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. If f(x) = 7x2 – 3x + 7, find f(2) + f(–1) + f(0).
f(x) = 7×2 – 3x + 7
Now, f(2) = 7(2)2 – 3(2) + 7 = 28 – 6 + 7 = 29
Also, f(–1) = 7(–1)2 – 3(–1) + 7 = 7 + 3 + 7 = 17
Also, f(0) = 7
∴ f(2) + f(–1) + f(0) = 29 + 17 + 7 = 53
OR
Factorise: 6x2 + 7x – 10.
Let p(x) = 6×2 + 7x – 10
By splitting the middle term, we get
p(x) = 6×2 + 12x – 5x – 10
= 6x (x + 2) – 5(x + 2) = (6x – 5) (x + 2)
2. One of the angles of a quadrilateral is 90° and the remaining three angles are in the ratio 2: 3: 4. Find the largest angle of the quadrilateral.
Let the quadrilateral be ABCD in which ∠A = 90°, ∠B = 2x, ∠C = 3x and ∠D = 4x.
Then, ∠A + ∠B + ∠C + ∠D = 360°
⇒ 90° + 2x + 3x + 4x = 360°
⇒ 9x = 270° ⇒ x = 30°
∴∠B = 60°, ∠C = 90°, ∠D = 120°
Hence, the largest angle is 120°.
3. Two cones have their heights in the ratio 1: 4 and radii of their bases in the ratio 4: 1. Find the ratio of their volumes.
Let the heights of the cones be h and 4h and radii of their bases be 4r and r and V1 and V2 are their respective volumes.
OR
If the height of a cylinder is 11 cm and area of curved surface is 968 sq. cm. Find the diameter of the cylinder
Height of cylinder = 11 cm
Curved surface area = 968 cm2
⇒ 2πrh = 968
⇒ 2 × 22/7 × π × 11 = 968
⇒ r = 14 cm
4. A survey of 700 families was conducted to know their opinion about a particular newspaper. If 325 families liked that newspaper and the remaining families disliked it, find the probability that a family chosen at random (i) likes the newspaper. (ii) does not like it.
Total number of families = 700
Number of families who likes newspaper = 325
(i) P(a family chosen at random likes the newspaper) = 325/700 = 13/28
(ii) P(a family does not like the newspaper) = 375/700 = 15/28
5. A letter is chosen at random from the letters of the word ′PROBABILITY′. Find the probability that the letter chosen is a
(i) consonant
There are 11 letters in the word ‘PROBABILITY’.
(i) Number of consonants = 7 ⇒ Required Probability = 7/11
(ii) vowel.
Number of vowels = 4
⇒ Required Probability = 4/11
6. In the given below figure, O is the centre of the circle. ∠CAB = 40°, ∠CBA = 110°, then find value of x.
In ∆ABC, 40° + 110° + ∠C = 180°
⇒ ∠C = 180° – 150° = 30°
Now, ∠AOB = 2 × ∠C
⇒ x = 2 × 30° = 60°.
SECTION B
7. If (2a + b) = 12 and ab = 15, then find the value of 8a3 + b3.
We have, (2a + b) = 12 … (i)
Cubing both sides of (i), we get (2a + b)3 = (12)3
⇒ (2a)3 + b3 + 3 × 2a × b(2a + b) = 1728
⇒ 8a3 + b3 + 3 × 2 × 15(12) = 1728 [Substituting ab = 15 and (2a + b) = 12]
⇒ 8a3 + b3 + 1080 = 1728
⇒ 8a3 + b3 = 1728 – 1080 = 648
OR
Factorise: (2y + x)2 (y – 2x) + (2x + y)2 (2x – y)
We have, (2y + x)2 (y – 2x) + (2x + y)2 (2x – y)
= (2y + x)2 (y – 2x) – (2x + y)2 (y – 2x)
= (y – 2x) [(2y + x)2 – (2x + y)2]
= (y – 2x) [(2y + x) + (2x + y)] [(2y + x) – (2x + y)]
= (y – 2x) (3x + 3y) (y – x)
= (y – 2x) 3(x + y) (y – x) = 3(y – 2x) (y + x) (y – x)
8. The dome of a building is in the form of a hemisphere. If its radius is 14 cm, find the cost of painting it at the rate of Rs. 3 per sq. cm.
Since the dome of building is in the shape of hemisphere
⇒ Curved surface area of dome = 2πr2
= 2 × 22/7 × 14 × 14 = 2 × 22 × 2 × 14 = 1232 cm2
Cost of painting the dome at the rate of 1 sq. cm = Rs. 3
∴ Cost of painting of 1232 cm2 dome = Rs. 3 × 1232 = Rs. 3696
9. Construct a DABC in which BC = 5.6 cm, AC – AB = 1.6 cm and ∠B = 45°
Steps of construction :
Step I : Draw BC = 5.6 cm.
Step II : At B, construct ∠CBX = 45°.
Step III : Produce XB to X′ to form line XBX′
Step IV : Along ray BX′, cut-off a line segment BD = 1.6 cm.
Step V : Join CD.
Step VI : Draw perpendicular bisector of CD which cuts BX – at A.
Step VII : Join CA to obtain required triangle BAC.
10. Find the factors of x3 + 9x2 + 23x + 15.
Let f(x) = x3 + 9×2 + 23x + 15
Here, constant term of f(x) is 15
Factors of 15 are ±1, ±3, ±5 and ±15
∴ f(–1) = (–1)3 + 9(–1)2 + 23(–1) + 15 = 0
f(–3) = (–3)3 + 9(–3)2 + 23(–3) + 15 = 0 f(–5)
= (–5)3 + 9(–5)2 + 23(–5) + 15 = 0
∴ (x + 1), (x + 3), (x + 5) are the factors of f(x).
SECTION C
11. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the below figure). Show that the line segments AF and EC trisect the diagonal BD.
ABCD is a parallelogram and E, F are the mid-points of sides AB and CD respectively
Since ABCD is a parallelogram, AB || DC
And AB = DC (Opposite sides of a parallelogram)
⇒ AE || FC and ½AB = ½DC ⇒ AE || FC and AE = FC
∴ AECF is a parallelogram
∴ AF || EC ⇒ EQ || AP and FP || CQ
In ΔBAP, E is the mid-point of AB and EQ || AP, so Q is the mid – point of BP.
(By converse of mid-point theorem)
BQ = PQ ……..(i)
Again, in ΔDQC,F is the mid-point of DC and FP || CQ, so P is the mid – point of DQ.
(By converse of mid-point theorem)
QP = DP ……. (ii)
From Equations (i) and (ii) , we get BQ = PQ = PD
Hence, CE and AF trisect the diagonal BD
OR
l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see the below figure). Show that l, m and n cut off equal intercepts DE and EF on q also.
We are given that AB = BC and have to prove that DE = EF.
Let us join A to F intersecting m at G.
The trapezium ACFD is divided into two triangles; namely ∆ACF and ∆AFD.
In ∆ ACF, it is given that B is the mid-point of AC (AB = BC) and BG || CF (since m || n).
So, G is the mid-point of AF (by using converse of midpoint theorem)
Now similarly, in ∆ AFD, G is the mid-point of AF, GE || AD and so by using converse of midpoint theorem,
E is the mid-point of DF, i.e., DE = EF.
In other words, l, m and n cut off equal intercepts on q also.
12. If x = 2 and x = 0 are zeroes of the polynomial 2x3 – 5x2 + ax + b, then find the values of a and b
Let p(x) = 2x3 – 5x2 + ax + b x = 2 and x = 0 are zeroes of p(x).
∴ p(2) = 0 and p(0) = 0
p(2) = 0 ⇒ 2(2)3 – 5(2)2 + 2a + b = 0
⇒ 16 – 20 + 2a + b = 0 ⇒ 2a + b = 4 …(i)
Now, p(0) = 0 ⇒ 2(0)3 – 5(0)2 + a ⋅ 0 + b = 0
⇒ 0 – 0 + 0 + b = 0 ⇒ b = 0
Put b = 0 in (i), we get 2a + 0 = 4 ⇒ a = 2
∴ a = 2 and b = 0.
CASE STUDY QUESTIONS
13. Aditya was doing an experiment to find the radius r of a green ball. For this he took a cylindrical container with radius R = 7 cm and height 10 cm. He filled the container almost half by water as shown in the below figure. Now he dropped the green ball into the container as in below figure. He observed that the water level in the container raised from P to Q i.e, to 3.4 cm.
(i). What is the approximate radius of the ball?
(ii). What is the volume of the cylinder?
(iii). What is the volume of the spherical ball?
Volume of ball = Increased volume of water
14. Sanjay visited a circular park with his father. He sees a triangular shaped pond and also observe that three shops are situated at P, Q, R as shown in the figure from where they have to purchase some food items according to their need. Distance between shop P and Q is 8 m, that of between shop Q and R is 10 m and between shop P and R is 6 m.
(i). Find the measure of ∠QPR.
(ii). Find the Area of ∆PQR
(iii). Find the Length of the longest chord of the circle.
(i) ∠QPR = 90° (Angle in semi-circle)
(ii) Area of ∆PQR = ½ × base × height
= ½ × PQ × PR = ½ × 8 × 6 = 24 m2
(iii) Since, longest chord of the circle is its diameter.
Length of the longest chord = QR = 10 m