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1

Class-IX CBSE

Latest Pattern Sample Paper {Mathematics}

Term-I Examination (SA I)

Time: 3hours Max. Marks: 90

Section-A

1. Let x be a rational number and y be an irrational number. It x + y necessarily an

irrational number?

Solution:

Yes, it is necessarily that sum of rational and irrational numbers is an irrational.

2. If an isosceles right angled triangle has an area of 12 cm2. Then find the length of its

base.

Solution:

Let equal sides of an isosceles triangle be x cm i.e. AB-BC= x cm.

Given, area of an isosceles triangle is 12 cm2.

General Instructions

(i) All questions are compulsory.

(ii) The question paper consists of 31 questions divided into 4 sections A, B, C

and D. Section A comprises of 4 questions of 1 mark each, Section B comprises of

6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks

each and Section D comprises of 11 questions of 4 marks each.

(iii) There is no overall choice. However, internal choice has been provided in 1

question of 2 marks, 3 questions of 3 marks each and 2 questions of 4 marks

each. You have to attempt only one of the alternatives in all such questions.

(iv)Use of calculator is not permitted.



2

Area = - x Base × Altitude

12

12 x x2

224 x

x 24 [taking positive square root]

Hence, length of the base of a triangle is 24 cm.

3. Find value of the polynomial p{x) = 5x + 3 – 4x2 at x = 2.

Solution:

Given, p(x) = 5x + 3 – 4x2

At x=2, then p(2) = 5 × 2 + 3 – 4 × (2)2

= 10 + 3– 4 ×4

= 13 – 16 = – 3

Hence, the value of p(x) is – 3 at x = 2.

4. Euclid divided his book 'Elements' into how many chapters?

Solution:

Euclid divided his book 'Elements' into 13 chapters.

Section-B

5. The polynomials kx3 + 3x2 – 8 and 3x3 – 5x + k are divided by x + 2. If the remainder in

each case is the same, then find the value of k.

Solution:

Let p(x) = Kx3+ 3x2 – 8

and q(x) = 3x3 – 5x + k.



3

When we divided p(x) and q(x) by x + 2, we get the remainder p(–2) and q(–2).

But according to the question, p(–2) = q(–2)

(–2)3 + 3(–2)2 – 8 = 3(–2)3 – 5(–2) + k

– 8k +12 – 8 = – 24 +10 + k

8k + 4 = –14 + k

–9k = – 18

k =2

6. The angles of a triangle are (x – 40)°), (x – 20)° and 0

1x 10

2

, find the angles of

triangle.

Or

In the given figure, AOC and BOC form a linear pair and a – b = 70°, find the values of

a and b.

Solution:

Let angles of a triangle be

A = (x – 40)°, S = (x – 20)°

and 0

1C x 10

2

We know that, A + B + C = 180°

[by angle sum property of a triangle]

x – 40° + x – 20° + – x –10° = 180°

0 012x x 70 180

2



4

012x x 250

2

04x x250

2

0250 2

x5

x = 1000

A = x – 40° = 100° - 40° = 60°

B = x – 20° = 100° – 20° = 80°

and C = 1

2 x – 10° =

1

2 × 100° – 10°= 50°-10° = 40°

Or

According to the question,

AOC + BOC = 180° [linear pair axiom]

a +b = 180° ...(i)

and a – b = 70° ...(ii) [given]

On adding Eqs. (i) and (ii), we get

a + b =180°

0

0

a b 70

2a 250

0

0250a 125

2

On putting the value of a in Eq. (i), we get

125° + b = 180°

b = 180° – 125°

b = 55°

7. Express 15.712 in the form p/q.

Solution:

Let x = 15.712



5

x = 15.71212...

On multiplying both sides by 10, we get

10x = 157.1212 ...(i)

On multiplying both sides by 100, we get

1000x = 157121212 ...(ii)

On subtracting Eq (i) from Eq (ii), we get

1000x – 10x = 15712.1212...– 1571212...

990x =15555

15555

x990

3111

x198

[dividing numerator and denominator by 5]

8. Two points with coordinates (4, 3) and (4, – 2) lie on a line, parallel to which axis?

Solution:

As x-coordinate of both points is 4.

So, both points lie on the line x = 4 which is parallel to y-axis.

9. Evaluate 103 × 107, without multiplying directly.

Solution:

103 × 107=(100 + 3)(100 + 7)

= (100)2 +100 × 7 + 3 ×100 + 3 × 7

= 10000+ 100(7+3)+21

= 10000 + 1000 + 21

= 11021

10. Find the factors of polynomial

4x2 + y2+4xy + 8x + 4y + 4

Solution:

4x2 + y2 + 4xy + 8x+4y+4



6

= (2x)2 + (y)2 + 2(2x)(y) +2(2x)(2) + 2(2)(y) + 22

= (2x)2+(y)2+(2)2 + 2(2x)(y) + 2(2)(y) + 2(2x)(2)

= (2x + y + 2)2

[ (a + b + c)2 =a2 + b2 +c2 + 2ab + 2bc + 2ca]

Section-C

11. Evaluate

(i) (104)3 (ii) (999)3

Solution:

(i) We have, (104)3 =(100 + 4)3

= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)

[ (a + B)3 = a3 + b3 +3ab (a + b)]

= 1000000 + 64 + 1200 × 104

= 1000000 + 64 + 124800

= 1124864

(ii) We have, (999)3 = (1000 – 1)3

= (1000)3 – (1)3 – 3 × 1000 × 1 × (1000–1)

[ (a – b)3 = a3 – (a – b)3 = a3 – b3 (a – b)]

= 1000000000 –1 – 3000 × 999

= 1000000000 – 1 – 2997000

= 997002999

12. Find the remainder when p(x) = x3 + 3x2 + 3x +1 is divided by 5 + 2x.

Or

Simplify

(x + y + z)2 – (x – y + z)2.

Solution:

By remainder theorem, when p(x) is divided by



7

5 + 2x = 25

x2

, then remainder is given by 5

p2

.

Now, p(x) = x3+3x2+3x+1

3 2

5 5 5 5p 3 3 1

2 2 2 2

= 125 25 15

3 18 4 2

= 125 75 15

18 4 2

= 125 150 60 8 27

8 8

Hence, the required remainder is 27

8 .

Or

(x + y + z)2 – (x – y + z)2

= [(x + y +z)2 – {x + (–y) + z}2]

= (x2 +y2 +z2 + 2xy + 2yz + 2zx) – [x2 + (–y)2 + (z)2 + 2(x) – (–y) + 2(–y) (z) + 2zx]

[ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

= (x2 + y2 + z2 + 2xy + 2yz + 2zx)

– (x2 + y2 + z2 – 2xy – 2yz + 2zx)

= x2 + y2 + z2 + 2xy + 2yz + 2zx – x2 – y2 – z2 + 2xy + 2yz – 2zx

= 4xy + 4yz

= 4y(x + z)

13. Evaluate

15

10 20 40 5 80 .

take

5 = 2.236 and 10 = 3.162)



8

Or

If x =9 4 5 , then find the value of 1

xx

.

Solution:

Now, 10 20 40 5 80

= 10 2 5 2 10 5 4 5

= 3 10 3 5 3( 10 5)

15

10 20 40 5 80

= 15 5

3( 10 5) 10 5

= 15 10 5

10 5 10 5

[multiplying numerator and denominator by 10 50 ]

= 2 2

5( 10 5) 5( 10 5)

10 5( 10) ( 5)

[ a2 – b2 = (a – b) (a + b)]

= 5( 10 5)

5

= 10 5 3.162 2.236 10 3.162

and 5 2.236

= 5.398

Or

Given x = 9 – 4 5

1 1

x 9 4 5



9

= 1 9 4 5

9 4 5 9 4 5

[multiplying numerator and denominator by 9 + 4 5 ]

= 9 4 5

9 4 581 80

[ 2 2a b a b a b ]

and 1

x 9 4 5 9 4 5 18x

..(i)

Now,

21 1 1

x x 2 x.xx x

[ (a – b)2 = a2 + b2 – 2ab]

= 1

x 2x

= 18 – 2 = 16 [from Eq. (i)]

1

x x 4x

14. Simplify

7/2 5/21 2 2 3

2 4 3 5

2 3 2 3

2 3 2 3

.

Or

If 2x = 3y = 6–z, then prove that 1 1 1

0.x y z

Solution:

7/2 5/21 2 2 3

2 4 3 5

2 3 2 3

2 3 2 3

=

7/22 4 3 5

2 1 3 2

3 3 3 3

2 2 2 2

m

m

1a

a



10

=

7/2 5/26 8

3 5

3 3

2 2

[ m n m na a a ]

=

7/2 5/26 5

7/2 83

3 2

32

[ m ma 1/a ]

= 6 7/2 5 5/2

3 7/2 8 5/2

3 2

2 3

nm mn[ a a ]

= 21 25/2

21/2 20

3 2

2 3

= 21 20 25/2 21/2

3 2

m

m n

n

aa

a

= 1 2

3 2 3 4 12

To prove, 1 1 1

0x y z

Let 2x = 3y = 6-z = k

Thus, 2= k1/x, 3 = k1/y = and 6 = k–1/z

Now, 2 × 3 = 6

1/x 1/y 1/zk k k

[put 2 = 1/x 1/y 1/zk , 3 k , 6 k ]

1/x 1/y 1/z m n m nk k [ a a a ]

On comparing the exponent, we get

1 1 1

x y z

1 1 1

0x y z

15. In the given figure, O is the mid-point of each of the line segments AB and CD. Prove that

AC = BD and AC BD.



11

Solution:

Given In the figure, AO = OB, OC = OD

To prove AC = BD and AC BD

Proof In AOC and BOD, we have AO = BO

[ O is the mid-point of AB]

AOC = BOD [vertical opposite angles]

and CO = DO [ O is the mid-point of CD]

AOC BOD [by SAS congruence rule]

Then, AC = BD [by CPCT]

and CAO = DBO [by CPCT]

CAB = DBA ...(i)

But CAB and DBA are alternate interior angles formed when transversal AB

intersects CA at A and DB at B.

AC BD

Hence, AC = BD and AC BD. Hence proved.

16. Write down Euclid's five postulates.

Solution:

First A straight line may be drawn from any one point to any other point.

Second A terminated line can be produced indefinitely.

Third A circle can be drawn with any centre and any radius.

Fourth All right angles are equal to one another.

Fifth For every line L and for every point P not lying on L, there exists a unique line M

passing through P and parallel to L.



12

17. In the given figure, if ,AB CD, EFCD and GED = 126°, then find AGE, GEF and

FGE.

Solution:

Given, AB CD, GE is a transversal line, then

AGE = GED

AGE =126° [alternate interior angles]

Now, GEF = GED – FED

GEF = 126° – 90° [ FED = 90°]

GEF =36°

Also, FGE + AGE = 180° [linear pair axiom]

FGE +126° =180°

FGE = 180°– 126° = 54°

Hence, AGE =126°, GEF =36°

and FGE = 54°.

18. Prove that two distinct lines cannot have more than one point in common.

Solution:

Given Two distinct lines l1 and l2

To prove Lines l1 and l2 have only one point in common.

Proof Suppose, lines l1 and 12 intersects at two distinct points, say P and Q.

Then, line l1 contains points P and Q.

Also, line 12 contains points P and Q.

So, two lines l1 and l2 pass through two distinct points P and Q.

But only one (unique) line can pass through two distinct points.



13

So, our assumption that two lines can pass through two distinct points is wrong.

Hence, two distinct lines cannot have more than one point in common.

19. If the sides of a triangle are produced in order, then prove that the sum of the exterior

angles so formed is equal to four right angles.

Let ABC be a triangle whose sides AB, BC and CA are produced in order, forming

exterior CBF, ACD and BAE.

In ABC, we have

CBF =1 + 3 ...(i)

[exterior angle is equal to the sum of opposite interior angles]

Similarly, ACD = 1 + 2 ...(ii)

and BAE = 2 + 3 ...(iii)

On adding Eqs. (i), (ii) and (iii), we get CBF + ACD + BAE =2[1 + 2 +3]

= 2 × 180° = 4 × 90°

[by angle sum property of a triangle is 180°]

CBF + ACD + BAE = 4 right angles

Thus, if the sides of a triangle are produced in order, then the sum of exterior angles so

formed is equal to four right angles.

20. Suppose E and F are the mid-points of the sides AB and AC of ABC. CE and BF are

produced to X and Y respectively, so that EX = CE and FY = BF.AX and AY are joined. Find

in figure, a triangle congruent to AEX and demonstrate the congruency. Prove that XAY

is a straight Line.



14

Given E and F are the mid-points of the sides AB and AC of AABC, CE and BF are

produced to X and Y, respectively such that EX = CE and FY = BF.

To prove (i) AEX = BFC

(ii) XAY is a straight line.

Construction Join AX and AY.

Proof In AEX and BEC, we have

AE = BE

[ E is the mid-point of AB]

AEX = BEC [vertically opposite angles]

and EX = EC [given]

AEX BEC [by SAS congruence rule]

XAE = CBE [by CPCT]

or XAB = CBA

[ XAE = XAB and CBE = CBA]

But XAB and CBA are alternate interior angles formed when a transversal AB meets

XA at A and BC at B.

XA BC ...(i)

Similarly, it can be proved that

AFY CFB and AY BC ...(ii)

From Eqs. (i) and (ii), we get

BC XA and BC Ay

Hence, XAY is a straight line. Hence proved.



15

Section-D

21. In the given figure, if AB = AC, then prove that AF > AE.

In the given figure, B < A and C < D. Prove that AD< BC.

Solution:

Given in the Figure, AB = AC

To prove AF > AE

Proof hi ABC, we have

AC = AB [given]

1 = 2 ...(i)

[angles opposite to equal sides are equal]

In DBE, whose side BE is extended to A, we have

5 > 1 ...(ii)

[exterior angle > each opposite interior angle]

From Eqs. (i) and (ii), we get

5 > 2 ...(iii)

Now, 3 = 4 ...(iv)

[vertical opposite angles]

Considering FDC, whose side DC is extended to B, we have

2 > 3 ...(v)



16

From Eqs. (iv) and (v), we get

2 >4 ...(vi)

From Eqs. (iii) and (vi), we get

5 > 4

AF > AE

Or

[sides opposite to greater angle is longer]

Hence, AF > AE

Or

Given In the figure B < A and C < D

To prove AD < BC

Proof In ABO, we have

B < A [given]

AO < BO . ..(i)

[sides opposite to smaller angle is shorter]

In COD, we have

C < D

OD < OC ...(ii)

[sides opposite to smaller angle is shorter]

On adding Eqs. (i) and (ii), we get

AO + OD < BO + OC

AD < BC

Hence, AD < BC

22. Draw the quadrilateral with vertices (– 4,4), (–6,0), (–4, –4)and (–2, 0). Name the type of

quadrilateral and find its area.

Solution:

Firstly, plot the points A(–4,4), B(–6,0),C(–4, –4) and D(–2,0) on a graph paper and join

all these points.



17

We obtained quadrilateral is a rhombus because its all sides are equal.

i.e., AB = BC = CD = DA

and diagonals are not equal.

Now, area of a rhombus = 1 2

1d d

2 1

2

d 4

and d 8

= 1

4 82

= 16 sq units

Hence, are a of quadrilateral is 16 sq units.

23. If x = 1 1

2 22 5 2 5 and 1

2y 2 5 2 5 , then evaluate x2 + y2.

Solution:

Given, 1/2 1/2

x 2 5 2 5

and 1/2 1/2

y 2 5 2 5

Now, 2 1/2 1/2x y [(2x 5) (2 5) + 1/2 2(2 5) (2 5)]

= 1/2 2[2(2 5) ]

= 1

224[2 5]

n

m mn[ a a ]

= 4(2 5) 8 4 5



18

and 1/2 1/22xy 2[(2 5) (2 5) ] × 1/2 1/2[(2 5) (2 5) ]

= 1/2 2 1/2 22[{(2 5) } {(2 5) } ]

[ 2 2a b a b a b ]

= 2[2 5 2 5] 2 4

= 8

2

x y 2xy 8 4 5 8

2 2x y 2xy 2xy 4 5

[ (a + b)2 = a2 + b2 + 2ab]

x2 + y2 = 4 5

24. In a class, a teacher conducted a small quiz to solve a question on blackboard. She needs

two students and a prize will be given to the students who solve the question first. For

this purpose she choose a boy and a girl. The problem is that in the given figure, AB CD.

Find the values of x, y and z.

Which of these values is depicted by the teacher in this question?

(i) Social value

(ii) Freedom

(iii) Truth value

(iv) Gender equality

Solution:

In ACO, AO AC [given]

Then, ACO = AOC



19


In ACO + AOC + CAO =180° [ by angle sum property of a triangle is 180° ]

ACO + ACO + 74°=180°

[ AOC = ACO]

ACO =180° – 74°

ACO = 106°

0

0106ACO 53

2

i.e., ACO = AOC = 53°

Again, AOC + COP = 180°

[linear pair axiom]

53° + x = 180° [ AOC = 53°]

x = 180° – 53°

x = 127° = POC

In A= POC, POC + OCP + CPO = 1800

[by angle sum property of a triangle is 1800]

127°+15°+ y =180°

142° + y =180°

y = 180° – 142°

= 380

Since, AB CD and AP is a transversal line.

Then, y=z [alternate interior angles]

z = 38°

Hence, x = 127°, y = 38° and z = 38°

Value depicted by the teacher is gender equality.

25. If in two right triangles, the hypotenuse and one side of one triangle are equal to the

hypotenuse and one side of the other triangle, then prove that the two triangles are

congruent.



20

Solution:

Given Two right angled ABC and DEF in which B = E = 90°, AC = DF and

BC = EF

To prove ABC DEF

Construction Produce DE to G such that GE = AB.

Join GF.

Proof In ABC and GEF, we have

BC = EF [given]

ABC = GEF = 90°

[by construction]

AB = GE [by construction]

ABC GEF

[by SAS congruence rule]

Then, A = G [by CPCT ]

and AC = OF ...(i) [by CPCT]

But AC = DP [given]

GF = DF

D = G


A = D

Now, in ABC and DEF,



21

A = D [ from Eq. (i)]

and B = E [given]

Remaining C = Remaining F

Now, in ABC and DEF, we have

BC = EF [given]

C = F

and AC = DE [given]

ABC DEF

[by SAS congruence rule] Hence proved.

26. Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a)

Solution:

To prove,

(a + b + c)3 –a3 –b3 –c3 = 3(a + b)(b + c)(c + a)

LHS = [(a + b + c)– a3]– [b3 + c3]

= [(a + b + c) – a][(a + b + c)2 + a(a + b + c) + a2] – (b + c) (b2 + c2 – bc]

[ x3 – y3 = (x – y) (x2 + y2 + xy)

and x3 + y3 =(x + y)(x2 +y 2–xy]

= (b + c)[a2 + b2 +c2 + 2ab + 2bc +2ca + a2 + ab + ac + a2] –[(b + c) (b2 + c2 – bc)

= (b + c)[3a2 + b2 +c2 + 3ab +2bc +3ca – b2 – c2 + bc]

= (b + c) [3a2 + 3ab + 3bc + 3ca]

= 3(b + c) [a2 + ab + bc + ca]

= 3(b + c) [a(a + b) + c(b + a)]

= 3(a + b) (b + c) (c + a) = RHS Hence proved.

27. Simplify

3/7 1/4 1/3

2

4 5 2.

2187 256 1331



22

Solution:

3/7 1/4 1/3

2

4 5 2

2187 256 1331

=

3/7 1/4 37 4

4 5 2

[ 11 ]2 1/33 4

=

3 1 2

4 5 2

3 4 11 [

nm mna a ]

= 3 1 2

4 3 5 4 2 11 m

m

1a

a

= 4 × 27 – 5 × 4 + 2 × 121

= 108 – 20 + 242

= 350 – 20 = 330

28. Factroise 3 2 2 31 1a a b ab b

3 27 .

Factorise 3

3

1x 14.

x

Solution:

3 2 2 31 1a a b ab b

3 27

= 2 3

3 2 1 1 1a 3 a b 3 a b b

3 3 3

= 3

1a b

3

[ (a – b)3 = a3 – b3 – 3ab (a – b)]

= 1 1 1

a b a b a b3 3 3

Hence, 3 2 2 31 1a a b ab b

3 27 =

1 1 1a b a b a b

3 3 3



23

OR

3 3

3 3

1 1x 14 x 8 6

x x =

333 1 1

x 2 3 x 2x x

= 2

221 1 1 1x 2 x 2 x 2 2 x

x x x x

[ x3 + y3 + z3 – 3xyz = (x + y + z)

(x2 + y2 + z2 – xy – yz – zx)]

= 2

2

1 1 2x 2 x 4 1 2x

x x x

= 2

2

1 1 2x 2 x 5 2x

x x x

29. Find the area of the cyclic quadrilateral AB CD by using Brahmagupta's formula, in

which AB = 9cm, BC = 12 cm, CD = 12 cm and DA = 15 cm.

Solution:

According to the given informations, the rough sketch of the cyclic quadrilateral ABCD

will be as shown alongside.

Let a = AB = 9cm, t = BC = 12cm,

C = CD = 12cm and d = AD=15cm

a b c d

s2

= 9 12 12 15

2

=

48

2 = 24 cm

Area of cyclic quadrilateral

= s a s b s c s d [by Brahmagupta’s formula]



24

= 24 9 24 12 24 12 24 15

= 15 12 12 9 = 3 5 12 12 3 3

= 12 3 3 5 = 236 15 cm

Hence, area of cyclic quadrilateral is 236 15 cm .

30. What is the maximum number of digits in the repeating block of digits in the decimal

expansion of 1

17? Perform the division to determine your answer.

Solution:

By long division method, we get



25

0.0588235294117647058

17 100

85

150

136

140

136

40

34

60

51

90

85

50

34

160

153

70

68

20

17

30

17

130

119

110

102

80

68

120

119

100

85

150

136

14



26

The remainders start repeating after 16 divisions.

10.0588235294117647

17

Hence, the maximum number of digits in the repeating block of digits in the decimal

expansion of 1

17 is 16.

31. A point O is taken inside an equilateral four sided figure ABCD such that its distances

from the angular points D and B are equal. Show that AO and OC are in the same straight

line.

Solution:

Let AB CD and transversal XY cuts AB and CD at P and Q, respectively. Let PR and QR be

the bisectors of BPQ and DQP respectively, meet at R.

AB CD and XPQY is the transversal. Then, BPQ + DQP = 180°

[cointerior angles]

01BPQ DQP 180

2

[dividing both sides by 2]

01 2 90 ...(i)

1BPQ 1

2

1DQP 2

2

In PRQ , we have

RPQ + PQR + PRQ = 180°

[ by angle sum property of a triangle is 180°]



27

1 + 2 + 3 = 180°

90° + 3 = 180°

[from Eq. (i), 1 + 2 = 90°]

3 = 180° – 90° = 90°

Hence, PRQ = 90°

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