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CBSEContinuous and Comprehensive Evaluation (CCE)

Sample

Question Papers

Term 2 (October to March 2014)

OSWAAL BOOKS

Class

Mathematics

Solutions

9



CONTENTS

l Solutions

q Sample Question Paper 6 1 - 6





ll



SAMPLE QUESTION PAPER-6

SECTION ‘A’

1. x = 3 y ...(1)

Put y = 0 in equation (1)

x = 3 × 0

x = 0 Yes, the equation x = 3 y passes through theorigin.

2. Let the two angles are 8 x and 15 x, than

50 + 80 + 8 x + 15 x = 360º

23 x = 360º – 130º

23 x = 230º

x = 10°

∴ 8 x = 8 × 10 = 80º

15 x = 15 × 10 = 150º

3. We know that

P( A) + P( B) = 1

0·32 + P( B) = 1

P( B) = 1 – 0·32

P( B) = 0·684. Equation 2 x + 2 y + 9 = 0

Put x = – 3

2 × (– 3) + 2 × y + 9 = 0

– 6 + 2 y + 9 = 0

2 y + 3 = 0

2 y = – 3

y =–32

No, the linear equation does not passesthrough the point (– 3, – 3).

SECTION ‘B’

5. (a) The two triangles on the same base ABand between the same paralles are :

∆ ADB and ∆ ACB 1

(b) A triangle ABC and ||gm ABCD on thesame base BC and between the sameparalles. 1

6. Given, ∠ MRS = 29º and RS is a diameter of the circle. NM || RS

In ∆ RMS,

∠ RMS = 90º (angle of semicircle)

∠ MRS = 29º (given)∴ ∠ MSR = 180º – (∠ RMS + ∠ MRS)

= 180º – (90º + 29º)

= 180º – 119º

= 61º 1

Now, ∠ NMS + ∠ MSR = 180º,( NM || RS)

∠ NMS + 61º = 180º

∠ NMS = 180º – 61º

= 119º 1

7.3 cm

3 cm

6 cm

Height of cylinder (h) = 6 cm

Radius of cylinder (r) = 3 cm

Curved surface area of cylinder

= 2πrh 1

= 2 × π × 3 × 6

= 36π cm2

8. Total no. of students = 50

No. of girls = 30

∴ No. of boys = 50 – 30 = 20

The mean of marks scored by girls = 73

∴ Total marks obtained by girls = 73 × 30

= 2190 ½

The mean of marks scored by boys = 71

∴ Total marks obtained by boys = 71 × 20

= 1420 ½

So, mean score of whole class =2190 1420

50+

=3610

50

= 72·2 marks 1

9. First seven multiples of 9 are

9, 18, 27, 36, 45, 54, 63 ½

Mean of first seven multiples

=Sum of first seven multiples of 9

7

= 2527

= 36 ½

10. Let the edge of the cube be a.

The length of a diagonal of a cube = 3 a 1

6 3 = 3 a

a =6 3

3

a = 6 cm. 1



2 | CBSE (CCE) Sample Question Papars (SA-II) Mathematics-IX

11. Let A (1, 2), B (– 1, – 16) and C (0, – 7) lie

on the graph of linear equation y = 9 x – 7,

then points A, B and C will satisfy the equa-

tion

y = 9 x – 7

OR y – 9 x = – 7Putting x = 1 and y = 2, as A lie on given

linear equation.

2 – 9 (1) = – 7

2 – 9 = – 7

– 7= – 7 1

Now, putting x = – 1 and y = – 16, as B lie

on given linear equation

– 16 – 9(– 1) = – 7

– 16 + 9 = – 7 1

– 7= – 7Now, putting x = 0 and y = – 7, as C lie on

given linear equation.

– 7 – 9(0) = – 7

– 7 – 0 = – 7 1

– 7= – 7

Hence, A (1, 2), B (– 1, – 16) and C (0, – 7) lie

on the graph of linear equation y = 9 x – 7

12. 7 y = 2 x

2 x – 7 y = 0

Equation 2 x – 7 y + 0 = 0 is of the form

ax + by + c = 0

Where a = 2, b = – 7 and c = 0 2

Put x = 0 in 2 x – 7 y + 0 = 0

2 × 0 – 7 y = 0

– 7 y = 0

y = 0

Yes, the graph of this linear equation passes

through origin.

13. Given : ∆ ABQ and parallelogram ABCD are

on the same base and between same paral-

lels, DC and AB. ½

DC R

A B

To prove :

ar (∆ ABQ) =12

ar (11gm ABCD). ½

Construction : Extend DC to R so that BR

|| AQ.

Proof : DCBA and QRBA are on the same

base and between same parallels.

∴ ar ( DCBA) = ar (QRBA) ...(i) ½

A diagonal divides a parallelogram into two

congruent triangles with equal area.

ar (QAB) =12

ar (QCBA). ...(ii) ½

From (i) and (ii), we get

ar (QAB) =

1

2 ar ( DCBA) 1

14. ∆ ABC is an isosceles triangle.

P

A D

B C

∴ ∠ ABC = ∠ BCA

∠ PAC = ∠ ABC + ∠ BCA

(exterior angle is the sum of two

opposite interior angle)

= 2 ∠ BCA ...(i)

AD bisects ∠ PAC.

∴ ∠ PAC = 2 ∠ DAC ...(ii) 1From (i) and (ii)

∠ BCA = ∠ DAC

These are alternative angles when lines BC

and AD are intersected by AC. 1

BC || AD

Also, BA || CD. (given)

∴ ABCD is a parallelogram 1

SECTION ‘C’



Solutions | 3

15. In ∆ APD and ∆ CQB,

A D

B C

P

Q

PD = BQ (given)

AD = BC (opposite sides of ||gm)

∠ ADP = ∠ QBC (alt. ∠ angles)

∴ ∆ APD ≅ ∆ CQB (by S.A.S) 1

∴ AP = CQ (by C.P.C.T )

In ∆ AQB and ∆ CPD

∠ ABC = ∠ CDP ( Alt.∠ S)

∴ ∆ AQB ≅ ∆ CPD (by SAS) 1

∴ AQ = CP∵ APCQ is a quadirlateral in which opposite

sides are equal.

∴ APCQ is a parallelogram 1

16. Edge of one cube = 12 cm

When we joined three cubes end to end.

Then length, breadth and height of resulting

Cuboid are 36 cm, 12 cm and 12 cm

respectively. 1

Volume of resulting cuboid = l × b × h

= 36 × 12 × 12= 5184 cm3. 2

17. Height of cone (h) = 16 cm

radius (r) = 12 cm

16 cm

20 cm

12 cm

Slant height, l = 2 2h r+

= 2 216 12+ = 256 144+

= 400 = 20 cm 1

Curved surface area of cone = πrl

= 3·14 × 12 × 20 = 753·6 cm2 1

Volume of cone =13

πr2h

=13

× 3·14 × 12 ×12 × 16

= 2411·52 cm3 1

18. First ten prime numbers are 2, 3, 5, 7, 11,

13, 17, 19, 23, 29. ½

Mean ( ) x =Sum of 10 prime numbers

10

x =+ + + + + + + + +2 3 5 7 11 13 17 19 23 29

10

=12910

= 12·9 1

Again prove ( )10

1

–i

i

x x

=

å = 0

L.H.S = ( )10

1

–ii

x x=

å

= ( )1 – x x + ( )2 – x x + ... + ( )10 – x x ½

= ( x1 + x2 + x3 + ... + x10) – 10 ( ) x

= (2 + 3 + 5 + ... + 29) – 10 × 12·9

= 129 – 129

= 0 = R.H.S Proved. 1

19. (a) Probability (obtained marks 60 or above)

=15 8

90+

=2390

1½

(b) Probability (obtained marks less than 40)

=+ +

=7 10 10 27

90 90

= =3

0·310 1½

20. (i) P (Earning ` 100 and more) =2

251

(ii) P (at least ` 60 but less than ` 80)

=2

25 1

(iii) P (less than ` 40) =5

25 =

15

1




21. 3 x + 4 y = 6

1 2 3 4 5 6–1

–2

0

5

4

3

2

1 ( 1, 3 / 4 )

(2, 0)

( 0, 3 / 2 )

–3

–4

–1–2–3–4

3 + 4 = 6

x y

(4, – 3/2)

2

y = 6 – 34 x ...(i)

(i) Put x = 0 in equation (i)

y =6 – 0

4 =

32

(ii) Put x = 2 in equation (i)

y =6 – 3 2

4´

= 0

(iii) Put x = 1 in equation (i)

y =6 – 3 1

4´

=34

(iv) Put x = 4 in equation (i)

y =6 – 3 4

4´

=– 6

4 = –

32

0 2 1 4

3 3 30 –2 4 2

x

y1

Point, where the given line cuts the x-axis

= (2, 0)

Point, where the given line cuts the y-axis

= ( )30,

2 1

22. Fixed charges for first two hours = ` 50

Let x be the hours ( x > 1) and y be the totalcharges for giving hours.

So, equation

y= 50 + 10( x – 2) ...(i) 1

(i) Charges for first two hours is fixed whichis ` 50

So charge for one hour = ` 50 1

(ii) For three hours put x = 3 in equation (i)

y = 50 + 10(3 – 2)

= 50 + 10 × 1

= ` 60 1

(iii) For six hours, put x = 6 in equation (i)

y = 50 + 10(6 – 2)

= 50 + 10 × 4 = 50 + 40 1

= ` 90

23. Steps of Construction :

(i) Draw a line AB = 11 cm (as XY + YZ + ZX = 11 cm)

(ii) Construct an angle ∠ PAB of 30º at point A and an angle ∠ QBA = 90º at point B.

1

S

A Y

T V

Z

30° 90°B

C

X P

2

(iii)Bisect ∠ PAB and ∠ QBA. These bisec-tors intersect each other at point X .

(iv)Draw perpendicular bisector ST of AX andUV of BX .

(v) ⊥ bisectors ST intersects AB at Y and UV

intersects AB at Z. join XY , and XZ.

(vi) ∆ XYZ is the required triangle. 1

24. Here

O y

x

100°

A

B C

D

E

x

y

∠ BAC = ∠ BDC = yº

(angle in the same segment) ½

∠ BAC + ∠ BEC = 180º

(∴ ABEC is a cyclic quadilateral) ½

SECTION ‘D’



Solutions | 5

y + 100º = 180º

y = 80º

In ∆ ABC, 1

AB = AC (given)

∴ ∠ ACB = ∠ ABC = xº ½

∴ x + x + y = 180º (angle sum property)

2 x + 80º = 180º ½

2 x = 100º

x = 50º 1

25. S

O

R

60°

65° 2 5 °

QP

T

(i) ∠ QRP = 90º (angle in the semi-circle)

∠ QPR = 180º – (∠ QRP + ∠ PQR)(angle sum property)

∠ QPR = 180º – (90º + 65º) = 25º 1

(ii) ∠ QPS = ∠ QPR + ∠ RPS = 25º + 25º = 50º

(∠ QRP + ∠ PRS) + ∠ QPS = 180º

( PQRS is a cyclic quadrilateral)

90º + ∠ PRS + 50º = 180º

∠ PRS = 40º 1

(iii) ∠ PSR + ∠

PQR = 180º

( PQRS is a cyclic quadrilateral)

∠ PSR + 65º = 180º∠ PSR = 115º 1

(iv) ∠ PTQ = 90º (angle in the semi-circle)

∠ PQT + ∠ PTQ + ∠ QPT = 180º

(angle sum property)

∠ PQT + 90º + 60º = 180º

∠ PQT = 30º 1

26. D ·C

4

A B2·5

DB is the transversal because DC || AB

because ∠ CDB = ∠ ABD = 90º

[form a pair of alternate ∠ s]

DC || AB and DC = AB 1

∴ ABCD is a ||gm 1

ar ( ABCD) = B × H 1

= 2·5 × 4

= 10 cm2. 1

27.

S R

A P B

ABCD is a parallelogram. so

AB = CD and AB || CD ...(1)

⇒12

AB =12

CD i.e., AP = CQ and AP||CQ 1

⇒ APCQ is a parallelogram ...(2)

(1) also implies that

12

AB =12

CD i.e., PB = DQ and PB || DQ

⇒ DPBQ is a parallelogram ...(3) 1

(2) ⇒ QS || PR and

(3) ⇒ SP || QR

⇒ PSQR is a parallelogram 2

28. A B

D C

Y

Given : ABCD is a trapezium with AB || BC

and diagonal AC || XY (point X on AB and Y

on BC)

To prove : ar ( ADX ) = ar ( ACY )

Construct : Join CX.

Proof : AB || DC (given) 1

∆ ADX and ∆ ACX are on the same base AX

and between the same parallels AB and DC.

∴ ar ( ADX ) = ar ( ACX ) ...(i)

Also AC || XY (given) 1

∆ ACY and ∆ ACX are on the same base AC

and between the same parallels AC and XY

∴ ar ( ACY ) = ar ( ACX ) ...(ii) 1


ar ( ADX ) = ar ( ACY ) 1




29.

l

r = 7 cm

h

Area of convas which is used to made aconicaltent = 551 – 1 = 550 m2

Again, radius of tent (r) = 7 m ½

Area of conical tent (surface area) = πrl ½

550 =227

× 7 × l

l = 25 m 1

We know that h = 2 2–l r

= −2 225 7

h = 625–49 = 576 = 24 m ½

Volume of conical tent =13

πr2h ½

=13

×227

× 7 × 7 × 24

= 1232 m3 1

30. Edge of cubical box = 10 cm

∴ lateral surface area of cubical box = 4a2

= 4 × 10 × 10 = 400 cm2 1

Dimension of cuboidal box is 12·5 cm long,10 cm wide and 8 cm high.

∴ lateral surface area of cuboidal box

= 2(l + b) h

= 2 (12·5 + 10) × 8= 2 × 22·5 × 8

= 360·0 cm2 1

Hence, cubical box has greater lateral sur-

face area.

∴ volume of cubical box = a3 = 103

= 10 × 10 × 10 = 1000 cm3. 1

31. (i) Probability (when 50 p coin will fall out)

=100

100 50 20 10+ + + =

100180

=59

1

(ii) Probability (when ` 50 coin will fell out)

=10

180 =

118

∴ Probability (when ` 50 coin will not fall out)

= 1 –1

18 =

18–118

=1718

1½

(iii) Probability ½

(iv) Economy is required everywhere. 1

ll




SECTION ‘A’

1. Equation x + a = 0 or x = – a will a lineparallel to y - axis and to the left of the y.

axis if and only if when a > 0.

2. 4 y = ax + 5 ...(i)

∵ Point (2, 3) lies on the line (i)

∴ 4 × 3 = a × 2 + 5

12 – 5 = 2a

2a = 7

=72

a

3. Bisector of an angle divides it in two equalparts.

4. Data : 144, 145, 147, 148, 149, 150, 152, 155,160

Here, N = 9 (odd)

∴ Median = ( )+ th12

N term

= ( )+th9 1

2term

= 5th term

= 149

SECTION ‘B’

5. Median QT and RT divide ∆ PQS and ∆ PRS

in two triangls of equal area.

∴ ar (QTS) =12

ar ( PQS) ...(i)

ar ( RTS) =12

ar ( RPS) ...(ii) 1

From (i) + (ii), we get

ar (QTS) + ar ( RTS) =12

[ar ( PQS) + ar ( RPS)]

ar (QTR) =12

ar ( PQR) 1

6. In ∆ ADC, P and Q are mid points of lines DA and DC respectively.

So, PQ || AC

∠ DPQ = ∠ PAC = 30º (corresponding) 1

∠ y = 30º

In ∆ PDQ ⇒ y + 120º + ∠ DQP = 180º

30º + 120º + ∠ DQP = 180º

∠ DQP = 30º

∠ x = 180º – ∠ DQP = 180º – 30º = 150º 1

7. The edge of cube = 10·5 mm =10·5

10

cm

= 1·05 cm ½

Total surface area of cube

= 6a2 = 6 × 1·05 × 1·05 1

= 6·615 cm2 ½

8. First 8 prime numbers are 2, 3, 5, 7, 11, 13,17, 19 ½

Mean ( ) x =Sum of 8 prime numbers

8½

=2 3 5 7 11 13 17 19

8+ + + + + + +

=778

1

9. Difference of weight = 38 – 33 = 5

half of difference =52

= 2·5 ½

∴ Table with class limits

ClassInterval

30·5 – 35·5 35·5 – 40·5 40·5 – 45·5of Weight(in kg)No. of

9 5 14Students

1½

10.

3 m

Height h = 3 mCircumference of base = 2πr = 22 1

Curved surface area of right

Circular cylinder = 2πrh

= 22 × 3 = 66 m2. 1

SECTION ‘C’

11. Equation

3 x – 5 y – 15 = 0

5 y = 3 x – 15

y = 3 – 155

x

y =35

( x – 5) 1



0 5 – 5– 3 0 – 6

A B C

x

y

5 4 3 2– – – ––1

–2

–3

–4

3 – 5

– 1 5

= 0

x

y–5

–6

y

y'

x'

x1–

A(0, –3)

B(5, 0)

C ( – 5,

– 6 )

1½

The graph of the line intersects x - axis at(5, 0) and y - axis at (0,– 3). ½

12. Solution x = 2, y = 3

( p + 1) × 2 – (2 p + 3) × 3 – 1 = 0

2 p + 2 – 6 p – 9 – 1 = 0 2

p = – 2

Equation – x + y – 1 = 0

y = 1 + x 1

13. D R

C

S Q

A P

B

Given : ABCD is a parallelogram and points P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively.

To prove : ar ( PQRS) =12

ar ( ABCD)

Construct : Join P to R.Proof : Here P and R are the mid points of sides AB and CD of parallelogram ABCD.

∴ PR || BC || AD

In parallelogram APRD,

ar ( PSR) =12

ar ( APRD) ...(i)

[∵ ∆ PSR and parallelogram APRD are onsame base PR and between same parallels

PR and AD] 1

Similarly in parallelogram PBCR

ar ( PQR) =12

ar ( PBCR) ...(ii)


as ( PSR) + ar ( PQR)

=1

2

[ar ( APRD) + ar ( PBCR)]

ar ( PQRS) =12

ar ( ABCD) Proved. 1

14. X A

D

60°

C B 1½

Steps of construction.

(i) Draw a line segment BC = 5 cm and atpoint C make an angle = 60º, ∠ XCB =

60º.

(ii) Cut the line segment CD = 15 cm (equal AC – AB = 1.5 cm) on ray CX.

(iii) Join DB and draw the perpendicular bi-

sector PQ of DB.(iv) The perpendicular bisector intersects CX

at point A.

(v) Joint AC.

(vi) ∆ ABC is the required triangle. 1½

15. A B E

1 2

D C

Through C draw CE || AD

∴ AECD is a parallelogram

∠ A + ∠ 2 = 180º ...(i)

AD = BC (given)

AD = CE (by construction)

∴ BC = CE

In ∆ BCE,

BC = CE ½



Solutions | 9

∠ 1 = ∠ 2

Also, ∠ B + ∠ 1 = 180º

∠ B + ∠ 2 = 180º ...(ii) ½


∠ A = ∠ B ½

Again, we get

∠ A + ∠ D = ∠ B + ∠ C = 180º

(Corresponding Angle)∠ C = ∠ D ½

16.

12 cm

15·5 cm

3·5 cm

Total height of toy = 15·5 cm

Radius of semi-sphere

= radius of cone = 3·5 cm

Height of cone (h) = 15·5 – 3·5

= 12 cm

l = 2 2h r+ = 2 212 3·5+

= 144 12·25+ = 156·25 1

= 12.5 cm

Total surface area of toy = πrl + 2πr2 1

= πr (l + 2r) =227 × 3·5 (12·5 + 2 × 3·5)

= 22 × 0·5 × 19·5

= 214·50 cm2. 1

17. Let the length and breadth of a rectangularhall are 13 x and 12 x.

Perimeter of rectangular hall = 2 (l + b)

250 = 2 (13 x + 12 x)

125 = 25 x

x = 5

∴ l = 13 × 5 = 65 m, b = 12 × 5 = 60 m 1

Area of four wall and ceiling

= 2 (l + b)h + l × b ½Cost of painting the four walls and ceiling atthe rate of 5 per m2

= [2 (l + b)h + lb] × 5 ½

27000 = [2 (65 + 60)h + 65 × 60] × 5

5400 = 250h + 3900

250h = 1500

h = 6 m 1

18. The mean age of 10 students of a class = 15years

∴ Total age of 10 students

= 15 × 10 = 150 years A 14 years student leaves the class,

then total age of 9 students

= 150 – 14 = 136 years ½

Let the age of teacher be x, than ½

mean =Sum of their ages

No. of person 1

16 =136

10 x+

136 + x = 160

x = 160 – 136 = 24 years 1

19. (a) Probabiltity (a number between 3 and 6)

=70 80

500+

=150500

=3

101

(b) Probability (an even number)

=50 70 115

500+ +

=235500

=47

1001

(c) Probability (an odd number)

=120 65 80

500+ +

=265500

=53

1001

20. Write the given data in ascending order

2, 3, 3, 4, 4, 5, 6, 6, 7, 7, 7, 7, 7, 8, 9Here n = 15 (odd)

Mean = xn

S

=

+ + + + + + + + + +

+ + + +

2 3 3 4 4 5 6 6 7 7 77 7 8 9

15

=8515

= 5·6 1

Mode = 7 1

Median = ( )12

thn +term = ( )15 1

2

th+

term

= 8th term = 6 1

SECTION ‘D’

21. Equation y – 5 x = 2 ... (i)

11 3 – 1 – 2 25

7 17 – 3 – 8 12 3

i

i

x

y

(i) Put x = 1 in equation (i)

y – 5 × 1 = 2 ⇒ = 7 y

(ii) Put y = 17 in equation (i)

17 – 5 x = 2




– 5 x = – 15 ⇒ = 3 x

(iii) Put y = – 3 in equation (i)

– 3 – 5 x = 2

– 5 x = 5 ⇒ = - 1 x

(iv) Put x = – 2 in equation (i)

y – 5 × (– 2) = 2

= - 8 y

(v) Put x = 2 in equation (i)

y – 5 × 2 = 2

= 12 y

(vi) Put y = 3 in equation (i)

3 – 5 x = 2

– 5 x = – 1 ⇒ =15

x

Put x = 0 in equation (i)

y – 0 = 2 ⇒ = 2 y

Line cuts the y- axis at (0, 2)

(½+½+½+½+½+½+½+½)

Put y = 0 in equation (i)

0 – 5 x = 2 ⇒ x =–25

Line cuts the x-axis at ( )–2 ,05

22. (a) 3 x + 15 = 0

–5 –4 –3 –2 –1 0 1 2 3 4 5 = – x

3 x = – 15

x = – 5 is a point on the number line. ½

(b) 3 x = – 15

y-ax s

x-axis

x = – 5

–6 –5 –4 –3 –2 –1 0 1 2 3 4 5(0,0)

x = – 5

is a line parallel to y-axis in two variables.

23. Steps of Construction :

(i) Draw a line segment BC = 4 cm.

(ii) Draw a ray BX such that ∠ CBX = 75º.

(iii) From ray BX , cut off BM = 10 cm.

(iv) Join MC.

(v) Draw perpendicular x bisector of MC.Intersecting BM at A.

X M

A

B C

75°

2

(vi) Join AC, then ∆ ABC is the required tri-angle.

24.

O D

A

N

C

M B

O is centre of circle.

AB is a chord with mid-point M.

To prove AB < CD 1

Join OM and ON ⊥ CD 1

∆ ONM is right angled triangle 1

∴ OM > ON, (OM is hypotenuse)

Chord CD is nearer to in comparison of AB, 1

⇒ CD > AB or AB < CD 1

25.

O

P

N

M Q

S

O is centre of the circle. Chord PQ and RQ

are equi-distance from the centre O.

Then OM = ON i.e., OM ⊥ PQ, ON ⊥ RQ 1

In ∆ OMQ and ∆ ONQ

OM = ON (given)

OQ = OQ (common) 1

∠ OMQ = ∠ ONQ = 90º

∆ OMQ ≅ ONQ, (by RHS.) 1

∴ ∠ OQM = ∠ OQN

i.e., diameter QS is a bisector of ∠ PQR. 1

26. In ∆ ABC,

3 x + 15 = 0



Solutions | 11

AB = AC

D

B

A

C

E

∴ ∠ C = ∠ B ...(i) (Angle opposite to equal

sides of a triangle are equal) 1

Again, ∠ ADE = ∠ C and ∠ AED = ∠ B

(Exterior angle of cyclic quadrilateral BCED) 1

∠ ADE = ∠ B and ∠ AED = ∠ C

[by equation (i)] 1

∴ BC || DE.

27. A P

T B

l1 3

S

4 2

C R D m

a

l || m (given)

∠ ATR = ∠ DRT (Alternate interior angle)

12

∠ ATR = 12

∠ DRT

∠ 1 = ∠ 2 (∵ TS and RQ arebisectors of interior angle) 1

But these are alternative interior angle

∴ ST || RQ and SR || TQ

∴

RST is a parallelogram 1

Again, ∠ ATR + ∠ BTR = 180º

(Linear pair of angle) 1

12

∠ ATR +12

∠ BTR =12

× 180°

∠ 1 + ∠ 3 = 90º∠ SQT = 90º

∴ QRST is a rectangle. 1

28. Length of wall (l) = 10 m = 1000 cm

Thikness of wall (b) = 36 cm

Height of wall (h) = 9 m = 900 cm 1

No. of bricks (for three fourth of this wall)

=

3 Area of wall4

Area of one brick

´

=34

×1000 36 900

36 15 9´ ´´ ´

1

= 5000 bricks 1

29. Inner radius of hemisphere ( x)

= 1 m = 100 cm

Outer radius of hemisphere ( R)

= 100 + 1 = 101 cm ½

Volume of hemisphere (used to make the

tank) =23

π ( R3 – r3) 1

=23

× 3·14 × (1013 – 1003) ½

= 63430·09 cm3 2

30. (i) Probability (getting a number 8)

=Number of outcomes

Total number of possibilities

=18

(ii) Probability (getting an odd number)

=( )4 1, 3, 5, 7

8 =

48

=12

(iii) Probability (getting a number greater

than 2 which is 3, 4, 5, 6, 7, 8) =68

=34

(iv) Probability (a number less than 9 which

is 8, 7, 6, 5, 4, 3, 2, 1) =88

= 1

(v) Probability (½+½+½+½+1+1)(vi) Equal distribution.

31. (i) Base BC = 120 m

Height AD = 90 m

Area of triangular plot

=12

× base × height

=12

× 120 × 90

= 5400 m2

(ii) In ∆ ABC they draw median AD on base BC and divide it into two equal areas ABD and ACD. Take any point E on AD

and join BE and CE

Two brothers get areas ar (∆ ABE) andar (∆ ACE) and ar (∆ BCE) is donated toschool.

(iii) Any positive value is acceptable. Bothbrothers know importance of educationlove their community.

ll

BD

C

E

A



1. ax + by + c = 0 ...(i)

Put x = – 2 and y = 3 in equation (i)

a × (–2) + b × 3 + c = 0

– 2a + 3b + c = 0

= -2 3c a b

2. 3 x + 2 y = 13 is a linear equation whosesolution is x = 3, y = 2.

3. An angle 40º is not possible to construct with

the help of rular and compass ?

4. The class mark of the class 130 – 150

=130 150

2+

=280

2 = 140


SECTION ‘A’

SECTION ‘B’

5. Arranging the data in increasing order.

40, 50, 65, 70, 75, 75, 95, 100

n = 8 (even)

Median = ( ) ( )+ +th th

obs. 1 obs.2 2

2

n n

1

=+th th4 obs. 5 obs.2

=70 75

2+

=145

2

= 72·5 1

6. Since opposite angles of parallelogram areequal.

∴ 3 x – 2 = 63 – 2 x

3 x + 2 x = 63 + 2

5 x = 65

x =652

= 13º 1

Angles of parallelogram

(3 × 13 – 2)º, (180º – 37º), (63 – 2 × 13)º,

(180º – 37º)

i.e. 37º, 143º, 37º, 143º 1

7.

= S =

4 5 206 10 609 10 90

10 7 7015 8 120

40 360

x f fx

n fx 1

Mean ( ) x = x

n

S =

36040

= 9 1

8. In ∆ AOB,

OA = OB (radii of circle)

∴ ∠ OAB = ∠ OBA = 30º

A

O

B

P

30º

Again, ∠ AOB = 180º – ∠ OAB – ∠ OBA

= 180º – 30º – 30º = 120º 1

Reflex∠ AOB= 360º – 120º = 240º

∠ APB =12

Reflex ∠ AOB

(Angle subtended by an arc at any pointon the remaining part of the circle is half the angle subtended by it at the centre) 1

=12

× 240º = 120º

9. Radius of the sphere =12

× the edge of cube

=72

cm ½

Volume of sphere =43

πr3

=43

×227

×72

×72

×72

½

= 179·66 cm3 ½

10. (i) Probability (it was correct)

=Number of outcomes

Total number of possibilitie

=175300

=7

121

(ii) Probability (it was not correct) = 1 –7

12

=12–7

12 =

512

1



Solutions | 13

11. Equation 2 x + 6 y + 1 = 0 ...(i)


2 × 0 + 6 y + 1 = 0

6 y = – 1

y = –1

6

Solution is ( )10,–6 1


2 x + 0 + 1 = 0

x = –12

Solution is ( )1– ,02 1

Again, put (– 3, 2) in equation (i)

L.H.S. = 2 × (– 3) + 6 × 2 + 1= – 6 +12 + 1

= – 6 + 13 = 7 ≅ R.H.S.

So, (– 3, 2) is not a solution of given equa-tion. 1

12. Equation 2 x + 3 y = 6 ...(i)


2 x + 0 = 6

x = 3

Point is (3, 0) 1


0 + 3 y = 6 y = 2

Point is (0, 2) 1

Hence, the line 2 x + 3 y = 6, cut the x - axisat (3, 0) and y - axis at (0, 2). 1

13. P Q

S RB

Through O, draw AB || PS

Also PA || BS

∴ PABS is a parallelogram

∴ ar ( POS) =12

ar ( PABS) ...(i)

(∆ POS and parallelogram PABS are onsame base and between same parallels) 1

Similarly, ar (QOR) =12

ar (QABR) ...(ii)


ar ( POS) + ar (QOR)

=12

[ar ( PABS) + ar (QABR)]

=12 ar ( PQRS) ½

14.

A

P

B CQ

Steps of construction :

(i) Draw a line BC = 6 cm.

(ii) With B as center and 5·2 cm draw an arc.

(iii) With C as centre and radius 4·8 cm drawan arc to cut the previous arc at A.

(iv) Join AB and AC.

(v) ABC is a required triangle.

(vi) Draw a perpendicular bisector PQS onside BC.

No, it does not passes through A. 1½15. PQ = QR = RS, ∠ PQR = 128º

∠1 + ∠2 =( )180º –128º

2 =

522

= 26º

∠

PTQ = ∠ QPR = 26º 1

∠ PTS = 3 ∠ PTQ = 3 × 26º = 78º 1

∠ ROS = 2 ∠ RTS = 2 × 26º = 52º 1

P T

1 2 8 °

O

R

S

1 43

2

16. External diameter = 16 cm

Radius ( R) =162

= 8 cm

Internal diameter = 12 cm

Radius (r) =122

= 6 cm

SECTION ‘C’

O




Total surface area is to be painted

= 2π R2 + 2πr2 + π R2 – πr2

= 3π R2 + πr2 = π[3 R2 + r2] 1

Cost of painting the vessel all over

= 2 × π[3 R2 + r2] ½

= 2 ×227

[3 × 82 + 62]

= 2 ×227

[192 + 36]

=447

× 228 = ` 1433·14 1

17. Hight of cone (h) = 24 cm

Slant height of cone (l) = 25 cm

Radius (r) = 2 2–l h = 2 225 – 24

= 625 – 576 = 49

= 7 cm 1

Area of metal sheet (to make two hollowcones)

= 2 × πrl = 2 ×227

× 7 × 25 1

= 1100 cm2 1

18. (i) No. of workers have salary below ` 3,000

= 10 + 30 = 40 1

(ii) No. of workers have salary between 3000 and ` 5,000

= 20 + 40 = 60 1

(iii) No. of workers have salary from ` 1000 to ` 5000

= 10 + 30 + 20 + 40 = 100 1

19. (i) Probability (a student was born in the month with 31 days)

=3 2 5 2 4 4

40+ + + + +

=2640

= 0·65 ½

(ii) Probability (a student was born in month

of Feb.) =4

40 =

110

= 0·1 1½

20.

+ +

S = = S = +

2 3 64 2 86 3 18

10 1 105 2 2 10

11 52 2

x f fx

n n

n fx n

Mean ( ) x = xn

S

6 =2 52

11n +

66 = 52 + 2n

2n = 66 – 52

2n = 14

n =142

n = 7

SECTION ‘D’21.

3

2

1

–3 –2 –1 0 1 2 3 4–1

–2

–3

–4

–5

–6

–7

–8

–9

2

Equation y = 9 x – 7 ...(i)

Put x = – 1 in equation (i)

y = – 9 – 7 = – 16


y = 0 – 7 = – 7


y = 9 × 1 – 7 = 9 – 7 = 2

Put x =12

in equation (i)

y = 9 ×12

– 7 =9 – 1 4

2

=–5

2

1–1 0 125

– 16 – 7 2 –2

x

y

Hence, point (– 1, – 16) and (0, – 7) lie on

graph, but point (– 1, –2) and (2, – 9) are notlie on graph. 2

22. Let the length and breadth of a rectangularfield be x and y metre.



Solutions | 15

Then according to question,

3

2

1

–3 –2 –1 0 1 2 3 4–1

–2

–3

–4

4

5

6

5

(20, 80)

(40, 60)

(60, 40)

(80, 20)

(0, 0)

Let 20 m = 1 cm

on -axis and

-axis

x

y

2

Perimeter of rectangular field = 2 ( x + y)100 = 2 ( x + y)

x + y = 100or y = 100 – x ...(i) 1

Put x = 20 in equation (i) y = 100 – 20 = 80




20 40 60 8080 60 40 20

x

y

Note : In this uestion the perimeter is 200 m instead of 100 m.

23.

D

B C

A X

Y

2

Steps of Construction :(i) Draw BC = 6 cm.(ii) Draw∠ CBX = 90º and cut off BD = 10 cm.(iii) Join CD and draw its perpendicular bisec-

tor meeting BD at A.(iv) Join AC, then ABC is the required trian-

gle.2

24. ABCD is a parallelogram∴ AB || DC

ar AR || PC ...(i)

P is the mid point of CD and AP || CR

(given) ...(ii) 1

D C

B A R

Q

From (i) and (ii),

opposite sides of a quadrilateral are parallel.

Hence ARCP is a parallelogram. 1

∴ AP = CR

In ∆ DQC,

P is the mid point of DC and AC || CQ.

∴ A is also the mid point of DQ(mid point theorem)

∴ DA = AQ 1

Again, by mid point theorem.

AP =12

CQ

CR =12

(CR + QR)

2 CR = CR + QR

CR = QR 1

25. A

D 3 4 E

B 1 2

C

We have to prove ∠ ECB + ∠ EDB = 180º

AB = AC ⇒ ∠ 1 = ∠ 2

AD = AE ⇒ ∠ 3 = ∠ 4 1

∠ A + ∠ 1 + ∠ 2 = 180º = ∠ A + ∠ 3 + ∠ 42∠ 1 = 2∠ 3 1

∠ 1 = ∠ 3 = ∠ 2 = ∠ 4 1

⇒ DE || BC (Corresponding angle)

∴ ∠ 1 + ∠ BDE = 180º

(∠ BDE = ∠ CED, as ∠ 3 = ∠ 4) 1

∠ 1 + ∠ CED = 180º,

∴ B, C, E, D are concyclic 1

26. Given : ABCD is a reclangle in which diago-nal AC bisects ∠ A as well as ∠ C.




D C4

3

21

A B

(i) Since AC bisects ∠ A as well as ∠ C in rectangle ABCD, therefore

∠ 1 = ∠ 2 = ∠ 3 = ∠ 4 (each 45º) 1

In ∆ ADC ⇒ ∠ 2 = ∠ 4 1

AD = CD 1

Thus, the rectangle ABCD is a square. 1

(ii) In a square, diagonals bisect the angles

So, BD bisects ∠ B as well as ∠ D. 1

27. Join A to C.

A B

D C

N

M

X

Draw AM ⊥ DC and CN ⊥ AX

∴ AB || DC

∴ AM = CN 1

ar ( ADC) =12

× base × height

= 12 × DC × AM ...(i) 1

Again, ar ( ABC) =12

× AB × CN ...(ii)

Add (i) and (ii), we get

ar ( ADC) + ar ( ABC) =12

[ DC × AM + AB × CN ]

1

ar ( ABCD) =12

[ DC × AM + AB × AM ]

Area of trapezium =12 × AM ( DC + AB)

=12

× height × sum of parallel side1

28. Radius of lead sphere (r) =62

= 3 cm

radius of beaker ( R) =182

= 9 cm

height of rises water level ( H ) = 40 cm 1

No. of lead spheres dropped in the water

=

Volume of raised water level incylindrical beaker

Volume of one lead Sphere 1

=2

34

3

H

r

p

p =

× × ×

× × ×

9 9 40 34 3 3 3

1

= 90. 1

29. Area of three adjacent face of cubiod are lb,bh and hl, where l, b and h are length,breadth and height of cubiod rspectively, then

lb = 15 cm2, bh = 20 cm2, hl = 12 cm2 1

lb × bh × hl = 15 × 20 × 12

(lbh)2 = 3 × 5 × 4 × 5 × 3 × 4 1

Volume of cubiod = lbh = 3 × 4 × 5 = 60 cm2 1

30.

+=

+=

+=

+=

+=

+=

+=

Number Team TeamClass Marks

of balls A B0 6

0 — 6 3 2 52

6 126 — 12 9 1 6

212 18

12 — 18 15 8 22

18 2418 — 24 21 9 10

224 30

24 — 30 27 4 52

30 3630 — 36 33 5 6

236 42

36 — 42 39 6 3

2

1

1 2 3 4 5 6 7 8 9 10 11 12 13 14

11

10

9

8

7

6

5

4

3

2

1

c o r e o

e m

n

e a

Class marks

Team B

T e a m A

Scale X-axis = 1 cm = 3 balls Y-axis = 1 cm = 1 score

2

(i) Statistics ½

(ii) Honesty ½



Solutions | 17

31. A B

O

DC

E

Let the plot be ABCD

Join AC, Draw BE || AC.

ar (∆ ADE) = ar (Quad. ABCD) 1

(i) Health centre can be constructed in triangu-lar plot (∆ AOB) and the farmer can have thetriangular plot ADE. 1

(ii) Helpful, wise, Good decision or any other.1

(iii) Yes, constructing a Health Centre is justi-fied and essential also.

Any positive points can be take into consid-eration as to why it is justified. 1

ll



1. Equation 4 x – 3 y = 12

Put x = 0 in equation

4 × 0 – 3 y = 12

0 – 3 y = 12

y = 12–3

= – 4

Hence, graph of linear equation cut the y-axis at (0, – 4).

2. In a circle the line joining the mid point of a chord of centre is perpendicular to thechord.

∴ ∠ OPA = 90º

3. Equation 2 x + y + 5 = 0

Put x = 1 in equation

2 × 1 + y + 5 = 0 y + 7 = 0

y = – 7

∴ Point = (1, – 7)

Again, put x = – 1 in equation

2 × (– 1) + y + 5 = 0

– 2 + y + 5 = 0

y + 3 = 0 y = – 3

∴ Point = (– 1, – 3)

4. First 8 prime numbers are 2, 3, 5, 7, 11, 13,17, 19

N = 8

Median =( ) ( )+

+th th1term term

2 22

N N

=+th th4 term 5 term

2

=7 11

2+

=182

= 9.


SECTION ‘A’

SECTION ‘B’

5. Since PQRS is a parallelogram

PQ || SR, PS || QR and QS is a transversal

4 y = 20º (Alternate interior angle)

y = 5º 1

10 x = 60º (Alternate interior angle)

x = 6º 16. Volume = 880 cm3, Area = 88 cm2

Volume of a cuboid = 1 × b × h = 880 cm3 ...(i)

Area = 1 × b = 88 cm2...(ii) 1

Substituting (ii) on (i)

88 × h = 880

h = 10 cm 1

7. Let the length of a cube is a then length of

a diagonal of a cube = 3a = 16 3

a = 16 cm 1

8. Observation : 3, 5, 7, 4, 7, 8, 3, 6, 7, 4, 7, 3

Mode of observation = 7 1 After adding,

New observation : 8, 10, 12, 9, 12, 13, 8, 11,12, 9, 12, 8

New mode = 12 1

9. Observation : 2, 3, x, x + 2, 11, 17

N = 6 (even)

Median =( ) ( )+

+th th1term term

2 22

N N

½

9 =( ) ( )+ +

th th6 6term 1 term2 2

2

18 = 3th term + 4th term ½

18 = x + x + 2 ½

2 x = 16

x = 8

∴ Value of 3 x + 1 = 3 × 8 + 1 = 25 ½

10. Let the angles of quadrilateral are 2 x, 3 x, 5 xand 8 x, then

2 x + 3 x + 5 x + 8 x = 360º

18 x = 360º x = 20º

Angles are 2 x, 3 x, 5 x and 8 x

2 × 20º, 3 × 20º, 5 × 20º and 8 × 20º

40º, 60º, 100º and 160º

SECTION ‘C’

11. (a) 2 x – 4 y = 32

Put (8, – 4) in equation

L.H.S. = 2 × 8 – 4 × (– 4) = 16 + 16

= 32 = R.H.S.

Hence, (8, – 4) is a solution of given linearequation. 1



Solutions | 19

(b) 4 x – 2 y = 10

Put (3, – 1) in equation

L.H.S. = 4 × 3 – 2 × (– 1) = 12 + 2

= 14 ≠ R.H.S.

Hence, (3, – 1) is not a solution of givenlinear equation. 1

(c) 2 x = 5

Put x = 0 in equationL.H.S. = 2 × 0 = 0 ≠ R.H.S.

Hence, (0, 5) is not a solution of given equa-tion. 1

12.

–1–2 O 1 2 3 4 5

Y

4

3

2

1

-1

-2

-3

(0,3)B

(3, 0) X´

X-axis

–3 X

A

Y´

XX ́ line represented the line y = 0

YY ́ line represented the line x = 0

Put x = 0 in equqation x + y = 3

0 + y = 3

y = 3 1

Again, put y = 0 in equation x + y = 3 x + 0 = 3

x = 3 1

ABC is the required triangle.

Co-ordinates of triangle ABC are (3, 0), (0, 3)

and (0, 0). 1

13. C

A BF

1½

Steps of construction :

(i) Draw any line seyment AB = 6·6 cm.

(ii) With A as centre and radius 6·6 cm drawan arc.

(iii) With B as centre and radius 6·6 cm draw

an arc to cut the previous arc at C.

(iv) Join AC and AB, then ABC is the re-quired triangle.

(v) Again, draw perpendicular bisector of AB

which cut AB at F .(vi) Join C to F which is median. 1½

14. ar (∆QER) = ar (∆QEP) [QE is median]

ar (∆QR) + ar (∆ ER)

= ar ( PFGE) + ar (∆GFQ) ...(i) 1

Also ar (∆ PER) = ar (∆QFR)

ar ( PFGE) + ar (∆ EGR)

= ar (∆GQR) + ar (∆GFQ)...(ii) 1

From (i) – (ii)

ar (∆GQR) – ar ( PFGE)

= ar ( PFGE) – ar (∆GQR)

ar (GQR) = ar ( PFGE) 1

15. A

B C

X

Y Z

In quadrilateral ABYX

AB = XY and AB || XY ∵ one pair of opposite side of a quadrilateralare equal and parallel.

∴ ABYX is a parallelogram ...(i)

Similarly, ACZX is a parallelogram ...(ii)

From (i) and (ii)

BY || CZ and BY || CZ

∴ BCZY is also a parallelogram. 1

BC = YZ and BC || YZ

In ∆ ABC and ∆ XYZ

AB = XY (given)

AC = XZ (given) BC = YZ (proved above)

∴ ∆ ABC ≅ ∆ XYZ

16. Radius of sphere (r) = 5 cm

Radius of cone ( R) = 4 cm

According to question

Surface area of sphere

= 5 × curved surface area of cone

4πr2 = 5 ×π Rl 1




l =24

5r R

=× ×

×

4 5 54 5

l = 5 cm 1

Height of cone, h = 2 2–l R = 2 25 – 4

= 25–16 = 9 = 3 cm 1

17. Let the radius of this ball = r cm

Volume of spherical ball =43

πr3

The total cost of making a solid spherical

ball = 7 ×43

πr3

33957 = 7 ×43

×227

× r3 1

r3 =×

×

33957 34 22

= 343 9 38´ ´ 1

r3 = ( )37 32´

r =212

= 10·5 cm 1

18. Data : 3, 21, 25, 17, ( x + 3), 19, ( x – 4)

Mean ( ) x =Sum of observations

Total number of observations1

18 =3 21 25 17 3 19 – 4

7 x x+ + + + + + +

126 = 88 – 4 + 2 x

2 x = 126 – 84 = 42

x =422

= 21 1

Data : 3, 21, 25, 17, (21 + 3), 19, (21 – 4) = 3, 21, 25, 17, 24, 19, 17

Mode = 17 1

19. (a) Probability (non occurrence of exactly 2 heads)

=216 270 130

1000+ +

=616

1000 = 0·616 1

(b) Probability (3 heads) =216

1000 = 0·216 1

(c) Probability (no head) =130

1000 = 0·13 1

20. (i) P (weight less than 65 kg)

=5 18 4 16 5

60+ + + +

=4860

=45

(ii) P (weight between 61 and 64)

=4 16

60+

=2060

=13

(iii) P (weight equal to or more than 64)

=5 12

60+

=1760

(1 + 1 + 1)

SECTION ‘D’

21. (a) Fixed charges = ` 1,000

Let the no. of days for which the food has

been availed = x

Let total charges = x

Then according to question

x = 1000 + 50 xy ...(i) 1

(b) Put y = 4 (days) in equation (i)

x = 1000 + 50 × 4

x = 1000 + 200

x = 1200 1

Again, put y = 6 (days) in equation (i)

x = 1000 + 6 × 50

= 1000 + 300

= 1300

Hence, (1200, 4) and (1300, 6) are the twosolutions.

(c) Again, put y = 21 (days)

x = 1000 + 50 × 21

= 1000 + 1050

x = ` 2,050 charges for 21 days. 1

22. x + y = 6

y = 6 – x ...(i)


y = 6 – 1 = 5


y = 6 – 2 = 4


y = 6 – 3 = 3

1 2 35 4 3

x

y 1½

2 x + 3 y = 16

y =1 6 – 2

3 x

...(ii)

Put x = 2 in equation (ii)

y =16 – 2 2

3´

=123

= 4

Put x = 5 in equation (ii)

y =16 – 2 5

3´

=63

= 2



Solutions | 21


y =( )16 – 2 –1

3

´ =

16 23+

= 6

2 5 – 14 2 6

x

y 1½

–1–2 0 1 2 3 4 5

4

3

2

1

–1

–2

–3

–3–4

6

5

x y+ = 6

2 + 3 = 16 x y

1

Intersecting point of two line = (2, 4).

23.

PB C

Q45°

30°

S

A

R

2½

Steps of Construction,

(i) Draw a line PQ = 16 cm.

(ii) At P, construct ∠ SPQ = 45º and at Qconstruct ∠ RQP = 30º.

(iii) Draw the bisectors of ∠ SPQ and ∠ RQP,intersecting at A.

(iv) Draw the right bisector of AP and AQintersecting PQ at B and C respectively.

(v) Join A to B and A to C.

ABC is the required triangle 1½

24.

D

E

B C

A

F

In ∆ BEC, BE || DF and D is the mid point of BC.

∴ F is the mid point of CE

∴ CF =12

CE ...(i) 1

As, BE is the median, E is the mid point of AC

CE = 12 AC ...(ii) 1


CF =12 ( )1

2 AC

CF =14

AC 1

25.

F E

B CD

AD is the median of ∆ ABC.

∴ ar (∆ ABD) = ar (∆ ACD) ...(i)

In ∆ GBC, GD is median

ar (GBD) = ar (GCD) ...(ii)


ar ( ABD) – ar (GBD) = ar ( ACD) – ar (GCD)

ar ( AGB) = ar ( AGC) ...(iii) ½

Similarly, we can prove thatar ( AGB) = ar ( BGC) ...(iv) ½

From (iii) and (iv), we get

ar ( AGB) = ar ( BGC) = ar ( AGC) ...(v) ½

Now, ar (∆ ABD)

= ar ( AGB) + ar ( BGC) + ar ( AGC)

= ar ( AGB) + ar ( AGB) + ar ( AGB) ½

= ar (∆ AGB)

ar (∆ AGB) =13

ar (∆ ABC) ½

Hence, ar ( AGB) = ar ( AGC) = ar ( BGC) =1

3ar ( ABC) ½

26. D2

4

3 1 5 A

EB

In trapezium ABCD,




AB || DC and AD = BC

Through C, draw CE || DA

DC || AE and CE is transverse

∴ ∠ 1 = ∠ 2 (alternate angles)

Also, ∠ 3 = ∠ 1 (Corresponding angles) 2

∴ ∠ 2 + ∠ 3 = ∠ 1

∠ 2 ≠ ∠ 3 = 2 ∠ 1

∴ ∠ A + ∠ C = ∠ 3 + ∠ 2 + ∠ 4

= 2 ∠ 1 + ∠ 4 ...(i)

Also, ∠ 1 = ∠ 5 (∵ EC = BC) 1

∠ A + ∠ C = ∠ 1 + ∠ 4 + ∠ 5 = 180º

Similarly, we can show that

∠ B + ∠ D = 180º.

Hence, the opposite angles of an isoscelestrapezium are supplementary. 1

27. Q P

O

A B 1

Chord AB = PQ (given)

In ∆ AOB and ∆ COD

OA = OP (radii of circle) 1

OB = OQ (radii of circle) 1

AB = PQ (given)

∆ AOB ≅ ∆ COD (by S.S.S.) 1

∴ ∠ APB = ∠ POQ (by C.P.C.T.) 1

28. Volume of water which is transferred into acylindrical vessel = lbh

= 6 m × 4 m × 1 cm

= 600 × 400 × 1 1½

= 240000 cm3

Let the height of water in cylindrical vessel

= h cm

Then volume of water

= volume of cylindrical vessel 1

240000 = πr2h =227

× 20 × 20 × h

h =240000 722 20 20

´´ ´

= 190·9 cm 1½

29. Here l = 30 cm, b = 25 cm, h = 25 cm

Area of glass = Total surface area

= 2 (lb + bh + hl) 1

= 2 [30 × 25 + 25 × 25 + 25 × 30]

= 2 [750 + 625 + 750]

= 2 × 2125

= 4250 cm2. 1

Top needed for all the 12 edges

= The sum of all the edges 1

= 4 (l + b + h) = 4 (30 + 25 + 25)

= 4 × 80

= 320 cm 1

30. (i) For histogram,

Y -axis = one square = 3

X -axis = one square = 50

(ii) For frequency polygon, first we obtain the class marks

C.I. Class Marks Frequency0 — 50 25 12

50 — 100 75 18100 — 150 125 27150 — 200 175 20200 — 250 225 17250 — 300 275 6

1

To obtain the frequency polygon we plot the

points (25, 12), (75, 18), (125, 27), (175, 20),(225, 17) and (275, 6) and join these points byline segment.

27

24

21

18

15

12

9

6

3

0 50 100 150 200 250 300

r e q u e n

c y

C.I.

2

(i) Statistics. ½

(ii) Sincerity. ½

31. Let ABCD be the plot and Naveen decided to

donate some portion to construct as home

for orphan girls from one corner say C of

plot ABCD. Now, Naveen also purchases equal

amount of land in lieu of land CDO, so that

he may have triangular form of plot. BD is

jointed. Draw a line through C parallel to

DB to meet AB produced to P.



Solutions | 23

Construction : Joint DP to intersect BC at O.

A B P

O

D

C

Proof : ∆ BCD and ∆ BPD are on the same

base and between same parallelsCP || DB.

⇒ ar (∆ BCD) = ar (∆ BPD)

⇒ ar (∆ COD) + ar (∆ DBO)

= ar (∆ BOP) + ar (∆ DBO)

⇒ ar (∆ COD) + ar (∆ BOP)

⇒ ar (quad. ABCD)

= ar (quad. ABOD) + ar (∆ COD) 1

= ar (quad. ABOD) + ar (∆ BOP)

[∴ ar (∆ COD) = ar (∆ BOP) proved above)

= ar (∆ APD)

Hence, Naveen purchased the portion ∆ BOP

to meet his requirement. 1

(ii) Area of parallelogram. ½

(iii) We should help the orphans. ½

ll



1. ax = by ...(i)

ay = bx

y =bxa ...(ii)

From equations (i) and (ii)

ax = b ×bxa

a2 x = b2 x

x(a2 – b2) = 0

x = 2 2

0

–a b = 0

From (ii), we get

y =b

a × 0 = 0

Hence, intersecting point is (0, 0).

2. 2 x + 3 y + c = 0

The line passes through origin (0, 0)

∴ 2 × 0 + 3 × 0 + c = 0


SECTION ‘A’

0 + 0 + c = 0

= 0c

3. Data : 11, 12, 14, 16, 18, x + 2, x + 4, 30, 32,35, 41

N = 11 (odd)

Median = ( )12

th N +term

22 = ( )11 12

th+

term

22 = 6th term

22 = x + 2

x = 22 – 2 = 20

4. In ||gm, the sum of adjacent angles is 180º∴ ∠ Q + ∠ R = 180º ...(i)

and opposite angles are equal.

∴ ∠Q = ∠ S

From (i), we get

∠ S + ∠ R = 180º.

SECTION ‘B’

5. Girls = 40, Boys = 100 – 40 = 60

Total marks for boys = 60 × 75 = 4500

Total marks for girls = 40 × 65 = 2600 1

Sum of class = 4500 + 2600 = 7100

Mean marks of the class =7100100

= 71% 1

6.

12 cm

21

Radius of cone = 242 = 12 cm

Total surface area of cone = πrl + πr2 1

= πr(l + r) =227

× 12 (21 + 12)

=227

× 12 × 33 = 1244·57 cm2 1

7. Class interval are 4 – 8, 8 – 12, 12 – 16,

16 – 20, 20 – 24, 24 – 28, 28 – 32

(a) class size = 8 – 4 = 4

(b) lower limit of second class = 8

(c) upper limit of last class = 32

(d) third class = 12 – 16 (½ + ½ + ½ + ½)8. Let the side of cugbe be a.

Then total surface area of cube = 6a2 = 864

a2 = 144

a = 12 1

Volume of cube = a3 = (12)3 = 1728 m3 1

9. D C

A B

Q

P

∆ APB and parallelogram ABCD are on samebase AB and between same parallel lines AB

and DC.

ar ( APB) =12

ar ( ABCD) ...(i) 1

Similarly, ar ( BCQ) =12

ar ( ABCD) ...(ii) ½

From (i) and (ii)

ar ( APB) = ar ( ABCD) ½

S R

P Q



Solutions | 25

10. In a cyclic quadrilateral

2 x + 4º + 4 x – 64 = 180º

6 x – 60 = 180º 1

6 x = 240º

x =240º

6

x = 40º 1

SECTION ‘C’

11. Equation y = mx + c

Put m = 2 and c = 1 y = 2 x + 1 ...(i)


y = 2 × 1 + 1 = 3


y = 2 × 2 + 1 = 5

Put x =32

in equation (i)

y = 2 ×32

+ 1 = 4

Put x =3

2−

in equation (i)

y = 2 ×–32

+ 1 1

= – 2

– 331 22 2

3 5 4 – 2

x

y

3

2

1

–3 –2 –1 0 1 2 3 4

–1

–2

–3

–4

4

5

5

-ax s y

6 x-axis

y x= 2 + 1

1½

Value of y, when x =32

is 4. ½

12. C =( )5 – 160º

9

F ...(i)

(i) if temperature = 104ºF

From (i), C =×5 104 – 160

9 =

520–160

9

C =360º

9 = 40º 1

(ii) Temperature = 35ºC

from (i) 35º =5 – 160

9 F

315 = 5 F – 60

5 F = 375º

F º = 75º 1

Put the same temperature x for both tem-peratures.

From (i) , xº =5 – 160º

9 x

9 x = 5 x – 160º

9 x – 5 x = – 160º

4 x = – 160º

x = – 40º

Yes, it is (– 40º). 1

13.

B C

l

m

90°

Here AB = 3 cm, BC = 5 cm, ∠ A = 90º

In right angled triangle ABC

BC2 = AB2 + AC2

(5)2 = 32 + AC2

AC2 = 25 – 9 = 16

AC = 4 cm 1

∆ ABC and ∆ DBC are on the same base BC

and between same parallels l and m. 1

∴ ar ( DBC) = ar ( ABC)

=12

× base height

=12

× AB × AC

=12

× 3 × 4

= 6 cm2 1




14. D C

12

3

A B

E

Here ∠ 1 = ∠ 2 ( AE is angle bisector)

But ∠ 1 = ∠ 3 (Alternate angle as AD || BC)

∴ ∠ 3 = ∠ 2 1

Hence, BE = AB (sides opposite to equal an-gles)

But BE =12

BC, ( E is the mid point of BC)

∴ AB =12

BC 1

and BC = AD (opposite sides of ||gm)

∴ AB =12

AD 1

15. D

A

C

BE

QP

According to question, E and F re the midpoints of sides AB and CD.

∵ AE =12 AB

and CF =1

2CD

∵ In the parallelogram opposite sides are

equal, so

AB = CD

∴ AE = CF 1

Again, AB || CD

So, AE || FC

Hence, AECF is a parallelogram

In ∆ ABP, E is the mid point of AB

EQ || AP 1

∴ Q is mid point of BP

Similarly, P is the mid point of DQ.

DP = PQ = QB

∴ Line segments AF and EC trisect the di-agonal BD. 1

16. Height of cone (h) = 3·5 m

Radius of base (r) = 12 m

Slant height (l) = 2 2h r+ = ( ) ( )2 23·5 12+

= 2 212·25 144+

= 156·25 = 12·5 cm 1

Area of canvas for making a conical tent

= πrl

=227

× 12 × 12·5

= 471·43 cm2 (app.) 1

17. Volume of sphere =43

πr3

90517

= 43

× 227

× r3

r3 = 63367

××

×

7 34 22

=×288 3

4

r3 = 216 = (6)3

r = 6 cm 1½Diameter of sphere = 2r = 2 × 6 = 12 cm ½

Surface Area = 4πr2 = 4 ×227

× 6 × 6

=3168

7 =

4452

7cm2 1

18. ( )Distance in km Tally Marks Frequency

40—6060—8080—100

100—1201200—140140—160

|

||||

|||| |||

||||||||

158

583

30 N

|||

|||

=

3

19. Total outcomes = 75 + 85 + 90 + 60 + 80 + 110

= 500

P(getting 1) =Number of outcomes

Total number of possibilities

=

75

500 =

3

20

P(getting 2) =85

500 =

17100

P(getting 3) =90

500 =

950

P(getting 4) =60

500 =

325

P(getting 5) =80

500 =

425



Solutions | 27

P(getting 6) =110500

=1150

2

Sum of all possibilities =3

20 +

17100

+9

50

+3

25 +

425

+1150

=15 17 18 12 16 22

100+ + + + +

=100100

= 1 1

20. Total no. of cars = 100

(i) P(exactly 5 occupants) =5

100 =

120

(ii) P (more than 2 occupants)

=23 17 5

100+ +

=45

100=

920

(iii) P (less than 5 occupants)

=29 26 23 17

100+ + +

=95

100=

1920

(1 + 1 + 1)

SECTION ‘D’

21. (a)3 x

+ 2 y = 5

2 y = 5 –3 x

=15 –

3 x

y =15 –

6 x

... (i)

is the expression of y in term of x ½

(b) Put x = 3 in equation (i)

y =1 5 – 3

6 = 2


y =1 5 – 6

6 =

96

= 1·5


y =( )15 – –3

6 =

15 36+

= 3 1

-3 6 3

2 1 3 x y

3

2

1

–3 –2 –1 0 1 2 3 4–1

–2

–3

–4

4

5

6

5

y-axis

x-axis6

–4

x y

/ 3 + 2

= 5

2

(c) Yes, (3, 2) is a solution of the given

equation. ½

22. Let Mahesh donated the money = ` x

All of denoted the money = ` y

According to question,

x + y = 95

y = 95 – x ...(i)


y = 95 – 20 = 75


y = 95 – 40 = 55


y = 95 – 60 = 35 1

20 40 6075 55 35

x

y

30

20

10

–30 –20 –10 0 10 20 30 40

–10

–20

–30

–40

40

50

60

50

-ax s

X -axis60

–40

x y+ = 95

70–50

70

80

90

On Y-axis 10 = 1 cm `

On X-axis 10 = 1 cm `

2




23.30° 30°

90°

S

X

Q

P

U

A Y

T

V

ZB

2

Steps of Construction :

(1) Draw a line segment AB = 18 cm ( XY +

YZ + ZX = 18 cm)

(2) Construct an angle ∠ PAB = 30º at point

A and an angle ∠ QBA = 90º at point B.

(3) Bisect ∠ PAB and ∠ QBA. These bisec-

tors intersect each other at point X .

(4) Draw perpendicular bisector ST of AX

and UV of BX .

(5) Perpendicular biesctor ST intersect AB

at Y and UV intersect AB at Z. Join XY,

XZ, then ∆ XYZ is the required triangle.

2

24. A

D F

BE C

D and E are mid points of AB and BC

respectively.

∴ DE || AC

Similarly, DF || BC and EF || AB

∴ ADEF , BDEF and DFCE are all

parallelogram 1

DE is the diagonal of parallelogram BDFE.

∴ ∆ BDE ≅ ∆ FED 1

and ∆ EFC ≅ ∆ FED

∴ All four triangles are concurrent. 1

25.

B C

A D

55°

45°

O

In ∆ ABC,

∠ BAC + ∠ ABC + ∠ BCA = 180º

(Angle sum property) 2

45º + 55º + ∠ BCA = 180º

∠ BCA = 180º – 100º

= 80º

26.

A E F

B

D

C

a

d

c

b

∠ BCD = 43º and ∠ BAE = 62º

In ∆ ACE,

43º + 62º + d = 180º

d = 180º – 105º

= °75d 1

a + d = 180º (opp. angles of

cyclic quad. are supplementary)

a + 75º = 180º

= °105a 1

In ∆ ABF,

62º + 105º + b = 180º

b = 180º – 167º

= °13b 1

∠ DEF = 180º – 75º = 105º

In ∆ DEF,

105º + 13º + c = 180º

118º + c = 180º

c = 180º – 118º

c = 62º 1

a = 105º, b = 13º, c = 62º, d = 75º



Solutions | 29

27. A P

B Q

C R

Given, BQ || CR

Therefore, ∆ BCQ and ∆ BQR are on thesame base and between the same parallels BQ and CR

So, ar (∆ BCQ) = ar (∆ BQR) ...(i) 1½

Also, AP || BQ (given)

Therefore, ∆ ABQ and ∆ PBQ are on thesame base BQ and between same parallels

BQ and AP∴ ar (∆ ABQ) = ar (∆ PBQ) ...(ii) 1½

Adding (i) and (ii), we get

(∆ BCQ) + ar (∆ ABQ)

= ar (∆ BQR) + ar (∆ PBQ)

ar (∆ AQC) = ar (∆ PBR). 1

28.

A P B

O

C Q D1

Since, perpendicular from the centre of thecircle to a chord bisects the chord.

∴ P and Q are the mid points of AB and CD

AP =12

AB =12

× 6 = 3 cm

CQ =12

CD =12

× 8 = 4 cm

In right triangle OAP

OA2 = OP2 + AP2

52 = OP2 + 32

OP2 = 25 – 9

OP2 = 16

OP = 4 cm

In right ∆ OCQ

OC2 = OQ2 + CQ2

52 = OQ2 + 42

OQ2 = 25 – 16

OQ2 = 9

OQ = 3 cm

∴ PQ = OP + OQ = 4 + 3 = 7 cm ½

29.

8 m

15 m

Heght of conical tent (h) = 15 cm

Radius (r) =162

= 8 cm

Slant height (l) = 2 2h r+

= 2 215 8+ = 225 64+

= 289 = 17 m 1

Area of canvas = surface area of conical tent 1

l × b = πrl

l × 2 = 3·14 × 8 × 17

l =× ×3·14 8 172

= 213·52 2

30. (i) Height of can (h) = 1·4 m = 140 cm

Radius (r) = 0·4 m = 40 cm

Volume of can = πr2h =227

× 40 × 40 × 140

= 2 × 352000 cm3 = 704000 cm3 1

Volume of milk =34

volume of cylindrical can

=34

× 704000 = 528000 cm3 ½

This milk is poured into some small cylindri-cal glasses whose height is 10 cm and radius5 cm.

∴ No. of small glasses

= Volume of milk

olume of one small glass

1

= 52800022 5 5 107

´ ´ ´ =

528000 722 5 5 10

´´ ´ ´

= 672 1½

31.

(i) For Histogram,

Y -axis = one square = one students

X -axis = one square = 10 marks




(ii) For frequency polygon, first we obtainthe class marks. ½

Marks Class Marks No. of students0 — 10 5 0

10 — 20 15 220 — 30 25 5

30 — 40 35 640 — 50 45 450 — 60 55 860 — 70 65 1070 — 80 75 5

1

To obtain the frequency polygon, we plot the

points (5, 0) (15, 2), (25, 5), (35, 6), (45, 4),(55, 8), (65, 10), (75, 5)

10

9

8

7

6

5

4

3

2

1

N o . o f S t u d e n t

10 20 30 40 50 60 70 80

Marks

1½

(i) Statistics. ½

(ii) Sincerity ½

ll

dancing solution

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