SET - B

Any method mathematically correct should be given full credit of marks.

SECTION - AQuestion number 1 to 6 carry 1 marks each.

1. an = a + (n – 1) d 0 = – 18 + (10 – 1) d 0 = – 18 + 9d 18 = 9d

d =189

d = 2 [1]

2. (A) 7cmJustification

In POQ,QPO = 900 ... (tangent at any point on the circle is to radius

through the point of contact) OQ² = OP² + PQ².......(by Pythagoras theorem) 25² = OP² + 24² 625 = OP² + 576 OP² = 625 – 576 OP² = 49 OP = 49 = 7 The radius of circle is 7cm. [1]

3. Let the radius of the bigger circle be r. Area of the bigger circle = r² As per the given condition,

r2 = (8)2 + (6)2

r2 = 82 + 62

r2 = 64 + 36 r2 = 100 r = 10 cm. Radius of the bigger circle = 10 cm. [1]

MT EDUCARE LTD.CBSE X

Date :

SUBJECT : MATHEMATICS

SET B

Marks : 80

Time : 3 hrs.

P

.Q O

24cm

25cm

QUEST - II (Semi Prelim II)MODEL ANSWER PAPER

SET - B... 2 ...

4. (x + 1)² = 2 (x – 3) x² + 2x + 1 = 2x – 6 x² + 7 = 0which is of the form ax² + bx + c = 0 The given equation is a quadratic equation. [1]

5. Curved surface area of hemisphere = 90517

=905 7 1

7

=6335 1

7

=6336

7Curved surface area of hemispherre = 2r2

6336

7 = 2 ×227 × r2

6336 77 2 22

= r2

r2 = 144 r = 12 cm

Volume of hemisphere =23 Tr3

=23 ×

227 × 12 × 12 × 12

=25344

7= 3620.57 cm3 [1]

6. Mean ( x ) = a +fidifi

= 17 +108100

= 17 + 1.08= 18.08 [1]

SECTION - BQuestion number 7 to 12 carry 2 marks each.

7. Let Rohan’s present Age = x years His mothers present age = (x + 26) years

SET - B... 3 ...

After 3 years, Rohan’s age = (x + 3) years

and, his mothers age = (x + 26 + 3) = (x + 29) yearsAccording to given condition,

(x + 3) (x + 29) = 360 x² + 29x + 3x + 87 = 360 x² + 32x – 273 = 0Rohan’s present age satisfies the equation x² + 32x – 273 = 0which is the required representation of problem mathematically. [2]

8. [2]

9. a = 3, d = 8 – 3 = 5, an = 78 an = a + (n – 1) d 78 = 3 + (n – 1) 5 78 – 3 = (n – 1) 5 75 = (n – 1) 5

755 = n – 1

n – 1 = 15 n = 15 + 1 n = 16 [2]

10. To Prove : BACO is a SquareProof :In BACO,

BAC = 90º [Given]OBA = 90 [Radius is perpendicular to

the tangent]OCA = 90º

BOC = 90º [Remaining angle of a Quadrilateral]

(Rough Figure)

M2.9 cmO

l

O M2.9 cm

l

SET - B... 4 ...

BACO is a rectangle But OB = OC [Radii of circle] BACO is a Square

(A rectangle with a pair of adjacent sides equal is a square) [2]

11. Angle between two ribs =3608

= 450

r = 45 cm Area between the consecutive ribs = Area of sector

=θ × r²

360

=45 22× × 45 × 45

360 7

=18

22 ×

4

11

× 45 × 457

Area of sector =22275

28 cm2 [2]

12. Let the height of the platform be h. Volume of Cylinder = Volume of Platform

7 7 202 2

= 22 × 14 × h

227

7

27

2

205 1

22

114

2

= h

h =5 m2 = 2.5 m

Height of the platform is 2.5 m. [2]

SECTION - CQuestion numbers 13 to 22 carry 3 marks each.

13. Let a be the first term and d be the common difference of the given A. P.Then, Sm = Sn

2m

{2a + (m – 1)d } = 2n

{2a + (n – 1)d}

450

SET - B... 5 ...

2a (m – n) + { m (m – 1) – n(n – 1)} d = 0 2a (m – n) + {(m2 – n2) – (m – n)} d = 0 (m – n) {2a (m + n – 1)d } = 0 2a + ( m + n – 1)d = 0 [ m – n 0] ....(i)

Now Sm+n =2

m n {2a + (m + n – 1)d }

Sm+n = 2m n

× 0

Sm+n = 0 [using (i)] [3]

14.

PQ = PR = 7.7 cm [3]

15. 2x2 – 7x + 3 = 0Here, a = 2, b = – 7, c = 3Using the quadratic formula, we get

x = 4

2–b b² – ac

a

10cmM

Q

O P

R

6 cm

SET - B... 6 ...

x =2– (–7) ± (–7) – 4 × 2 × 3

2 × 2 = 7 49 – 24

4 = 7 25

4 =

7 54

x =7 5

4

or x =7 5

4

x =124

or x =24

x = 3 or x =12

The roots of the given quadratic equation are 3 and12 . [3]

16. (i) r = 5 cm, = 900

A (A –MXN) =θ × r²

360

=90 × 3.14 × 5 × 5360

=78.5

4= 19.625 m2

Area of the grazed part = 19.625 m2

(ii) If r = 10 cm

Area of sector =θ × r²

360

=90360 × 3.14 × 10 × 10

=3144

= 78.5 m2

Increase in the grazing area = 78.5 – 19.625Increase in the grazing area = 58.875 m2 [3]

OR

16. A(O - AXB) =θ

360 × (21)2

A(O - CYD) =θ

360 × (7)2

Area of shaded portion = A(O - AXB) A(O - CYD)

D C

BA

5

15 m

XM

N

SET - B... 7 ...

=θ

360 × (21)2 –360

θ × (7)2

=2 230 × (21 7 )

360

=1

12 ×227

× (21 + 7) (21 – 7)

=112

22 ×7

3

× 28

1

41

× 14 =22 ×14

3

Area of shaded portion =3083 cm2 [3]

17. Given : A parallelogram ABCD iscircumscribing a circle.Prove that : ABCD is a rhombus.Proof :

AP = AS ... (I) Length of the tangentsBP = BQ ... (II) drawn from an externalCR = CQ ... (III) point to the circle areDR = DS ... (IV) equalAP + BP + CR + DR = AS + BQ + CQ + DS ...adding (I), (II), (III) and (IV)

(AP + BP) + (CR + DR)= (AS + DS) + (BQ + CQ) AB + CD = AD + BC ...(V)ABCD is a parallelogram ...(given)

AB = CD ...(VI)...(opposite sides of a gm are equal)

AD = BC ...(VII) 2AB = 2BC ...[from (V), (VI) and (VII)] AB = BC

ABCD is a rhombus....(A parallelogram is a rhombus, if its adjacent sides are equal) [3]

18. r1 = 10 cmr2 = 4 cml = 15 cmOne part of the cap is open, so cloth is not needed there. Area of material = l (r1 + r2) + r1

2 + r22

P

Q

R

S

D C

BA

SET - B... 8 ...

= 22 22×15 10 + 4 + 0 + × 4²7 7

( The cap is open form

the base)

=227

× 15 × 14 +227

× 16

= 660 +3527

=4972

7

Area of material used for making it = 22710 cm7 [3]

OR

18. Let the radius of the upper part be r1 & of lower part be r2. r1 = 18

r1 =

182

r1 =9

cm

and r2 = 6

r2 =

62

r2 =3

cm

Curved surface area of the frustum = (r1 + r2) l

= 9 3+

× 4

Curved surface area of the frustum = 48 cm² [3]

19. Class Frequency c.ƒ0 - 10 5 5

10 - 20 x 5 + x c.f20 - 30 20 f 25 + x30 - 40 15 40 + x40 - 50 y 40 + x + y50 - 60 5 45 + x + y

60

4 cm

10 cm

15 cm

SET - B... 9 ...

Here, 45 + x + y = 60 x + y = 15 .......(1)

Since the median is 28.5 which lies in the class 20 - 30 Median class is 20 - 30.n = 60, l = 20, h = 10, f = 20, c.f. = (5 + x)

Median =– . .

2+ ×

n c fl h

f

28.5 =30 – (5 + )20 + × 10

20x

28.5 – 20 =(30 – 5 – )

2x

2 (8.5) = 25 – x 17 = 25 – x x = 8 units.Substituting. x = 8 in eqn (1) 8 + y = 15 y = 7

Hence, x = 8 and y = 7. [3]

20. Let Shefali’s marks in Mathematics be = x Her marks obtained in English = (30 – x)Had she got two marks more in Mathematics = x + 2

Had she got three marks less in English = 30 – x – 3= (27 – x)

According to given condition,Product of their marks = 210i.e. (x + 2) (27 – x) = 210 27x – x2 + 54 – 2x = 210 – x2 + 25x – 156 = 0 x2 – 25x + 156 = 0Here a = 1, b = – 25, c = 156Using the quadratic formula, we get.

x =– ± – 4

2b b² ac

a

x =2– (– 25) ± (– 25) – 4×1×156

2 × 1

x =25 ± 625 – 624

2 × 1

SET - B... 10 ...

x =25 ± 1

2

x =25 + 1

2 or x =25 – 1

2

x =262 or x =

242

x = 13 or x = 12 When x = 13, 30 – x = 30 – 13 = 17If she got 13 marks in Maths, then she scored 17 marks in English. When x = 12, 30 – x = 30 – 12 = 18If she got 12 marks in Maths, then she scored 18 marks in English. [3]

21. QPR = 900 ... (diameter always subtends a right angle.)

A (PQR) =12 × PR × QP

=12 × 7 × 24

= 84 cm2

In PQR,QP2 + PR2 = QR2 ... (by Pythagoras theorem)

(24)2 + (7)2 = QR2

QR2 = 576 + 49 QR2 = 625 QR = 25 cm

radius =252 cm

Area of semicircle = 21 r2

= 12

22×

1125 25× ×

7 2 2

=687528 cm2

Area of the shaded portion = Area of semicircle – A (PQR)

=687528 – 84

=6875 – 2352

28

Area of the shaded portion =452328 cm2 [3]

Q

O

RP

SET - B... 11 ...

22. Given : Two concentric circles with centre O and radii5cm and 3cm. Chord AB of larger circle touches thesmaller circle at P.To Find : ABConstruction : Join OP and OB.Solution :OP = 3cm ....(radius of smaller circle)OB = 5cm ....(radius of larger circle)In OPB,OPB = 900 ....(I) ....(tangent at any point on the circle is to

radius of the circle through the point of contact)

OB2 = OP2 + PB2 ....(by Pythagoras theorem) 52 = 32 + PB2

25 = 9 + PB2

PB2 = 25 – 9 PB2 = 16 PB = 16 PB = 4cm

OP AB ....[from (I)]

PB =12 AB ....( from centre of circle to the chord bisects

the chord)

4 =12 AB

AB = 8cm [3]

OR

22. Given : A ABCD is drawn to circumscribe a circle.Prove that : AB + CD = AD + BCProof :

AP = AS ...(I)BP = BQ ...(II) Length of the tangents drawn from an externalCR = CQ ...(III) point to the circle are equalDR = DS ...(IV)

AP + BP + CR + DR = AS + BQ + CQ + DS ...adding (I), (II), (III) & (IV)(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC [3]

O

A P B3

5

SET - B... 12 ...

SECTION - DQuestion numbers 23 to 30 carry 4 marks each.

23. For the cylindrical bucket,h = 32 cmr = 18 cmVolume of the cylindrical bucket = r2h

= × 182 × 32 cm³For the cone, h = 24 cm

Volume of the cone =13 r

2h

=13 r2 × 24 cm³

But, Volume of the cylinder = Volume of the cone

× 18 × 18× 32 =13 × × r2 × 24

183

×18 × 328

×3

2441

= r2

r² = 9 × 18 × 8 r2 = (3 ×3) × (3 × 3 × 2) × (2 × 4) r = 3 × 3 × 2 × 2 r = 36 cm

Radius of the heap = 36 cm

Now l² = r2 + h2

l² = 362 + 242

l² = 1296 + 576 l² = 1872 l = 12 13 cm

Slant height of the heap = 12 13 cm [4]

24. Given : (i) A circle with centre O.(ii) The circle touches the sides

AB, BC, CD and AD of ABCDat points P, Q, R and S respectively.

To Prove : (i) mAOB + mCOD = 1800

(ii) mAOD + mBOC = 1800

A

S

P

B Q C

O

D

R

SET - B... 13 ...

Construction : Draw seg OP, seg OQ, seg OR and seg OS.Proof : In APO and ASO,

mAPO = mASO = 900 [Radius is perpendicular to the tangent]

hypotenuse OA hypotenuse OA [Common side]seg OP seg OS [Radii of the same circle]

APO ASO [By RHS rule] AOP AOS [c.p.c.t.]Let, mAOP = mAOS = a0 ........(i)Similarly,

mBOP = mBOQ = b0 ........(ii)mCOQ = mCOR = c0 .......(iii)mDOR = mDOS = d0 .......(iv)

mAOB + mBOC + mCOD + mAOD= 3600

[ Sum of the measures of all angles at a point is 3600] a + b + b + c + c + d + a + d = 360 2a + 2b + 2c + 2d = 360 2(a + b + c + d) = 360 (a + b) + (c + d) = 180 mAOB + mCOD = 180

(a + d) + (b + c) = 180

mAOD + mBOC = 180 [4]

25. Now, Sn = 4n – n² S1 = 4 (1) – (1)2

= 4 – 1

S1 = 3 a = 3 S2 = 4 (2) – (2)2

= 8 – 4

S2 = 4Now a2 = S2 – S1

= 4 – 3

a2 = 1Now d = a2– a1

= 1 – 3= – 2

a3 = a2 + d= 1 + (– 2)

SET - B... 14 ...

a3 = – 1Now a10 = 3 + (10 – 1) (–2)

= 3 + 9 × (–2)= 3 – 18

a10 = – 15 an = a + (n – 1) d

= 3 + (n – 1) (–2)= 3 + (–2n) + 2

an = 5 – 2n [4]

26. Let the time taken by the tap of smaller diameter be x hrs The time taken by the tap of larger diameter = (x – 10) hrs.

In 1 hour the smaller tap will fill1x

part of the tank, and the larger tap

will fill110x

part of the tank.

In 1 hour the two taps together will fill1 1+

10x x –

part of the tank

The tank is filled by the two taps in398 hours =

758 hrs.

In 1 hour the two taps together will fill =8

75 of the total volume

1x +

1– 10x =

875

– 10 +

– 10x xx x =

875

2

(2 – 10 )– 10

xx x =

875

75 (2x –10) = 8 (x² – 10x) 150x – 750 = 8x2 – 80x – 8x² + 150x + 80x – 750 = 0 8x2 – 230x + 750 = 0Dividing throughout by 2 4x2 – 115x + 375 = 0 4x2 – 100x – 15x + 375 = 0 4x (x – 25) – 15 (x – 25) = 0

SET - B... 15...

(x – 25) (4x – 15) = 0 x – 25 = 0 or 4x – 15 = 0

x = 25 or x =154

If x =154

then x – 10 =154

– 10

=15 – 40

4

=– 254

< 0

x –10 =– 254

which is not possible as time cannot be negative

x = 25 and (x – 10) = 25 – 10 = 15

Time taken by larger tap = 15 hrs.and time taken by smaller tap = 25 hrs. [4]

OR

26. 3

7 15 3

xx – 4

5 37 1xx = 11

21 35 3

xx –

20 127 1

xx = 11

21 3 7 1 20 12 5 3

5 3 7 1

x x x xx x

= 11

2 2

2

147 21 27 3 100 60 60 36

35 5 21 3

x x x x x xx x x

= 11

2 2

2

147 42 3 100 120 3635 16 3

x x x xx x

= 11

2

2

47 162 3335 16 3

x xx x

= 11

47x2 + 162x – 33 = 385x2 – 176x – 33.

385x2 – 47x2 – 176x – 162x – 33 + 33 = 0

SET - B... 16 ...

338x2 – 338x = 0

338x (x – 1) = 0

338x = 0 or x – 1 = 0

x = 0 or x = 1

Hence, x = 0 or 1. [4]

27. Volume of water = Volume of cone =13 r

2h

=13 × × 5 × 5 × 8

=2003 cm3

Now,14

of the volume of water =14

×2003

=503 cm3

Lead is in the form of sphere, r = 0.5 cm

Volume of one lead shot (sphere) = 343

r

=34

× × (0.5)3

=0.53 cm3

Number of lead shots =Volume of water

Volume of one shot

=

503

0.53

=5010

330×

1

10

5

Number of lead shots = 100 [4]

OR

5 cm

8 cm

SET - B... 17 ...

27. For well, diameter = 3m r = m32

For embankment, diameter = 3 + 4 + 4= 11 m

R =112

m

Volume of the earth taken out= r2h

=222 3 14

7 2

....(i)

Also volume of earth taken = R2h – r2h= (R2 – r2 ) × h

=2 222 11 3

7 2 2h

.....(ii)

But2 222 11 3

7 2 2h

=222 3 14

7 2

...[From (i) and (ii)]

h =

227

3 3 142 2

227

2 211 32 2

=

3 3 72121 9

4 4

=63 42 112

=89

h = 1.125 m Height of the embankment is 1.125 m [4]

1 4 m

4m

3 m3m4m 4m

11m

SET - B... 18 ...

28.

AnalysisIn ABC,B = 450, A = 1050

C = 1800 – (A + B)= 1800 – (1050 + 450)= 1800 – 1500

C = 300

Justification : AC || AC .... (by construction)

ABC ABC .... (AA similarity)

A BAB

=BCBC =

A CAC

.... (corresponding sides of similar triangles)

ButBCBC =

4

3

BBBB =

43

BCBC =

43

A BAB

=BCBC =

A CAC

=43 [4]

29. Radius of the circle =802

= 40 cm.Distance travelled in one revolution = Circumference

= 2r

Q

C

B1

B2

B3

B4

7 cmBC45

0 300

A

A

1050

SET - B... 19 ...

= 2 × × 40= 80 cm

Distance travelled in 1hr. = 66 km

Distance travelled in 10 minutes =6660 × 10

= 11 km.= 11 × 1000 × 100cm

... (1km = 100000 cm)

Total no. revolution =Total distance covered

Distance covered in one revolution

=11 ×100000

2280 ×7

=11 × 1000 × 100 × 7

1760

=77 × 1000 × 100

1760 Total no. revolutions = 4375 [4]

OR

29. A (ABD) =12 × AD × AB

=12 × 8 × 8

= 32 cm2

Area of sector A - BXD =θ

360 × r2

=90360 ×

227

× 8 × 8

=3527

cm2

Area of minor segment BXD = A (A - BXD) A (ABD)

=3527

– 32

=352 – 224

7

SET - B... 20 ...

=128

7 cm2

Similarly,Area of minor segment BYD =128

7 cm2

Area of the shaded portion = 2 ×128

7

Area of the shaded portion =256

7 cm2 [4]

30. Production Number of Production yield Cumulative Points to beyield(in kg/ha.) farms Frequency plotted

50 - 55 2 50 or more than 50 100 (50, 100)

55 - 60 8 55 or more than 55 98 (55, 98)

60 - 65 12 60 or more than 60 90 (60, 90)

65 - 70 24 65 or more than 65 78 (65, 78)

70 - 75 38 70 or more than 70 54 (70, 54)

75 - 80 16 75 or more than 75 16 (75, 16)

SET - B

10

20

30

40

50

60

X50 55 60 65 70 75

Y

No.

of f

arm

s

SCALE : X-axis, 1cm = 5 kg/haY-axis, 1cm = 10 farms

Y

70

80

90

100

110

80

• •

•

•

•

•

Production yield (in kg/ha)

X 0

(50, 100)

(55, 98)

(60, 90)

(65, 78)

(70, 54)

(75, 16)

•••••••

•

•

•

•

•

•

•

•

•

•

•

... 21 ...