Summative Assessment Paper-1

Mathematics IX (Term - I) 1

SECTION A

(Question numbers 1 to 8 carry 1 mark each. For each question, four alternative choices

have been provided of which only one is correct. You have to select the correct choice).

1. When x3 – 6x2 + 9x + 3 is divided by (x – 1), the remainder is :

(a) 20 (b) 7 (c) 13 (d) –7

Sol. (b) Let p(x) = x3 – 6x2 + 9x + 3

∴ p(1) = 1 – 6 + 9 + 3 = 13 – 6 = 7

2. The coefficient of y in (x + y + z)2 is :

(a) 2x (b) 2z (c) x + z (d) 2x + 2z

Sol. (d) (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

= x2 + y2 + z2 + y (2x + 2z) + 2zx

Thus, the coefficient of y is (2x + 2z).

3. On factorising x2 + 8x + 15, we get :

(a) (x + 3) (x – 5) (b) (x – 3) (x + 5) (c) (x + 3) (x + 5) (d) (x – 3) (x – 5)

Sol. (c) x2 + 8x + 15 = x2 + 3x + 5x + 15

= x (x + 3) + 5 (x + 3) = (x + 3) (x + 5)

4. The length of the sides of a triangle are 5 cm, 7 cm and 8 cm. Area of the triangle is :

(a) 300 cm2 (b) 100 23 cm (c) 50 23 cm (d) 10 23 cm

Sol. (d) s = 5 7 8

cm =10 cm2

+ +

∴ Area of the triangle = 210 (10 – 5) (10 – 7) (10 – 8) cm

= 2 210 5 3 2 cm 10 3 cm× × × = .

5. To locate n , where n is a positive integer, on the number line, we first locate :

(a) n (b) 1n − (c) 1n + (d)2

n

Sol. (b) We can locate n , for any positive integer n, after 1n − has been located.

6. Lines AB and CD intersect each other at O. If ∠AOC : ∠BOC = 2 : 7, then ∠BOD

equals :

(a) 20° (b) 40° (c) 60° (d) 80°

Sol. (b) Let ∠AOC = 2x and ∠BOC = 7x, then

2x + 7x = 180° [Linear pair]

⇒ 9x = 180°

⇒ x = 20°

MODEL TEST PAPER – 2 (SOLVED)

Maximum Marks : 90 Maximum Time : 3 hours

General Instructions : Same as in CBSE Sample Question Paper.

2 Mathematics IX (Term - I)

∴ ∠AOC = 2 × 20° = 40°

Also, ∠BOD = ∠AOC = 40° [Vertically opposite angles]

7. In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be

congruent by SAS axiom if :

(a) BC = EF

(b) AC = DE

(c) BC = DE

(d) AC = EF

Sol. (b) In ∆ABC and ∆DEF

AB = FD [Given]

∠A = ∠D [Given]

The two triangles will be congruent by SAS rule, if AC = DE.

8. An isosceles right triangle has area 8 cm2.The length of its hypotenuse is :

(a) 32 cm (b) 16 cm (c) 48 cm (d) 24 cm

Sol. (a) We have, 1

2× x × x = 8 ⇒ x = 4

∴ Hypotenuse of the triangle = 2 24 4 cm = 32 cm+

SECTION B

(Question numbers 9 to 14 carry 2 marks each)

9. Using remainder theorem, find the value of k so that (4x2 + kx – 1) leaves the remainder

2 when divided by (x – 3).

Sol. Let f(x) = 4x2 + kx – 1

⇒ f(3) = 4(3)2 + k(3) – 1 = 2

⇒ 4 × 9 + 3k – 1 = 2 ⇒ 3k = 2 – 35 ⇒ 3k = –33 ⇒ k = – 11

10. Evaluate 103 × 107, without multiplying directly.

Sol. We have, 103 × 107 = (100 + 3) (100 + 7)

= (100)2 + (3 + 7) × 100 + 3 × 7

[Using (x + a) (x + b) = x2 + (a + b) x + ab]

= 10000 + 1000 + 21

= 11021

11. Angles X, Y, and Z of a triangle are equal. Prove that ∆XYZ is equilateral.

Sol. In ∆XYZ, ∠X = ∠Y ⇒ YZ = ZX ... (i)

[Sides opposite to equal angles are equal]

Also, ∠Y = ∠Z ⇒ ZX = XY ... (ii)

From (i) and (ii), we have XY = YZ = ZX

So, triangle XYZ is equilateral. Proved.


OR

In the figure, PQ = PR and ∠Q = ∠R. Prove that ∆PQS ≅ ∆PRT.

Sol. In ∆PQS and ∆PRT,

∠Q = ∠R [Given]

PQ = PR [Given]

∠P = ∠P [Common]

∴ By ASA axiom, ∆PQS ≅ ∆PRT. Proved.

12. Where will you find all points with negative abscissa and positive ordinate?

Sol. In the II quadrant we can find all points with negative abscissa and positive ordinate

because the sign of all points in the II quadrant are (–, +).

13. Find the value of (13 + 23 + 33)–3/2

Sol. We have, (13 + 23 + 33)–3/2 = (1 + 8 + 27)–3/2

= (36)–3/2 = (62)–3/2 = 62×(–3/2) = 3

3

1 16

6 216

−= = .

14. Two adjacent angles on a straight line are in the ratio 2 : 7. Find the measure of the

greater angle.

Sol. Let the angles are 2x and 7x.

∴ 2x + 7x = 180° ⇒ 9x = 180° ⇒ x = 20°

∴ Greater angle = 7 × 20° = 140°.

SECTION C


15. Simplify ( )

2

2 3 1

3 1 4

−

− −

and then express it with a rational denominator...

Sol. We have, ( )

2

2 3 1

3 1 4

−

− −=

2 3 1

3 1 2 3 4

−

+ − −=

2 3 1

2 3

−

−

= ( ) 3

3

2 3 1

2 3

−×

−[Rationalising the denominator]

= 6 3

6

−

−=

3 6

6

−

OR

Show how 5 can be represented on the number line.

Sol. Draw a number line as shown in the given figure. Let the point O represent 0 (zero)

and point A represent 2. Draw perpendicular AX at A on the number line and cut-off

AB = 1 unit. Join OB.

We have, OA = 2 units and AB = 1 unit

Using Pythagoras theorem, we have


OB2 = OA2 + AB2 ⇒ OB2 = (2)2 + 12 = 5 ⇒ OB = 5

Taking O as the centre and OB = 5 as radius, draw an arc cutting the number line

at C. Clearly, OC = OB = 5 .

Hence, C represents 5 on the number line.

16. Express 3

3 2 2 5+ with a rational denominator...

Sol. We have, 3

3 2 2 5+

Here, rationalising factor of denominator is 3 2 2 5− . So, multiply both numerator

and denominator by 3 2 2 5−

3

3 2 2 5+=

( )( ) ( )

( )

( ) ( )2 2

3 3 2 2 5 3 3 2 2 5

3 2 2 5 3 2 2 5 3 2 2 5

× − −=

+ × − −

=( ) ( )3 3 2 2 5 3 3 2 2 5

18 20 2

× − − −=

−.

17. In the figure, in ∆ABC, ∠DAC = ∠ECA and AB = BC.

Prove that ∆ABD ≅ ∆CBE.

Sol. In ∆ABC, AB = BC

⇒ ∠C = ∠A

⇒ ∠BCE + ∠ECA = ∠BAD + ∠DAC

⇒ ∠BCE = ∠BAD ... (i) [∵ ∠ECA = ∠DAC]

In ∆ABD and ∆CBE, we have

∠BAD = ∠BCE [From (i)]

AB = CB [Given]

∠ABD = ∠CBE [Common]

∴ By ASA axiom, ∆ABD ≅ ∆CBE. Proved.

18. Two lines AB and CD intersect at a point O such that ∠BOC + ∠AOD = 280°, as

shown in the given figure. Find all the four angles.

Sol. We have, ∠BOC + ∠AOD = 280°

But, ∠BOC = ∠AOD [Vertically opposite angles]


∴ 2∠BOC = 280° ⇒ ∠BOC = 140°

Now, ∠COA + ∠BOD = 360° – 280° = 80°

But, ∠COA = ∠BOD [Vertically opposite angles]

∴ 2∠COA = 80° ⇒ ∠COA = 40°

So, measures of all 4 angles are 140°, 40°, 140°, 40°.

19. Find the remainder when the polynomial f(x) = x4 + 2x3 – 3x2 + x – 1 is divided by(x – 2).

Sol. x – 2 = 0 ⇒ x = 2By the remainder theorem, we know that when f(x) is divided by (x – 2), the remainderis f(2).Now, f(2) = 24 + 2 × 23 – 3 × 22 + 2 – 1 = 16 + 16 – 12 + 2 – 1 = 21

Hence, the required remainder is 21.

20. Factorise : 27p3 – 1216

9

2

1

42− +p p

Sol. We have, 27p3 – 1216

9

2

1

42− +p p = (3p)3 –

1

6

3⎛⎝⎜

⎞⎠⎟ – 3 × (3p)2 ×

1

6

⎛⎝⎜

⎞⎠⎟ + 3 × (3p)

1

6

2⎛⎝⎜

⎞⎠⎟

= 31

6

3

p −⎛⎝⎜

⎞⎠⎟ [∵ (a – b)3 = a3 – b3 – 3a2b + 3ab2]

= 31

6p −⎛

⎝⎜⎞⎠⎟ 3

1

6p −⎛

⎝⎜⎞⎠⎟ 3

1

6p −⎛

⎝⎜⎞⎠⎟ .

OR

Factorise : a b3 32 2−( )Sol. We have, a b3 32 2−( ) = ( )a b3

32− ( )⎡

⎣⎢⎤⎦⎥

= a b a a b b−( ) + ( ) + ( )⎡⎣⎢

⎤⎦⎥

2 2 222

= a b a ab b−( ) + +( )2 2 22 2

∴ a b3 32 2−( ) = a b a ab b−( ) + +( )2 2 22 2

21. In the given figure, AB ⊥ BC, AD || BC,

∠CDE = 75° and ∠ACD = 22°. Findthe values of x and y.

Sol. In the given figure, AB ⊥ BC,AD || BC, ∠CDE = 75° and ∠ACD = 22°.Now, ∠BCD = ∠CDE [Alternate angles]⇒ y + 22° = 75°⇒ y = 75° – 22° = 53°In ∆ABC, x = 180° – (90° + 53°) [Angle sum property of a triangle]

= 180° – 143° = 37°Hence, x = 37° and y = 53°.

C B

A D

O


22. A triangle ABC is right angled at A. L is a point on BC such that AL ⊥ BC. Prove that

∠BAL = ∠ACB.

Sol. In right ∆BAC, we have

⇒ ∠BAC = ∠ABC + ∠ACB ...(i)

Again, in right ∆ALB, we have

⇒ ∠ALC = ∠ABL + ∠BAL

[Exterior angle property]

⇒ ∠ALC = ∠ABC + ∠BAL ...(ii)

[∵ ∠ABL = ∠ABC]

From (i) and (ii), we get

∠ABC + ∠ACB = ∠ABC + ∠BAL [∵ ∠BAC = ∠ALC = 90°]

⇒ ∠ACB = ∠BAL

Or, ∠BAL = ∠ACB. Proved.

OR

If the altitudes from two vertices of a triangle to opposite sides are equal, prove that

triangle is isosceles.

Sol. ABC is a triangle in which altitude CD = altitude BE.

In ∆CBD and ∆BCE, we have

Hypotenuse BC = Hypotenuse BC [Common]

CD = BE [Given]

∠BDC = ∠CEB = 90°

∴ ∆CBD ≅ ∆BCE [RHS]

⇒ ∠CBD = ∠BCE [CPCT]

⇒ AB = AC [Side opposite to equal angles are equal]

Hence, the triangle is isosceles. Proved.

23. The difference between the legs of a right-angled triangle is 14 cm. The area of the

triangle is 120 cm2. Calculate the perimeter of the triangle.

Sol. Let the sides containing the right angle be x cm and (x – 14) cm.

Then, the area of the triangle = 1

2× x × (x – 14) cm2

But, area = 120 cm2 [Given]

∴ 1

2× x (x – 14) = 120 ⇒ x2 – 14x – 240 = 0

⇒ x2 – 24x + 10x – 240 = 0 ⇒ x(x – 24) + 10(x – 24) = 0

⇒ (x – 24) (x + 10) = 0 ⇒ x = 24 [Neglecting x = – 10]

∴ One side = 24 cm, other side = (24 – 14) cm = 10 cm.

Hypotenuse = ( ) ( )2 2

24 10+ cm = 576 100+ cm = 676 cm = 26 cm

∴ Perimeter of the triangle = (24 + 10 + 26) cm = 60 cm

24. In the figure, ∠Q > ∠R. If QS and RS are bisectors of ∠Q and ∠R respectively, then

show that SR > SQ.

Sol. We have, ∠Q > ∠R


⇒1

2∠Q >

1

2∠R ⇒ ∠SQR > ∠SRQ

⇒ SR >SQ [Side opposite to greater angle is greater]Proved.

SECTION D


25. Factorise : 2 5 1

26 12

x x− +

Sol. We have, 2 5 1

26 12

x x− + =( )224 10 1

12

− +x x =

1

12(24x2 – 10x + 1)

=1

12(24x2 – 6x – 4x + 1) =

1

12[6x(4x – 1) – 1(4x – 1)]

=1

12(4x – 1) (6x – 1)

Hence, 2 5 12

6 12x x− + =

1

12(4x – 1) (6x – 1)

OR

If p = 2 – a, prove that a3 + 6ap + p3 – 8 = 0.

Sol. We have, p = 2 – a ⇒ a + p – 2 = 0

Now, a3 + 6ap + p3 – 8 = a3 + p3 + (–2)3 – 3ap(–2)

= {a + p + (–2)}{a2 + p2 + (–2)2 – ap – p(–2) – a(–2)}

= (a + p – 2) (a2 + p2 + 4 – ap + 2p + 2a)

= 0 × (a2 + p2 + 4 – ap + 2p + 2a) = 0.

26. If a + b = 10 and a2 + b2 = 58, find the value of a3 + b3.

Sol. We know that (a + b)2 = a2 + b2 + 2ab

Putting a + b = 10 and a2 + b2 = 58, we get

102 = 58 + 2ab ⇒ 100 = 58 + 2ab ⇒ 2ab = 100 – 58 = 42 ⇒ ab = 21

Thus, we have a + b = 10 and ab = 21

Now, (a + b)3 = a3 + b3 + 3ab (a + b)

⇒ (10)3 = a3 + b3 + 3 × 21 × 10 [Putting a + b = 10 and ab = 21]

⇒ 1000 = a3 + b3 + 630

⇒ a3 + b3 = 1000 – 630 = 370.

27. Factorise : x8 – y8

Sol. We have, x8 – y8 = {(x4)2 – (y4)2} = (x4 – y4)(x4 + y4)

= {(x2)2 – (y2)2} (x4 + y4)

= (x2 – y2) (x2 + y2) (x4 + y4)

= (x – y)(x + y) (x2 + y2) (x4 + y4)

= (x – y)(x + y) (x2 + y2){(x2)2 + (y2)2 + 2x2y2 – 2x2y2}

= (x – y) (x + y)(x2 + y2) {(x2 + y2)2 – ( 2 xy)2}

= (x – y) (x + y)(x2 + y2)(x2 + y2 – 2 xy)(x2 + y2 + 2 xy)


28. (i) Plot each of the points

A (–2, 4), B (–2, –3),

C(4, – 3) and D(4, 4).

(ii) Draw the segments AB, BC,

CD and DA. What is the name

of the figure ABCD?

(iii) What are the coordinates of the

point where the segment AD

cuts the y-axis?

(iv) What are the coordinates of the

points where the segment CD

cuts the x-axis?

Sol. (i) Points A(–2, 4), B(–2, –3),

C(4, –3) and D(4, 4) are plotted

in the adjoining figure.

(ii) From the graph, we observe that

AB = CD = 7 units and BC = AD = 6 units.

Therefore, the figure ABCD is a rectangle.

(iii) From the graph, it is clear that the segment AD cuts the y-axis at (0, 4).

(iv) From the graph, it is clear that segment CD cuts the x-axis at (4, 0)

29. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that

AD = AB (see figure). Show that ∠BCD is a right angle.

Sol. We have, AB = AC [Given]

∠ACB = ∠ABC ...(i)

[Angles opposite to equal sides are equal]

Also, AB = AD [Given]

∴ AD = AC [∵ AB = AC]

∴ ∠ACD = ∠ADC ...(ii)

[Angles opposite to equal sides are equal]

Adding (i) and (ii), we get

∠ACB + ∠ACD = ∠ABC + ∠ADC

⇒ ∠BCD = ∠ABC + ∠ADC ...(iii)

Now, in ∆BCD, we have

∠BCD + ∠DBC + ∠BDC = 180° [Angle sum property of a triangle]

∴ ∠BCD + ∠BCD = 180° ⇒ 2∠BCD = 180° ⇒ ∠BCD = 90°. Proved.

30. Prove that the perimeter of a triangle is greater than the sum of its three altitudes.

Sol. We have, ∆ABC in which AD, BE and CF are its altitudes.

In ∆ABD, AD < AB ...(i)

[Perpendicular line segment is the shortest]

Similarly, in ∆ADC, AD < AC ...(ii)

Adding (i) and (ii), we get

2AD < AB + AC ...(iii)

Similarly, we can prove


2BE < BC + AB ...(iv)

And, 2CF < AC + BC ...(v)

Adding (iii), (iv) and (v), we get

2(AD + BE + CF) < 2 (AB + BC + CA)

⇒ AB + BC + CA > AD + BE + CF. Proved.

31. In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray

lying between rays OP and OR. Prove that

∠ROS = 1

2(∠QOS – ∠POS)

Sol. We have, ∠ROS = ∠ROP – ∠POS ...(i)

And ∠ROS = ∠QOS – ∠QOR ...(ii)

Adding (i) and (ii),

∠ROS + ∠ROS = ∠QOS – ∠QOR + ∠ROP – ∠POS

⇒ 2∠ROS = ∠QOS – ∠POS

[∵ ∠QOR = ∠ROP = 90°]

⇒ ∠ROS =1

2(∠QOS – ∠POS). Proved.

32. In the given figure, AB || CD. Find the value of x

Sol. Through E, draw a line GEH || AB || CD.

Now, GE || AB and EA is a transversal.

∴ ∠GEA = ∠EAB = 50° [Alt. Int. ∠s]

Again, EH || CD and EC is a transversal.

∴ ∠HEC + ∠ECD = 180° [Co. Int. ∠s]

⇒ ∠HEC + 100° = 180° ⇒ ∠HEC = 80°

Now, GEH is a straight line.

∴ ∠GEA + ∠AEC + ∠HEC = 180° [Straight angle]

⇒ 50° + x° + 80° = 180° ⇒ x° = 50°

Hence, x = 50.

33. If x = 5 21

2

−, find the value of

2

2

1x

x+ .

Sol. We have, x + 1

x=

5 21 2

2 5 21

−+

−

=

( )( )( )

2 5 215 21

2 5 21 5 21

+−+

− +


= ( )2 5 215 21

2 25 21

+−+

− =

5 21 5 21

2 2

− ++ = 5

Now,

2

2

2

1 1x x

x x

+ = +

– 2 = 52 – 2 = 25 – 2 = 23.

OR

Simplify : 3 2 3 2

3 2 3 2

+ −+ − +

Sol. We have, ( ) ( )

( )( )

( ) ( ){ }( )

2 2

2 2

3 2 3 2 3 2 3 23 2

3 23 2 3 2 3 2 3 2

+ + + ++= × = =

−− − + −

= ( ) ( ) ( )2 2 2

3 2 3 2 2 3 2+ = + + × × = ( )3 2 2 6 5 2 6+ + = +

And, ( )( )

( )( )

3 2 3 23 2

3 2 3 2 3 2

− −−= ×

+ + − =

( )( )

2

23 23 2

3 2

−= −

−

= ( ) ( )2 2

3 2 2 3 2+ − × × = 3 + 2 – 2 6 = ( )5 2 6−

∴ 3 2 3 2

3 2 3 2

+ −+

− + = ( ) ( )5 2 6 5 2 6+ + − = 10.

34. Prove that :

–1 –1 2

–1 –1 –1 –1 2 2

2

– –

a a b

a b a b b a+ =

+

Sol. We have,

–1 –1

–1 –1 –1 –1–

a a

a b a b+

+

= 1/ 1/ 1/ 1/

1 1 1 1 ––

+ = ++

+

a a a a

b a b a

a b a b ab ab

= 1 1

. .– –

ab ab b b

a b a a b a b a b a+ = +

+ +

=

2 2

2 2

( – ) ( ) –

( )( – ) –

b b a b b a b ba b ab

b a b a b a

+ + + +=

+=

2

2 2

2

–

b

b a. Proved.

Summative Assessment Paper-1

Education