ICSE Class 9 Maths Sample Question Paper 10 with Answers

ICSE Class 9 Maths Sample Question Paper 10 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Evaluate without using tables :
sin 38° sin 52° – cos 38° cos 52°.
Answer:
sin 38° sin 52° – cos 38° cos 52° = sin 38° sin (90° – 38°) – cos 38° cos (90° – 38°)
= sin 38° cos 38° – cos 38° sin 38°
= 0.

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(b) In the figure, CD is a diameter which meets the chord AB at E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle
ICSE Class 9 Maths Sample Question Paper 10 with Answers 1
Answer:
Given : CD is a diameter, AE = BE = 4 cm, CE = 3 cm.
∠OEB = 90° (∵ E is the mid-point of BA).
Let the radius OB = OC = r cm.
Then, OE = OC – CE = (r – 3) cm.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 7
In ΔOBE, by Pythagoras theorem,
OB2 = OE2 + BE2
⇒ r2 = (r – 3)2 + 42
⇒ r2 = r2 – 2.r.3 + 9 + 16
⇒ 6r = 25
⇒ \(r=\frac{25}{6}=4 \frac{1}{6}\)
The required radius = \(4 \frac{1}{6}\) cm.

(c) If a = 1 – √3, find the value of \(\left(a-\frac{1}{a}\right)^{3}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 8
ICSE Class 9 Maths Sample Question Paper 10 with Answers 9

ICSE Class 9 Maths Sample Question Paper 10 with Answers

Question 2.
If \(a=\frac{1}{a-5}\),Find
(i) \(a-\frac{1}{a}\)
(ii) \(a+\frac{1}{a}\)
(iii) \(a^{2}-\frac{1}{a^{2}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 10

(b) The water bills (in ₹) of 32 houses in a locality are given below.
80, 48, 52, 78, 103, 85, 37, 94, 72, 73, 66, 52, 92, 85, 78, 81, 64, 60, 75, 78, 108, 63, 71, 54, 59, 75, 100, 103, 35, 89, 95, 73.
Construct a frequency distribution table with a class size of 10.
Answer:
Here, minimum value = 35
Maximum value = 108
Class size =10.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 11

(c) Factorize : (a2 – b2) (c2 – d2) – 4 abcd.
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 12

Question 3.
(a) Show that:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 14
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 13

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(b) In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that:
(i) ΔACE ≅ ΔDBE
(ii) AE = DF.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 2
Answer:
Given: AB = CD,CE = BF and ∠ACE= ∠DBF.
AB = CD
AB + BC = BC + CD (Adding BC on both sides)
⇒ AC = BD.
(i) Now, in ΔACE and ΔDBF
AC = BD (Proved above)
∠ACE = ∠DBF (Given)
CE = BF (Given)
ΔACE ≅ ΔDBF (SAS axiom)
Hence Proved.

(ii) AE = DF (c.p.c.t.)
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(c) If log10y = x, find the value of 102x in terms of y.
Answer:
Given:
log10 y = x
⇒ y = 10x
Or 10x = y
∴ 102x = (10x)2 = y2

Question 4.
(a) Given, 5 cos A -12 sin A = 0, find the value of \(\frac{\sin A+\cos A}{2 \cos A-\sin A}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 47
ICSE Class 9 Maths Sample Question Paper 10 with Answers 46

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(b) Solve by the substitution method :
2x – \(\frac{3}{4}\)y – 3 = 0; 5x – 2y – 7 = 0
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 45

(c) What sum of money will amount to ₹ 3630 in 2 years at 10% p. a. compound interest ?
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 44

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

Question 5.
(a) If x= \(7-4 \sqrt{3}\),find the value of \(\sqrt{x}+\frac{1}{\sqrt{x}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 18
ICSE Class 9 Maths Sample Question Paper 10 with Answers 42
ICSE Class 9 Maths Sample Question Paper 10 with Answers 43

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(b) In the given figure, X and Y are mid-points of sides AB and AC respectively of ΔABC. If
= 6 cm, AB = 7.4 cm and AC = 6.4 cm, then find the perimeter of the trapezium XBCY.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 3
Answer:
Given : BC = 6 cm, AB = 7.4 cm and AC = 6.4 cm X,
∴ Y are mid-points of AB and AC, respectively
∵ By mid-points theory.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 22

(c) The mean weight of 8 students is 46.5 kg. Three more students having weight 41.7 kg, 52.8 kg and 51.8 kg joined the group. What is the new mean weight of the students ?
Answer:
Given : No. of students = 8, mean weight = 46.5 kg.
∴ Total weight of 8 students = 8 x 46.5 kg = 372 kg
∵ 3 more students joined the group,
∴ Total weight of 11 students = 372 + 41.7 + 52.8 + 51.8 = 518.3 kg.
∴ New mean of 11 students = \(\frac{518.3}{11}\) = 47.12 kg.

ICSE Class 9 Maths Sample Question Paper 10 with Answers

Question 6.
(a) Factorize : (x2 – 4x) {x2 – 4x – 1} – 20.
Answer:
(x2 – 4x) (x2 – 4x – 1) – 20
Let x2 – 4x = a
⇒ a (a – 1) – 20 = a2 – a – 20
⇒ a2 – (5 – 4) a – 20
⇒ a2 -5a + 4a -20
⇒ a (a – 5) + 4 (a – 5)
⇒ (a – 5) (a + 4)
Putting the value of a = x2 – 4x, we get
(x2 – 4x – 5) (x2 – 4x + 4)
= (x2 – 5x + x -5) (x2 – 1. x .2 +22)
= {x {x – 5) + 1 (x – 5)} (x – 2)2
= (x -5) (x + 1) (x – 2)2.

(b) The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them saves ₹ 200 per month, find their monthly incomes.
Answer:
The ratio of incomes = 9 : 7 and the ratio of expenditures = 4 : 3
Let their incomes be 9x and 7x and their expenditures be 4y and 3y.
:. Their savings are 9x – 4y and 7x – 3y respectively.
According to the question,
9x – = 200 ———- (i)
7x – = 200 ———- (ii)
Multiplying equation (i) by 3 and equation (ii) by 4, we get
ICSE Class 9 Maths Sample Question Paper 10 with Answers 24
Their incomes are ₹ 1800 and ₹ 1400.

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(c) The centre of a circle is C (2α -1,3α +1) and it passes through the point A (- 3, – 1). If the diameter of the circle is of length 20 units, find the value (s) of a.
Answer:
Given points are C(2α – 1, 3α+1) and A(-3,-1)
Diameter 20 units.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 25
ICSE Class 9 Maths Sample Question Paper 10 with Answers 26

Question 7.
(a) Insert 3 rational numbers between \(\frac{1}{4}\) and \(\frac{1}{2}\).
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 27

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(b) In the figure, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm and
∠ABD = ∠BCD = 90°. Calculate the length of AB.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 4
Answer:
Given : AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90°. (Pythagoras theorem)
In ΔBCD, BD2 = BC2 + CD2
= 32 + 122 = 9 + 144 = 153
In ΔBCD, AD2 = BD2 + AB2
132= 153 + AB2
169 = 153 + AB2
AB2 = 169 – 153 = 16
AB = √16 = 4 cm.

(c) Prove that \(: \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1\)
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 28

ICSE Class 9 Maths Sample Question Paper 10 with Answers

Question 8.
(a) A sum amounts to ₹ 9680 in 2 years and to ₹ 10648 in 3 years compounded annually. Find the sum and the rate of interest per annum.
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 29
ICSE Class 9 Maths Sample Question Paper 10 with Answers 30

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(b) A triangle is formed by the lines x + 2y – 3 = 0, 3x – 2y + 7 = 0 and y + 1=0. Find graphically :
(i) The coordinates of vertices of the triangle.
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 31
ICSE Class 9 Maths Sample Question Paper 10 with Answers 32
ICSE Class 9 Maths Sample Question Paper 10 with Answers 33
Let the lines intersect at the points A, B and C.
(i) The coordinates of vertices of triangle are A (- 1, 2), B (- 3, – 1) and C (5, – 1).
(ii) From A, draw AM ⊥ BC.
Area of Δ ABC
= \(\frac{1}{2}\) × BC × AM
= \(\frac{1}{2}\) × 8 × 3
= 12 sq. units.

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(c) Solve : \(\frac{8}{x+3}-\frac{3}{2-x}=2\)
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 34

Question 9.
(a) In the figure, ABCD is a parallelogram and P is any point on BC. Prove that area of ΔABP + area of ΔDPC = area of ΔAPD.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 5
Answer:
Given : ABCD is a parallelogram
We know, triangles on same base and between same parallels are equal.
∴ Area of AAPD = Area of AABD ……. (i)
and Area of AABP = Area of ABDP ……. (ii)
Also, since diagonal of a parallelogram divides it into two triangles of equal area
∴ Area of ΔBCD = Area of ΔABD …… (iii)
⇒ Area of ΔBDP + Area of ΔDPC = Area of ΔABD
⇒ Area of ΔABP + Area of ΔDPC = Area of ΔAPD [Using (i) and (ii)]
Hence Proved.

(b) Factorize : 1+a+b+c+ab+bc+ca+abc.
Answer:
1+a+b+c+ab+bc+ca+abc=(1+a)+(b+ab)+(c+ca)+(bc+abc)
=(1+a)+b(1+a)+c(1+a)+bc(1+a)
=(1+a)(1+b+c+bc)
=(1+a)((1+b)+c(1+b)
= (1+ a) (1+ b) (1+ c).

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(c) The area of cross-section of a pipe is 5.4 cm2 and water is pumped out of it at the rate of 27 km/h. Find in litres the volume of water which flows out of the pipe in one minute.
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 35

Question 10.
(a) If 4 cos2 x° – 1 = 0 and 0 ≤ x°≤ 90°,find the value of cos2 x° sin2 x°.
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 36

(b) Draw the parallelogram ABCD in which AB = BC = 4.8 cm and AC = 7.5 cm. Find the angle between the two diagonals. What special name can you give it to this parallelogram?
Answer:
Given : AB = BC = 4.8 cm, AC = 7.5cm
ICSE Class 9 Maths Sample Question Paper 10 with Answers 37
Steps of construction:
(1) Draw BC = 4.8 cm.
(2) From B and C, draw arcs of length 4.8 cm and 7.5 cm, respectively which intersect at A.
(3) From A and C, draw arcs each of length 4.8 cm which intersect at D.
(4) Join BA, AD and CD to complete the required parallelogram.
(5) Join AC and BD which intersect at O.
∴ ∠AOD=90°
The special name of this parallelogram is rhombus as all sides are equal.

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(c) Elavuate \(\frac{5^{10+n} \times 25^{3 n-4}}{5^{7 n}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 38

Question 11.
(a) Prove that: \(x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}=1\)
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 39

(b) In the figure, ABCD is a rhombus in which the diagonal DB is produced to E. If ∠ABE = 160°, then find x, y and z.
ICSE Class 9 Maths Sample Question Paper 10 with Answers 6
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 40

ICSE Class 9 Maths Sample Question Paper 10 with Answers

(c) If a + b – c = 4 and a2 + b2 + c2 = 38, find the value of ab – bc – ca.
Answer:
ICSE Class 9 Maths Sample Question Paper 10 with Answers 41

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 9 with Answers

ICSE Class 9 Maths Sample Question Paper 9 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Find xy, if x + y = 6 and x – y = 4.
Answer:
Given: x + y =6,x – y=4.
Now, 4xy =(x+y)2-(x-y)2
= 62 – 42
=20
xy=\(\frac{20}{4}\)=5.

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(b) Find the mean of first 10 prime numbers.
Answer:
First lo prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
∴ ∑ =129,n=10
Mean = \(\frac{\sum x}{n}=\frac{129}{10}=12.9\)

(c) If x = acosθ + bsinθ and y = asinθ – bcosθ, prove that x2 + y2 = a2 + b2.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 10

ICSE Class 9 Maths Sample Question Paper 9 with Answers

Question 2.
(a) Simplify : \(\left(\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\right)\left(\frac{1}{2+\sqrt{3}}+\frac{1}{2-\sqrt{3}}\right)\)
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 11
(b) The area of a trapezium is 540 cm2. If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.
Answer:
Given : Area of trapezium = 510 cm2
Perpendicular distance between parallel lines = 18 cm
Ratio of parallel sides = 7 : 5
Let the length of parallel sides be 7x and 5x.
Area of trapezium = \(\frac{1}{2}\) (Sum of parallel sides) x Perpendicular distance
ICSE Class 9 Maths Sample Question Paper 9 with Answers 12
7x = 7 x 5 = 35
and 5x = 5 x 5 = 25
Hence, the length of parallel sides are 35 cm and 25 cm.

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(c) Solve : (3x + 1) (2x + 3) = 3.
Answer:
(3x + 1) (2x + 3) = 3
6 x 2 + 9x + 2x + 3 – 3 = O
6 x 2 +11x = 0
x (6x + 11) = 0
x=0 or 6x+11=0
x = 0 or x=\(\frac{-11}{6}\)
x = 0 or \(\frac{-11}{6}\)

Question 3.
(a) Express \(\log _{10}\left(\frac{a^{3} c^{2}}{\sqrt{b}}\right)\)
A ladder 13 cm long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the in terms of log10 a, log10 b and log10 c.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 14

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(b) A ladder 13 cm long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.
Answer:
Let AC be the ladder and BC be the height of the wall.
Then, AC = 13m, AB= 15m, ∠B=90°
∴ By Pythagoras theorem,
ICSE Class 9 Maths Sample Question Paper 9 with Answers 15

(c) Factorize : a4 + b4 – 11a2 b2.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 16

Question 4.
(a) Simplify: \(\sqrt{\frac{1}{4}}+(0.04)^{-1 / 4}-(8)^{2 / 3}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 17

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(b) In the figure, ABCD is a trapezium in which DA || CB. AB has been produced to E. Find the angles of the trapezium.
ICSE Class 9 Maths Sample Question Paper 9 with Answers 1
Answer:
Given: DA || CB
x + + 100 = 180° (Co-interior angles) … (i)
⇒ x+2y = 170°
Also, x + 25° = y (Corresponding angles) … (ii)
x – y= – 25°
Subtracting equation (ii) from equation (i), we get
3y = 195°
= \(y=\frac{195^{\circ}}{3}=65^{\circ}\)
Putting y = 65° in equation (ii), we get
X – 65° = – 25°
x = 65 – 25°=40°
∠A = x + 25° = 40°+ 25° = 65°
∠B = 180° – y= 180 – 650= 115°
∠C = 2y + 10° =2 x 65° + 10° = 1400
∠D = x = 40°

(c) Calculate the compound interest for the second year on ₹ 8000 when invested for 3 years at 10% p.a.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 18

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

Question 5.
(a) Solve for x : 25x – 1 = 52x – 1 – 100.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 19
ICSE Class 9 Maths Sample Question Paper 9 with Answers 20

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(b) In a ΔABC, ∠A = 80°, ∠B = 40° and bisectors of ∠B and ∠C meet at O. Find ∠BOC.
Answer:
\frac{\sum x}{n}=\frac{129}{10}=12.9
ICSE Class 9 Maths Sample Question Paper 9 with Answers 21
ICSE Class 9 Maths Sample Question Paper 9 with Answers 22

(c) Without using tables, evaluate :
ICSE Class 9 Maths Sample Question Paper 9 with Answers 3
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 23

ICSE Class 9 Maths Sample Question Paper 9 with Answers

Question 6.
(a) If x = 3 + 2√2 , find the value of \(x^{3}-\frac{1}{x^{3}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 24

(b) Construct a rectangle each of whose diagonals measures 6 cm and the diagonals intersect at an angle of 45°.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 25
Given: Each diagonal = 6 cm, diagonals intersect at 45°.
Steps of construction:
(1) Draw diagonal AC = 6 cm.
(2) Bisect AC at O
(3) At O, draw ∠COX = 45° and extend XO to Y.
(4) From O, cut-off XY at D and B such that OD = OB = 3 cm,
i.e., BD = 6 cm.
(5) Join A, B, C, D to get the required rectangle ABCD.

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(c) KM is a straight line of 13 units. If K has the coordinates (2, 5) and M has the coordinates (x, – 7), find the possible values of x.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 26

Question 7.
(a) If \(x=\frac{\sqrt{7}+1}{\sqrt{7}-1} \text { and } y=\frac{\sqrt{7}-1}{\sqrt{7}+1}\)find the value of \(\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 27
ICSE Class 9 Maths Sample Question Paper 9 with Answers 28

(b) In ΔABC, AC = 3 cm. If M is the mid-point of AB and a straight line through M parallel to AC cuts BC at N, what is the length of MN ?
ICSE Class 9 Maths Sample Question Paper 9 with Answers 4
Answer:
Given : AC = 3 cm, M is the mid-point of AB, MN || AC.
Since, M the mid-point of AB and MN || AC
Therefore, by mid-point theorem, we have
MN = \(\frac{1}{2}\)AC= \(\frac{1}{2}\) x 3cm = 1.5cm

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(c) If θ is an acute angle and tan θ = \(\frac{5}{12}\), find the value of cosθ + cot θ.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 29
ICSE Class 9 Maths Sample Question Paper 9 with Answers 30

Question 8.
(a) Factorize : 1 – 2ab – (a2 + b2)
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 31

ICSE Class 9 Maths Sample Question Paper 9 with Answers
(b) In the given figure, AOC is a diameter of a circle with centre O and arc A×B = \(\frac{1}{2}\) arc BYC. Find ∠BOC.
ICSE Class 9 Maths Sample Question Paper 9 with Answers 5
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 32

(c) In a class of 90 students, the marks obtained in a weekly test were as under.
ICSE Class 9 Maths Sample Question Paper 9 with Answers 6
Construct a combined histogram and frequency polygon.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 33
ICSE Class 9 Maths Sample Question Paper 9 with Answers 34

ICSE Class 9 Maths Sample Question Paper 9 with Answers

Question 9.
(a) Divide ₹ 1,95,150 between A and B so that the amount that A receives in 2 years is the same as that of B receives in 4 years. The interest is compounded annually at the rate of 4% p.a.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 35
ICSE Class 9 Maths Sample Question Paper 9 with Answers 36
A’s share = 1,01,400.
and B’s share = (1,95,150 – 1,01,400) = ₹ 93,750.

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(b) Solve simultaneously : \(\frac{3}{x+y}+\frac{2}{x-y}=3 ; \frac{2}{x+y}+\frac{3}{x-y}=\frac{11}{3}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 37
ICSE Class 9 Maths Sample Question Paper 9 with Answers 38
ICSE Class 9 Maths Sample Question Paper 9 with Answers 39

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(c) If a = cz, b = ax and c = by, prove that xyz = 1.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 40

Question 10.
(a) If sin (A + B) = 1 and cos (A – B) = \(\frac{\sqrt{3}}{2}\), 0° < A + B ≤ 90°, A > B, then find A and B.
Answer:
sin (A + B) = 1 = sin 90°
A + B = 90°
cos (A – B) =\(\frac{\sqrt{3}}{2}\) = cos 30°
A – B = 30°
Adding (i) and (ii), we get
A + B + A – B = 90° + 30°
⇒ 2A = 120°
A = \(\frac{120^{\circ}}{2}\) = 60°
Putting A= 60° in (i), we get
60° + B = 90°
⇒ B = 90° – 60° = 30°
∴ A = 60°, B = 30°

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(b) The sum of the digits of a two-digit number is 5. The digit obtained by increasing the digit in ten’s place by unity is one-eighth of the number. Find the number.
ICSE Class 9 Maths Sample Question Paper 9 with Answers 7
Answer:
Let the digit at ten’s place be  and that at unit’s place be y.
∴ The number = 10x + y
By 1st condition,
x + y = 5
By 2nd condition,
x + 1 = \(\frac{1}{8}\) (10x + y)
⇒ 8 (x + 1) = 10x + y
⇒ 8x + 8 = 10x + y
⇒ 2x + y = 8
Subtracting (i) from (ii), we get
x = 3.
Putting x = 3 in (i), we get
3 + y = 5
⇒ y = 5 – 3 = 2
The required number = 10x + y = 10 x 3 + 2 = 32

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(c) Given, log10 x = a and log10 y =
(i) Write down 10a-1 in terms of x
(ii) Write down 102b in terms of y.
(iii) If log10 P = 2a-b, express P in terms of x and y.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 42

Question 11.
(a) Draw the graph of the equations 2x – 3y = 7 and x + 6y = 11 and find their solutions.
Answer:
ICSE Class 9 Maths Sample Question Paper 9 with Answers 43ICSE Class 9 Maths Sample Question Paper 9 with Answers 44
The two lines intersect each other at the point (5, 1).
∴ x = 5, y = 1

ICSE Class 9 Maths Sample Question Paper 9 with Answers

(b) In the figure, area of ΔABD = 24 sq. units. If AB = 8 units, find the height of ΔABC.
Answer:
Given : Area of A ABD = 24 sq. units, AB = 8 units, DC || AB.
Area of ΔABC = Area of ΔABD
(∵ They are on same base and between same parallels)
= 24 sq. units.
⇒ \(\frac{1}{2}\) x AB x Height of ΔABC = 24
⇒ \(\frac{1}{2}\) x 8 x Height of ΔABC = 24
⇒ Height of Δ ABC = \(\frac{24}{4}\) = 6 units.

(c) If x – y = 8 and xy = 20, evaluate : (i) x + y (ii) x2 – y2.
Answer:
Given x – y – 8, xy = 20.
(i) (x + y)2 = (x – y)2 + 4 xy = 82 + 4 x 20 = 64 + 80 = 144
x + y = ± √144 = ± 12
x2 – y2 = (x + y) (x – y)
= (± 12) x 8
= ± 96.

ICSE Class 9 Maths Question Papers with Answers

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) 25-30 30-35 35-40 40-45 45-50
No. of doctors 40 60 50 35 20

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 Interval 40 – 48 48-56 56-64 64-72 72 – 80
Frequency 15 25 35 30 10

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.

Marks Below 45 Below 60 Below 75 Below 90 Below 105 Below 120
No. of Students 0 8 23 48 85 116

Answer:

Marks No. of Students Class Interval Frequency
Below 45 0 0-45 0
Below 60 8 45 – 60 8 (8-0)
Below 75 23 60 – 75 15 (23 – 8)
Below 90 48 75 – 90 25 (48 – 23)
Below 105 85 90 -105 37 (85 – 48)
Below 120 116 105 – 120 31 (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

X 6 3 0
y 0 1 2

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

X 6 3 0
y 0 1 2

(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) 40-44 45-49 50-54 55-59 60-64 65-69
No. of students 2 8 12 10 6 4

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.

Marks 40 – 50 50-60 60 – 70 70-80 80 – 90 90 – 100
No. of Students 5 8 13 9 7 5

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]

Marks 40-50 50-60 60-70 70-80 80-90 90-100
No. of students 7 18 26 37 20 6

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

X 3. 0 -3
y 0 -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

X 3 0 -3
y 0 2 4

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 9 Maths Sample Question Papers with Answers 2021-2022

ICSE Class 9 Maths Sample Question Paper with Answers 2021-2022

ICSE Sample Papers for Class 9 with Answers