Practice Paper – TERM II (2021 – 2022)
Class – IX
Mathematics
Time: 2 hours Maximum Marks: 40
General Instructions:
1. The question paper consists of 14 questions divided into 3 sections A, B, C.
2. Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
3. Section B comprises of 4questions of 3 marks each. Internal choice has been provided in one question.
4. Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question. It contains two case study based questions.
SECTION – A
1. Factorise: 2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz.
Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
2x2 + y2 + 8z2 – 2√2xy + 4√2yz – 8xz
= (-√2x)2 + (y)2 + (2√2z)2 + (2×-√2x×y) + (2×y×2√2z) + (2×2√2×−√2x)
= (−√2x + y + 2√2z)2 = (−√2x + y + 2√2z)(−√2x + y + 2√2z)
OR
Expand (–2x + 5y – 3z)2, using suitable identities.
Using identity, (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
(–2x + 5y – 3z)2 = (–2x)2 + (5y)2 + (–3z)2 + (2×–2x × 5y) + (2× 5y×– 3z) + (2×–3z ×–2x)
= 4x2 + 25y2 + 9z2 – 20xy – 30yz + 12zx
2. The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral.
We know that the sum of the interior angles of the quadrilateral = 360°
Now, 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°
⇒ x = 12°,
Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°
3. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7)
Height of cylinder, h = 14cm
Let the diameter of the cylinder be d
Curved surface area of cylinder = 88 cm2
So 2πrh =88 cm2 (r is the radius of the base of the cylinder)
2×(22/7)×r×14 = 88 cm2
2r = 2 cm
d = 2 cm
Therefore, the diameter of the base of the cylinder is 2 cm.
OR
The height of a cone is 15cm. If its volume is 1570cm3, find the diameter of its base. (Use π = 3.14)
Height of the cone, h = 15 cm
Volume of cone =1570 cm3
Let r be the radius of the cone
As we know: Volume of cone, V = (1/3) πr2h
So, (1/3) πr2 h = 1570
(1/3)×3.14×r2 ×15 = 1570 ⇒ r2 = 100 ⇒ r = 10
Radius of the base of cone 10 cm.
4. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? (ii) white?
The Total no. of balls = 5+8+4 = 17
P(E) = (Number of favourable outcomes/ Total number of outcomes)
(i) Total number of red balls = 5
P (red ball) = 5/17 = 0.29
(ii) Total number of white balls = 8
P (white ball) = 8/17 = 0.47
5. An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?
Number of integers between 0 and 100 = 99
(i) Number of integers divisible by 7 = 14 (7, 14, 21,…., 98)
Hence, the required probability = 14/99
(ii) Number of integers not divisible by 7 = 99 – Number of integers divisible by 7
= 99-14 = 85
Hence, the required probability = 85/99
6. In the below figure, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.
We know that angles in the segment of the circle are equal
so, ∠BAC = ∠BDC
Now in the in ΔABC, the sum of all the interior angles will be 180°
So, ∠ABC+∠BAC+∠ACB = 180°
Now, by putting the values,
∠BAC = 180° – 69° – 31° So, ∠BAC = 80° ∴ ∠BDC = 80°
SECTION B
7. Factorise: (i) 2x2 + 7x + 3
2x2 + 7x + 3
= 2x2 + 6x + x + 3
= 2x (x + 3) + 1(x + 3)
= (2x + 1)(x + 3)
(ii) 3x2 – x – 4
3x2 – x – 4
= 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)
OR
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area: 25a2 – 35a + 12
Area : 25a2 –35a+12
= 25a2 –15a−20a+12
= 5a(5a–3)–4(5a–3)
= (5a–4)(5a–3)
Possible expression for length = 5a–4
Possible expression for breadth = 5a –3
(ii) Area: 35y2 + 13y – 12
Area : 35y2+13y–12
= 35y2 –15y+28y–12
= 5y(7y–3)+4(7y–3)
= (5y+4)(7y–3)
Possible expression for length = (5y+4)
Possible expression for breadth = (7y–3)
8. The diameter of a metallic ball is 4.2cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3 ? (Assume π=22/7)
Diameter of a metallic ball = 4.2 cm
Radius(r) of the metallic ball, r = 4.2/2 cm = 2.1 cm
Volume formula = 4/3 πr3
Volume of the metallic ball = (4/3)×(22/7)×2.1 cm3
Volume of the metallic ball = 38.808 cm3
Now, using relationship between, density, mass and volume,
Density = Mass/Volume
Mass = Density × volume
= (8.9×38.808) g
= 345.3912 g
Mass of the ball is 345.39 g (approx).
9. Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.
Construction Procedure:
The steps to draw the triangle of given measurement is as follows:
1. Draw a line segment of base BC = 7 cm
2. Measure and draw ∠B = 75° and draw the ray BX
3. Take a compass and measure AB+AC = 13 cm.
4. With B as centre and draw an arc at the point be D
5. Join DC
6. Now draw the perpendicular bisector of the line DC and the intersection point is taken as A.
7. Now join AC 8. Therefore, ABC is the required triangle.
10. Factorise: x3 – 23x2 + 142x – 120
Let p(x) = x3 – 23x2 + 142x – 120
By trial, we find that p(1) = 0. So x – 1 is a factor of p(x).
Now we see that x3 – 23x2 + 142x – 120
= x3 – x2 – 22x2 + 22x + 120x – 120
= x2(x –1) – 22x(x – 1) + 120(x – 1)
= (x – 1) (x2 – 22x + 120)
We could have also got this by dividing p(x) by x – 1.
x2 – 22x + 120 = x2 – 12x – 10x + 120
= x(x – 12) – 10(x – 12) = (x – 12) (x – 10)
So, x3 – 23x2 – 142x – 120 = (x – 1)(x – 10)(x – 12)
SECTION C
11. 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 the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB
(i) In ΔACB, M is the midpoint of AB and MD || BC ,
D is the midpoint of AC (Converse of mid-point theorem)
(ii) ∠ACB = ∠ADM (Corresponding angles)
also, ∠ACB = 90° ,
∠ADM = 90° and MD ⊥ AC
(iii) In ΔAMD and ΔCMD,
AD = CD (D is the midpoint of side AC)
∠ADM = ∠CDM (Each 90°)
DM = DM (common) ,
ΔAMD ≅ ΔCMD [SAS congruency]
AM = CM [CPCT]
also, AM = ½ AB (M is midpoint of AB)
Hence, CM = MA = ½ AB
OR
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Given that ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.
Join AC and BD.
In ΔABC
P and Q are the mid-points of AB and BC respectively,
PQ || AC and PQ = ½ AC (Midpoint theorem) — (i)
In ΔADC,
SR || AC and SR = ½ AC (Midpoint theorem) — (ii)
So, PQ || SR and PQ = SR
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram. ,
PS || QR and PS = QR (Opposite sides of parallelogram) — (iii)
Now, In ΔBCD,
Q and R are mid points of side BC and CD respectively. ,
QR || BD and QR = ½ BD (Midpoint theorem) — (iv)
AC = BD (Diagonals of a rectangle are equal) — (v)
From equations (i), (ii), (iii), (iv) and (v),
PQ = QR = SR = PS
Hence, PQRS is a rhombus.
12. Factorise each of the following:
(i) 27y3 + 125z3
The expression, 27y3 + 125z3 can be written as (3y)3 + (5z)3
27y3 + 125z3 = (3y)3 + (5z)3
We know that, x3 + y3 = (x + y)(x2 – xy + y2)
27y3 + 125z3 = (3y)3 + (5z)3 = (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2)
(ii) 64m3 – 343n3
The expression, 64m3 – 343n3 can be written as (4m)3 – (7n)3
64m3 – 343n3 = (4m)3 – (7n)3
We know that, x3 – y3 = (x – y)(x2 + xy + y2)
64m3 – 343n3 = (4m)3 – (7n)3
= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)(16m2 + 28mn + 49n2 )
CASE STUDY QUESTIONS
13. In an exhibition, a stall is installed for soft drink. A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm.
(i) What is the Capacity of tin can in litres?
tin can will be cuboidal in shape
Dimensions of tin can are
Length, l = 5 cm
Breadth, b = 4 cm
Height, h = 15 cm
Capacity of tin can = l×b×h= (5×4×15) cm3
= 300 cm3 = 300/1000 litres = 0.3 litre
(ii) Which container has greater capacity and by how much?
Plastic cylinder will be cylindrical in shape.
Dimensions of plastic can are:
Radius of circular end of plastic cylinder,
r = 3.5cm Height , H = 10 cm
Capacity of plastic cylinder = πr2H
Capacity of plastic cylinder = (22/7)×(3.5)2×10 = 385
Capacity of plastic cylinder is 385 cm3
From results of (i) and (ii), plastic cylinder has more capacity.
Difference in capacity = (385-300) cm3 = 85cm3
14. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. The distance between Reshma and Salma and between Salma and Mandip is 6m each. In the given below figure Reshma’s position is denoted by R, Salma’s position is denoted by S and Mandip’s position is denoted by M.
(i) Find the area of triangle ORS.
(ii) What is the distance between Reshma and Mandip?
Area of ∆ORS = ½ x X x 5 = 3x/2 …(i)
(⸪in ∆ORM, RM is a chord \ OX ^ RM)
In right angled triangle, ∆ORN OR2 = RN2 + NO2
⇒ 52 = 32 + NO2
⇒ 25 = 9 + NO2
⇒ NO2 = 25 – 9 = 16
⇒ NO = 4 cm
So, Area of = 1/2 × RS × ON = 1/2 × 6 × 4 = 12 cm2 …(ii)
From Equations (i) and (ii), we get,
5x/2 = 12
x = 24/5 = 4.8 cm
Since, P is the mid- Point of RM,
Thus, RM = 2RP = 2 × 4.8
RM = 9.6m
Hence, the distance between Reshma and Mandip is 9.6m.