ICSE Class 9 Maths Sample Question Paper 1 with Answers

Max Marks :80
[2 Hours]

General Instructions

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

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

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

X3.0-3
y0-1-2

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

X30-3
y024

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

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

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

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

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

ICSE Class 9 Maths Sample Question Paper 1 with Answers

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

ICSE Class 9 Maths Question Papers with Answers

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