Class 9 Exam  >  Class 9 Notes  >  Mathematics (Maths) Class 9  >  Class 9 Math: Sample Question Paper Term II - 1 (With Solutions)

Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 PDF Download

Class IX Mathematics


Time: 120 Minutes


Max. Marks: 40

General Instructions :

  1. The question paper consists of 14 questions divided into sections A, and C.
  2. All questions are compulsory.
  3. Section comprises of questions of marks each. Internal choice has been provided in two questions.
  4. Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
  5. Section C comprises of 4 questions of marks each. An internal choice has been provided in one question. It contains two case study based questions.

Section - A

(2 Marks each)

Q.1: Two coins are tossed simultaneously 500 times, following are the outcomes
No head= 100 times
One head = 200 times
Two heads = 200 times
If the two coins are simultaneously tossed again, compute the probability of obtaining:

(i) One Head
(ii) Two Heads

Total number of outcomes, n(S) = 500
Let E1 and E2 be the events of one head and two heads respectively.
(i) Favourable outcomes n(E1) = 200
Then, P(one head) = n(E1)/n(S)
P(one head) = 200/500= 2/5
(ii) Favourable outcomes n(E2) = 200
Then, P(Two heads) = n(E2)/n(S)
P(E2) = 200/500 = 2/5

Q.2: Factorize: 64a3 – 27b– 144a2b + 108ab2

OR

Find the value of k, so that polynomial x3 + 3x2 – kx – 3 has one factor as x + 3.

64a3 – 27b– 144a2b + 108ab2
= (4a)3 – (3b)3 – 3 × (4a)× (3b) + 3 × (4a) × (3b)2 
[Using identity, x3 – y3 – 3x2y + 3xy2 = (x – y)3]
= (4a – 3b)3

OR

Let f(x) = x3 + 3x2 – kx – 3
Since, (x + 3) is a factor of f(x).
Then, f(– 3) = 0
or, (– 3)3 + 3(– 3)2 – k(–3) – 3 = 0
or, – 27 + 27 + 3k – 3 = 0
or, 3k – 3 = 0
or, k = 1.

Q.3: In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angle.

Let ABCD be a parallelogram, then
∠ADC + ∠BCD = 180° [Co-interior angles]
or, 1/2 ∠ADC + 1/2 ∠BCD = 90° [Divide by 2]
or ∠2 + ∠1 = 90°
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
In Δ ODC,
∠1 + ∠2  + ∠DOC = 180° [Angle sum property of a triangle]
∴ ∠DOC = 90°.
Hence, the angle bisectors of two adjacent angles intersect at 90°.

Q.4: The angles of a quadrilateral are 4x°, 7x°, 15x° and 10x°. Find the smallest and largest angles of the quadrilateral.

Sum of the angles of a quadrilateral is 360°.
∴ 4x° + 7x° + 15x° + 10x° = 360°
or, 36x° = 360°
or, x° = 10°
∴ Smallest angle = 4x° = 4 × 10° = 40°
Largest angle = 15x° = 15 × 10°
= 150°

Q.5: A coin is tossed 1200 times with the following outcomes:

Head: 455, Tail: 745
Compute the probability for: (i) getting head, (ii) getting tail.

(i) Number of favorable outcomes n(A) = 455

Total outcomes n(S) = 455 + 745 = 1200
Probability of getting head = n(A) / n(S)

= 455/1200 = 91/240
(ii) Number of favourable outcomes n(B) = 745
Total outcomes n(S) = 1200
Probability of getting tail = n(B)/n(S)
= 745/1200 = 149/240

Q.6: A chord of length 10 cm is at a distance of 12 cm from the centre of a circle. Find the radius of the circle.

OR

In the given figure, find the value of x.
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9

Let AB be the chord of a circle and ON be the distance of the chord from the centre.
Given, AB = 10 cm
ON = 12 cm
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
Also, ON ⊥ AB
and AN = BN
[Q Perpendicular drawn from the centre of the circle to chord of circle bisects the chord]
In Δ ONB,
OB= ON2 + NB2[By Pythagoras theorem]
∴ OB2 = 122 + 52 [∵ BN = 5 cm]
= 144 + 25 = 169
∴ OB = 13 cm
Hence, the radius of the circle is 13 cm.

OR

Here, ABCD is a cyclic quadrilateral.
In a cyclic quadrilateral,
∠A + ∠C = 180° [opposite angles of cyclic quadrilateral  are supplementary]
or, 2x + 4° + 4x – 64° = 180°
or, 6x – 60° = 180°
or, 6x = 180° + 60° = 240°
or, x = 240°/6
∴ x = 40°

Section - B

(3 Marks each)

Q.7: Find the value of k, if x – 2 is a factor of f(x) = x2 + kx + 2k. Also find the factorise of f(x), when putting the value of k.

Given, (x – 2) is a factor of f(x).
∴ f(2) = 0
or, (2)2 + k(2) + 2k = 0
or, 4 + 2k + 2k = 0
or, 4 + 4k = 0
or, k = – 1
So, f(x) = x+ (–1)x + 2(–1)
= x2 – x – 2
= x– 2x + x – 2
= x(x – 2) + 1(x – 2)
= (x – 2)(x + 1)

Q.8: Verify if – 2 and 3 are zeroes of the polynomial 2x3 – 3x2 – 11x + 6. If yes, factorize the polynomials.

Let p(x) = 2x3 – 3x2 – 11x + 6
For, x = – 2
p(– 2) = 2(– 2)3 – 3(– 2)2 – 11(– 2) + 6
= – 16 –  12 + 22 + 6
= – 28 +  28 = 0
For, x = 3
p(3) = 2(3)3 – 3(3)2 – 11(3) + 6
= 54 – 27 – 33 + 6
= 60 – 60 = 0
So, – 2 and 3 are zeroes of the given polynomial.
Now, p(x) = 2x3 – 3x2 – 11x + 6
(x + 2)(x – 3) = x– x – 6 is a factor of p(x).
∴ 2x3 – 3x2 – 11x + 6
= 2x3 + 4x2 – 7x2 – 14x + 3x + 6
= 2x2(x + 2) – 7x(x + 2) + 3(x + 2)
= (x + 2)(2x2 – 7x + 3)
= (x + 2)(2x2 – 6x – x + 3)
= (x + 2)[(2x(x – 3) – 1(x – 3)]
= (x + 2)(x – 3)(2x – 1)

Q.9: If ab + bc + ca = 0, then find the value of Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9

ab + bc + ca = 0       ...(i)
⇒ – bc = ab + ca      ...(ii)
– ca = ab + bc           ...(iii)
and – a b = bc + ca ...(iv)
Now, Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 
= Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 [Using (i), (iii) & (iv)]
= Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
= Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
= Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 [Using (i)]
= 0

Q.10: Find the radius of the base of a right circular cylinder whose curved surface area is 2/3 

of the sum of the surface areas of two circular faces. The height of the cylinder is given to be 15 cm.

OR

The radius and slant height of a cone are in the ratio 4: 7. If its curved surface area is 792 cm2, find its radius.

Let the radius and height of the cylinder be r and h respectively, then
h = 15 cm [given]
C.S.A. of the cylinder = 2/3 (Sum of areas of 2 circular faces)
2πrh = 2/3(2πr2)
h = 2/3r
15 = 2/3r
or, r = 22.5 cm.

OR

Let the radius of cone be r = 4x
and slant height l = 7x
∵ CSA of a cone = 792 cm2
∴ πrl = 792
or, 22/7 x 4x x 7x  = 792
x2 = Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
= 9
or, x = 3
∴ radius = 4 × 3
= 12 cm.

Section - C

(4 Marks each)

Q.11: A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

According to the question,
OA = AB = OB
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
∴ ΔOAB is an equilateral triangle
i.e., ∠AOB = 60°
∠ACB = 1/2 ∠AOB
[angle subtended by an arc at the circumference is half of the angle at the centre of circle]
∠ACB = 1/2 x 60°
∠ACB = 30°
∠ACB + ∠ADB = 180°
[opposite angles of cyclic quadrilateral are supplementary]
or, ∠ADB = 180° – ∠ACB
∠ADB = 180° – 30°
= 150°

Q.12: What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m ? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).

OR

Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius R' of the new sphere, (ii) ratio of S' and S.

Given, Conical tent: height (h) = 8 m
base radius (r) = 6 m
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
l2 = r2 + h2
l2 = 82 + 62
l = √64 +36 = √100 = 10 cm
C.S.A of tent = πrl unit2
= 3.14 × 6 × 10 m2
= 188.4 m2
Area of Tarpaulin = C.S.A of tent width × length of tarpaulin = 188.4 m2
3 × length of tarpaulin = 188.4 m2
length of tarpaulin = 188.4/3
= 62.8 m
Extra length required for stitching and wastage of cutting
= 20 cm = 0.20 m
∴ Total length of tarpaulin
= 62.8 + 0.2
= 63 m

OR

Given: radius of each sphere = r
Volume of 1 solid iron sphere = 4/3πr3
Volume of 27 solid iron spheres = 4/3πr3 x 27
Volume of new sphere = 4/3 x  27πr3
Let radius of now sphere be R, then according to given condition,
Volume of new sphere made after melting 27 spheres = Volume of 27 spheres
4/3πR3 = 4/3 x  27πr3
R3 = 27r3
(i) R = 3r unit
(ii) Surface area of new sphere = 4πR2
S' = 4π × (3r)2
S' = 4π × 9r2 unit2
Surface area of Sphere (S) = 4πr2
Now,
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
S' : S = 9 : 1

Case Study-1
Q.13: Read the following text and answer the following questions on the basis of the same: Beti Bachao, Beti Padhao (BBBP) is a personal campaign of the Government of India that aims to generate awareness and improve the efficiency of welfare services intended for girls.
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9In a school, a group of (x + y) teachers, (x2 + y2) girls and (x3 + y3) boys organised a campaign on Beti Bachao, Beti Padhao.
(i) If in the group, there are 10 teachers and 58 girls, then what is the number of boys?
(ii) If x  – y = 23, then find x2 – y2.

(i) No. of teachers= x + y = 10
⇒ (x + y)2 = (10)2
⇒ x+ y2 + 2xy = 100        ...(i)
No. of girls = (x2 + y2) = 58
⇒ 58 + 2xy = 100    ...using equation (i)
⇒ 2xy = 100 – 58
⇒ 2xy = 42
⇒ xy = 42/2
⇒ xy = 21
Now, since (x + y)3 = [x3 + y3 + 3xy(x + y)]
⇒ (10)3 = [x3 + y3 + 3 × 21(10)]
⇒ 1000 = (x3 + y3 + 630)
⇒ 1000 – 630 = (x3 + y3)
⇒ (x3 + y3) = 370
Hence, no. boys = 370
(ii) Given x – y = 23
Also, x + y = 10 [Taking from part (i)]
x– y2 = (x + y) (x – y)
= 10 × 23 = 230
Hence, the value of x2 – y2 is 230.

Case Study-2
Q.14:  Read the following text and answer the questions given below: National Association for the Blind (NAB) aimed to empower and well-inform visually challenged population of our country, thus enabling them to lead a life of dignity and productivity.
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9Ravi donated ₹ Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 to NAB. When his cousin asks to tell the amount donated by him, he just gave ,the hint. x + 1/x = 10
(i) Find the amount donated by Ravi.
(ii) Find the amount donated by Ravi if  x + 1/x =7

(i) Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 = 1000
[Use formula : (a + b)3 = a3 + b2 + 3ab(a + b)]
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 = 1000
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 = 1000 - 30 = 970
Hence, amount donated by Ravi = ₹ 970.
(ii) Given: x + 1/x =7
Taking cube on both sides, we get
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 = 343
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 = 343
Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 = 343 – 21 = 322.
Hence, amount donated by Ravi = ₹322.

The document Class 9 Math: Sample Question Paper Term II - 1 (With Solutions) | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
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Ans. The duration of the Class IX Mathematics exam is 120 minutes.
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Ans. There are three sections in the Class IX Mathematics exam - Section A, Section B, and Section C.
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Ans. The questions in the Class IX Mathematics exam are designed to match the complexity of the textbook and syllabus.
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Ans. Section C in the Class IX Mathematics exam is designed to test the application and problem-solving skills of the students.
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