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

Class IX Mathematics

Max. Marks: 40

General Instructions :

1. The question paper consists of 14 questions divided into 3 sections A, B and C.
2. All questions are compulsory.
3. Section A comprises of 6 questions of 2 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 4 marks each. An internal choice has been provided in one question. It contains two case study based questions.

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 – 27b³ – 144a²b + 108ab²

OR

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

64a³ – 27b³ – 144a²b + 108ab²
= (4a)³ – (3b)³ – 3 × (4a)² × (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°

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.

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

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,
OB² = 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°

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) = x² + (–1)x + 2(–1)
= x² – x – 2
= x² – 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) = x² – 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

ab + bc + ca = 0       ...(i)
⇒ – bc = ab + ca      ...(ii)
– ca = ab + bc           ...(iii)
and – a b = bc + ca ...(iv)
Now,  
=  [Using (i), (iii) & (iv)]
=
=
=  [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 =
= 9
or, x = 3
∴ radius = 4 × 3
= 12 cm.

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

∴ 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

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,

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.
In 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
⇒ x² + 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)]
x² – 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.
Ravi donated ₹  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) 
 = 1000
[Use formula : (a + b)³ = a³ + b² + 3ab(a + b)]
 = 1000
 = 1000 - 30 = 970
Hence, amount donated by Ravi = ₹ 970.
(ii) Given: x + 1/x =7
Taking cube on both sides, we get
 = 343
 = 343
 = 343 – 21 = 322.
Hence, amount donated by Ravi = ₹322.