mathematics sample paper cbse 2015 x

7/26/2019 Mathematics Sample Paper CBSE 2015 X

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Section A

1. If the area of a circle is 301.84 cm2 , then find the circumference of this circle.

Sol. Given, area of the circle =301.84 cm 2

r2 =301.84

r2 = 301.84 7

22 r2 =96.04 r= 9.8 cm

Circumference= = 2 2 22

7r 9.8 =61.6 cm

2. A letter is chosen at random from the word PROBABILITY. Find the probability thatit is a vowel.

Sol. In the word PROBABILITY, there are 11 letters out of which, 4 are vowels (O, A, I, I).

Total possible outcomes, n S( )=11Number of favourable outcomes, n E( )=4

P(getting a vowel)=n E

n S

( )

( )=

4

11

STAGE

Mathematics

A Highly Simulated Practice Question Paper forCBSE Class XTerm II Examination (SA II)

SampleQuestionPaper 1Fully Solved (Question-Solution)

Time : 3 Hours Max. Marks: 90

General Instructions

1. All questions are compulsory.

2. The question paper consists of 31 questions divided into four 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 of10 questions of 3 markseach and Section D comprises of 11 questions of 4 markseach.

3. 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.

4. Use of calculator is not permitted.

1

1


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3. If the length of the shadow of a pole is

3 times the height of the pole, then

calculate the angle of elevation of the Sun.Sol. LetAB= hbe the height of the pole.

Then, length of the shadow= =BC h3

Let denotes the angle of elevation of the Sun.Now, inABC,

tan = = =AB

BC

h

h3

1

3= tan30

= 30

4. What is the mid-point of the line segmentjoining the pointsA(2, 8) andB ( 6, 4)?

Sol. Let Cbe the mid-point ofAB.

Then, coordinates of C

=

=

( ),( )

( , )2 6

2

8 4

242

Qmid- point ofAB x x y y

= + +

1 2 1 2

2 2,

Section B

5. In an AP, if d= 2, n= 5andTn= 0, thenfind the value ofa.

Sol. Given, d n= =2 5, and Tn =0We know that,T a n d n = + ( )1Q Tn =0 a n d+ =( )1 0 a+ =( )( )5 1 2 0

a =8 0 a=8

6. If 1/2 is a root of the equation

x px2 5

40+ = , then find the other root of

the quadratic equation.

Sol. We knowthat, ifx1is a root of thequadraticequation

x ax b 2 0+ + = , then it will satisfy the givenequation.

On puttingx=1

2

in given equation x px2 5

4

0+ = ,

we get

1

2

1

2

5

40

2

+ =p

1

4 2

5

40

1 2 5

40+ =

+ =

p p

1 2 5 0+ = p 2 4p= p=2

Now, putting the value ofpin given equation,

we get

x x2 2 54

0+ =

4 8 5 02x x+ =

4 10 2 5 02x x x+ =

2 2 5 1 2 5 0x x x( ) ( )+ + = ( ) ( )2 5 2 1 0x x+ = 2 5 0x+ = or 2 1 0x =

x= 1

2

5

2,

Hence, the other root of the given equation is52

.

7. The rootsand of the quadratic equationx x k

2 5 3 1 0 + =( ) , are such that = 11. Find the value ofk.

Sol. Given, and aretheroots of thequadratic equationx x k2 5 3 1 0 + =( ) .

Then, + =

=

5

15 Q + =

b

a...(i)

and = = 3 113 3( )k k Q =

ca

...(ii)

Also given, =11 ...(iii)

On solving Eqs. (i) and (iii), we get

2 16 8 = = (iv)

On putting =8 in Eq. (i), we get8 5+ =

= = 5 8 3 ...(v)

Now, putting the values ofandin Eq. (ii),

we get( )( )8 3 3 3 = k = 24 3 3k

3 24 3k= +

14 www.arihantbooks.com

1

1

h

3h

A

B C

1/2

1/2

1

1

1

1/2

1/2


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3 21k= k= 7

8. In the given figure, PQL and PRM aretangents to the circle with centre O atthe points Q and R, respectively. If S is apoint on the circle, such that = SQL 50and = SRM 60 , then find the value ofQSR.

Sol. In the given figure,Ois the centre of the circle.

Therefore, = = OQL ORM 90[Qradius is perpendicular to the tangent at the

point of contact]

Now, = OSQ OQS

[Qangles corresponding to equal sides are equal,here OS OQ = =radii of circle]

= 90 50

= 40Similarly, = = = RSO SRO 90 60 30

= + QSR OSQ RSO = + 40 30= 70

9. Three unbiased coins are tossed together.Find the probability of getting atleast twoheads.

Or

Two dice are thrown simultaneously. Findthe probability of getting a doublet of evennumber.

Sol. For atleast two heads, we will also consider threeheads i.e. for favourable outcomes, the eventscontaining two heads and three heads will beconsidered. Here, three unbiased coins are tossed.

Sample space, S= {HHH,HHT, HTH, THH, HTT,

THT, TTH, TTT}

Total possible outcomes, n S( )=8Outcomes of getting atleast two heads

={HHH, HHT, HTH, THH}

Number of favourable outcomes, n E( )=4

P(getting atleast two heads)= = =n E

n S

( )

( )

4

8

1

2

Or

Here, two dice are thrown simultaneously.

Then, sample space,

S= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Total possible outcomes when two dice are thrown

= =6 6 36

Let A be the event of getting a doublet of even

number. Doublets of even number are (2, 2), (4, 4) and

(6, 6).

Number of favourable outcomes=3

Hence,P(doubletsof evennumber) = =3

36

1

12

10. Find the value ofk, for which 2k + 7 , 6k 2and 8k + 4 are 3 consecutive terms of an

AP.

Sol. We know that, if three terms a, b and c are

consecutive terms of an AP, then 2b=a+c.Thus, 2k + 7, 6k 2 and 8k + 4 will be threeconsecutive terms of an AP, if

2 6 2 2 7 8 4( ) ( ) ( )k k k= + + + 12 4 10 11k k = + 12 10 11 4k k = + 2 15k=

k=15

2

Sample Question Papers MathematicsClassXth (Term II) 15

OS

60

50

L

M

Q

R

P

1

1

1

1

1

1/2

1/2

1

1

1


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Section C

11. Two concentric circles are of radii 7cm and5 cm. Find the length of the chord of thelarger circle which touches the smallercircle.

Sol. Let Obe thecommon centre of twoconcentric circleswhose radii are OB = 7 cm and OP = 5 cm,respectively.

LetABbe a chord of the largercircle touching the smaller circle atP. JoinOP. Since,OPis the radius ofthe smaller circle andABis a

tangent to the smaller circle at apointP.

OPAB

In right angledOPB,we haveOB OP PB 2 2 2= +

PB OB OP 2 2 2= = ( ) ( )7 52 2 = 49 25

PB= 24 PB=2 6cmWe know that, the perpendicular drawn from thecentre of a circle to any chord of the circle bisectsthe chord.

AP=PBLength of the chord of the larger circle,

AB=AP+PB=2PB [QAP BP = ]

= =2 2 6 4 6( ) cm

12. Find the length of the median drawnthrough A on BC of a ABC, whosevertices areA(7, 3), B(5, 3) andC(3, 1).

Or

Determine the points (1, 5), (2, 3) and

( , ) 2 11 are collinear.Sol. The median from a vertex of a triangle bisects theside oppositeto that vertex. So,let Dbe themid-pointof the sideBC. Then,

Coordinates ofD= +

5 3

2

3 1

2

,

Qmid-point= + +

x x y y 1 2 1 2

2 2,

=( , )4 1

Therefore, length of medianADis given by

AD x x y y = + ( ) ( )1 22

1 22

= + ( ) ( )7 4 3 12 2 = + 3 42 2( )

= + =9 16 5 units

Or

Three pointsA,BandCare said to be collinear, ifAB BC AC + = . LetA(1, 5),B(2, 3) andC(2,11)be the given points. Then,

AB= + ( ) ( )2 1 3 52 2

[Qdistance= + ( ) ( )x x y y 2 12

2 12 ]

= + 1 22 2( ) = + =1 4 5 units

BC= + ( ) ( )2 2 11 32 2

= + ( ) ( )4 142 2 = + =16 196 212

=2 53units

andAC= + ( ) ( )2 1 11 52 2 = + ( ) ( )3 162 2

= +9 256

= 265units

AC AB BC +Hence,A,BandCare non-collinear.

13. In a circle of radius 10 cm, an arc subtendsan angle of 90 at the centre. Find the areaof major sector.

Sol. The area of major sector is

obtained by subtracting the area

of minor sector from area of

circle. It also can find by the

formula

r2

360, where is theangle subtended by the major

arc of the centre.

Area of minor sectorOAPBO=

r2

360

=

22

7

10 10 90

360

=550

7cm 2

Area of major sector=Area of circle

Area of minor sectorOAPBO

= r2 550

7=

22

710 10

550

7

1

1

A(7, 3)

DB

(5, 3)C

(3, 1)

1

1

1

1

1

1

1

1

O

A

P

B

A B

P

O

1


5/12

= 2200

7

550

7=

2200 550

7=

1650

7cm 2

Alternate Method

Angle subtended by the major arc at the centre

= 360 Angle subtended by the minor arc= 360 90 = 270

Area of major sector=

r2

360=

22

710 10

270

360

=1650

7cm 2

14.A paper is in the form of a rectangleABCD, in which AB = 16 cm andBC= 12 cm. A semi-circular portion withBCas diameter is cut off. Find the area ofremaining part.

Sol. The area of the remaining part is obtained bysubtracting the area of semi-circle from the area ofrectangle.

Given,Length of the rectangle,AB=16 cmBreadth of the rectangle,BC=12 cm Area of the rectangle= LengthBreadth

= =( )16 12 192 cm 2

Radius of the semi-circle=6cmArea of the remaining part

= Area of rectangleABCD Area of semi-circle with radius 6 cm

= 192 1

2

2r = 192 1

2

22

762

= 192 22 187

= 192 56.57=135.43 cm 2

15. Find the value of k for which the givenequation 2 10 02x x k + = has real andequal roots.

Or

Solve forx,

2 2 3

325

3

2 35

x

x

x

x

+

+

= .

Sol. If the quadratic equation has real and equal roots,then discriminant,D=0.The given quadratic equation is 2 10 02x x k + =Here, a b= = 2 10, and c k=

D b ac k = = 2 24 10 4 2( )

= 100 8k

The equation will have real and equal roots, ifD=0 100 8 0 =k

k= =100

8

25

2

Or

Let 2 3

3

x

xy

+

= (i)

Then, x

x y

+

=3

2 3

1

Therefore, the given equation reduces to

2 25

1

5 =y y

2 25 52y y =

2 5 25 02y y =

2 10 5 25 02y y y + =

2 5 5 5 0y y y( ) ( ) + = ( )( )y y + =5 2 5 0

y=5 andy=5

2

Now, putting y=5 in Eq. (i), we get2 3

3

5

1

x

x

+ = 5 15 2 3x x = +

3 18x= x=6

Again, putting y= 5

2in Eq. (i), we get

2 3

3

5

2

x

x

+

=

+ = +5 15 4 6x x

9 9x= x=1Hence, the values ofxare 1 and 6.

16.What is the probability of having53 Sunday in a non-leap year?

Sol.In a non-leapyear, there are 365 daysor52 weeksand1 day.

This extra one day can be Mon, Tue, Wed, Thu, Fri,Sat or Sun.

Total possible outcomes,n S( )=7Number of favourable outcomes,

n E( )=1 [only Sunday]

P(having 53 Sunday)= =n E

n S

( )

( )

1

7

17. Find the value of p for which the points

A B p( , ), ( , ) 1 3 2 and C ( , )5 1 are collinear.Also, find the ratio in whichB divides AC.


1

C

B

D

A

12cm

16 cm

1

1

1

1

1

1/2

1

1/2

1

1

2

1

1

1

1


6/12

Sol. If the given pointsA B p( , ), ( , )1 3 2 and C( , )5 1 arecollinear, then area of triangle formed by thesepoints is 0.

+ + =12

01 2 3 2 3 1 3 1 2[ ( ) ( ) ( )]x y y x y y x y y

x y y x y y x y y 1 2 3 2 3 1 3 1 2 0( ) ( ) ( ) + + = + + + =1 1 2 1 3 5 3 0( ) ( ) ( )p p + =p p1 8 15 5 0 + =6 6 0p p=1Thus, forp=1, the given points are collinear.Now, let B p( , )2 =B( , )2 1 divides the line segmentjoiningthe points A andCinternally in theratiok: 1.

By using section formula,

mx nx

m n

my ny

m n

2 1 2 1++

++

=, (2, 1)

Here, m k n= =and 1

5 1

1

3

12 1

k

k

k

kB

+

++

=, ( , )

5 1

12

k

k

+

=

5 1 2 2k k = + 5 2 2 1k k = + 3 3k= k=1Hence, the ratio is 1 : 1. So,Bis the mid-point ofAC.

18. If A B a C b( , ), ( , ), ( , ) 2 1 0 4 and D( , )1 2 arethe vertices of a parallelogram, then findthe values ofa andb.

Or

The line segment joining the pointsA( , )3 3

and B( , )6 6 is trisected at the points P andQ such thatP is near to A. IfP also lies onthe line given by2 0x y k+ + = , then findthe value ofk.

Sol. We know that, the diagonals of a parallelogrambisect each other. So, the mid-point ofACwill be thesame as that ofBD.

Mid-point ofAC=Mid-point ofBD

+

+

=

+

+

2 4

2

1

2

1

2

0 2

2, ,

b a

1 1

2

1

21, ,

b a

=

+

On comparing both sides, we get

1 1

2=

+aand

b=

1

21

2 1= +a and b =1 2 a=1 and b=3

Or

Let P and Q be the points which bisect the linesegment joining the pointsA ( , )3 3 andB( , ),6 6 suchthatPis near toA.Then,AP PQ QB = =

AP

PB

AP

PQ QB

AP

PQ

AP

AP=

+ = = =

2 2

1

2

Thus,Pdivides the line segment joining the pointsA( , )3 3 andB( , )6 6 in the ratio 1 : 2.Then, the coordinates ofP

=

+

+

+

+

mx nx

m n

my ny

m n

2 1 2 1

,

= +

+ +

+

1 6 2 3

1 2

1 6 2 3

1 2,

( )

[here, m n= =1 2, ]

= + +

=

6 6

3

6 6

34 0, ( , )

Now, if point P(4, 0) lies on the line2 0x y k+ + = ,then it will satisfy it.

On puttingx=4 and y=0 in 2 0x y k+ + = , we get2 4 0 0( )+ + =k

k= 8

19. The angles of a triangle are in AP. If thegreatest angle equals to the sum of theother two, then find the angles.

Sol. Consider the angles of the triangle as three numbersin AP. So, let the three angles of a triangle in AP be

( ),a d a and ( )a d+ .Sum of three angles of a triangle= 180 a d a a d + + + = 180 3 180a=

a= 60So, three angles of a triangle which are in AP, are

60 60 d, and 60 + d.


1

k 1( 1, 3)A C(5, 1)B

(2, 1)

1

1

D C

A B

O

(1, 2) (4, )b

( 2 , 1 ) ( , 0)a1

1

1

1 2:

A(3, 3) B(6, 6)P Q 1

1

1

1


7/12

According to the question,

Greatest angle=Sum of two smaller angles

60 60 60 + = + d d 2 60d= d= 30Hence, the required angles of a triangle are

60 30 60 60 30 + , , i.e. 30 60 90 , , .

20.A circle touches the side BCof a ABCat Pand the extended sides AB and AC at QandR, respectively. Prove that

AQ BC CA AB= + +1

2( ).

Sol. Given Acircle touches the sideBCofABCat Pand extended sides AB and AC at Q and R,respectively.

To prove

AQ=1

2( )BC CA AB + +

Proof Here,BQ BP =

[since, length of tangents drawn from an externalpoint to a circle are equal]

Similarly, CP CR = and AQ AR =

Now, 2AQ AQ AR = + [QAQ AR = ]

2AQ AB BP AC CP = + + +

[ , ]Q

BP BQ CP CR = = 2AQ BP CP AC AB = + + +( ) ( )

2AQ BC CA AB = + +

AQ BC CA AB = + +1

2( )

Hence proved.

Section D

21.At a point on level ground, the angle ofelevation of a vertical tower is found to be

such that its tangent is 5

12.

On walking 192 m towards the tower, the

tangent of the angle of elevation is3

4. Find

the height of the tower.

Or

A ladder rests against a vertical wall at an

inclination to the horizontal. Its foot ispulled away from the wall through adistance p, so that its upper end slides adistance q down the wall and then theladder makes an angle to the horizontal.Show that

p

q=

cos cos

sin sin

.

Sol. LetAB= hm be the heigt of tower and the angle ofelevation of its top at pointCbe .Again, letDbe a point at a distance of 192 m fromC,such that the angle of elevation of the top of thetower atDbe andAD x= m.

Given, tan = 5

12

and tan =3

4

InBAC,

tan = =+

AB

AC

h

x

5

12 192(i)

InDAB,

tan = AB

AD

3

4=

h

x

x h=43

(ii)


1

1

A

B CP

Q R

1

1

C A

B

h m

xm192 mD

1

1

1

1


8/12

On putting the value ofxfrom Eq. (ii) in Eq. (i),

we get

5

12192

4

3

=+

h

h

5 192 4

312+

=

hh

5 576 4 36( )+ =h h 2880 20 36+ =h h 16 2880h= h=180 mHence, the height of the tower is 180 m.

OrLet BC H= m be the height of the wall and AC l= mbe the length of the ladder, which rests against thevertical wall at =BAC . When the ladder pulledawayfrom the wall, its new position will be DE l= mand thenAD p= m,EC q= m and =BDE .

Let AB x=In right angledABC,

sin = =BC

CA

H

l

and cos = =BA

CA

x

l(i)

and in right angledDBE,

sin = = = BEDE

BC CE DE

H ql

(ii)

and cos = = +

= +BD

DE

BA AD

DE

p x

l

Now, RHS=

cos cos

sin sin

=

+

p x

l

x

lH

l

H q

l

= +

p x x

H H q( )

= =pq LHS

Hence proved.

22.A round table cover has six equal designs

as shown in the figure. If the radius of the

cover is 28 cm, then find the cost of

making the design at the rate of

` 0.35 per cm2 .(take, 3 1.73= )

Sol. Here, six equal designs represent six identical

segments. Area of one segment is obtained bysubtracting the area of triangle formed by that chordand radius from the area of the sector formed by thatchord and radius.We know that, central angle of a circle is 360.

Angle of each sector=

= 360

660

[Qsix designs divide the circle in six sectors also]

Now, we determine the area of one segment ofcircle. Here,ACis a chord of the circle. Draw,OD AC , it bisects ACatD.

AD DC =Thus, AOD COD ~= [by SSS congruency] = = COD AOD 30 [Q = AOC 60 ]In right angledODC,

sin 30 = DCOC

1

2 28=

DC

DC=28

2

DC=14 cmandcos 30 =OD

OC

3

2 28=

OD

OD

=

28 3

2 OD= = =14 3 14 1.73 24.22 cm AC CD = = =2 2 14 28 cm


1

1

p x

DB

C

A

H q

q

H

l

lE

1

1

1

1

60O

28cm

D

C

B

A

1

1


9/12

Area ofAOC AC OD = 1

2

=

1

2 28 24.22=339.08 cm2

Now, area of sector OABCO=

r2

360

=

60

22 28

7 360

2( )

= 22 4 28

6

=2464

6

=410.67 cm 2

Area of segmentABCA=Area of sector OABCO Area ofAOC

= 410.67 339.08=71.59 cm 2

Now, area of six segments= =6 71.59 429.54 cm2

Since, the cost of making the design at the rate of

`0.35 per cm 2 .

Total cost= =464.82 0.35 150.34`

23.A cone of maximum size is cut out from acube of edge 14 cm. Find the surface area

of the cone and of the remaining solid leftout after the cone is cut out.

Sol.The cone of maximum size that is cut out from a cube

of edge 14 cm will be of base diameter 14 cm and the

height 14 cm.

Radius of the base of the cone= =14

27 cm

Total surface area of the cone= + rl r2

= + + 22

77 7 14

22

772 2 2

[ ]Ql r h= +2 2

= + + 22

77 49 196

22

749

= +22

77 245 154= +22 7 5 154

= +154 5 154 = +154 5 1( ) cm 2

Now, surface area of cube= = 6 14 6 1962( )

=1176 2cm

So, surface area of the remaining solid left out afterthe cone is cut out

=Surface area of cubeSurface area of cone= +1176 154 5 154( ) = ( )1176 154 154 5

= ( )1022 154 5 cm 2

24. In a violent storm, a tree got bent by thewind. The top of the tree meets the groundat an angle of 30, at a distance of 30 mfrom the root. At what height from thebottom did the tree get bent? What was theoriginal height of tree? [take, 3 173= . ]

Sol. LetABbe the original height of the tree, got bent atpointCsuch that CB CD = .Then, AD=30m, = ADC 30

LetAC a= m andBC b= m

In right angledDAC,

tan30 =AC

AD

1 =3 30a

a= =

=30

3

30 3

3 310 3m

Also,sin30 =AC

CD

1

2

10 3=

b

b=20 3m

AB a b = + = +10 3 20 3

= = =30 3 30 1 73. 51.90 mHence, the tree got bent from the height of

10 3 10= =1.73 17.3mand the original height of the tree is 51.90 m.

25.A canal is 300 cm wide and 120 cm deep.The water in the canal is flowing with aspeed of 20 km/h. How much area will itirrigate in 20 min, if 8 cm of standing

water is desired?

Sol. Here, required area will be obtained by dividing the

volume of water in the canal in 20 min by the desireheight of water in 20 min.

Given, width of canal =300cm=3m


1

1

1

1

1

a

m

C

A

B

D

30

30 m

b

m

1

1

1

1


10/12

Depth of canal=120cm=1.2mSpeed of canals water=20km/h= 20 1000m/h

Volume of water that flows in the canal in one hour=Width of the canalDepth of the canal

Speed of the canals water

= 3 1.2 20 1000 =72000 3m

In 20 min, the volume of the water in the canal

= 72000 20

60=24000 3m

Area irrigated in 20 min, if 8 cm or 0.08 m standing

water is required= =24000

008300000 2

.m

=30 hec [Q1 hec=10000 2m ]

26.A train travels at a certain average speedfor a distance of 63 km and then travels adistance of 72 km at an average speed of6 km/h more than its original speed. If ittakes 3 h to complete the total journey,then find its original average speed.

Or

Atakes 6 days less than B to finish a pieceof work. If bothA and B together can finish

the work in 4 days, then find the timetaken byB to finish the work.

Sol. We know that,

Speed Distance

Time= Time

Distance

Speed=

Let the original average speed of the train bexkm/h.


63 72

63

x x+

+ =

On dividing by 9 both sides, we get7 8

6

1

3x x+

+ =

7 6 8

6

1

3

( )

( )

x x

x x

+ ++

=

21 6 24 6( ) ( )x x x x + + = +

21 126 24 62x x x x + + = +

x x2 39 126 0 =

( )( )x x+ =3 42 0 x= 3 andx=42

Since, xis the average speed of the train, soxcannot be negative.

Hence, the original average speed of the train is42 km/h.

Or

LetBalone takesxdays to finish the work. Then,Aalone can finish it in( )x6 days.

Bs one day work=1

x

As one day work=1

6x

Now,As one day work+Bs one day work

= +

1 1

6x x


( )A B+

s one day work=

1

4

1 1

6

1

4x x+

=

x x

x x

+

=6

6

1

4( )

8 24 62x x x =

x x2 14 24 0 + =

x x x2 12 2 24 0 + =

x x x( ) ( ) =12 2 12 0

( ) ( )x x

=12 2 0

x=2, 12Butxcannot be less than 6.

x=12Hence,Balone can finish the work in 12 days.

27. Find the length of a tangent drawn to acircle with radius 5 cm, from a point 13 cmaway from the centre of the circle.

Sol. Given, radius of the circle, OT=5cm and OP=13cm

We know that, the radius is perpendicular to thetangent at the point of contact.

OT PT Then, = OTP 90Now, in right angledOTP,

OP OT TP 2 2 2= + [by Pythagoras theorem]

TP OP OT 2 2 2= TP2 2 213 5=


1

1

1

1

1

1

1

1/2

1

1

1

1/2

1

O

T

P

5 cm

13 cm

1

1

1


11/12

TP2 169 25 144= =

TP2 144=

TP=12 cmHence, length of the tangent is 12 cm.

28. Draw aABC with sides BC= 6 cm,AB = 5 cm and = ABC 60, then

construct a triangle whose sides are 3

4times, the correspording sides ofABC.

Sol. Steps of construction

(i) Draw a line segment AB=5cm.(ii) At pointB, make = ABX 60 .

(iii) With Bas centre and radius BC= 6 cm, draw anarc intersectingBXat C.

(iv) Join AC. Then, ABC is the required giventriangle.

(v) Draw a rayAYmaking an acute angle with ABon the opposite side of the vertex C.

(vi) Locate 4 points the greater of 3 and 4 in3

4

A A A1 2 3, , and A 4 on AY, such thatA A A A A A A A1 1 2 2 3 3 4= = = .

(vii) Join A B4 and draw a line throughA 3 the 3rd

point, 3 being smaller of 3 and 4 in3

4parallel to

A B4 intersectingABat B.(viii) Draw a line through B parallel to the line BCto

intersectACat C .Thus,AB C is the required triangle.

29. 216 logs are stacked in the followingmanner as 21 logs in the bottom row, 20 inthe next row, 19 in the row next to it and soon (see figure). In how many rows are the216 logs placed and how many logs are inthe top row?

Sol. Here, logs are stacked in each row form a series21 20 19 18 17+ + + + +

Clearly, it is an AP with first term, a=21 andcommon difference, d= = 20 21 1.Let number of rows be n, then Sn =216.

2162

2 21 1 1= + n

n[ ( ) ( )]

QS n

a n dn = +

2

2 1{ ( ) }

432 42 1= +n n( ) 432 42 2= +n n n n n2 43 432 0 + =

n n n2 27 16 432 0 + =

n n n( ) ( ) =27 16 27 0 ( ) ( )n n =27 16 0 n=16, 27Hence, the number of rows is 16.

[Q27 is not possible]

T a n d 16 1 21 16 1 1= + = + ( ) ( ) ( )= =21 15 6

Hence, in sixteenth row, there are six logs.

30.A container is in the form of a frustum of acone of height 30 cm with radii of its lower

and upper ends as 10 cm and 20 cm,respectively. Find the capacity and surfacearea of the container. Also, find the cost ofmilk which can completely fill thecontainer at the rate of ` 25 per litre.(take, = 314. )

Sol. Capacity of the container means the volume ofcontainer. Its surface area is equal to the sum ofcurved surface area of container and area of bottomof the container.Volume (capacity) of the container

= + +h

r r r r

3 1

222

1 2( )

Given, h =30 cm, r1 20= cm and r2 10=So, the capacity of the container

= + + 314 30

320 10 20 102 2. [( ) ( ) ]

=21980 3cm

=21980. L

[Q1 L=1000 cm 3 ]Q Cost of 1 L milk= `25Cost of 21.980 L milk= 21980 25.

=`549 50.

Now, slant height of container,

l h r r = + 2 1 22( )


C

6 cm

C

60

5 cm B B

X

A1A2

A3A4 Y

1

1

1

1

11

2

11

2

1

11

2

1/2

1


12/12

l= +900 100 =3162.

Surface area of the container

=Curved surface area of the container+Surface area of the bottom= + + l r r r ( )1 2 2

2

= + + 3.14 31.62 (20 10) 3.14 ( )10 2

= +3.14 (948.60 100)= 314. 1048.60=3292.60 cm 2

31. In sports day activities of Meerut PublicSchool, the lines were drawn with chalkpowder in rectangular shape OBCD. Each

line is at 12

m distance from each other.

60 flower pots have been placed at a

distance of1

2m from each other along OD.

Yamini runs1

4th of the distance OD on the

3rd line and makes a red flower. Kamla

runs1

5th of the distanceOD on the 7th line

and makes a yellow flower.

(a) Find the position of red and yellowflowers.

(b) Find the distance between these flowers.

(c) Who makes the flower closer to theorigin?

(d) Write the values associated with sportsactivities in day-to-day life.

Sol. Given, the distance between two flower pots is1

2m

and there are 60 pots along OD. Also, distance

between two lines is 12

m. So, the x-coordinate of

position of any student in the field will obtained bymultiply the distance between two lines by numberof lineand y-coordinatewillobtained by multiplythedistance cover by number of flower pots.

(a) From the above figure, the position of the redflower made by Yamini is

A A3

2

1

460, (

= 1.5, 15)

and the position of the yellow flower made byKamla is

H H7

2

1

560 12, ( , )

= 3.5

(b) Now, the distance between these flowers is

AH x x y y = + ( ) ( )2 12

2 12

= + ( ) ( )3.5 1.5 2 212 15

= + ( ) ( )2 32 2

= +4 9 = 13 units

(c) Now, the distance between OandA is

OA= + ( ) ( )1.5 0 15 02 2

= +2.25 225= 227.25 units

and the distance between OandHis

OH= + ( ) ( )3.5 0 12 02 2

= +12.25 144= 156.25units

Hence,Kamla makes flowercloser to the origin.

(d) Values associated with sports activities inday-to-day life are, they

(i) give break from regular life.

(ii) increase physical strength of individual.(iii) relax mind and body.

(iv) help in increasing the concentration inother works.


11

2

O 1 2 3 4 5 6 7 8 9 10 11

D

B

C

1/5 th1/4 th

X

Y

Flowerpots

A H

1/2

1

1

1

1

mathematics sample paper cbse 2015 x

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