ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

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

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

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

ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

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

ICSE Class 9 Maths Question Papers with Answers

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