a-b; a = a-b;
System.out.println(“a=”+a+””b=”+b);
}
Answer:
a=50 b=30

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

(d) Rewrite the following statement using if-else statement amount = (x!=50)?((x<50)?(4.0/100*x):(l 0.0/100*x)):500;
Answer:

if(x!=50)
{
if(x<50)
amount = 4.0/100*x; 
else
amount = 10/100*x;
}
else
amount = 500;

(e) Name any two tokens of java.
Answer:
variables and constants.

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 3
(a) What are the different keywords that checks the visibility of a member of the class? What are they called?
(b) Determine how many times loop will be executed and print the output.
int a=1, b=2; while(++b<6)a*b;
System.out.println(a);
(c) Attempt the following:
(i) A package that is involved to manipulate character as well as String?
(ii) Name a data type that can hold 16 bit Unicode characters.
(iii) What is the output of code below:
charc = ‘A’; int m = 5;
System.out.println(char(c+m));
System.out.println(c+m);
(iv) Write statements to show how finding the length of a character array char[ ] differs from finding the lengths of string object str.
(v) Give the output of the following functions System.out. println(“MALAYALAN”.indexOf(‘A,)+”Sidharth”.lastlndexOf(‘h’));
(vi) double a = Math.pow(“200”.index0f(‘0’),2);
System.out.println(a);
(vii) Differentiate between linear search and binary search.
(viii) Evaluate the following expression:
int p, k= 8,m=11,r=7; p = (r++%7)+(-m%5)+k*(++k-8);
Answer:
(a) private, public, protected
(b) 3 times 60
(c) (i) java.lang
(ii) byte
(iii) F 70
(iv) array.length and str.length( )
(v) 8
(vi) 1.0
(vii)

Linear SearchBinary Search
It compares each element of the array with rest of the elements in the array.It’s based on divide and conquer rule. Array is sorted in this search. Element is searched only in the selected halves.

(viii) 8

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 4
Write a program in java to input a number and check whether it is a pronic number or Heteromecic number or not.
Pronic number: A pronic number, oblong number, rectangular number or heteromecic number is a number which is the product of two consecutive integers i.e n(n-t-l)
The first few Pronic numbers are:
0,2,6,12.20,30,42,56,72.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.print("lnput a number:"); 
int n = sc.nextlnt( ); 
int result = 0; for(int i=0; i<n; i++)
{
if(i*(i+1) == n)
{
result = 1; break;
}
}
if(result == 1)
System.out.println(''Pronic Number."+n); 
else
System.out.println("Not a Pronic Number."+n);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 5
Write a program to input a word from the user and remove the duplicate characters present in it.
Example:
input: crcricicket
output: criket
Answer:

import java.util.*; 
class Main
{
public static void main(String[ ] args)
{
Scanner sc=new Scanner(System.in); 
System.out.printlnC'Enter a sentence:"); 
String str=sc.nextLine();
String word =
char c = str.charAt(O);
word+=c;
for(int i=0;i<str.iength();i++)
{
char ch=str.charAt(i);
boolean flag=false;
for(int j = 0; j<word.length();j++)
{
if(ch==word.charAt(j))
{
flag = true;
}
}
if(flag == false) 
word+=ch;
}
System.out.println(word);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 6
Write a program to accept a sentence and print only the first letter of each word of the sentence in capital letters separated by a full stop.
e.g.
INPUT Sentence: Sid is a cricket output: S.I.A.C
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc=new Scanner(System.in);
System.out.println(''Enter a sentence: ");
String str=sc.nextLine( );
str = "" +str;
str = str.toUpperCase( );
String word =
for(int i = 0; i < str.length( ); i++)
{
char ch = str.charAt(i); 
if(ch =='')
{
word+=str.charAt(i+1
}
}
System.out.println(word);
} //end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 7
Write a program to accept name and corresponding age in two different single dimensional array. Display the records in descending order of age using bubble sort.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
int temp;
String temp 1;
String a[ ] = new String[15];
int age[ ] = new int[15];
Scanner sc = new Scanner(System.in); 
System.out.printlnC'Enter a 15 names and their ages:"); 
for(int i =0; i<a.length; i++)
{
System.out.print("Enter name"+ (i+1) 
a[i] =sc.next( );
System.out.print("Enter age"+ (i+1) +":"); 
age[i]=sc.nextlnt( );
}
for(int i = 0;i<age.length; i++)
{
for(int j =0; j<age.length - i - 1 ;j++)
{
if(age[j]<(age[j+1]))
{
temp = age[j]; tempi = a[j]; 
age[j]=age[j+1]; 
a[j] = a[j+1 ]; 
age[j+1] = temp; 
a[j+1]=temp1;
}
}
}
for(int i =0; i <a.length;i++)
{
System.out.println(a[i]+"\t"+age[i]);
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 8
Write a menu driven program to find area of an Equilateral triangle, an isosceles triangle and a scalene triangle as per the users choice.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in); int n;
float a,c,s,b; double area;
System.out.priintln("1 .Area of equilateral triangle"); System.out.priintln("2.Area of isosceles triangle");
System.out.priintln("3.Area of scalene triangle");
n=sc.nextlnt( );
switch(n)
{

case 1:
System.out.printlnC'Enter side of an equilateral triangle");
s=sc.nextFloat( );
area=Math.sqrt(3.0*s*s)/4.0;
System.out.println("Area="+area);
break;

case 2:
System.out.printlnC'Enter the side and base of isosceles triangle”);
a=sc.nextFloat( );
b=sc.nextFloat( );
area=b/4.0*(Math.sqrt(4.0*a*a-b*b));
System.out.println("Area="+area);
break;

case 3:
System.out.printlnC'Enter the 3 sides of scalene triangle");
a=sc.nextFloat( );
b=sc.nextFloat( );
c= sc.nextFloat( );
s= (a+b+c)/2;
area=Math.sqrt(s*(s-a)*(s-b)*(s-c));
System.out.println(”Area="+area);
break-
default:
System.out.println("Wrong choice");
}
}//end of main 
}//end of class

Question 9
Define a class to overload a function Sum() as follows:
(i) int Sum(int a, int b) – with integer arguments a and b.
Calculate and return sum of all even numbers in the range of a and b.
Sample input: a = 4, b = 16
Sample output: sum = 4 + 6 + 8 + 10 + 12 + 16

(ii) double Sum(double n) – with one double argument n. Calculate and return the product of the following series:
sum = 1.0*1.2*1.4*…’*n

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

(iii) int Sum(int n) – with one integer argument n. Calculate and return sum of only odd digits of the number n.
Sample input: n = 43961
Sample output: sum = 3 + 9+1=13
Answer:

import java.util.*;
class Main
{
public int sum(int a, int b)
{
int sum= 0; for(int i=a; i<=b;i++)
{
if(i%2==0)
sum+=i;
}
return sum;
Wend of sum
public double sum(double n)
{
double prod=1.0;
for(double i =1.0;i<=n; i+=0.2)
prod*=i; 
return prod;
}
public int sum(int n)
{
int sum = 0; 
int a = n; 
int digit; 
while(a>0)
{
digit = a%10; 
if(digit%2!=0) 
sum+=digit; 
a=a/10;
}
return sum;
}
public static void main(String args[ ])
{
Main s = new Main( );
System.out.println(s.sum(1,5));
System.out.println(s.sum(10.0));
System.out.println(s.sum(10935));
}//end of main
} //end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

VariableData TypeDescription
nintUsers input integer value
resultintFlag to check pronic number
ccharTo store first character
flagbooleanTo check character is present in word or not
strStringUser input string.
wordStringModified string
a[ ]stringTo store 15 names
age[ ]intTo store integer number
tempintTemporary variable for swapping
tempiStringTemporary variable for swapping
nintUsers choice
a c s bfloatS = side of equilateral triangle A b side and base of issocless triangle Abe sides of scalene triangle
areadoubleTo store area
proddoubleStore product of decimal number in series
digitintTo store digit of number

ICSE Class 10 Computer Applications Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Section- A (40 Marks)
Attempt all questions

Question 1
(a) Give one similarity and one difference between do while and while loop.
(b) Write the java statement to find largest of 3 numbers using conditional operator.
(c) What is polymorphism?
(d) Write a function to round off any fractional number? Give one example.
(e) What is difference between infinite loop and fall through statement?
Answer:
(a) Similarity – Both while and do while loop is used when the no. of iterations are not specified. Difference: while loop is entry controlled loop and while do while loop is exit controlled.

(b) int a = (b>c)?b:c; int d = (a>e)?a:e;

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(c) Polymorphism is the identity to represent an object in more than one form.

(d) Math.round( )
double k = Math.round(2.9);

(e)

Infinite LoopFall Through
Occurs for iteration statementOccurs for conditional statement.
Occurs when conditions is always satisfied.Occurs when the break statement is missing in every case statement.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 2
(a) Compare local, instance and class variables on the basis of the given criteria.
(i) identification in program
(ii) Usage of access specifier with them.
(iii) provision of initial values.
(iv) Number of copies for each object of class.
Answer:
(a) (i) Local variables are usually used in a block of statement. Instance variables are used by a method while class variable has only one copy throughout out the class.

(ii) Local variable scope remains inside the compound statement. Inside the curly braces.
Instance variable scope remains inside the method. All access specifiers can be used in class variables. Scope is throughout the class.

(iii) Local variables is initialized in its scope.
Instance variables initialized in the method where it is used in.
Class variables are initialized after the class declaration.

(iv) local variables multiple copies can exists.
Instance variables multiple copies can exists.
Class variables single copy exists.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(b) Which OOPs principle is implemented when you:
(i) use the keyword extends
(ii) create an interface
(iii) create various functions
(iv) use private data members?
Answer:
(i) Inheritance
(ii) Abstraction
(iii) Polymorphism
(iv) Encapsulation

(c) What are functions? Give one advantage using functions.
Answer:
Functions are a set of statements used to perform certain tasks. Advantage: It organizes the code into a blocks and helps in performing specific task.

(d) Differentiate between classes and objects. Why is object referred as instance of class?
Answer:
(i) Object is an identifiable entity with some characteristics and behaviour. Class is a collection of object with similar characteristics.

(ii) Object is called instance of class as object helps in accessing the class members and functions to get the required output.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(e) Differentiate between linear search and binary search.
Answer:

Linear SearchBinary Search
The search element is compared with each element of the array.The array is sorted and divided into two halves and then search element is found accordingly.
Array need not be sorted.Array has to be sorted.

Question 3
(a) What is other name of java interpreter? Explain it.
Answer:
JVM (Java virtual machine): It is used to convert java byte code to machine code which can be understood by the computer running on any operating system.

(b) Name the java keyword
(i) used to finish the execution of the method.
(ii) used to implement the concept of inheritance.
Answer:
(i) exit or return
(ii) extends

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(c) Attempt the following:
(i) Assign the constant value pi as 3.14 to variable using suitable data type?
(ii) Give difference between unary and binary operators.
(iii) Give examples of each
(i) Composite date type.
(ii) Escape sequence.
(iii) Comment lines
(iv) Wrapper class
(iv) Write any two rules of naming variable
(v) Differentiate between type conversion and coericion.
(vi) State the difference between final and finally.
(vii) a. State the use of toString( ) and valueOf( ).
b. Write a statement to extract the last word of the string str.
(viii) a. What is called byte code?
b. Explicit and implicit type conversion.
Answer:
(i) double const pi =3.14;
(ii) Unary operator is is used for one operand.
Binary operator is used for two operands.
(iii) (i) Array, class
(ii) “\n”‘;\\’
(iii) //or/*java*/
(iv) Character, Integer
(iv) Variables should not be keyword and it should not start with digit or special character except for the $ and & and _. It must start with alphabet.

Type conversionCoercion
Assigning one data type to anotherPromotion of data type
It’s called implicit type castingAlso called type promotion.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(v) Final makes a variable constant throughout the program. Finally is a block which is always executed even if exception occurred or not.

(vi) a.

toString( )valueOf( )
Converts integer to string.Converts strings to integer.

b. String word = str.substring(sr.lastlndexOf(“)+1);

(vii) a. Byte code is an specific set of instruction to be executed by the JVM.
b. Implicit type conversion: Java compiler will automatically convert one data type to another when data types are compatible.
E.g. int 1 = 10;
double b = I;
Explicit type conversion: Java compiler is to be told explicitly to do the conversion form one data type to another.
E.g.
double d = 10.0;
float f = (float)d;

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 4
Using switch case write a menu driven program to print the patterns.
(a)
0 0 0 0 0
2 2 2 2
6 6 6
12 12
20

(b)
X
YY
XXX
YYYY
XXXXX
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter the choice:");
int a1 = sc.nextlntO;
switch(al)
{
case 1: int a =0; int b = 2;
for(int i =5;i>= 1;i—)
{
for(int j=1; j<=i;j++)
System.out.print(a+""); -
a+=b;
b+=2;
System.out.println( );
}
break;
case 2: for(int i =1 ; i<=5; i++)
{
for(intj=1;j<=i;j++)
{
if(j%2 != 0)
System.out.printC'X");
else
System.out.printC'Y");
}
System.out.printlnO;
}
break;
default: System.out.println("lnvalid Choice");
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 5
Write a program to print the frequency of the digits occurring in number.
Eg.2566773
Output:
2 1
5 1
6 2
7 2
3 1
Answer:

import java.util.*; 
class Main {
static int frequency(int number, int digit)
{
int count = 0;
while (number > 0)
{
if (number % 10 == digit) count++;
number = number / 10;
}
return count;
}
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in);
System.out.println(''Input the number:");
int n = sc.nextlnt( );
int a = n;
int search = 0;
int countdigit = 0;
int []arr = new int[10];
while(a >0)
{
search = a%10;
countdigit = frequency(n,search);
if(countdigit>0)
{
arr[search]=countdigit;
countdigit=0;
}
a=a/10;
}
for(int i =0; 
i<arr.length;i++)
{
if(arr[i]>0)
System.out.println(i+"\t"+arr[i]);
}
}//end of main
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 6
Write a class with a special member function to input a multi-digit number(max 9 digit) and print the following:
(i) total number of even and odd digits
(ii) the reverse number
(iii) total number of zeros present.
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in);
System.out.priintln("lnput the number:");
long a = sc.nextlntQ;
int sumeven =0;
int sumodd =0, sum3 = 0;
String str - long a1 = a;
while(a1 <=999999999&&a1 !=0)
{
int k= (int)al %10;
str+=lnteger.toString(k);
if(k%2 == 0)
sumeven++;
else if(k%2!=0)
sumodd++;
else if (k==0)
sum3++;
else;
a 1 =a 1/10;
}
System.out.println("No. even: "+sumeven+"\t"+ "odd:"+sumodd+"\ t"+"zero: "+sum3+"\t"+"Reverse no:"+str);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 7
Write a class to input a string (combination of letters and digits) and replace the repecated characters with star(*) sign. Then display both old and newly created string.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc=new Scanner(System.in); 
System.out.println("Enter a sentence:"); 
String str=sc.nextLine( );
for(int i=0;i<str.length( );i++)
{
char ch=str.charAt(i); if(ch!='*')
{
for(int j=str.length( )-1;j>i;j—)
{
char ch 1 =str.charAt(j); jf(ch—chi)
{
StringBuffer strl =new StringBuffer(str);
str1.setCharAt(j,'*');
str=""+strl;
}
}
}
}
System.out.println(str);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 8
Write a program to print the following pattern by taking the input n from the user.
Note: n determines number of rows.
Eg. n = 4
Output:
* * * *
* * *
* * * *
* * *
Answer:

import java.util.*;
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in); 
System.out.println("lnput the number:"); 
int n = sc.nextlnt( ); 
for(int i = 1; i <=n;i++)
{
if(i%2==0)
{
for(int j =1; j<=3;j++) 
System.out.printC *");
}
else
{
for(int j =1; j<=4;j++)
System.out.printC*");
}
System.out.println( );
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 9
Write a java class to calculate and print the electricity bill to be paid by a customer.
Assume that the customer pays a rent of Rs.350.00
No. of units — Charge per unit
Upto 100 units — Rs. 1.50
For the next 100 units — Rs. 2.00
For next 50 units — Rs. 2.50
Beyond 250 units — Rs. 4.00
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc=new Scanner(System.in); 
double fixed_price = 350.0; 
double amt = 0.0;
System.out.println("Enter a unit of electricity consumption: ");
int n=sc.nextlnt( );
if(n<=100)
amt+= amt+1.5*n;
else if(n>100&&n<=200)
amt+= 100*1,5+(n-100)*2.0;
else if(n>200&&n<=250)
amt+= 100*1.5+0 00)*2.0 + (n-200)*2.5;
else
amt+= 100*1.5+0 00)*2.0 + (50)*2.5+(n-250)*4.0;
System.out.println("Electricity bill amount: "+amt);
}//end of main 
}//end of class
VariableData TypeDescription
alintStore users choice
a bintFor printing pattern
ijintFor looping
numberintNumber in which digits to be searched
digitintDigit which is to be searched.
countintCounting number of digits
nintUsers input number
arr[]intArray of 10 integers.
search, countdigitintDigit to be searched and count of that digit
sumeven,sumodd,

sum3

intSum of even digits, sum of odd digits and count of zero.
strStringUser input string
strlString BufferTo store modified string
chcharTo store character
fixed „price, amtdoubleFor fixed price and calculating amount

ICSE Class 10 Computer Applications Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Section – A
Attempt all questions

Question 1
(a) What do you mean by abstraction.
(b) Using an example explain the term object.
(c) What is wrapper class.
(d) State the use of new operator.
(e) Write a java expression for square root of sinx.
Answer:
(a) Abstraction is the act of representing the essential features of the program without involving in its complexity.

(b) Eg. While using computer user is not concerned about the different parts of the computer only needs to know about usage. Characteristics (big, small, flat etc)

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

(c) Wrapper class is a class which helps to convert a primitive data type to object type. E.g. Integer, Short etc.

(d) New operator supports instantiations. It is used to allocate memory space to the newly created object.
E.g. Scanner sc = new Scanner(System.in)

(e) double d = Math.sin(x*3.14/180);

Question 2
(a) Differentiate between local variable and global variable.
Answer:

Local VariableGlobal Variable
Scope remains only inside a block or a function.Scope remains throughout the class.
Multiple copies of the variable are used throughout the class.Only a single copy of the variable is used throughout the class.

(b) What is the purpose of default in switch statement?
Answer:
When none of the cases of the switch statement are satisfied, the default case is used and executed in the program.

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

(c) Define an impure function.
Answer:
An impure function is a function where values of the arguments keep changing during the execution of the program.

(d) What is called default constructor?
Answer:
A default constructor is a constructor without any parameters. It is used to assign a default value to the class variables.

(e) Explain the concept of constructor overloading using an example.
Answer:

class lol
{
int a; int n;
lol( )
{
a=0;
}
lol(int c)
{
a=1;
}
public static void main( )
{
lol obj = new lol( );
 lol obj1=new lol(2);
}
}

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 3
(a) How java make use of formal and actual parameter?
Answer:
Formal parameters are used in the function definition of the program while actual parameters are used in the function call statement.

(b) State the java concept that is implemented into smaller groups
(i) Dividing a long set of instruction into smaller groups and modules.
(ii) Wrapping up of data and its associated function into a class.
Answer:
object oriented programming Encapsulation

(c) Attempt the following:
(i) What is role of access specifier?
(ii) Why is java platform independent?
(iii) Differentiate between public and private visibility label.
(iv) Int a = 7,b = 6; b = a>b?a!=0?3:b<10?9:4:8; System.out.println(b);
(v) Int a = 7,b = 6; b=++a+ ++a/ ++a; System.out.println(b);
(vi) What are functions? Give one advantage of using functions.
(vii) How do objects communicate with each other?
(viii) What are two types of methods? Give two concrete differences between them and include example of each type of function from java.lang.Math class.
Answer:
(i) Access specifiers are used to moderate the accessibility of the data according to user’s choice.
(ii) Java is platform-independent because the JVM can interpret the byte code which can run on any platform.
(iii) Public: It can be accessed anywhere.
Private: It can be accessed only inside the class.
(iv) 3
(v) b=8+9/10=8+0=8
(vi) A function is block of statements in a java program used to perform certain tasks. It organizes the data into segments and helps in performing tasks easily.
(vii) Objects communicate with each other using methods.
(viii)

Pure FunctionImpure Function
Functions where the value of variables don’t changeFunction where the values of the variables changes.
Depends on the actual parameters passed in the function call.Depends on where created object calls the functions.
E.g. Math.sqrt( )E.g. Math.random(i)

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 4
A Smith number is a composite number, the sum of whose digits is the sum of the digits of its prime factors obtained as a result of prime factorization(excluding 1).
The first few such numbers are 4,22,27,58,85,94,121
eg.666
Prime factors are 2,3,3 and 37
Sum of the digits are (6+6+6) = 18
Sum of the digits of the factors (2+3+3+3+7)=18
Write a program to input a number and display whether the number is a Smith number or not.
Answer:

import java.util.*; 
class Main {
int sumDigfint n) 
{
int sum=0; while(n>0)
{
sum+=n%10;
n=n/10;
}
return sum;
} 
int sumPrimeFact(int n)
{
int i=2, sum=0; while(n>1)
{
if(n%i-=0)
{
sum=sum+sumDig(i);
n=n/i;
}
else
i++;
}
return sum;
}
public static void main(String[ ] args)
{
Main ob=new Main( );
Scanner sc = new Scanner(System.in);
System.out.print("Enter a Number:");
int n=sc.nextlnt( );
int a=ob.sumDig(n);
int b=ob.sumPrimeFact(n);
if(a==b)
System.out.print("lt is a Smith Number"); else
System.out.print("lt is Not a Smith Number"); 
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 5
Write a program to assign a full path and file name as given below. Using library functions, extract and output the file path, file name and file extension separately as shown.
Input: C:\users\sid\pictures\cricket.jpg
output: C:\users\sid\pictures
File name: Cricket
Extension: jpg
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in); 
System.out.priintlnflnput the file path:");
String str = sc.nextLine( );
String strl = str.substring(0,str.lastlndexOf('\V)); 
String str2 = str.substring(str.lastlndexOf('\V)+1); 
String str3 = str2.substring(str2.indexOf(' ’)+1); 
System.out.println("Path:" +str1); 
System.out.println("Filename:"+str2); 
System.out.println("Extension:"+str3);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 6
Write a program to input a word from user and remove the consecutive repeated characters by replacing the sequence of repeated characters by its single occurrence.
input: ssiiddaarrtth
Output: sidarth
Answer:

import java.util.*;
class Main {
public static void main(String args[ ])
{
String word ="";
Scanner sc = new Scanner(System.in);
System.out.println("lnput the sentence:");
String str = sc.nextLine( ); char chprev='';
for(int i =0;i<str.length( );i++)
{
char ch = str.charAt(i); 
if(chprev!=ch)
{
word+=ch; chprev = ch;
}
else
continue;
}
System.out.println("word without successive repeated characters is:"+word);
}//end of main
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 7
Write a program in java to input a number and check whether it is Duck number or not.
Note: A Duck number is a number which has zeroes present in it, but there should be no zero present in the beginning of the number. For example 3210, 7056, 8430709 are all duck numbers whereas 08237,04309 are not.
Answer:

import java.util .*;
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in);
System.out.printlnC'Input the number:");
int a = sc.nextlnt( );
int a 1 = a;
int count =0;
String str = Integer.toString(a);
while(a1!=0)
{
if (str.charAt(0)==0)
break;
else
{
int k = a 1 %10; 
if(k==0) count++; a1/=10;
}
}
if(count>=1)
System.out.println("Duck no."); else
System.out.println("Not Duck no.");
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 8
Write a program in java to input a number in Decimal number system and convert it into its equivalent number in the octal number system.
Note: Octal number system is a number system which can represent a number in any other number system in terms of digits ranging from 0 to 7 only. This number system consists of only eight basic digits i.e. 0,1,2,3,4,5,6 and 7.
eg. 25 in the decimal number system can be represented as 31 in the octal number system.
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in); 
System.out.println("lnput the decimal number:"); 
int a = sc.nextlnt( ); 
int a1 =a%8; 
int a2 = a/8;
String str=" "; 
int a3=a2%8;
String strl = Integer.toString(al);
String str2 = lnteger.toString(a3); 
str=str2+str1;
int ah = Integer.parselnt(str); 
System.out.print(ah);
}//end of main
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 9
Using Scanner class, write a program to input a string and display all those words of the strings which begins with capital letter and end with a small letter.
Sample Input: We all love Java for School Students because of its Uniqueness
Sample Output: We Java School Students Uniqueness
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
String word =
Scanner sc = new Scanner(System.in);
System.out.println("lnput the sentence:");
String str = sc.nextLine( ); str+="";
for(int i =0;i<str.length( );i++)
{
char ch = str.charAt(i);
if(ch!='') word+=ch; 
else {
jf(Character.isUpperCase(word.charAt(0))&&Character.isLowerCase(word. charAt(word.lengthO-1)))
{
System.out.print(word+"");
}
word =
}
}
} //end of main 
}//end of class
VariableData TypeDescription
sumintTo store sum of digits of number
nintNumber whose digits are to be added
sumintTo store sum
iintTo start with 2
bintStore sum of prime factors
aintStore sum of digits
strStringStore the string
strlStringTo store the path
str2StringTo store filename
str3StringStore file extension
wordStringTo store modified word
chcharto store character
chprevcharTo store previous character
a al countintUsers input number, modified number and count number of 0
a1intTo store octal number last digit
a2intQuotient after dividing by 8
a3intQuotient after dividing by 8
strlStringTo store a 1 as string
str2StringTo store a2 as string
ahintStore converted integer from string.

ICSE Class 10 Computer Applications Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Section- A (40 Marks)
Attempt all questions

Question 1
(a) What is the return of following function.
(i) equals( )
(ii) rint( ).
(b) What do you mean by compound statement. When do you need it.
(c) Explain difference between character constant and string constant.
(d) State any two features of constructor.
(e) Define abstraction.
Answer:
(a) (i) boolean
(ii) double

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

(b) A set or a block of statement in a program is called compound statement. It can be used to divide the program into segments and organise the data accordingly.

(c)

Character ConstantString Constant
Enclosed in single quotes E.g. ‘A’, ‘e’Enclosed in double quotes
E.g. “hello”, “1 like JAVA”

(d) It has same name as that of class. It has no return type.
(e) Abstraction is the act of representing the essential features of a program without involving in its complexity.

Question 2
(a) Explain the term object using example?
(b) What is the wrapper class. Justify with example?
(c) State the purpose of new operator?
(d) Write java expression for
\(\sin x+\sqrt[2]{a x^{2}-b x+c}\)
Answer:
(a) Object is an identifiable entity with some particular characteristics and behaviour.
E.g.TV, characteristics: Big, small etc.
Behaviour: Show different channels, off/on etc.
(b) The class which helps in converting primitive data types to object type is called wrapper class. Eg. Character, Integer etc.
(c) The ‘new’ keyword supports instantiation. It allocates memory for the newly created object.
(d) double z = Math.sin(x)+Math.sqrt(a*x*x + b*x + c);.
(e) class lol

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 3
(a) Explain the concept of constructor overloading with example.
(b) Differentiate between equals( ) and compareTo( ).
(c) Attempt the following:
(i) Differentiate between call by value and call by reference.
(ii) What is an identifier. Give example
(iii) Name the package that contains the scanner class. Which unit or class gets called when object is created.
(iv) What is meant by encapsulation.
(v) What is meant by inheritance.
(vi) char a[ ]={‘a’,’b’,,c’}; System.out.println(a); What will be the output.
(vii) char a[ ]={‘a’,’b’,’c’}; System.out.println(a[1]++); System.out. println(++a[1]);
(viii) inta[ ]={1,2,3};System.out.println(a[1]++);System.out.println(++a[1]);
Answer:
(a) Same as above.

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

(b)

Equals( )CompareTo( )
It returns boolean valueIt returns integer value
Compares the whole string at once.It compares the ASCII values of each character of the string.

(c) (i)

Call by ValueCall by Reference
Actual parameters are copied into the formal parameters.Actual parameters alias are created in the formal parameters.
Changes made by the formal parameters is not reflected back to the actual parameterChanges made by the formal parameter directly affect the actual parameters.

(ii) Identifiers are those quantities which change their values during the execution of the program. E.g. Int a, b;

(iii) java.util and constructor.

(iv) Encapsulation: Wrapping of data and its associated functions into a single unit.

(v) Inheritance is the method of inheriting the properties of one class from another class. The class which derives its own properties is called base class and the class which derives the properties of another class is called sub class.

(vi) Address of array a[ ]

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

(vii) b d

(viii) 2

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

Question 4
Write a program to print the sum of prime digits from given number.
E.g. 134667
Output: 1+3+7=11
Answer:

import java.util.*;
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.println("enter the number:"); 
int a = sc.nextlnt( ); 
int a 1 = a; int sum = 0; while (a 11=0)
{
int count = 0; int k = a1%10;
for(int i=1; i<=k; i++)
{
if(k%i==0) count++;
}
if(count==2) sum+=k; a 1 /=10;
}
System.out.println("sum = "+sum); 
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 5
Write a program to print the longest word from the sentence and print the number of characters of the longest word.
Answer:

import java.util.*; 
class Main
{
public static void main(String [ ] largs)
{
int count = 0; String temp;
Scanner sc = new Scanner(System.in); 
System.out.println("Enter a sentence:"); 
String str = sc.nextLine( ), 
word =""; str - str+" ";
for(int i = 0; i<str.length( ); i++)
{
char ch = str.charAt(i);
if(ch=='') count++;
}
String a[ ] = new Stringfcount]; 
for(int i = 0; i<count;i++)
{
for(int j = 0;j< str.length( ); j++)
{
char ch = str.charAt(j);
if(ch!='') word+=ch; 
else {
a[i] =word; word ="
}
}
}
for(int i = 0;i<a.length; i++)
{
for(int j = 0;j<a.length-i-1;j++)
{
if(a[j].length() > a[j+1].length())
{
temp = a[j]; a[j]=a[j+1]; a[j+1]=temp;
}
}
}
System.out.printlnfLONGESTWORD: "+a[a.length-1]); 
System.out.println("No. of characters:"+a[a. length-1 ].length()); 
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 6
Write a program to find sum of negative numbers, sum of positive odd numbers and sum of positive even numbers entered by user and list terminates when user enters 0.
Answer:

import java.util class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
int sum = 0, sum1 = 0, sum2 = 0;
System.out.println("Enter the number to calculate and 0 for exit:"); 
int n = sc.nextlnt( ); int a = n; while(a!=0)
{
if(a>0)
{
if(a%2 == 0)
sum1+=a;
else
sum2+=a;
}
else
{
sum+=a;
}
System.out.println(''Enter the next number and 0 to exit:"); 
a = sc.nextlnt( );
}
System.out.println("Positive even sum: "+suml+"\t"+"Positive odd sum:"+sum2+"\t"+"Negative sum:"+sum);
} //end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 7
Using switch case write a menu driven program to print the patterns.
(a)
0 0 0 0 0
2 2 2 2
6 6 6
12 12
2 0

(b)
X
Y Y
X X X
Y Y Y Y
X X X X X
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.println(''Enter the choice:");
intal = sc.nextlnt( ); switch(al)
{
case 1: inta = 0; int b = 2;
for(int i =5;i>= 1;i—)
{
for(int j=1; j<=i;j++)
System.out.print(a+" ");
a+=b;
b+=2;
System.out.println( );
}
break; case 2:
for(int i =1; i<=5; i++)
{
for(int j=1;j<=i;j++)
{
if(j%2 != 0)
System.out.printC'X");
else
System.out.printC'Y'');
}
System.out.println( );
}
break;
default: System.out.println("lnvalid Choice");
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 8
Write a program to find using binary search method from list of roll numbers entered by user in ascending order. If the search is successful print “you are selected to go” else print “Try next time”.
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
int first, last, middle, n, search, array! ];
Scanner sc = new Scanner(System.in);
System.out.println(''Enter total number of students");
n = sc.nextlntO; array = new int[n];
System.out.printlnC'Enter" + n + " integers"); 
for (int i = 0; i<n; i++) arrayti] = sc.nextlnt( );
System.out.printlnC'Enter value to find");
search = sc.nextlnt( ); first = 0; last = n - 1;
middle = (first + last)/2; while(first <= last)
{
if (array[middle] < search)
first = middle + 1;
else if (array[middle] == search)
{
System.out.println(search + " you are selected to go " + (middle + 1) + "•"); break;
}
else
last = middle - 1; 
middle = (first + last)/2;
}
if (first > last)
System.out.println(search + " Try next timeAn");
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 9
Write a program to input sentence and print in the lowercase letters and replace all the words like “is” and “are” with “were” and “had” and “has” with “had”.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in); 
System.out.println("Enter the sentence:"); 
String str = sc.nextLine( );
String str1 = str.toLowerCaseQ;
String word = strl +=" ";
String str2 =" ";
for(int i =0;i<str1 Jength(); i++)
{
char ch = strl .charAt(i);
if(ch!='') word+=ch;
else {
if(word ,equals("is")) str2=str2+"were"+"";
else if(word.equals("are")) str2=str2 + "had"+""; 
else if(word.equals("has")) str2=str2+"had" +""; else
str2=str2+word+" "; word =
}
}
System.out.println(str2);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

VariableData TypeDescription
aintStore input integer from user
a1,sumintFor modifying digits and to calculate sum of digits
strStringStore input string from user
chcharStore the character
countintCount number of blank spaces
wordStringTo store new word
tempStringTemporary storage for swapping strings.
a[ ]StringArray of strings.
nintStore integer from user
aintFor modifying value
sum1,sum2,sum3intEven number sum Positive odd sum Negative sum
bintFor printing even pattern
‘jintFor looping
first last middle n search arrayj]intFirst last and middle are Index of array N is total number of elements Search is number to be searched in array. Array[ ] array of sorted roll numbers
strlStringStore lowercase word
str2StringTo store the modified sentence.

ICSE Class 10 Computer Applications Question Papers with Answers