General Instructions :

i. All questions are compulsory.

ii. The question paper consists of 30 questions divided into four

sections A, B, C and D.

iii. Section A contains 6 questions of 1 mark each. Section B

contains 6 questions of 2 marks each. Section C contains 10

questions of 3 marks each. Section D contains 8 questions of 4

marks each.

iv. There is no overall choice. However, an internal choice has been

provided in four questions of 3 marks each and three questions

of 4 marks each. You have to attempt only one of the

alternatives in all such questions.

v. Use of calculators is not permitted.

SECTION A

1. Evaluate (25)1/3× (5)1/3.

Solution:

(25)1/3× (5)1/3=(23× 5) 1/3=(5³)1/3=5

2. Find the total surface area of a cone whose radius is 2r and slant

height is 𝑙

𝑟.

Solution:

Total surface area of cone is given as,

S=𝜋2𝑟 (𝑙

2+ 2𝑟) = 𝜋𝑟(𝑙 + 4𝑟)𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠

CLASS IX – CBSE BOARD

Mathematics

Solved Practice Paper Total Marks: 90

Time: 3 hours

3. Find the radius of largest sphere that is carved out of the cube of

side 8 cm.

Solution:

The largest sphere can be carved out from a cube, if we take

diameter of the sphere equal to edge of the cube.

Diameter of the sphere = 8 cm

Thus, radius of the sphere =8

2= 4 𝑐𝑚

4. An angle is 14o more than its complement. Find its measure.

Solution:

Let the measure of the angle be x.

Now, two complementary angles have sum equal to 90o

So, measure of its complement = 90 – x

From the given condition, we have:

x = 14 + (90 − x)

⟹ 2x = 104

⟹ x = 52o

5. Calculate the mean of first five multiples of 3.

Solution:

The first five multiples of 3 are 3, 6, 9, 12 and15

Now, mean of ungrouped data is sum of the data divided by

number of quantities in data.

Mean=3+6+9+12+15

5=

45

5= 9

6. If P(E)= 0.25 what is the value of P(not E).

Solution:

P(E) + P(not E) = 1 [∵ Sum of the probabilities of all the

elementary events is 1]

∴ 0.25 + P(not E) = 1

P(not E) = 1 − 0.25

P(not E) = 0.75

SECTION B

7. A, B and C are three points on a circle with centre O such that

∠BOC = 300 and ∠AOB = 60°. If D is another point on the circle

other than the arc ABC, find ∠ADC.

Solution:

Consider the following diagram:

Here, ∠AOC = ∠AOB+ ∠BOC

= 60°+30°=90°

Here, ∠AOC and ∠ADC are the angles subtended by the same arc

ABC at the centre and the circumference of circle.

But we know that the angle subtended by an arc at the centre is

double the angle subtended by it at any point of the remaining part of

the circle.

∠ADC=1

2∠𝐴𝑂𝐶 =

1

2× 90° = 45°

8. For the following distribution, find the value of a and the frequencies

of 30 and 70,if the mean of the distribution is 50.

X F

10

90

30

70

50

17

19

5a+3

7a-11

32

Solution:

Mean of the given data can be found as:

Mean=10×17×90×19+30(5𝑎+3)+70(7𝑎−11)+50×32

17+19+5𝑎+3+7𝑎−11+32

50=170+1710+150𝑎+90+490𝑎−770+1600

12𝑎+60

50(12a+60)=2800+640a

600a+3000=2800+640a

= 40a=200

a =5

Hence, the value of a is 5 and frequencies of 30 and 70 are 28 and 24

respectively.

9. Some wooden crates, each measuring 1.5 m x 1.25 m x 0.5 m, have

to be stored in a go down that measures 40 m x 25 m x 10 m. Find

the maximum number of wooden that can be stored in the god own.

Solution:

Volume of the go down = 40 × 30 × 10 m3

Volume of each wooden crate = 1.5 × 1.25 × 0.5 m3

Volume of go down Maximum number of wooden crates

= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑔𝑜𝑑𝑜𝑤𝑛

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑐𝑟𝑎𝑡𝑒

Hence, the maximum number of wooden crates that can be stored in

the go down are 12800.

SECTION C

10. Prove that equal chords of a circle subtends equal angles at the

centre.

Given: AB and CD are the two equal chords of a circle C (O, r)

To Prove: ∠AOB =∠COD

Solution:

Proof: In ∆ABO and ∆COD

AB=CD [ Given]

OA=OD = r

And, OB = OC = r

ABO and ∆COD [By SSS criteria]

Thus, ∠AOB = ∠COD [CPCT]

11. Prove that a diagonal of a parallelogram divide it into two congruent

triangles.

Given : A parallelogram ABCD and AC is its diagonal.

To Prove: ∆ABC = ∆CDA

Solution:

Proof: In ∆ABC and ∆CDA, we have:

∠DAC=∠BCA [alt, int, angles, since AD‖BC]

AC=AC

And ∠BAC = ∠DCA [alt, int, angles, since AD‖BC]

By ASA congruence axiom, we have

∆ABC≅ ∆ CDA

12. In 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)

Solution:

∠QOS- ∠POS =( ∠QOR+ ∠ROS)- ∠POS

= 90°+ ∠ROS- ∠POS

= (90°- ∠POS)+ ∠ROS

= (∠ROP- ∠POS)+ ∠ROS

= ∠ROS+ ∠ROS=2ROS

Hence , ∠ROS =1

2(∠QOS − ∠POS)

SECTION D

13. Convert the given frequency distribution into a continuous grouped

Frequency distribution.

Class-

Intervals

Frequency

150-153

154-157

158-161

162-165

166-169

170-173

7

7

15

10

5

6

In which intervals would 153.5 and 157.5 be included.

Solution:

We convert the given distribution into a continuous grouped

frequency distribution as follows: Here, difference between the

lower limit of a class and the upper limit of the preceding class,

d=154 – 153 = 1

Adjustment factor=1

2= 0.5

So, we have to subtract 0.5 from the upper and lower limits of each

class which will give us a continuous frequency distribution as

below:

Class- Intervals Frequency

149.5-153.5

153.5-157.5

157.5-161.5

161.5-165.5

165.5-169.5

169.5-173.5

7

7

15

10

5

6

Thus, 153.5 is included in the class-interval 153.5 − 157.5 and 157.5

is included in the class-interval 157.5 − 161.5.

14. The diagonals AC and BD of parallelogram ABCD intersect at the

point O. If ∠DAC = 34o and ∠AOB = 75o, then what is the measure

of ∠DBC?

Solution:

ABCD is a parallelogram

∴AD‖BC ∠ACB=∠DAC=34°

Now ∠AOC is an exterior angle of ∆BOC

∠OBC+ ∠OCB= ∠AOB

[∵ ext ∠=sum of two ini.opp. ∠s]

∠OBC+34°=75°

∠OBC=75°-34°=41°

OR

∠DBC = 41°

15. Express 2.4178̅̅ ̅̅ ̅̅ ̅ in the form 𝑝

𝑞

Solution:

Let 𝑝

𝑞= 2. 4178̅̅ ̅̅ ̅̅ ̅

Or

𝑝

𝑞= 2.4178178178

Multiplying by 10,

10𝑝

𝑞= 24.178178178 … … (𝑖)

Multiplying by 1000,

1000𝑝

𝑞= 24178.178178 … … (𝑖𝑖)

Subtracting (i)from (ii), we get:

10000𝑝

𝑞− 10

𝑝

𝑞= 24178.178178 − 24.178178178

9990𝑝

𝑞= 24154

𝑝

𝑞=

24154

9990

𝑝

𝑞=

12077

4995

16. Factorise(9𝑥 −1

5)

2+ (𝑥 +

1

3)

2

Solution:

We have, (9𝑥 −1

5)

2+ (𝑥 +

1

3)

2

= [(9𝑥 −1

5) − ( 𝑥 +

1

3)] [(9𝑥 −

1

5) + (𝑥 +

1

3)] [∵a²-b² =(a+b)(a-b)]

= (9𝑥 −1

5− 𝑥 −

1

3) (9𝑥 −

1

5+ 𝑥 +

1

3)

= (8𝑥 −8

15) (10𝑥 −

2

15)

= ( 120𝑥 − 8

15) (

150𝑥 − 2

15)

17. If a, b, c are all non-zero and a + b + c = 0, prove that

𝑎2

𝑏𝑐+

𝑏2

𝑐𝑎+

𝑐2

𝑎𝑏= 3

Solution:

Given, 𝑎 + 𝑏 + 𝑐 = 0

Now, 𝑎3 + 𝑏3 + 𝑐3 − 3𝑎𝑏𝑐

= (𝑎 + 𝑏 + 𝑐)(𝑎2 + 𝑏2 + 𝑐2 − 𝑎𝑏 − 𝑏𝑐 − 𝑐𝑎)

𝑎3 + 𝑏3 + 𝑐3 − 3𝑎𝑏𝑐 = 0 [∵𝑎 + 𝑏 + 𝑐 = 0]

𝑎3 + 𝑏3 + 𝑐3 − 3𝑎𝑏𝑐

Now, we have, 𝑎2

𝑏𝑐+

𝑏2

𝑐𝑎+

𝑐2

𝑎𝑏= 3

LHS =𝑎2

𝑏𝑐+

𝑏2

𝑐𝑎+

𝑐2

𝑎𝑏=

𝑎3+ 𝑏3+𝑐3

𝑎𝑏𝑐

3𝑎𝑏𝑐

𝑎𝑏𝑐= 3 = 𝑅𝐻𝑆

18. A triangle ABC is right-angled at A. AL is drawn perpendicular to

BC. Prove that

∠𝐵𝐴𝐿 = ∠𝐴𝐶𝐵

Solution:

Consider the ∆ABC as given below:

In ∆ABC, we have

∠A+∠B+∠C = 180o

900 + ∠B + ∠ = 1800

∠B+∠C =90°

∠C = 90o− ∠B ……(i)

Now, in ∆ABL, we have:

∠ALB + ∠BAL + ∠B = 180o

90o+ ∠BAL +∠B = 180o

∠BAL +∠B =90o

∠BAL =90o− ∠B ...(ii)

From equations (i) and (ii), we get:

∠BAL = ∠ACB

19. If x = 2 + √3, then find the value of 𝑥2 +1

𝑥2

Solution:

Given, 𝑥 = 2 + √3

1

𝑥=

1

2+√3

1

𝑥=

1

2+√3×

2−√3

2−√3

1

𝑥=

2−√3

4−3= 2 − √3

𝑥 +1

𝑥= 2 + √3 + 2 − √3

𝑥 +1

𝑥= 4 … . . (𝑖)

Now 𝑥2 +1

𝑥2 = 𝑥2 +1

𝑥2 + 2. 𝑥1

𝑥− 2. 𝑥

1

𝑥

𝑥2 +1

𝑥2 = (𝑥 +1

𝑥)

2− 2

𝑥2 +1

𝑥2= (4)2 − 2 [𝑈𝑠𝑖𝑛𝑔(𝑖)]

= 16-2=14

20. The volume of a rectangular block of stone is 10368 dm3and its

dimensions are in the ratio of 3:2:1. (i) Find the dimensions (ii) Find

the cost of polishing this rectangular block’s entire surface at Rs. 2

per dm2 ?

Solution:

Let the length of the block = 3 x dm,

Width of the block = 2x dm and

Height of the block = x dm

Given, volume of the block = 10368 dm3

3𝑥 × 2𝑥 × 𝑥 = 10368

6𝑥3 = 10368

𝑥3 =10368

6= 1728

𝑥 = √17283

𝑥 = √12 × 12 × 123

=12

length of the block = 3x = 3 × 12 = 36 dm,

Width of the block = 2x = 2 × 12 = 24 dm and

Height of the block = x = 12 dm

Thus dimensions of the block are 36dm, 24dm and 12dm.

Now, Surface area of the block = 2(lb + bh + hl)

=2(36 × 24 + 24 × 12 × 36)𝑑𝑚3

=2(864 + 288 + 432)

=2 × 1584 = 3168 𝑑𝑚3

Thus, cost of polishing the surface = Rs 2 × 3168 = Rs 6336

21. Construct a triangle ABC in which BC = 7cm, ∠B = 75oand AB +

AC = 9cm Ans.

Solution:

Steps of construction:

a. Draw BC = 7cm

b. Draw ∠DBC=75o

c. Cut a line segment BD = 9cm

d. Join DC and make ∠DCY = ∠BDC

e. Let CY intersect BD at A

f. Triangle ABC is required triangle

22. In the given figure, BC and CO are bisectors of ∠ DBC and ∠ ECB

respectively. If ∠BAC = 70° and ∠ABC = 40o, find the measure of

∠BOC?

Solution:

∠DBC = 180o− 40° = 140° (Linear pair)

∠CBO = 1

2∠DBC =

1

2× 140o= 70°

∠ACB = 180o− (70o+ 40o) = 70o

∠BCE = 180° − 70° = 110° (Linear pair)

∠BCO = 1

2× 110° = 55° (Angle bisector)

∠BOC = 180° − (∠CBO + ∠BCO)

= 180° − (70° + 55°) = 55°

SECTION E

23. The diagonals of a quadrilateral ABCD are perpendicular to each

other. Show that the quadrilateral formed by joining the mid-points

of its sides is a rectangle.

Solution:

Given: A quadrilateral ABCD whose diagonals AC and BD are

perpendicular to each other at O. P, Q, R and S are mid-points of

sides AB, BC, CD and DA respectively and are joined are form a

quadrilateral PQRS. To Prove: PQRS is a rectangle.

Proof : In ∆ABC, P and Q are mid-points of AB and BC

respectively.

PQ‖AC and PQ =1

2 AC ……(i) [mid point theorem]

Further in ∆ACD, R and S are mid-points of CD and DA repectively.

SR‖AC and SR=1

2𝐴𝐶 …..(ii)[mid point theorem]

From (i) and(ii), we have PQ ‖ SR and PQ = SR

Thus, one pair of opposite sides of quadrilateral PQRS are parallel

and equal.

PQRS is a parallelogram.

Since PQ‖ AC PM‖NO

In ∆ABD, P and S are mid- point of AB and AD respectively

PS‖BD [mid point theorem]

PN‖MO

Opposite sides of quadrilateral PMON are parallel.

PMON are parallel.

∠MPN =∠MON [opposite angles of ‖gm are equal]

But ∠MON = 90° [given]

∠MPN =90° ∠QPS =90°

Thus , PQRS is a parallelogram whose one angle is 90°

PQRS is a rectangle.

24. Without actual division, prove that(2𝑥4 − 6𝑥3 + 3𝑥2 + 3𝑥 − 2) is

exactly divisible by (x²-3x+2)

Solution:

Let 𝑓(𝑥) = 2𝑥4 − 6𝑥3 + 3𝑥2 + 3𝑥 − 2

And 𝑔(𝑥) = x2 − 3x + 2

= 𝑥2 − 2𝑥 − 𝑥 + 2

= 𝑥(𝑥 − 2) − 1(𝑥 − 2) = (𝑥 − 1)(𝑥 − 2)

Now if x²-3x+2 is a factor of 𝑓(𝑥)

Then (x-1) and (x-2) both should be factors of 𝑓(𝑥)

Then, 𝑓(1) = 𝑓(2) = 0

So,𝑓91) = 2.14 − 3.22 + 3.2 − 2

= 2-6+3+3-2=0

𝑓(2) = 2.24 − 6.23 + 3.22 + 3.2 − 2

= 32 − 48 + 12 + 6 − 2 = 0

Since 𝑓(1) = 𝑓(2) = 0

𝑓(𝑥) = 2𝑥4 − 6𝑥3 + 3𝑥2 + 3𝑥 − 2 is exactly divisible by (x – 1)

and (x – 2).

∴ x2– 3x + 2 is a factor of𝑓(𝑥).

25. AC and BD are chords of a circle that bisect each other. Prove that

AC and BD are diameters and ABCD is a rectangle.

Solution:

Let AC and BD bisect each other at point O.

Then, OA = OC and OB = OD ... (i)

In triangles AOB and COD we have,

OA = OC

OB = OD

And ∠AOB = ∠DOC (Vertically opposite angles)

∆AOB≅ COD (SAS criterion)

AB= CD (CPCT)

AB≅ CD ... (i)

Similarly, BC = DA

BC= DA ... (ii)

From (i) and (ii), we get:

AB+ BC ≅CD +DA

ABC≅CDA

AC divides the circle into two equal parts.

Thus, AC is the diameter of the circle.

Similarly, we can prove that BD is also a diameter of the circle.

Now, since AC and BD are diameters of the circle.

∠ABC =∠BCD, ∠CDA=∠DAB=90° [As all are the angles in

semicircles]

Also, AB = CD and BC = DA (Proved above)

Hence, ABCD is a rectangle.

26. Three girls (Anita, Payal and Riya) are sitting at equal distance on

the boundary of a circular bed of radius 2m as shown in the figure

below. Each of the three girls is having a to y telephone in their

hands to talk to each other. Find the length of the string of each

telephone.

Solution:

Let the length of the string of each telephone wire be a. Then,

AR =AP= RP= a

We know that perpendicular from the centre to a chord bisects the

chord.

PN =AN=𝑎

2

Let us join OA, then in ∆ANO

𝑂𝐴2 = 𝑂𝑁2 + 𝐴𝑁2

22 = 𝑂𝑁2 + (𝑎

2)

2

4 −a2

4= ON2

Also, in ∆ANR

𝐴𝑅2 = 𝑅𝑁2 + 𝐴𝑁2

𝑎2 = 𝑅𝑁2 + (𝑎

2)

2

a2 −a2

4= RN2

3a2

4= RN2

Now consider,

OR= RN- ON

2= √3𝑎2

4− √4 −

𝑎2

4

2=√3𝑎

2−

√16−𝑎2

2

4=√3𝑎 − √16 − 𝑎2

√16 − 𝑎2 = √3𝑎 − 4

16 − 𝑎2 = 3𝑎2 + 16 − 8√3𝑎

8√3𝑎 = 4𝑎2

𝑎 = 2√2

Thus, length of the string of each telephone is 2√3m.

27. Water in a canal, 30 dm wide and 12 dm deep is flowing with a

velocity of 20 km per hour. How much area will it irrigate in 30 min,

if 9 cm of standing water is desired? ( 10 dm= 1m )

Solution:

Width of the canal = 30 dm = 300 cm

Depth of the canal = 12 dm = 120 cm

The velocity of the flow of the water in the canal = 20 km/hr

= 20 × 1000 × 100 = 2000000 cm/hr

Volume of water irrigating in 1 hour = 300 × 120 × 2000000 cm3

Volume of water irrigating in 30 min

= 300 × 120 × 2000000 ×30

60= 300 × 60 × 2000000 𝑐𝑚3

The area irrigated by this water if 9 cm of standing water is desired

= 300 × 60 × 2000000 = 400000000 𝑐𝑚2 =4000000000

10000

= 400000 𝑚2

Therefore, the area irrigated by this water if 9 cm of standing water

is desired =400000

1000000= 0.4 𝑘𝑚2

28. Factorise: 1

27(2𝑥 + 5𝑦)2 + (−

5

3𝑦 +

3

4𝑧) − (

3

4𝑧 +

2

3𝑥)

2

Solution:

We can write the above expression as below:

[1

3(2𝑥 + 5𝑦)]

3

+ [−5

3𝑦 +

3

4𝑧]

3

+ [−3

4𝑧 −

2

3𝑥]

3

Put 1

3(2𝑥 + 5𝑦) = 𝑎, −

5

3𝑦 +

3

4𝑧 = 𝑏 𝑎𝑛𝑑 −

3

4𝑧 −

2

3𝑥 = 𝑐

Now 𝑎 + 𝑏 + 𝑐+=2

3𝑥 +

5

3𝑦 −

5

3𝑦 +

5

4𝑧 −

2

3𝑥 = 0

𝑎3 + 𝑏3 + 𝑐3 = 3𝑎𝑏

Thus,

[1

27(2𝑥 + 5𝑦)3] + [−

5

3𝑦 +

3

4𝑧]

3

+ [−3

4𝑧 −

2

3𝑥]

3

=3[1

392𝑥 + 5𝑦) (−

5

3𝑦 +

3

4𝑧) (−

3

4𝑧 −

2

3𝑥)]

= −(2𝑥 + 5𝑦) (−20𝑦 + 9𝑧

12) (

9𝑧 + 8𝑥

12)

=1

144(2𝑥 + 5𝑦)(−20𝑦 + 9𝑧)(9𝑧 + 8𝑥)