&%6( 35(9,286

3$3(56�ZLWK

62/87,216�

&%6(35(9,286��<($56

�

&/$66�;&/$66�;&/$66�;0DWKHPDWLFV

��

YYY�RCFJNG�KP #RCFJNG�KP

YYY�RCFJNG�KP #RCFJNG�KP

CONTE

NTS

TA

BLE

O

F154 2 0 1 6

Answers Begin at page 161

122 2 0 1 7Answers Begin at page 128









CBSE Ma hema ic 2020 Ge e a I c : ( ) A e a e c . ( ) T e a e c f 30 e d ded f ec A, B, C a d D.

( ) Sec A c a 6 e f 1 a eac . Sec B c a 6

e f 2 a eac , Sec C c a 10 e f 3 a eac . Sec D c a 8 e f 4 a eac . ( ) T e e e a c ce. H e e , a e a c ce a bee

ded f e f 3 a eac a d 3 e f 4 a eac . Y a e a e e f e a e a e a c

e . ( ) U e f ca c a e ed.

Downloaded from www.padhle.in

Q e i n

Sec i n-A (1 Ma k Each)

1. If one of the eroes of the quadratic pol nomial x2 + 3x + k is 2, then thevalue of k is:

(a) 10 (b) 10 (c) 7 (d) 2

2. The total number of factors of a prime number is:(a) 1 (b) 0 (c) 2 (d) 3

3. The quadratic pol nomial, the sum of whose eroes is 5 and theirproduct is 6, is

(a) x2 + 5x + 6 (b) x2 5x + 6 (c) x2 5x 6 (d) x2 + 5x + 6 4. The value of k for which the s stem of equations x + 4 = 0 and 2x + k=3 has no solution, is

(a) -2 (b) 2 (c) 3 (d) 2

5. The HCF and the LCM of 12, 21, 15 respectivel are


(a) 3140 (b) 12,420 (c) 3,420 (d) 4,203

6. The value of x for which 2x,(x + 10) and ( 3x + 2 ) are the three consecutiveterms of an AP, is(a) 6 (b) 6 (c) 18 (d) 18

7. The first term of an AP is p and the common difference is q, then its 10thterm is(a) q + 9p (b) p 9q (c) p + 9q (d) 2p + 9q

8. The distance between the points (a cos + b sin , 0) and (0, a sin bcos ), is(a) a2 + b2 (b) a2 b2 (c) a2 +b2 (d) a2 - b2

9. Find the ratio in which the segment joining the points (1, -3) and (4, 5) isdivided b x-axis? Also, find the coordinates of this point on the x-axis.

(a) 1 (b) 2 (c) 2 (d) 1

10. The value of p, for which the points A(3, 1), B(5, p) and C(7, 5) arecollinear, is(a) 2 (b) 2 (c) 1 (d) 1

11. In Fig. 1, ABC is circumscribing a circle, the length of BC is _______cm.


12. Given ABC PQR, if AB/PQ = ⅓ , then =a ( PQR ) a ( ABC )

13. ABC is an equilateral triangle of side 2a, then length of one of its altitudeis __________ 14. + c 59 c ec 31 = _______. 10°

c 80° 15. The al e f ( in 2 + 1/ 1 + an2 ) = _______.

OR The al e f (1 + an2 ) (1 in ) (1 + in ) =_______. 16. The ratio of the length of a vertical rod and the length of its shadow is 1:3 . Find the angle of elevation of the sun at that moment?


17. Two cones have their heights in the ratio 1:3 and radii in the ratio 3:1.What is the ratio of their volumes? 18. A letter of English alphabet is chosen at random. What is the probabilitthat the chosen letter is a consonant.

19. A die is thrown once. What is the probabilit of getting a number lessthan 3?

OR

If the probabilit of winning a game is 0.07, what is the probabilit of losingit?

20. If the mean of first n natural number is 15, then find n.

Sec i n-B (2 Ma k Each)

21.Sh ha (a b) 2 , (a 2 + b2) and (a + b) 2 a e in AP.

22. In the given Fig. DE AC and DC AP. Prove that = BEEC CP

BC


OR In he gi en Fig. , angen TP and TQ a e d a n a ci cle i h cen e O f m an e e nal in T. P e ha ∠PTQ = 2∠OPQ.

23. The rod AC of a TV disc antenna is fixed at right angle to the wall AB anda rod CD is supporting the disc as shown in Fig. 4. If AC = 1.5m long and CD= 3m, find (i) tan (ii) sec + cosec .


24. A iece f i e 22 cm l ng i ben in he f m f an a c f ci cle b ending an angle f 60 a i cen e. Find he adi f he ci cle. (

U e = 22/7 ) 25. If a n mbe i ch en a and m f m he n mbe 3, 2, 1, 0, 1, 2, 3. Wha i he babili ha 2 4?

26. Find the mean of the following distribution:

Sec i n- C (3 Ma k Each)

27. Find the quadratic pol nomial whose eroes are reciprocal of the eroesof the pol nomial f (x) = ax2 + bx + c, a 0, c 0.

OR


Divide the pol nomial f(x) = 3x2 x3 3x + 5 b the pol nomial g(x) = x 1 x2 and verif the division algorithm.

28. Determine graphicall the coordinates of the vertices of a triangle, theequations of whose sides are given b 2 x = 8, 5 x = 14 and 2x = 1.

OR

If 4 is the ero of the cubic pol nomial x2 3x2 10x + 24, find its other twoeroes.

29. In a flight of 600 km, an aircraft was slowed due to bad weather. Itsaverage speed for the trip was reduced to 200 km/hr and time of flightincreased b 30 minutes. Find the original duration of flight.

30. Find the area of triangle PQR formed b the points P( 5, 7), Q( 4, 5)and R(4, 5).

OR

If the point C( 1, 2) divides internall the line segment joining A(2, 5) andB(x, ) in the ratio 3 : 4, find the coordinates of B.

31. In Fig.5, ∠D = ∠E and AD/DB = AE/EC , e ha BAC i an i cele iangle.


32. In a iangle, if a e f ne ide i e al he m f he a e f he he ide , hen e ha he angle i e he fi ide i a igh angle.

33. If in + c = 3 , hen e ha an + c = 1 34. A c ne f ba e adi 4 cm i di ided in a b d a ing a

lane h gh he mid in f i heigh and a allel i ba e. C m a e he l me f he a .

Sec i n- D (4 Ma k Each)

35. Sh ha he a e f an i i e in ege cann be f f m (5 + 2) (5 +3) f an in ege .

OR P e ha ne f e e h ee c n ec i e i i e in ege i di i ible b 3.


36. The m f f c n ec i e n mbe in AP i 32 and he a i f d c f he fi and la e m he d c f middle e m i

7:15. Find he n mbe . 37. D a a line egmen AB f leng h 7 cm. Taking A a cen e, d a a ci cle f adi 3 cm and aking B a cen e, d a an he ci cle f adi 2 cm. C n c angen each ci cle f m he cen e f he he ci cle.

38. A e ical e and n a h i n al lane and i m n ed b a

e ical flag- aff f heigh 6 m. A a in n he lane, he angle f ele a i n f he b m and f he flag- aff a e 30 and 45 e ec i el . Find he heigh f he e . (Take 3 = 1.73)

39. A b cke in he f m f a f m f a c ne f heigh 30 cm i h adii

f i l e and e end a 10 cm and 20 cm e ec i el . Find he ca aci f he b cke . Al find he al c f milk ha can c m le el fill he b cke a he a e f R . 40 e li e. ( U e = 22/7 ) 40. +The f ll ing able gi e d c i n ield e hec a e (in in al )

f hea f 100 fa m f a illage:


An 1). (b) 10

An 2). (c) 2

An 3). (a) 2 + 5 + 6

An 4). (d) 2

An 5). (c) 3,420

An 6). (a) 6

An 7). (c) + 9

An 8). (c) a 2 + b2

An 9). (d) 1

An 10). (a) 2

An 11). 10cm i he leng h f BC.

An 12). We kn ha a ea f imila iangle a e e al

he i n f he a e f i nal Side .

S , a ABC/a PQR= = 1/9


An 13).

Gi en : ABC i an e ila e al iangle f ide 2a. In e ila e al iangle, △ADB = △ADC

& ∠ADB = ∠ADC (B h 90 deg ee a AD ⊥ BC)

Th , △ ADB ≅ △ ADC (B R. H. S. C ng enc )

A e CPCT i.e. he a e c e nding a f c ng en iangle,

BD = DC

U ing P hag a he em,

Hence,


Th , he al i de f he gi en e ila e al iangle i e al

An Hence, leng h f ne f he al i de i a 3

An 14).

+ 10 c (90 10) c 59

c ec 31

+ 10 c (90 10) c 59

c ec 31

+ 10 10 c 59

c ec 31

1+c 59 / in(90-31)

1+c 59/c 59

1+1 = 2

Hence ed.

An 15).


= ( in2 + c 2 )

= 1

An 16).

GI en: The a i f he leng h f a e ical d and he leng h f i had i 1: 3 .

S l i n: an = 1/ 3

= 30

An 17). Gi en: T c ne ha e hei heigh in he a i 1:3 and adii in he a i 3:1.

S l i n: = =. 2 1

.1 3

.3 1

Ra i f l me : 12h1 / 22h2 = 3:1

An 18). The e a e 26 le e . S , he babili ha he ch en le e i a c n nan i :


P(c n nan ) = .26 21

An 19).

We kn in a die, he al n . f c me d e n e ceed 6.

S ,

P (f n mbe le han 3) = T a . f c e Fa ab e c e

P (f n mbe le han 3) = = 6 2

OR An 20).

P (l ing) = T a . f c e Fa ab e c e

P (l ing) 1 0.07 = 0.93

An 20).

Gi en : Mean f fi n na al n mbe i 15

We kn , S m f fi n na al n mbe = 2 [ ( +1) ]


Mean f fi n na al n mbe = S m f b e a i n / T al n mbe f

b e a i n = 2 [ ( +1) ]

= (n + 1)/2 = 15

n + 1 = 30

n = 29

An 21).

Gi en: (a-b) 2, (a 2 + b2) and (a+b) 2 a e in A.P.

S , (a 2 + b2) - (a-b) 2 = 2ab

(a+b) 2 - (a 2 + b2) = 2ab

A e can ee, C mm n diffe ence i ame.

An Hence, he gi en e m a e in AP

An 22). Gi en: In ABC, DE AC

T P e: BE/EC = BC/CP

S l i n:

In ABC, DE AC, = DA BD

EC BE

In ABP, DC AP, = DA BD

CP BC


N , f m (i) & (ii), = EC BE

CP BC

An Hence, ed.

OR Le ∠OPQ =

∠TPQ = ∠TQP = 90

In TPQ, 2(90 ) + ∠PTQ = 180

∠PTQ = 2 = 2∠OPQ

An Hence, P ed.

An 23). Gi en: AC = 1.5m l ng and CD = 3m

I ACD , CD 2 = AC 2 +AD 2

Tan

Tan

(ii) sec + cosec


An Hence, he al e f Tan = 1/ 3 and he al e f ec + c ec i

2 ( )31 + 3

An 24). Gi en: Leng h f he i e = 22 cm, b end an angle f 60 a i cen e

S l i n: 2 ( ) = 22722 60°

360°

2 ( ) = 22722

61

= 21cm

An The adi f he ci cle i 21cm.

An 25).

T al n mbe f c me = 7


Fa able c me a e 2, 1, 0, 1, 2, i.e., 5

P( 2 4) = 5/7

An 26).

Cla F e enc (f) X X*f

3 - 5 5 4 20

5 - 7 10 6 60

7 - 9 10 8 80

9 - 11 7 10 70

11 - 13 8 12 96

f = 40 X*f = 326

Mean f he da a, = = = 8.15 ∑ f∑ f

40326

OR Gi en acc ding he able:

M dal cla : 60 80

F m la f m de: l + hf1 f02f1 f0 f2

On ing he al e : 60 + 2012 1024 10 6


= 60 + 5

= 65

An The m de f he f ll ing da a i 65.

An 27).

Le and a e he e e f he l n mial f( ) = a 2 + b + c.

Acc ding he e i n, 1/ and 1/ a e he e e f he e i ed ad a ic l n mial


E a i n f he e i ed ad a ic l n mial

O (c 2 + b +a )

An 28).

Le 2 - =8 .. e n 1

5 - =14 .. e n 2

2 - = -1 .. e n 3

On l ing 1 and 2, e ge = -4 and = 2

C dina e f B = (-4,2)


On l ing 2 and 3:

We b ain =1 and =3

C dina e f C = (1,3)

On l ing 1 and 3:

=2 and =5

Hence C dina e f A=(2, 5)

An S , C dina e f he e ice f he iangle a e A( 4, 2), B(1, 3) and C(2, 5)


OR Gi en:

1). Gi en ha he c bic l n mial i

2). 4 i a e f he gi en l n mial

T find : I he e e .

S l i n: Since, 4 i a e f he gi en l n mial, ( - 4) i i fac .

On f he l ing, 2 + + 6 = ( + 3)( - 2)

An The he e e f he l n mial a e 3 and 2.

An 29). Le he eed f ai c af be km/h


_ = 30/60600 200

600

2 - 200 - 240000 = 0

( 600) ( + 400) = 0

= 600,

Since, 400 cann be acce ed 1/2 S eed f ai c af = 600 km/h ∴ D a i n f fligh = 1 h

An 30).

S l i n:

P = (-5 , 7)

Q = ( -4 , - 5)

R = ( 4 , 5)

A ea f T iangle PQR

= (1/2) -5( -5 - 5 ) -4( 5 - 7 ) + 4(7 + 5) . ni

= (1/2) 50 + 8 + 48

= (1/2) 106

= 106/2

= 53

A ea f T iangle PQR = 53 ni

An 31).


C dina e f C a e ( , ) = ( 1, -2) 73 + 8

73 + 20

= 5, = 2

S , C dina e f B a e ( 5, 2)

An 31).

Gi en: ∠D = ∠E , hich mean AE = ED

= DB = ECAD DB

AE EC

AD + DB = AE + EC

AB = AC

A , AB = AC , BAC i an i cele iangle.

An BAC i an i cele iangle.


An 32). Gi en:- ABC i a iangle

AC 2 =AB 2 +BC 2

T e:- ∠B=90

C n c i n:- C n c a iangle PQR igh angled a Q ch ha , PQ=AB and QR=BC

P f:- In △PQR PR 2 =PQ2 + QR 2

(B hag a he em) PR 2 =AB 2 +BC 2 .....(1) a AB=PQ and QR=BC

AC 2=AB 2+BC 2....(2)(Gi en)

F m e a i n (1)&(2), e ha e AC 2 = PR 2

AC=PR.....(3) N , in △ABC and △PQR AB=PQ BC=QR AC=PR(F m (3))

△ABC≅△PQR(B SSS c ng enc ) The ef e, b C.P.C.T., ∠B=∠Q ∠Q=90 ∠B=90

Hence ed.


An 32). Gi en in + c = 3

T e: an + c = 1. S l i n: in + c = 3

( in + c ) 2 = ( 3) 2

in2 + c 2 + 2 in c = 3 in c = 1

L.H.S = an + c =

+ = = 1 RHS c

c

1 c

Hence, P ed.

An 33).

Le heigh f he c ne i h

N B i he mid in f AC.


Le V1 and V2 be he l me f e and l e a f he c ne e ec i el

In ABE and ACD

∠B=90 0 =∠C

∠A=∠A (C mm n angle)

ABE∼ ACD

AB/BE = AC/CD

= h/4BEh/2

BE = 2 cm

Ra i f l me f a :

= 4/28 = 1/7 Hence, he a i f l me f a i 1:7.

An 35). Le a be an i i e in ege . Take b = 5 a he di i .


a = 5m + , = 0,1,2,3,4 1 Ca e-1 : a = 5m ⇒ a2 = 25m2 = 5(5m2) = 5 Ca e-2 : a = 5m+1 ⇒ a2 = 5(5m2 + 2m) + 1 = 5 + 1 Ca e-3 : a = 5m+2 ⇒ a2 = 5(5m2 + 4m) + 4 = 5 + 4 Ca e-4 : a = 5m+3 ⇒ a2 = 5(5m2 + 6m + 1) + 4 = 5 + 4 Ca e-5 : a = 5m+4 ⇒ a2 = 5(5m2 + 8m + 3) + 1 = 5 + 1

An Hence ed,

a e f an i i e in ege cann be f he f m (5 + 2) (5 + 3) f an in ege .

OR

Le n be an i i e in ege . Di ide i b 3.

n = 3 + = 0, 1, 2 1

Ca e-1 : n = 3 (di i ible b 3)

n + 1 = 3 + 1 n + 2 = 3 + 2

Ca e-2 : n = 3 + 1 n + 1 = 3 + 2 n + 2 = 3 + 3 (di i ible b 3)

Ca e-3 : n = 3 + 2 n + 1 = 3 + 3 (di i ible b 3) n + 2 = 3 + 4


An 36).

GI en: S m f f c n ec i e n mbe in AP i 32 Le he f c n ec i e n mbe in AP be (a 3d),(a d),(a+d) and (a+3d) a 3d+a d+a+d+a+3d=32 4a=32 a=32/4 a=8 ----> e n (1) N , (a 3d)(a+3d)/(a d)(a+d)=7/15

15(a 9d )=7(a d ) 15a 135d =7a 7d 15a 7a =135d 7d

8a =128d

P ing he al e f a=8 in ab e e ge . 8(8) =128d 128d =512 d =512/128 d =4 d=2 S , he f c n ec i e n mbe a e 8 (3 2) 8 6=2 8 2=6 8+2=10 8+(3 2)


8+6=14 F c n ec i e n mbe a e 2,6,10and14

OR

Gi en : a = 1, d = 3

We kn ha ,

S(n) = n/2 [2a + (n - 1)d]

P ing all he al e , e ge

287 = n/2[2 1 + (n - 1) (3) ]

287 = n/2[2 + (n - 1) 3]

574 = 3n - n

3n - n - 574 = 0

3n - 42n + 41n - 574 = 0

3n(n - 14) + 41(n - 14) = 0

n = 14, - 41/3 ( Since, n can' be nega i e)

n = 14

We kn ha ,

a + (n - 1)d =

1 + (14 - 1) (3) = 3

1 + 13 (3) = 3

= 40.


An The al e f i 40.

An 37). F ll ing a e he e f c n c i n:

1. Take AB = 7 cm.

2. Wi h A a cen e and 3 cm a adi , d a a ci cle.

3. Simila l , i h B a cen e and 2 cm a adi , d a a ci cle.

4. N , d a he e endic la bi ec f AB and ma k he in f in e ec i n O.

5. Wi h O a cen e and OA a adi , d a a ci cle. Ma k he 2 in he e he ci cle i h cen e O and A mee a Q and R. Simila l , ma k

he in he e he ci cle i h cen e O and B mee a S and T e ec i el .

6. J in BR and BQ a ell a AS and AT. N , BR, BQ, AS and AT a e he e i ed angen .


An 38).

Le BC be he heigh f he e and DC be he heigh f he flag - aff.

In .△ABC,

AB = BC c 30

AB = BC 3 -----> e n (i)

In .△ABD,

AB = BD c 45

AB = (BC + CD) c 45

AB = (BC + 6) -----> e n (ii)

E a ing (i) and (ii)

(BC + 6) = BC 3

BC( 3-1) = 6


BC = 6/0.73 = 9.58 m

The ef e he heigh f he e i 9.58 m.

An 39). Gi en: Radi f he e end f he f m f c ne = R = 20 cm

adi f he l e end f he f m f c ne = = 10 cm H = 30 cm

S l i n: V l me = h[R + + R* ]31

= 30 [20 + 10 + 20*10]31

722

= 660/21 [400 + 100 + 200] = (660 700)/21 = 22000 cm3 Hence he ca aci = 22L S , he C f milk = ₹ 40 ₹ 22 = ₹880

An 40).

P d c i n ield / hec a e N . f f m



CBSE Ma hema ics 2019 Ge e a I c i : (i) A e i a e c . (ii) Thi e i a e c i f 30 e i di ided i f ec i A, B, C a d D.

(iii) Sec i A c ai 6 e i f 1 a each. Sec i B c ai 6

e i f 2 a each, Sec i C c ai 10 e i f 3 a each. Sec i D c ai 8 e i f 4 a each. (i ) The e i e a ch ice. H e e , a i e a ch ice ha bee

ided i f e i f 3 a each a d 3 e i f 4 a each. Y ha e a e e f he a e a i e i a ch

e i . ( ) U e f ca c a i e i ed.


Q es ions

Sec ion-A (1 Mark Each)

1. Find he coo dina e of a oin A, he e AB i diame e of a ci cle ho ecen e i (2, -3) and B i he oin (1, 4). 2. Fo ha al e of k, he oo of he e a ion 2 + 4 + k = 0 a e eal? [1]

ORFind he al e of k fo hich he oo of he e a ion 3 2 10 + k = 0 a eeci ocal of each o he .

3. Find A if an 2A = co (A 24 )

ORFind he al e of ( in2 33 + in2 57 ) 4. Ho man o digi n mbe a e di i ible b 3? 5. In Fig., DE BC, AD = 1 cm and BD = 2 cm. ha i he a io of he a( ABC) o he a ( ADE)


6. Find a a ional n mbe be een 2 and 3.

Sec ion-B (2 Marks Each) 7. Find he HCF of 1260 and 7344 ing E clid algo i hm.

ORSho ha e e o i i e odd in ege i of he fo m (4 + 1) o (4 + 3), he e

i ome in ege .

8. Which e m of he A.P. 3, 15, 27, 39, ill be 120 mo e han i 21e m? [2]

ORIf Sn, he m of fi e m of an A.P. i gi en b Sn = 3n2 4n, find hen h e m.

9. Find he a io in hich he egmen joining he oin (1, -3) and (4, 5) idi ided b -a i ? Al o, find he coo dina e of hi oin on he -a i . 10. A game con i of o ing a coin 3 ime and no ing he o come eachime. If ge ing he ame e l in all he o e i a cce , find he

obabili of lo ing he game.

11. A die i h o n once. Find he obabili of ge ing a n mbe hich(i) i a ime n mbe(ii) lie be een 2 and 6. 12. Find c if he em of e a ion c + 3 + (3 c) = 0, 12 + c c = 0ha infini el man ol ion ?


Sec ion- C (3 Marks Each) 13.P o e ha 2 i an i a ional n mbe . 14. Find he al e of k ch ha he ol nomial 2 (k + 6) + 2(2k 1) ha

m of i e o e al o half o hei od c 15. A fa he age i h ee ime he m of he age of hi o child en.Af e 5 ea hi age ill be o ime he m of hei age . Find he

e en age of he fa he .

OR

A f ac ion become hen 2 i b ac ed f om he n me a o and i

become hen 1 i b ac ed f om he denomina o . Find he f ac ion. 16. Find he oin on -a i hich i e idi an f om he oin (5, -2) and(-3, 2).

ORThe line egmen joining he oin A(2, 1) and B(5, -8) i i ec ed a he

oin P and Q ch ha P i nea e o A. If P al o lie on he line gi en b 2 + k = 0, find he al e of k.

17. P o e ha ( in + co ec )2 + (co + ec )2 = 7 + an2 + co 2 . [3]

OR

P o e ha (1 + co A co ec A) (1 + an A + ec A) = 2.


18. In Fig. PQ i a cho d of leng h 8 cm of a ci cle of adi 5 cm and cen eO. The angen a P and Q in e ec a oin T. Find he leng h of TP.

19.In Fig. ∠ACB = 90 and CD AB, o e ha CD2 = BD AD.

OR

If P and Q a e he oin on ide CA and CB e ec i el of ABC,igh -angled a C, o e ha (AQ2 + BP2 ) = (AB2 + PQ2).

20. Find he a ea of he haded egion in Fig. if ABCD i a ec angle i h

ide 8 cm and 6 cm and D i he cen e of he ci cle.

21. Wa e in a canal, 6 m ide and 1.5 m dee , i flo ing i h a eed of 10km/ho . Ho m ch a ea ill i i iga e in 30 min e , if 8 cm anding a ei needed? 22. Find he mode of he follo ing f e enc di ib ion.


Sec ion- D (4 Marks Each)

23. T o a e a oge he can fill a ank in 1 7/8 ho . The a i hlonge diame e ake 2 ho le han he a i h a malle one o fill heank e a a el . Find he ime in hich each a can fill he ank e a a el .

24. If he m of fi fo e m of an A.P. i 40 and ha of fi 14 e m i280. Find he m of i fi n e m .

25. 26. A man in a boa o ing a a f om a ligh ho e 100 m high ake 2min e o change he angle of ele a ion of he o of he ligh ho e f om60 o 30 . Find he eed of he boa in me e e min e. [U e 3 = 1.732] 27. Con c a ABC in hich CA = 6 cm, AB = 5 cm and ∠BAC = 45 . Thencon c a iangle ho e ide a e ⅗ of he co e onding ide of ABC.


28. A b cke o en a he o i in he fo m of a f m of a cone i h aca aci of 12308.8 cm3. The adii of he o and bo om of ci c la end ofhe b cke a e 20 cm and 12 cm e ec i el . Find he heigh of he b cke

and al o he a ea of he me al hee ed in making i . 29. P o e ha in a igh -angle iangle, he a e of he h o en e i e alhe m of a e of he o he o ide .

30.If he median of he follo ing f e enc di ib ion i 32.5. Find he

al e of f1 and f2.

ORThe ma k ob ained b 100 den of a cla in an e amina ion a egi en belo .


D a a le han e c m la i e f e enc c e (ogi e). Hencefind he median.

Ans ers

An 1). Le he coo dina e of oin A be ( , ) and oin O (2, -3) be oin hecen e, henB mid oin fo m la,


The coo dina e of oin A a e (3, -10)

Ans 2). The gi en e a ion i 2 + 4 + k = 0On com a ing he gi en e a ion i h a 2 + b + c = 0, e gea = 1, b = 4 and c = k

Fo eal oo , D 0o b2 4ac 0o 16 4k 0o k 4

Fo k 4, e a ion 2 + 4 + k ill ha e eal oo .

OR The gi en e a ion i 3 2 10 + k = 0

On com a ing i i h a 2 + b + c = 0, e gea = 3, b = -10, c = kLe he oo of he e a ion a e and 1/P od c of he oo = c/

. 1/ = k/3


o k = 3

An 3).

Gi en, an 2A = co (A 24 )

o co (90 2A) = co (A 24 ) [ an = co (90 )]o 90 2A = A 24o 3A = 90 + 24o 3A = 114A = 38

OR in233 + in257

= in2 33 + co 2(90 57 )= in2 33 + co 2 33 [ in2 + co 2 = 1]= 1

An 4). The o-digi n mbe di i ible b 3 a e 12, 15, 18, 99 Thi i an A.P. in hich a = 12, d = 3, an = 99 an = a + (n 1) d

99 = 12 + (n 1) 387 = (n 1) 3o n 1 = 29o n = 30So, he e a e 30 o-digi n mbe di i ible b 3.

An 5).

Gi en, AD = 1 cm, BD = 2 cmAB = 1 + 2 = 3 cm


Al o, DE BC (Gi en)∠ADE = ∠ABC (i) (co e onding angle )

In ABC and ADE∠A = ∠A (common)∠ABC = ∠ADE [b e a ion (i)]

ABC ADE (b AA le)

No ,

= (AB/AD)2ar(ADE) ar(ABC)

ABC ADE

= (3/1)2 = 9:1ar(ADE) ar(ABC)

An 6). We kno , 2=1.414 and 3=1.732

so,ra ional n mber be een 2 and 3 =1.432 =1.563 =1.576 =1.657 =1.711 e c.

An 7).


S ep-b -s ep e plana ion: In e clid's di ision algori hm,

7344 = 1260 5 + 1044

Al o, 1260 = 1044 1 + 216

1044 = 216 4 + 180

216 = 180 1 + 36

180 = 36 5 + 0

We ill follo he follo ing s eps,

S ep 1 : di ide 7344 b 1260,

Q o ien = 5 remainder = 1044,

S ep 2: di ide di isor 1260 b remainder,

Q o ien = 1 remainder = 216,

S ep 3: Repea he abo e s eps n il e ge remainder 0.

Th s, he las q o ien hich gi es he remainder ero is 36.

OR An posi i e in eger is of he form 4q+1or4q+3

As per E clid s Di ision lemma. If a and b are o posi i e in egers, hen,

a=bq+r Where 0 r<b.

Le posi i e in egers be a.and b=4 Hence,a=bq+r


Where, (0 r<4) R is an in eger grea er han or eq al o 0 and less han 4 Hence, r can be ei her 0,1,2and3

No , If r=1 Then, o r be eq a ion is becomes a=bq+r a=4q+1 This ill al a s be odd in eger.

No , If r=3 Then, o r be eq a ion is becomes a=bq+r a=4q+3 This ill al a s be an odd in eger. Hence pro ed.

Ans 8). The gi en A.P. i 3, 15, 27, 39,He e a = 3, d = 12

a21 = a + 20d = 3 + 20 12 = 3 + 240 = 243No , an = a21 + 120 = 243 + 120 = 363an =a + (n 1) d363 = 3 + (n 1) 12o 360 = (n 1) 12o n 1 = 30

n = 31Hence, he e m hich i 120 mo e han i 21 e m ill be i 31 e m.


OR

Gi en, Sn = 3n2 4nWe kno ha

an = Sn Sn-1= 3n2 4n [3 (n 1)2 4 (n 1)]= 3n2 4n [3 (n2 2n + 1) 4n + 4]= 3n2 4n (3n2 6n + 3 4n + 4)= 3n2 4n 3n2 + 10n 7= 6n 7So, n h e m ill be 6n 7

An 9). Le he gi en oin be A (1, -3) and B (4, -5) and he line- egmenjoining b he e oin i di ided b he -a i , o he co-o dina e of he

oin of in e ec ion ill be P( , 0)


A 10). The possible o comes of 3 imes ossing a coin 1s = HHH 2nd= HHT or HTH or THH 3rd= TTHor THTor HTT 4 h=TTT

To al n mber of e en s=8 No of fa o rable e en s= 6

P(E)= o al o come no. of fa o rable o come


P(E)= 6/8 = 3/4 Ans Hence, he obabili of lo ing he game i 3/4.

Ans 11). Possible n mbers of e en s on hro ing a dice = 6 N mbers on dice = 1,2,3,4,5 and 6 (i) Prime n mbers = 2, 3 and 5 Fa o rable n mber of e en s = 3 Probabili ha i ill be a prime n mber

= = 3/6 = 1/2o al o come no. of fa o rable o come

(ii) N mbers l ing be een 2 and 6 = 3, 4 and 5 Fa o rable n mber of e en s = 3 Probabili ha a n mber be een 2 and 6 = 3/6 =

Ans 12). Gi en S s em of eq a ions are

-c + 3 + ( 3 - c ) = 0 and 12 - c - c = 0

To find: Val e of c hen s s em of eq a ion has infini el man sol ion

Condi ion for infini el man sol ion is gi en b ,

From gi en s s em of eq a ion,

P ing hese al e in condi ion e ge ,

= = 12 c

c c c


To find al e of c

consider firs eq ali .i.e.,

-c/12 = 3/-c

c = 36

c = 6

Here c = -6 is rejec ed as i does no sa isf he 2nd eq ali .

Therefore, c = 6

Ans 13). Le 2 i a a ional n mbe .

So, 2 = he e a and b a e co ime in ege and b 0

o 2 b = a

S a ing on bo h ide , e ge

2b2 = a2

The efo e, 2 di ide a2

o 2 di ide a (f om heo em)

Le a = 2c, fo ome in ege c

F om e a ion (i)

2b2 = (2c)2

o 2b2 = 4c2

o b2 = 2c2


I mean ha 2 di ide b2 and o 2 di ide b

The efo e a and b ha e a lea 2 a a common fac o .

B hi con adic he fac ha a and b a e co- ime.

Thi con adic ion i d e o o ong a m ion ha 2 i a ional.

So, e concl de ha 2 i i a ional.

Hence P o ed.

An 14). - (k+6) + 2(2k-1)

As s m of eros

= -b/a

= -(-(k+6))/1

= k+6

Prod c of eros

= c/a

= 2(2k-1)/1

= 2(2k-1)

Acc di g he e i

S m of eros =1/2 Prod c of eros


-b/a = 1/2 c/a

k + 6 = 1/2 2(2k -1)

k + 6 = 2k -1

6 + 1 = 2k - k

7 = k

So Val e of k = 7

Ans 15).

Le he e en age of fa he be ea and m of age of hi ochild en be eaAcco ding o e ion

= 3 (i) Af e 5 ea

Fa he age = ( + 5) eaS m of age of o child en = ( + 5 + 5) ea = ( + 10) ea

In 2nd ca eAcco ding o e ion

+ 5 = 2 ( + 10)o + 5 = 2 + 20o 2 = 15o 3 2 = 15 (U ing e a ion (i))

= 15No f om e a ion (i)

= 3 (P = 15)


o = 3 15 = 45

An P e en age of fa he i 45 ea .

OR

Le he f ac ion be /

Acco ding o e ion = 1/3o 3( 2) = o 3 = 6 (i)again, Acco ding o e ion

= 1/2o 2 = 1o 2 = -1 (ii)On ol ing e a ion (i) and (ii), e ge

= 7, = 15

An The e i ed f ac ion i 7/15

An 16).We kno ha a oin on he -a i i of he fo m (0, ).So, le he oin P(0, ) be e idi an f om A (5, -2) and B (-3, 2)

Then AP = BP o AP2 = BP2

o (5 0)2 + (-2 )2 = (-3 0)2 + (2 )2

o 25 + 4 + 2 + 4 = 9 + 4 + 2 48 = -16

= -2So, he e i ed oin i (0, -2)


OR

The line egmen AB i i ec ed a he oin P and Q and P i nea e oASo, P di ide AB in he a io 1 : 2

P lie on he line 2 + k = 0I ill a i f he e a ion.

On ing = 3 and = -2 in he gi en e a ion, e ge2(3) (-2) + k = 06 + 2 + k = 0k = -8Hence, k = -8

An 17).

We kno he rigonome ric iden i ies ,

1 ) sin A + cos A = 1


2 ) sec A = 1 + an A

3 ) cosec A = 1 + co A

4 ) sinA cosecA = 1

5 ) cosA secA = 1

No ,

LHS = (sinA+cosecA) + (cosA+secA)

= sin A+cosec A + 2sinAcosecA+ cos A + sec A + 2cosAsecA

= (Sin A + cos A ) + 2 + 2 + cosec A + Sec A

= 1 + 2 + 2 + ( 1 + co A )+( 1 + an A )

= 7 + co A + an A

Hence Pro ed.

OR


Ans 18).

Le TR =

Since OT i e endic la bi ec o of PQ.


The efo e, PR=QR=4cm

In igh iangle OTP and PTR, e ha e:

TP2=TR2+PR2

Al o, OT2=TP2+OP2

OT2=(TR2+PR2) + OP2

( +3)2= 2+16+25 (OR = 3, a OR2 = OP2 - PR2)

6 =32

=

TP2=TR2+PR2

TP2= +42 = +16 =

TP= cm

Ans 19).

To pro e CD = BD AD In CAD, CA = CD + AD .... (1)

Also in CDB, CB = CD + BD .... (2) (1) + (2) e ge :


CA + CB = 2CD + AD + BD

AB = 2CD + AD + BD AB - AD = BD + 2CD (AB + AD)(AB - AD) - BD = 2CD (AB + AD)BD - BD = 2CD BD(AB + AD - BD) = 2CD BD(AD + AD) = 2CD BD 2AD = 2CD

CD = BD AD Hence pro ed.

OR

Gi en, ABC i a igh -angled iangle in hich ∠C = 90

To o e : AQ2 + BP2 = AB2 + PQ2

Con c ion: Join AQ, PB and PQP oof: In AQC, ∠C = 90AQ2 = AC2 + CQ2 (i) (U ing P hago a heo em)In PBC, ∠C = 90BP2 = BC2 + CP2 (ii) (U ing P hago a heo em)Adding e a ion (i) and (ii)AQ2 + BP2 = AC2 + CQ2 + BC2 + CP2 = AC2 + BC2 + CQ2 + CP2


oAQ2 + BP2 = AB2 + PQ2

Hence P o ed.

Ans 20).

Gi en, ABCD i a ec angle i h ide AB = 8 cm and BC = 6 cmIn ABCAC2 = 82` + 62 (B P hago a Theo em) AC2 = 64 + 36 AC2 = 100 AC = 10 cm

The diagonal of he ec angle ill be he diame e of he ci cle

adi of he ci cle = = 5 cmA ea of haded o ion = A ea of ci cle A ea of Rec angle= 2 l b= 3.14 5 5 8 6= 78.50 48= 30.50 cm2

Hence, A ea of haded o ion = 30.5 cm2

An 21).

Le b be he id h and h be he de h of he canalb = 6 m and h = 1.5 mWa e i flo ing i h a eed = 10 km/h = 10,000 m/h

Leng h of a e flo ing in 1 h = 10,000 m

Leng h (l) of a e flo ing in h = 5,000 mVol me of a e flo ing in 30 min. = l b h = 5000 6 1.5 m3

Le he a ea i iga ed in 30 min ( h ) be m2

Vol me of a e e i ed fo i iga ion = Vol me of a e flo ing in 30 min.


= 5000 6 1.5 = 562500 m2 = 56.25 hec a e . (∵ 1 hec a e = 104 m2)

Hence, he canal ill i iga e 56.25 hec a e in 30 min.

An 22). He e, he ma im m cla f e enc i 16Modal cla = 30-40 Lo e limi (l) of modal cla = 30 Cla i e (h) =10 F e enc (f1) of he modal cla = 16 F e enc (f0) of eceding cla = 10 F e enc (f2) of cceeding cla = 12

Mode = l + ( ) h f1 f0

2f1 f0 f2

30 + ( ) 10 1 10 2 10 12

= 30 + 6/10 10 Ans Hence, Mode is 36.

Ans 23). Le he a A i h longe diame e ake ho and he a B i h

malle diame e ake ( + 2) ho o fill he ank.

Po ion of ank filled b he a A in 1 h . =

and Po ion of ank filled b he a B in 1 h . =

Po ion of he ank filled b bo h a in 1 h . = + =

Time aken b bo h a o fill he ank = 1 7/8 h = h


Po ion of he ank filled b bo h in 1 h . = 8/15Acco ding o e ion,

=

=

15 + 15 = 4 2 + 8 4 2 12 + 5 15 = 0 4 ( 3) + 5 ( 3 ) = 0 (4 + 5)( 3) =0 4 + 5 = 0 o 3 = 0 = -5/4

Since, ime can no be nega i e hence, neglec ed hi al e i ; = 3Hence, he ime aken i h longe diame e a = 3 hoand he ime aken i h a malle diame e a = 5 ho .

OR

Le he speed of he boa in s ill a er be km/h and he speed of he s ream be km/h.

Speed ps ream= ( - ) km/h

Speed do ns ream= ( + ) km/h

On s b rac ing, e ge ,


On sol ing, e ge ,

= 8 and =3

Speed of boa in s ill a er = 8 km/hr

And, Speed of s ream = 3 km/hr

Ans 24). GI en:

S 4 = 40 S 14 =280

Sn=n/2[2a+(n-1)d] s=4/2[2a+(4-1)d] 40=2[2a+3d] 40/2=2a+3d 2a+3d=20 (1) similarl , S14=14/2[2a+(14-1)d] 280=7[2a+13d] 280/7=2a+13d 2a+13d=40. .(2) B s b rac ing eq(2) from(1)

e ge ,


-10d =-20 10d = 20 d = 20/10 d = 2

No s bs i e d=2 in eq(1) a=7

We ha e ,a=7 & d=2

Sn= n/2 [2a+(n-1)d] Sn= n/2 [2(7)+(n-1)2] Sn= n/2 [14+2n-2] Sn= n/2 [12+2n] Sn= n/2 [2(6+n)] Sn= n (6+n) Sn= 6n + n2

Ans 25).


Ans 26).

Consider ABC Where AB is Heigh of To er BC is dis ance from boa o o er

TanC = AB/BC = 3 No , as heigh of he o er is 100m, AB=100 100/BC = 3 BC = 57.73

No le D be he poin he boa ra elled TanD = AB/BD = 1/ 3 = 100/BD = 173.2

Dis ance ra elled b boa CD=BD BC= 173.2 57.73 = 115.47m


Speed of he boa = 115.47/2 = 57.73 m/min. Ans Speed of he boa is 57.73 m/min.

Ans 27). S e of Con c ion a e a follo :

1. D a AB = 5 cm2. A he oin , A d a ∠BAX = 453. F om AX c off AC = 6 cm4. Join BC, ABC i fo med i h gi en da a.5. D a AY making an ac e angle i h AB a ho n in he fig e.

6. D a 5 a c P1, P2, P3, P4, and P5 i h e al in e al .7. Join BP5.


8. D a P3B P5B mee ing AB a B .9. F om B , d a B C BC mee ing AC a C

AB C ABCHence AB C i he e i ed iangle.

An 28). Gi en: Vol me of he b cke - 12308.8cm3

1 = 20cm and 2 = 12cm


An Hence, heigh of he b cke = 15cm and A ea of he me al hee edi 2160.32 .cm.

An 29).

Gi en: A righ angled ABC, righ angled a B

To Pro e- AC =AB +BC

Cons r c ion: Dra perpendic lar BD on o he side AC .


Proof: We kno ha if a perpendic lar is dra n from he er e of a righ angle of a righ angled riangle o he h po en se, hen he riangles on bo h sides of he perpendic lar are similar o he hole riangle and o each o her.

We ha e , ADB ABC. (b AA similari )

Therefore, AD/ AB=AB/AC

(In similar Triangles corresponding sides are propor ional)

AB =AD AC (1)

Also, BDC ABC

Therefore, CD/BC=BC/AC (in similar Triangles corresponding sides are propor ional)

Or, BC =CD AC (2)

Adding he eq a ions (1) and (2) e ge ,


AB +BC =AD AC+CD AC

AB +BC =AC(AD+CD)

( From he fig re AD + CD = AC)

AB +BC =AC . AC

AC =AB +BC

Ans Hence, Pro ed.

Ans 30).


Gi en: Median=32.5, hen 30 40 is class in er al.

l = 30 , cf = 14+f 1

f=12 and h = 40 30 = 10.

N = 40 and N/2 = 20

31+ f1 +f2 = 40. So, f1 + f2 =9


Median = hf cf2 N

32.5 = 30 + 1012 20 ( 1 + f1)

2.5 = 5 f1

15 = 30 5f1

5f 1 = 15

and on s bs i ing al es, f1 =3 and f2 = 6

OR

To d a a le han ogi e,

e ma k he e -cla limi of he cla in e al on he -a i and heic.f. on he -a i b aking a con enien cale.

He e, n = 100 ⇒ n/2 = 50


To ge median f om g a h F om 50,

D a a e endic la , he oin he e hi e endic la mee on he-a i ill be he median.

Median = 29


CBSE Ma hema ic 2018 Ge e a I c : ( ) A e a e c . ( ) Th e a e c f 30 e d ded f ec A, B, C a d D.

( ) Sec A c a 6 e f 1 a each. Sec B c a 6

e f 2 a each, Sec C c a 10 e f 3 a each. Sec D c a 8 e f 4 a each. ( ) The e e a ch ce. H e e , a e a ch ce ha bee

ded f e f 3 a each a d 3 e f 4 a each. Y ha e a e e f he a e a e a ch

e . ( ) U e f ca c a e ed.


Q e i

Sec i -A (1 Ma k Each)

1. If = 3 i e f he ad a ic e a i 2 2k 6 = 0, he fi d he al e f k. 2. Wha i he HCF f malle ime mbe a d he malle c m i e mbe ? 3. Fi d he di a ce f a i P( , ) f m he igi . 4. I a AP, if he c mm diffe e ce (d) = 4 a d he e e h e m (a7) i 4, he fi d he fi e m. 5. Wha i he al e f (c 267 i 2 23 ) ?

6. Gi e ABC PQR, if = 1/3 , he fi d ?ABP Q

a ABCa P QR

Sec i -B (2 Ma k Each) 7. Gi e ha i i a i al, e ha (5 + 3 ) i a i a i al2 2

mbe


8. I fig. 1, ABCD i a ec a gle. Fi d he al e f a d .

9. Fi d he m f fi 8 m l i le f 3. 10. Fi d he a i i hich P (4, m) di ide he li e egme j i i g he

i A (2, 3) a d B (6, 3). He ce fi d m. 11. T diffe e dice a e ed ge he . Fi d he babili : (i) f ge i g a d ble . (ii) f ge i g a m 10, f he mbe he dice. 12. A i ege i ch e a a d m be ee 1 a d 100. Fi d he

babili ha i i : (i) di i ible b 8 (ii) di i ible b 8.

Sec i - C (3 Ma k Each) 13. Fi d HCF a d LCM f 404 a d 96 a d e if ha HCF LCM = P d c f he gi e mbe . 14. Fi d all e e f he l mial (2 4 9 3 + 5 2 + 3 1) if f i e e a e (2 + ) a d (2 ).3 3


15. If A ( 2, 1), B (a, 0), C (4, b) a d D (1, 2) a e he e ice f a a allel g am ABCD, fi d he al e f a a d b. He ce fi d he

le g h f i ide .

OR

If A ( 5, 7), B ( 4, 5), C ( 1, 6) a d D (4, 5) a e he e ice f a ad ila e al, fi d he a ea f he ad ila e al ABCD.

16. A la e lef 30 mi e la e ha i ched led ime a d i de each he de i a i 1500 km a a i ime, i had i c ea e i

eed b 100 km/h f m he al eed. Fi d i al eed. 17. P e ha he a ea f a e ila e al ia gle de c ibed e ide

f he a e i e al half he a ea f he e ila e al ia gle de c ibed e f i diag al.

OR If he a ea f imila ia gle a e e al, e ha he a e c g e . 18. P e ha he le g h f a ge d a f m a e e al i a ci cle a e e al.

19. If 4 a = 3, e al a e 4 c 14 c ( 1)

OR

If a 2A = c (A 18 ), he e 2A i a ac e a gle, fi d he al e f A.


20. Fi d he a ea f he haded egi i Fig. 2, he e a c d a i h ce e A, B, C a d D i e ec i ai a mid- i P, Q, R a d S f he ide AB, BC, CD a d DA e ec i el f a a e ABCD f ide

12 cm. [U e = 3.14] 21. A de a icle a made b c i g a hemi he e f m each e d f a lid c li de , a h i he fig e. If he heigh f he c li de i 10 cm a d i ba e i f adi 3.5 cm. Fi d he al

face a ea f he a icle.

OR A hea f ice i i he f m f a c e f ba e diame e 24 m a d heigh 3.5 m. Fi d he l me f he ice. H m ch ca a cl h i e i ed j c e he hea ?

22. The able bel h he ala ie f 280 e :

Sala ( i Th a d ₹ ) N . f Pe

5-10 49

10-15 133

15-20 63

20-25 15

25-30 6

30-35 7

35-40 4


40-45 2

45-50 1 Calc la e he media ala f he da a.

Sec i - D (4 Ma k Each)

23. A m b a h e eed i 18 km/h i ill a e ake 1 h m e g 24 km eam ha e d eam he ame

. Fi d he eed f he eam.

OR A ai a el a a ce ai a e age eed f a di a ce f 63 km a d he a el a a di a ce f 72 km a a a e age eed f 6 km/h

m e ha i igi al eed. If i ake 3 h c m le e a al j e , ha i he igi al a e age eed ? 24. The m f f c ec i e mbe i a AP i 32 a d he a i

f he d c f he fi a d he la e m he d c f middle e m i 7 : 15. Fi d he mbe . 25. I a e ila e al ABC, D i a i ide BC ch ha BD = 1/3 BC. P e ha 9(AD)2 = 7(AB)2 .

OR P e ha , i a igh ia gle, he a e he h e e i e al

he m f he a e he he ide .


26. D a a ia gle ABC i h BC = 6 cm, AB = 5 cm a d ∠ ABC = 60 . The c c a ia gle h e ide a e 3/4 f he c e di g

ide f he ABC. 27. P e ha : (Si A - 2Si 3A) / (2C 3A - C A) = Ta A 28. The diame e f he l e a d e e d f a b cke i he f m

f a f m f a c e a e 10 cm a d 30 cm e ec i el . If i heigh i 24 cm, fi d : (i) The a ea f he me al hee ed make he b cke . (ii) Wh e h ld a id he b cke made b di a la ic ? [U e = 3.14] 29. A b e ed f m he f a 100 m high ligh h e f m he

ea-le el, he a gle f de e i f hi a e 30 a d 45 . If e hi i e ac l behi d he he he ame ide f he ligh h e,

fi d he di a ce be ee he hi . [U e = 1.732]3 30. The mea f he f ll i g di ib i i 18. Fi d he f e e c f

f he cla 19 21. Cla 11 13 13-15 15-17 17-19 19-21 21-23 23-25

F e e c 3 6 9 13 f 5 4

OR The f ll i g di ib i gi e he dail i c me f 50 ke f a fac :


Dail I c me (i ₹) 100 120 120 140 140 160 160 180 180 200

N mbe f ke 12 14 8 6 10

C e he di ib i ab e a le ha e c m la i e f e e c di ib i a d d a i gi e.

A e

A 1). Gi e ad a ic e a i i : 2 2k 6 = 0

: = 3 i a f ab e e a i The , (3)2 2k (3) 6 = 0 9 6k 6 = 0 3 6k = 0 3 = 6k k =

A 2). Smalle ime mbe = 2

Smalle c m i e mbe = 4 P ime fac i a i f 2 i 1 2 P ime fac i a i f 4 i 1 2 2

A : HCF (2, 4) = 2

A 3). Di a ce f m P ( , ) igi


P i f igi Q (0,0)

B a l i g di a ce f m la

( 2 - 1) + ( 2 - 1)

(0 - ) + (0 - )

+ i .

A The di a ce f a i ( , ) f m he igi i + i .

A 4). Gi e : c mm diffe e ce (d) = 4

he e e h e m (a7) = 4 T fi d: The fi e m (a) S ,

d = 4 a7 = 4

a + 6d = 4 a + 6( 4) = 4

a = 28

A The fi e m (a) i 28

A 5). S l i :

c 2 67 i 2 23 = c 2 67 c 2 (90 23 )


We k , [ i (90 ) = c ]

= c 2 67 c 2 67 = 0

A he al e f (c 267 i 223 ) i 0

A 6). Gi e , ABC PQR & = P Q AB

3 1

S , = a (P QR) a (ABC)

= ( )2 = 1/9

A = 1/9a (P QR) a (ABC)

A 7). Gi e , i i a i al mbe . 2 Le = m S e, 5 + 3 i a a i al mbe . 2 2 S , 5 + 3 = a/b 2 3 = a/b - 5 2

3 = 2 b a 5b

= 2 3b

a 5b

= m3b a 5b


B i a i al mbe a d ha be m, b hi3b a 5b

c adic he fac ha m = hich i i a i al 2 A He ce, 5 + 3 i al i a i al. 2

A 8).

Since ABCD is a rectangle. its opposite sides are equal.

AD = BC and AB = DC

x - y = 14

x + y = 30

Solving both the equations by elimination method,

Add the equations to get:

2x = 44

x = 22

Put the value of x in any equation, we get

x + y = 30


22 + y = 30

y = 8

Hence, x = 22 and y = 8

Ans 9).We know , m f e m f m 1 e m i

I he gi e e i ,

Fi e m = 3

C mm diffe e ce = 3

The , a l i g f m la f m ab e.

A mi g he al e f a 8 a d a a 3


Ans The sum of first 8 multiples of 3.

Ans 10). Let P divides line segment AB in the ratio k : 1. Le he a i be k : 1

B ec i f m la : , 1 2 1 2 2 1

1 2 1 2 2 1

( , ) = [ (k 2 + 1) / (k + 1) , (k 2 + 1) / (k + 1) ] (4, m) = [ k(6) + 2 / (k + 1) , k( 3) + 3 / (k + 1) ] 4 = (6k + 2) / (k + 1) a d m = ( 3k + 3) / (k + 1)

4k + 4 = 6k + 2

6k 4k = 4 2 2k = 2

k = 1

N , b i i g k = 1 i m' al e m = [ 3(1) + 3] / (1 + 1)

3+3 / 2 0 / 2

0


A 11). T al mbe a e 2, 3, 4, ........., 99 Le E be he e e f ge i g a mbe di i ible b 8. E = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96 = 12

P(E) = T a O c e F a ab e O c e

= 0.12248 12

Le E be he e e f ge i g a mbe di i ible b 8. The , P(E ) = 1 P(E) = 1 0.1224 = 0.8756

A 12).

HCF f 404 a d 96 404 = 2*2*101 96 = 2*2*2*2*2*3 C mm Fac = 2*2 HCF f 404 a d 96 = 4 LCM f 404 a d 96 404 = 2*2*101 96 = 2*2*2*2*2*3


LCM = 2*2*2*2*2*3*101 = 9696 LCM f 404 a d 96 i 9696 HCF LCM = P d c f he mbe

4*9696 = 404*96

38784 = 38784 LHS = RHS

A 14). Gi e : T e e f he l mial a e: (2 + ) a d (2 ). 3 3

Q ad a ic l mial i h e i gi e b : (2 + ) . (2 - ) 3 3

( 2 ) ( - (2 + ) 3 3 ( -2)2 - ( ) 2 3

2 - 4 + 4 - 3 2 - 4 + 1 = g( )

N , g( ) ill be a fac f ( ) g( ) ill be di i ible b ( )


F he e e , 2 2 1 = 0 2 2 2 + 1 = 0

2 ( 1) + 1 ( 1) = 0 ( 1) (2 + 1) = 0

1 = 0, 2 + 1 = 0

= 1, = -

He ce, Ze e f ( ) = 2 4 9 3 + 5 2 + 3 1 a e 1, - , 2 + , 2 - 3 3

A 15).


Gi e ide f a allel g am A ( 2,1), B ( ,0), C (4, ), D (1,2) We k ha diag al f Pa allel g am bi ec each he . Mid- i le a f diag al AC i gi e b

= ( ) a d = ( )2 1 + 2

2 1 + 2

O( ) ,2 2 + 4 2

1+ b

Mid- i le a P f diag al BD i gi e b

P( ) ,2 a + 1 2

0 + 2

P i O a d P a e ame

E a i g he c e di g c di a e f b h mid i , e ge

2 2 + 4 2

a + 1

a = 1 a d


2 1 + b 2

0 + 2

b = 1 a d N he Gi e c di a e f he a allel g am a e i e a A( 2,1), B(1,0), C(4,1), D(1,2)

B di a ce f m la, ( 2 1)2 ( 2 1)2 + We ca fi d he le g h f each ide AB = ( 1 )2 (1 0)2 2 + AB = = (3)2 (1)2 + 10

AB = CD ( ai f i e ide f he a allel g am a e a allel a d e al) BC = = (4 )2 (1 )2 1 + 0 10 BC = = (3)2 (1)2 + 10 BC=AD ( ai f i e ide f he a allel g am a e a allel a d e al )

AB=BC=CD=AD = 10

ABCD i a Rh mb


OR Gi e ABCD i ad ila e al.

B j i i g i A a d C, he ad ila e al i di ided i ia gle .

N , A ea f ad. ABCD = A ea f ABC + A ea f ACD A ea f ABC

A ea f ABC

= [ 1 2 - 3 2 3 - 1 3 1- 2 ]

= [ 5 5 6 4 6 7 1 7 5 ]

= [ 5(1) 4( 13) 1(12) ] = ( 5 52 12 ) = (35) . i

A ea f ADC

= [ 1 2 - 3 2 3 - 1 3 1- 2 ]

= [ 5 5 6 4 6 7 ( 1) 7 5 ]

= [ 5 5 6 4 6 7 ( 1) 7 5 ]


= [ 5(11) 4( 13) 1(2) ] = ( 55 52 12) = 109/2 . i . S , A ea f ad ila e al ABCD =

(35) (109) = 72 . i .

A 16). Le he al eed f he la e be km/h .

I c ea ed eed = 100 km/h

Time ake c e 1500 km = 1500/ h . Time ake c e 1500 km i h i c ea ed eed = 1500/ + 100 h Acc di g he Q e i ,

1500/ - 1500( + 100) = 30/60

+ 100 - 300000 = 0

+ 600 - 500 - 300000 = 0

( + 600)( - 500) = 0


= -600 500 (Neglec i g ega i e ig a eed ca be

ega i e)

= 500 A He ce, he al eed f he la e i 500 km/h .

A 17).

I ABC, AC2 = AB 2 + BC 2

a 2 + a 2 = 2a 2

AC = a 2 = a 2 2 A ea f e ila e al BEC (f med ide BC f a e ABCD)

= ( ide)24 3

= a24 3


A ea f e ila e al ACF (f med diag al AC f a e ABCD)

= ( )24 3 a 2

= 24

3 a 2

= 2 a 24 3

F m e . (i) a d (ii), a ACF = 2 a BCE O ,

a ( BCE) = a ( ACF)

He ce P ed, a ea f ia gle de c ibed e ide f a e i half he a ea f ia gle de c ibed i diag al.

OR


Gi e , ABC PQR A ( ABC) = a ( PQR) T P e: ABC ≅ PQR

= AB 2/PQ2 = BC 2/QR 2 = AC 2/PR 2a (P QR) a (ABC)

: A Ra i f a ea f imila ia gle i e al he a e f c e di g ide

B , a gi e , = 1a (P QR) a (ABC)

AB 2/PQ2 = BC 2/QR 2 = AC 2/PR 2 = 1

AB 2 = PQ2 / AB = PQ

BC 2 = QR 2 / BC = QR

AC 2 = QR 2 / AC = PR : - B SSS c g e ce a i m ABC PQR , he ce P ed.


A 18).

Le a ge PT a d QT a e d a ci cle f ce e O a h i fig e. B h he gi e a ge PT a d QT ch he ci cle a P a d Q e ec i el .

We ha e e :

le g h f PT = le g h f QT

C c i :- d a a li e egme ,f m ce e O e e al i T chi g i f a ge .

N POT a d QOT

We k a ge make he igh a gle i h he adi f he ci cle.

He e, PO a d QO a e adii . S , ∠OPT = ∠OQT = 90


POT a d QOT ∠OPT = OQT = 90

C mm h e e OT A d OP = OQ [ OP a d OQ a e adii]

S , R - H - S le f imila i POT QOT

He ce, OP/OQ = PT/QT = OT/OT PT/QT = 1

PT = QT [ he ce ed]

A 19). Gi e , 4 a = 3

Ta = = 4 3

B P

N , P = 3K, B = 4K,

We k , a = ( e e dic la / ba e)

, af e c m a i g , e e dic la = 3 a d ba e = 4

H e e = (3 + 4 ) = 5

N , i = e e dic la / h e e = 3/5 c = ba e / h e e = 4/5

N , (4 i - c + 1)/(4 i + c - 1) = (4 3/5 - 4/5 + 1)/(4 3/5 + 4/5 - 1) = (12 - 4 + 5)/(12 + 4 - 5)


= 13/11

A 20). Gi e a 2A = c (A 18 )

c (90 2A) = c (A 18 )

[ a =c (90 )] C m a i g a gle e ge , 90 2A = A 18

90+18=A+2A

3A=108

A= 108/3

A = 36

A 20).


Gi e , ABCD i a a e f ide = 12 cm.

P, Q, R a d S a e he mid i f ide AB, BC, CD a d AD e ec i el .

A ea f haded egi = A ea f a e 4 A ea f ad a =

a 2 4 1/4 2

= (12)2 3.14 (6)2

=144 3.14 36 =144 113.04 = 30.96 cm2

A 21).


Le he igi al heigh f c li de = 10cm Ba e adi f c li de = 3.5cm ba e adi f hemi he e = 3.5cm( ame a ha f c li de )

The al face a ea ld be he m f he c ed face a ea f

c li de a d he face a ea f 2 hemi he e .

face a ea f c li de = 2 h face a ea f e hemi he e = 2

OR

V l me f ice = l me f c e = He e , = 12 [ ∵ diame e = 24 m ] a d h = 3.5m

N , l me f ice = 1/3 22/7 (12) 3.5 m = 1/3 22/7 12 12 7/2 m =22 24 = 528 m Agai , a ea f ca a cl h = c e face a ea f c e = l

He e l = = 12 + 3.5 = 144 + 12.25 = 156.25 = 12.5 m


∴ A ea f ca a cl h = 22/7 12 12.5 m = 471.43 m

= = 1402 N

2 280

The c m la i e f e e c j g ea e ha 140 i 182.

Media cla i 10 15. l = 10, h = 5, N = 280, c.f. = 49 a d f = 133

MEDIAN = l + hf c.f . 2

N

= 10 + 5133

140 4 = 10 + 133

1 5 * = 10 + 455/133 = 13.42 A 13.42


A 23).

Gi e , eed f m b a i ill a e = 18 km/h .

Le eed f eam = km/h

S eed f b a d eam=(18 + ) km/h . A d eed f b a eam=(18 ) km/h .

Time f he eam j e = 24 18

Time f he eam j e = 24

18 + N , Acc di g he e i ,

- = 1 24 18

24 18 +

= 1(18 ) (18+ ) 24( 18 + ) 24 (18 )

= 1324 2 24 18 + 24 24 18 24 * *

48 / 324 2 = 1

48 = 324 2

2 + 48 324 = 0


2 + 54 6 324 = 0

( + 54) 6 ( + 54) = 0 ( + 54) ( 6) = 0

Ei he + 54 = 0 , = 54 N , a he eed ca be ega i e, - 6 = 0 , = 6 A S , he eed f he eam i 6 km/h

OR

Le igi al eed f he ai be km/h.

The , ime ake a el 63 km = 63/ h

Ne eed = ( + 6) km/h Time ake a el 72 km = 72/( + 6) h

Acc di g he e i ,

+ = 3 63 72

+6

O l i g, 135 +378 = 3 2 + 18 2 - 39 - 126 = 0


( - 42)( + 3) = 0

= -3 = 42 A he eed ca be ega i e, = 42

Th , he a e age eed f he ai i 42 km/h . A The a e age eed f he ai i 42 km/h .

A 24). Le he f c ec i e mbe i AP be (a 3d),(a d),(a+d) a d (a+3d) S , acc di g he e i .

a 3d+a d+a+d+a+3d=32 4a=32 a=32/4

a=8......(1) N , (a 3d)(a+3d)/(a d)(a+d)=7/15

15(a 9d )=7(a d ) 15a 135d =7a 7d 15a 7a =135d 7d 8a =128d

P i g he al e f a=8 i ab e e ge . 8(8) =128d 128d =512 d =512/128


d =4 d=2

S , he f c ec i e mbe a e 8 (3 2) 8 6=2 8 2=6 8+2=10 8+(3 2) 8+6=14 F c ec i e mbe a e 2,6,10 a d 14.

A 25).


ABC i a e ila e al ia gle , he e D i ide BC i ch a a ha BD = BC/3 . Le E i he i ide BC i ch a a ha

AE BC .

N , ABE a d AEC ∠AEB = ∠ACE = 90 AE i c mm ide f b h ia gle , AB = AC [ all ide f e ila e al ia gle a e e al ]

F m R - H - S c g e ce le , ABE ACE

∴ BE = EC = BC/2 N , f m P hag a he em ,

ADE i igh a gle ia gle ∴ AD = AE + DE ------(1) ABE i al a igh a gle ia gle ∴ AB = BE + AE ------(2)

F m e a i (1) a d (2) AB - AD = BE - DE = (BC/2) - (BE - BD) = BC /4 - (BC/2) - (BC/3) = BC /4 - (BC/6) = BC /4 - BC /36 = 8BC /36 = 2BC /9

AB = BC = CA


S , AB = AD + 2AB /9

9AB - 2AB = 9AD

He ce, 9AD = 7AB

OR

Gi e : A igh a gled ABC, igh a gled a B T P e- AC =AB +BC C c i : d a e e dic la BD he ide AC . P f: We k ha if a e e dic la i d a f m he e e f a igh a gle

f a igh a gled ia gle he h e e, ha ia gle b h ide f he e e dic la a e imila he h le ia gle a d each he .

We ha e,

ADB∼ ABC. (b AA imila i )


The ef e, AD/ AB=AB/AC (I imila T ia gle c e di g ide a e i al) AB =AD AC ..(1) Al , BDC∼ ABC The ef e, CD/BC=BC/AC (i imila T ia gle c e di g ide a e i al) O , BC =CD AC ..(2) Addi g he e a i (1) a d (2) e ge , AB +BC =AD AC+CD AC AB +BC =AC(AD+CD)


( F m he fig e AD + CD = AC) AB +BC =AC . AC The ef e, AC =AB +BC

A 26). S e f c c i :- (i) D a a li e egme BC i h mea eme f 6 cm.

(ii) N c c a gle 60 f m i B a d d a AB = 5 cm.

(iii) J i he i C i h i A. Th ABC i he e i ed ia gle.

(i ) D a a li e BX hich make a ac e a gle i h BC a d i i e f e e A.

( ) C f e al a f li e BX amel BB1, BB2, BB3, BB4.

( i) N j i B4 C. D a a li e B3C' a allel B4C.

( ii) A d he d a a li e B'C' a allel BC.

He ce AB'C' i he e i ed ia gle.


A 27). L.H.S. = ( Si A - 2Si 3A ) / ( 2c 3A - c A)

= (1 - 2 i 2A ) / [ 2 ( 1- i 2A) - 1] Ac A

= (1 - 2 i 2A ) / [ 2 - 2 i 2A - 1] Ac A

= (1 - 2 i 2A ) / [ 2 - 2 i 2A - 1] Ac A

= (1 - 2 i 2A ) / (1 - 2 i 2A ) Ac A

= Ta A ( He ce P ed )


A 28). Given:

Diameter of upper end of bucket = 30 cm

Radius of the upper end of the frustum of cone, r1 = 15 cm

Diameter of lower end of bucket = 10 cm

Radius of the lower end of the frustum of cone, r2 = 5 cm

Height of the frustum of Cone, h = 24 cm

Slant height of bucket, L = √[(h² + (r1 - r2)²]

L = √[24² + (15 - 5)²]

L = √(576 + 10²)

L = 26 cm

Area of metal sheet = Curved Surface Area of bucket + area of lower end

= π(r1 + r2)L + πr2²

= 3.14(15 + 5) × 26 + π(5)²

= 3.14 × 20 × 26 + 25 × 3.14

= 1711.3 cm²

Hence, the Area of metal sheet used to make the bucket is 1711.3 cm².

ii). We should avoid bucket made by ordinary plastic because it is non-biodegradable. It is harmful for our environment.


Ans 29).

Heigh f ligh h e = 100 m

The a gle f de e i f 2 hi a e 30 a d 45 deg ee.

T Fi d: If e hi i e ac l behi d he he he ame ide f he ligh h e, he fi d he di a ce be ee he 2 hi

S l i :

Refe he a ached fig e

I ABC

AB = Heigh f e = Pe e dic la = 100 m

BC = Ba e

∠ACB = 45

We ill e ig me ic a i fi d he le g h f ba e :

I ABD

AB = Heigh f e = Pe e dic la = 100 m

BD = Ba e


∠ADB = 30

We ill e ig me ic a i fi d he le g h f ba e :

U e 3 = 1.732

BD= 100 * 1.732

BD = 173.2 m

N e a e e i ed fi d he di a ce be ee he 2 hi i.e. CD = BD-BC.

S , CD = BD-BC.

CD = 173.2 m - 100 m

CD= 73.2 m

Th he di a ce be ee he 2 hi i 73.2 m.

A 30).


⟹ 720 704 20f 18f

⟹ 2f 16

⟹ f 8

OR



CBSE Ma he a ic 2017

G a I c : ( ) A a c . ( ) T a c 30 d d d

c A, B, C a d D. ( ) S c A c a 6 1 a ac . S c B c a 6

2 a ac , S c C c a 10 3 a ac . S c D c a 8 4 a ac .

( ) T a c c . H , a a c c a b

d d 3 a ac a d 3 4 a ac . Y a a a a a c

. ( ) U a ca c a d.


Q e i

Sec i -A (1 Ma Each)

1. Wha i he c diffe e ce f a A.P. i hich a 21 a 7 = 84 ? 2. If he a g e be ee a ge d a f a e e a i P a ci c e f adi a a d ce e O, i 60 , he fi d he e g h f OP 3. If a e 30 high, ca a had 10 g he g d,3 he ha i he a g e f e e a i f he ?

4. The babi i f e ec i g a e a e a d f a hea f 900 a e i 0 18. Wha i he be f e a e i he hea ?

Sec i - B (2 Ma Each)

5. Fi d he a e f , f hich e f he ad a ic e a i 2 14 + 8 = 0 i 6 i e he he . 6. Which e f he g e i 20, 19 , 18 , 17 , ... i he fi4

121

43

ega i e e ? 7. P e ha he a ge d a a he e d i f a ch d f a ci c e a e e a a g e i h he ch d


8. A ci c e che a he f ide f a ad i a e a ABCD. P e ha AB + CD = BC + DA

9. A i e i e ec he -a i a d -a i a he i P a d Q e ec i e . If (2, 5) i he id i f PQ

10. If he di a ce f P( , ), f A(5, 1) a d B( 1, 5) a e e a , he

e ha 3 = 2 .

Sec i - C (3 Ma Each)

11. If ad bc, he e ha he e a i (a 2 + b2 ) 2 + 2 (ac + bd) + (c 2 + d2 ) = 0 ha ea . 12. The fi e f a A.P. i 5, he a e i 45 a d he f a i e i 400. Fi d he be f e a d he c diffe e ce

f he A.P 13. O a aigh i e a i g h gh he f f a e , i C a d D a e a di a ce f 4 a d 16 f he f e ec i e . If he a g e f e e a i f C a d D f he f he e a e

c e e a , he fi d he heigh f he e . 14. A bag c ai 15 hi e a d e b ac ba . If he babi i f d a i g a b ac ba f he bag i h ice ha f d a i g a hi e ba , fi d he be f b ac ba i he bag. 15. I ha a i d e he i ( 24/11 , ) di ide he i e eg e j i i g he i P(2, 2) a d Q(3, 7) ? A fi d he a e f .


16. Th ee e ici c e each f dia e e 3 c , a ci c e f dia e e 4 5 c a d a e ici c e f adi 4 5 c a e d a i he gi e fig e. Fi d he a ea f he haded egi .

17. I he gi e fig e, c ce ic ci c e i h ce e O ha e adii 21 c a d 42 c . If ∠ AOB = 60 , fi d he a ea f he haded egi . [U e ] 22/7

18. Wa e i a ca a , 5 4 ide a d 1 8 dee , i f i g i h a eed f 25 /h . H ch a ea ca i i iga e i 40 i e , if

10 c f a di g a e i e i ed f i iga i ?


19. The a heigh f a f f a c e i 4 c a d he e i e e f i ci c a e d a e 18 c a d 6 c . Fi d he c ed face a ea f he f .

20. The di e i f a id i c b id a e 4 4 2 6 1 0 . I i e ed a d eca i a h c i d ica i e f 30 c i e adi a d hic e 5 c . Fi d he e g h f he i e.

Sec i - D (4 Ma Each)

21. S e f :

+ = , 1, , 41+1

35 +1

5+ 4 5

1

22. T a i g ge he ca fi a a i 3 h . If e a113

a e 3 h e ha he he fi he a , he h ch i e i each a a e fi he a ?

23. If he a i f he f he fi e f A.P. i (7 + 1) : (4 + 27), he fi d he a i f hei 9 h e 24. P e ha he e g h f a ge d a f a e e a

i a ci c e a e e a . 25. I he gi e fig e, XY a d X'Y' a e a a e a ge a ci c e

i h ce e O a d a he a ge AB i h i f c ac C, i e ec i g XY a a d X'Y' a B. P e ha ∠ AOB = 90 .


26. C c a ia g e ABC i h ide BC = 7 c , ∠ B = 45 , ∠ A = 105 . The c c a he ia g e h e ide a e 3 4 i e he c e di g ide f he ABC 27. A ae a e i f i g a a heigh f 300 ab e he g d. F i g a hi heigh , he a g e f de e i f he ae a e f

i b h ba f a i e i i e di ec i a e 45 a d 60 e ec i e . Fi d he id h f he i e . [U e = 1 732]3 28. If he i A( + 1, 2 ), B(3 , 2 + 3) a d C(5 1, 5 ) a e c i ea , he fi d he a e f . 29. T diffe e dice a e h ge he . Fi d he babi i ha he be b ai ed ha e (i) e e , a d (ii) e e d c

30. I he gi e fig e, ABCD i a ec a g e f di e i 21 c 14 c . A e ici c e i d a i h BC a dia e e . Fi d he a ea a d he

e i e e f he haded egi i he fig e.


31. I a ai - a e ha e i g e , he ai - a e f a f f 22 20 d ai i a c i d ica a ha i g a dia e e f ba e 2

a d heigh 3.5 . If he a i f , fi d he ai fa i c . W i e ie a e c e a i .

A e

A 1). Gi e , a21 a7 = 84

(a + 20d) (a + 6d) = 84

a + 20d a 6d = 84

20d 6d = 84

14d = 84

d = 6

S , he c diffe e ce i 6


A 2).

Gi e , ∠ A B = 60 a d ∠ A = 30

I igh a g e A , = c ec30OAOP

= 2OP

OP = 2a

He ce, he e g h f OP i 2a.

A 3). Gi e : Heigh f he e - 30

Le g h f had he g d - 0 1 3

S i :

I ABC , a = ABBC

Ta = 3010 3

Ta = Ta 60


= 60

He ce, he a g e f e e a i i 60

A 4).

Gi e : P babi i f e ec i g a e a e a d f a hea f 900 a e i 0 18

T a a e = 900

P(E) = 0.18

= 0.18T . N .

= 0.18900 N .

N . f e a e = 900 0 18 = 162

A : He ce, he . f e a e i 162.

A 5). Gi e : Q ad a ic e a i - 2 14 + 8 = 0

S , e e =

he i be = 6

S f =

+ 6 = ( 14)


7 = 14

= 2

We , P d c f =

* 6 = 8

6 2 = 8

O P i g a e f a f e . (i)

6 )2 = ( 2 8

6* 4/ 2 = 8

24 = 8 2

8 2 24 = 0

8 ( 3) = 0

Ei he 8 = 0 = 0

O

3 = 0 = 3

F = 0, he gi e c di i i be a i fied

SO, a e i be = 3


A 6).

Gi e , A.P. i 20, 19 , 18 , 17 .4 1

2 1

4 3

S0, hi i e a 20, , , .4 77

2 37

4 71

He e, a = 20, d = - 20 = = 4 77

4 77 80

4 3

Le a i fi ega i e e

a + ( 1)d < 0

20 + ( 1) ( ) < 04 3

20 ( ) + < 04 3

4 3

20 + < 4 3

4 3

< 4 83

4 3

> 4 83

4 3

> 27.663 83

( SI ce e ca be i f ac i , d i ff)

A : 28 h e i be he fi ega i e e f he gi e A.P.


A 7). Gi e , a ci c e f adi OA a d ce ed a O i h ch d AB a d a ge PQ & RS a e d a f i A a d B e ec i e .

Le AB be a ch d f a ci c e i h ce e O a d e AP a d BP be he a ge a A a d B.

Le he a ge ee a P.J i OP S e OP ee AB a C.

T e : ∠PAC = ∠PBC

P f : I PAC a d PBC

PA = PB [Ta ge f a e e a i a ci c e a e e a ]

∠APC = ∠BPC [PA a d PB a e e a i c i ed OP]

PC = PC [C ]

PAC PBC [SAS C g e ce]

∠PAC = ∠PBC [C.P.C.T]

He ce P ed.

A 8). Gi e : A ci c e che a he f ide f a ad i a e a ABCD

T P e: AB + CD = BC + DA


A i ' ha he e g h f a ge d a f a e e a i he ci c e a e e a , hich ea , AP=AS

BP=BQ

CR=CQ

DR=DS

(AP+PB)+(CR+RD)=(AS+SD)+(BQ+QC)

AB+BC=AD+BC

A 9). Le he i P( ,0) a d Q(0, ) (beca e a i , =0 a d a i , =0 ) a d id i (2, )

N , fi d id i

= 22 0+

= -52 0 +


S , =4, = 10

A - The c di a e f P: (4,0) : (4,0) a d (0, 10)

A 10).

Gi e : Di a ce f P( , ), f A(5, 1) a d B( 1, 5) a e e a T e: 3 = 2

A i i gi e , PA=PB

= ( ) ( )5 2 + 1 2 ( ) ( )+ 1 2 + 5 2 O a i g b h he ide :

= ( ) ( )5 2 + 1 2 ( ) ( )+ 1 2 + 5 2

2 + 25 -10 + 2 +1 - 2 = 2 + 1 + 2 + 2 + 25 - 10

10 2 = 2 10 10 2 = 10 + 2

12 = 8 3 = 2

He ce, P ed!

A 11) Gi e , ad bc (a 2 + b2) 2 + 2(ac + bd) + (c 2 + d2) = 0 D = b2 4ac


= [2 (ac + bd)] 2 - 4(a 2 + b2)(c 2 + d2) = 4 [a 2 c 2 + b2 d2 + 2abcd] 4 (a 2 c 2 + a 2 d2 + b2 c 2 + b2 d2 ) = 4 [a 2 c 2 + b2 d2 + 2abcd - a 2 c 2 - a 2 d2 - b2 c 2 - b2 d2 ] = 4 [ a 2d2 b 2c 2 + 2abcd] = 4 [a 2d2 + b2c 2 2abcd] = 4 [ad bc] 2

He ce, D i ega i e!

A 12). Gi e : a = 5, a = 45, S = 400 S , a e , S = /2 [a + ] he e i he a e

400 = /2 [5 + 45 ] 400 = /2 [50] 25 = 400 = 400/25 = 16

N , a e , a = a + ( 1) d

45 = 5 + (16 1) d 45 5 = 15d 15d = 40 d = 8/3

He ce, = 16 a d d = 8/3

A 13). Gi e : Di a ce f 4 a d 16 Le heigh AB f e = h


N , a e ca ee, i ABC

= = a (90 )AC AB

= c 4

I ABD,

= aAC AB

= a 16

M i e . (i) a d (ii)

= c a 4

16

h2/64 = 1

[ c a = 1/ a a = 1]

h2 / 64 h = 8

Heigh f e = 8


A 14). Gi e , . f hi e ba = 15

Le . f b ac ba =

T a ba = (15 + )

Acc di g he e i , P (B ac ba ) = 3 P (Whi e ba )

= 3 (15 + )

15(15 + )

= 45

He ce, f b ac ba i he bag = 45

A 15). Le i R di ide PQ i he a i : 1

We ha , R = ( , )1 + 2 1 2 + 2 1

1 + 2 1 2 + 2 1

(24/11 , ) = ( , ) + 1 (3) + 1(2)

+ 1 (7) + 1( 2)

= , + 1 3 + 2

+ 1 7 + 2

= 3 + 2 = 24

11 (3 + 2) = 24 ( + 1)

33 + 22 = 24 + 24

33 24 = 24 22

9 = 2

= 2/9


S , :1 = 2:9

= = + 1 7 2

+ 19 2

7 ( ) 2 9 2

O i g a d i g he a e f

= 11 4

A : Li e PQ di ide i he a i 2 : 9 a d a e f = 11 4

A 16). Gi e : Radi f a ge e i-ci c e = 4 5 c

A ea f a ge e i-ci c e = R 2 = 4 5 4 52 1

2 1

7 22

Dia e e f i e ci c e = 4 5 c , = 4.5/2 c

A ea f i e ci c e = 2 = 7 22

2 4.5

2 4.5

Dia e e f a e i-ci c e = 3 c , = 2 3

A ea f a e i-ci c e = 22 1

2 1

7 22

2 3

2 3

2 3

A ea f haded egi = A ea f a ge e i ci c e + A ea f 1 a e ici c e A ea f i e ci c e A ea f 2 a e i ci c e


( 4.5 4.5) + ( - 2 1

7 22

2 1

7 22

2 3

2 3

7 22

2 4.5

) - ( 2 )2 4.5

2 1

7 22

2 3

2 3

O i g, = = 28 990 643.5

28 346.5

A - 12.37 c 2 a i a e

A 17).

S i : A g e f haded egi = 360 60 = 300

A ea f haded egi

= (R 2 - 2) 360

= [42 2 - 21 2]7 22

360 300

= 63 217 22

6 5


He ce, he a ea f he haded egi = 3465 c 2

A 18).

Wid h f ca a =5.4

De h f ca a =1.8

I 60 i ., 25 f a e f h gh i .

Which ea . i 40 i , he e i be

40 = , i be a ai ab e f i iga i .60 25

3 50

He ce, he e f a e i he ca a = e f a d i iga ed.

V e f ca a = b h

. 1. 33 50000

A ea f a d i iga ed if 10cm a di g a e i e i ed

= . 1. 33 50000 1

0.1

= 16,20,000 2

A 19). S i :

S a heigh f f ( ) = 4 c


Pe i e e f e = 18 c 2 R = 18 c ; R = 9 c

Pe i e e f e b = 6 c

2 = 6 ; = 3/ c

C ed S.A. f f = [R + ]

= 4 [ ] 9 + 3

= 4 = 48 c 2 12

He ce, he c ed face a ea f he f i 48 c 2

A 20). Gi e :

I e adi f i e = 30 c Thic e f i e = 5 c O e adi = 30 + 5 ;R = 35 c

N , acc di g he e i , a id i c b id i e ed a d eca i a h c i d ica i e.

S , V . f h i e = V . f c b id

h(R 2 2 ) = b h

h[ 35 2 - 30 2 ] = 4.4 2.6 1 100 100 1007 22

h 65 5 = 44 26 1 100 1007 22

O i g h,

h = = 11200 c / 11222 65 5 44 26 100 100 7

A 21).


Gi e : + = 1 + 1

3 5 +1

5 + 4

- = - 1 + 1

5 + 4

3 5 +1

= -( + 1)( + 4) ( + 4) 5( + 1) 3

5 +1 = -( 2 + 5 +4)

( + 4 5 5) 3 5 +1

= - ( 4 1) ( 2 + 5 +4)

3 5 +1

(4 + 1) (5 + 1) = 3( 2 + 5 + 4)

20 2 + 4 + 5 + 1 = 3 2 + 15 + 12 17 2 6 11 = 0 17 2 17 + 11 11 = 0 17 ( 1) + 11 ( 1) = 0 ( 1) (17 + 11) = 0

A Ei he = 1 = 11/ 17

A 22). Le a fi b e a = h Le a fi b he a = ( + 3) h

A gi e , ge he he fi b h13 40

Which ea + = 1

1 +3

1340

= + 3 + ( ) ( +3)

1340

(2 + 3) / ( 2 + 3 ) = 1340


13 2 + 39 = 80 + 120 13 2 41 120 = 0

13 2 65 + 24 120 = 0 13 ( 5) + 24( 5) = 0 ( 5) (13 + 24) = 0

Ei he 5 = 0 13 + 24 = 0

= 5, = 24/13 ca be 24/13 i ce e a e a i g ab i e. S , = 5

O e a fi he a i 5 h he a fi he a i 5 + 3 = 8 h

A 23). We , Ra i f f fi e f A.P. a e:

= 2 + ( 1)2 2A + ( 1)D2

7 + 14 + 27

N , P = 17

= = 24/19 2 + 16 2A + (16)D 95

120

O i if i g,

= 24/19 + 8 A + 8D


A Ra i f 9 h e f A.P. i 24 : 19

A 24).

S a e e : The a ge d a f a e e a i a ci c e a e e a .

T e : PA = PB C : J i adi OA a d OB a j i O P.

P f : I OAP a d OBP OA = OB (Radii)

∠A = ∠B (Each 90 ) OP = OP (C ) AOP BOP (RHS c g.)

PA = PB (c c ) A He ce, he a e e i ed.

A 25).

Gi e :


XY a d X'Y' a e a a e a ge a ci c e i h ce e O a d a he a ge AB i h i f c ac C, i i e ec i g XY a A a d X'Y' a B

Le j i i O C

I OPAa d OCA

OP=OC (Radii f he a e ci c e) AP=AC (Ta ge f i A)

AO=AO (C ide) OPA OCA (SSS c g e ce c i e i )

S , ∠POA=∠COA.........(1) Si i a ,

∠QOB ∠OCB ∠QOB=∠COB.........(2)

Si ce,POQ i he dia e e f he ci c e, i i a aigh i e. The ef e, ∠POA+∠COA+∠COB+∠QOB=180


∘

S , f e a i (1) a d e a i (2)

2∠COA+2∠COB=180∘ ∠COA+∠COB=90∘

∠AOB=90 0

He ce, P ed.

A 26). G : BC = 7 c , ∠ B = 45 , ∠ A = 105 .

∠A + ∠B + ∠C = 180

[A g e e ]


105 + 45 + ∠C = 180 150 + ∠C = 180 ∠C = 180 - 150

∠C= 30

S e f C c i :

D a a i e eg e BC= 7 c .

Ma e a a g e f 45 a i B a d 30 a i C, hich i e ec

each he a i A. Th , ABC i a gi e T ia g e.

N f B d a a a BY b a i g a ac e ∠CBY i h ba e BC he ide i e he e e A.

N , ca e 4 i B1 ,B2 , B3 ,B4 BY ch ha BB1= B1B2=

B2B3 =B3B4.

J i B3C, a d d a a i e h gh B4M B3 C i e ec i g he

e e ded i e BC a M.

F M d a OM AC i e ec i g he e e ded i e BA a O.

A Th , OBM i he e i ed ia g e h e ide a e ( ) i e

f he c e di g ide f ABC.


A 27). Le ae a e i a A, 300 high f a i e . C a d D a e i e ba f i e .

S , I igh ABC, = c 60 BC

AB

= 300 / 1 3

= = 100 3 300 3

3 3

= 100 1 732 = 173 2


I igh ABD,

= c 45BD AB

= 1 300 = 300 Wid h f i e = + = 173 2 + 300 = 473 2 A Wid h f i e i 473 2 .

A 28). Gi e : A( + 1, 2 ), B(3 , 2 + 3) a d C(5 1, 5 ) a e c i ea

T fi d: he a e f .

if A ( 1, 1) , B( 2, 2), C( 2, 2) a e h ee c i ea i he

A ea f ia g eABC = 0

[ 1( 2- 1) + ( 3- 1) + 3 ( 1- 2) ] =01 2

Gi e A( +1,2 ) ,B(3 ,2 +3) ,C(5 -1,5 ), he ef e

( +1)[2 +3-5 ]+3 [5 -2 ]+(5 -1)[2 -(2 +3)] =01 2

( +1)[3-3 ]+3 *3 +(5 -1)(-3)=0


3 -3 +3-3 +9 -15 +3=0

6 -15 +6=0

Di ide each e i h 2

2 -5 +2=0

2 -4 - +2=0

2 ( -2)-( -2)=0

( -2)(2 -1)=0

-2 = 0 2 -1 =0

=2 = 1 2

A 29). Whe diffe e dice a e h ge he

T a c e = 6 6 = 36

(i) F e e Fa ab e c e a e (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)

N . f fa ab e c e = 18

P (e e ) = = = T 18

36 1 2

(ii) F e e d c Fa ab e c e a e (1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

N . f fa ab e c e = 27


P (e e ) = = = T 27

36 34

A 30).

A ea f haded egi = A ea f ec a gle - A ea f emi-ci cle

= (21 14) 7 7

= 294 77

= 217 cm2

Pe ime e f haded egi = 21 + 14 + 21 + 7

= 56 + 22 = 78 cm

A A ea f haded egi = 78 cm

A 31). Gi e : Ba e 2 m a d Heigh 3 5 m.


Rec a g a f : =22 ,b=20

C i d ica a : =1 ,h=3.5

The e f a = 2h

= 22/7 1 1 3.5

= 11 3

A ea f f = b = 22 20 = 440 2

Rai fa =

A

Rai fa = 11 3/440 3

= 0.025 25 c


CBSE Ma he a ic 2016

G I : ( ) A . ( ) T 30

A, B, C D. ( ) S A 6 1 . S B 6

2 , S C 10 3 . S D 8 4 .

( ) T . H ,

3 3 4 . Y

. ( ) U .


Q e i

Sec i -A (1 Ma k Each)

1. I ABC, D a d E a e i AC a d BC e ec i el ch ha DE AB. If AD = 2 , BE = 2 1, CD = + 1 a d CE = 1, he fi d he al e f .

2. I A, B a d C a e i e i a gle f ABC, he e ha :

= 2Sin (B+C)

2CosA

3. If = 3 i a d = 4 c , fi d he al e f (16 2 + 9 2)

4. If e i ical ela i hi be ee ea , edia a d de i e e ed a ea = k(3 edia de), he fi d he al e f k.

Sec i -B (2 Ma k Each)


5. E e 23150 a d c f i i e fac . I i i e ? 6. S a e he he he eal be 52.0521 i a i al . If i i a i al e e i i he f / , he e , a e c - i e, i ege

a d 0. Wha ca a ab i e fac i a i f ? 7. Gi e he li ea e a i 2 6 = 0, i e a he li ea e a i i he e a iable , ch ha he ge e ical e e e a i f he ai f ed i : (i) c i cide li e (ii)

i e ec i li e 8. I a i cele ABC igh a gled a B, e ha AC 2 = 2AB 2 . 9. P e he f ll i g ide i :

[ ] 2 = a 2A ∠A i ac e1 cotA1 tanA

10. Gi e bel i a c la i e f e e c di ib i able. C e di g i , ake a di a f e e c di ib i able

cf

M e ha e al 0 45

M e ha e al 10 38

M e ha e al 20 29

M e ha e al 30 17

M e ha e al 40 11

M e ha e al 50 6


Sec i -C (3 Ma k Each) 11. Fi d LCM a d HCF f 3930 a d 1800 b i e fac i a i

e h d. 12. U i g di i i alg i h , fi d he ie a d e ai de di idi g f( ) b g( ) he e f( ) = 6 3 + 13 2 + 2 a d g( ) = 2 + 1. 13. If h ee e e f a l ial 4 3 3 2 +3 a e 0, a d - ,3 3 he fi d he f h e .

14. S l e he f ll i g ai f e a i b ed ci g he a ai f li ea e a i :

- = 2x1 4

+ = 9x1 3

15. ABC i a igh a gled ia gle i hich ∠B = 90 . D a d E a e a i AB a d BC e ec i el . P e ha AE2 + CD 2 = AC 2 + DE2 . 16. I he gi e fig e, RQ a d TP a e e e dic la PQ, al TS ⊥ PR e ha ST.RQ = PS.PQ.


17. If ec A = , fi d he al e23

f:

+ tanAcosA tanA

1+ sinA 18. P e ha : ec 2 c 2(90 ) = c 2(90 ) + c 2 . 19. F he h f Feb a , a cla eache f Cla IX ha he f ll i g ab e ee ec d f 45 de . Fi d he ea be f da , a de a ab e . N be f da f ab e 0 4 4 8 8 12 12 16 16 20 20 24

N be f de 18 3 6 2 0 1

20. Fi d he i i g f e e c ( ) f he f ll i g di ib i , if

de i 34.5 :

C.I. F e e c


0 10 4

10 20 8 = f0

l = 20 30 10 = f1

30 40 = f2

40 50 8

Sec i -C (4 Ma k Each)

21. P e ha i a i a i al be . He ce h ha 3 + 2 i5 5 al a i a i al be . 22. Ob ai all he e e he l ial 4 + 6 3 + 2 24 20, if

f i e e a e + 2 a d 5. 23. D a g a h f f ll i g ai f li ea e a i : = 2( 1) 4 +

= 4 Al i e he c di a e f he i he e he e li e ee -a i a d -a i .

24. A b a g e 30 k ea a d 44 k d ea i 10 h . The a e b a g e 40 k ea a d 55 k d ea i 13 h . O hi i f a i e de g e ed he eed f he b a i ill a e a 8.5 k /h a d eed f he ea a 3.8 k /h. D

ag ee i h hei g e ? E lai ha d e lea f he i cide ? 25. I a e ila e al ABC, E i a i BC ch ha BE = 1/4 BC. P e ha 16 AE2 = 13 AB 2 .


26. I he fig e, if ∠ABD = ∠XYD = ∠CDB = 90 . AB = a, XY = c a d CD = b, he e ha c (a + b) = ab. 27. I he ABC ( ee fig e), ∠A = igh a gle, AB = a d BC =x

E al a e i C. c C. a C + c 2C. i Ax + 5

28. If = a d = , he h ha ( 2 + 2 ) c 2A = 2 .SinAcosB

cosAcosB

29. P e ha : = ( )2secA 1secA + 1

sinA1 + cosA

30. F ll i g able h a k ( f 100) f de i a cla e :

Ma k N . f de

M e ha e al 0 80

M e ha e al 10 77

M e ha e al 20 72

M e ha e al 30 65

M e ha e al 40 55

M e ha e al 50 43

M e ha e al 60 28

M e ha e al 70 16

M e ha e al 80 10

M e ha e al 90 8


M e ha e al 100 0 D a a ' e ha e' gi e. F he c e, fi d he edia . Al , check he al e f he edia b ac al calc la i .

A e A 1): Gi e :

1. DE BC 2. AD = 3. DB = - 2 4. AE = + 2 a d EC = - 1.

T fi d: Val e f I ABC, e ha e

DE BC , b Thale' he e

= ADDB

AEEC

AD EC = AE DB

( -1) = ( -2)( +2)


2 - = 2 - 4 = 4 A e : The al e f i 4.

A 2): Gi e : A,B a d C a e i e i a gle f ABC

N , ∠A+ ∠B+ ∠C = 180 f i e i a gle f ia gle ABC S , ∠B+∠C= 180 - ∠A.

N , M l i l b h ide b 21

( ∠B+∠C) = ( 180 - ∠A) = ( 90 - A/2) [ e ed he b acke ]21

21

N , ake i e f 2(∠B+∠C)

= [ Si ce, = ]2

Sin (∠B+∠C)2

Sin (90 A)2

(∠B+∠C)2

(90 A)

N , e k Si (90- ) = C

S ,

= 2Sin (90 A)

2Cos A

He ce, P ed.


A 3) Gi e :

= 3 i a = 4c a

T fi d, N , b i e 3 i a a he lace f a d 4c a a he lace f

N , e k ha Si 2 + C 2 = 1 O a l i g hi i he l i

( i ce i 2a + C 2a bec e 1)


A 4). Gi e : ea = k(3 edia de) We , e E ca f a

M d = 3 M d a - 2 M a

S ,

2 Mea = 3 Med a - M de

Mea = 23Median Mode

We a e gi e , ea = k(3 edia de) S , b i i g he al e

k(3 edia de) = 23Median Mode

k =

He ce, he al e f k i

A 5). F da e al he e f a i h e ic a e ha he e a i f i e fac i a i f a be i i e. S , he e a i f i e fac i a i f 23150: 23150 = 2 5 5 463


A d 6). Ye , 52.0521 i a i al a i ca be e e ed i he / f

52.0521 = 10000520521

A 7).

The li e a1 + b1 + c1 a d a2 + b2 + c2 = 0 a e c i cide if

= = a2a1

b2b1

c2c1

S , a he li e ca be 2 -4 -12=0

C di i f li e be i e ec i g :

a2a1

b2b1

S , a he li e ca be -4 -6=0

A 8).

Gi e : I ia gle ABC a gle B= 90

T P e: AC 2 = 2AB 2

We k , B P hag a he e

AB 2 + BC 2 = AC 2

A d, b i cele ia gle e ,


AB = BC

S , AB 2 + AB 2 = AC 2 ( Si ce, AB =BC)

The ef e, 2AB 2 = AC 2

He ce, ed 2AB2 = AC 2

A 9). T e, [ ] 2 = a 2A1 cotA1 tanA

[ ] 2 (We k , C A = 1/Ta A)1 cotA1 tanA

S , [ ] 21 tanA1 1/tanA

[ ] 2 tanAtanA 1

1 tanA

[ ] 2tanA 1tanA (1 tanA)

F de i a , ake (-)

[ ] 2 (1 tanA)tanA (1 tanA)

Red ce he i le f

[ - a A ]2

= a A

He ce, P ed!


A 10). We f d e d a f e e c (f )

45 38=7

38 29=9

29 17=12

17 11=6

11 6=5

6=6

TABLE

cf fi

M e ha e al 0

45 7

M e ha e al 10

38 9

M e ha e al 20

29 12

M e ha e al 30

17 6

M e ha e al 40

11 5

M e ha e al 6 6


50

A 11). HCf(3930,1800)= 30 LCM(3930,1800)=235800

S e -b - e e la a i :

We ha e ,

2 3930

________

3 1965

________

5 655

________

* 131

2 1800

__________

2 900

__________

2 450

__________

5 225

__________

5 45


__________

3 9

__________

3

N ,

3930= 2 3 5 131

a d

1800= 2 2 2 3 3 5 5

= 2 3 5

HCF(3930,1800)=2 3 5 = 30

P d c f he alle e f each c i e fac f he be

LCM(3930,1800)= 2 3 5 131

= 235800

A 12). Di ide d = Di i =

Di ide d = (Di i *Q ie ) + Re ai de

6 3 + 13 2 + - 2 = (2 + 1 * 3 2) + (10 2 + - 2)

6 3 + 13 2 + - 2 = (2 + 1 * 3 2 + 5 ) + (-4 - 2)

6 3 + 13 2 + - 2 = (2 + 1 * 3 2 + 5 -2) + 0


He ce, he e ai de c e be 0 a d ie i 3 2 + 5 -2

A 13). Gi e :

P l ial: 4 3 3 2 +3

R : 0, a d - 3 3

T fi d: 4 h

If 0, a d - a e f he l ial, ,( - 3) a d ( + 3) d be e3 3

fac .

Le a e f f e e a , c ea , ( -a) d be e fac

S ,

( - 3)( + 3)( -a) = 4 3 3 2 +3

( -3)( -a) = 4 3 3 2 +3 [ (a+b)(a-b) = a 2-b2 ]

4-a 3-3 2+3a = 4 3 3 2 +3

O c a i g b h he ide , e ge a=1

He ce, he f h i a = 1.

A 14).


A 15). A ABE , ,

(AE) 2 = (AB) 2 +(BE)2 ---------------> E (1)

S , DBC , ,

(CD)2 =(BD)2 + (BC) 2 -----------------> E (2)

O (1) (2), (AE) 2 + (CD) 2 = (AB) 2 +(BE)2 + (BD) 2 + (BC) 2

(AE)2 + (CD) 2 = (AB 2 + BC 2) + (BE 2 + BD 2) -----------------> E (3)

I ABC,

AC 2 =AB 2 +BC 2

I DBE

DE2 = BE 2 + BD 2 -----------------> E (4)


O E (4) E (3) (AE) 2 + (CD) 2 = (AC 2) + (DE 2) -------- H P

A 16).

Gi e : RQ a d TP a e e e dic la PQ, al TS ⊥ PR T P e: ST.RQ = PS.PQ.

S l :B a gle e

R + P + Q = 180

1 + 2 + 3 = 180

Gi e ha , 3 = 90 . S ,

1 + 2 + 90 = 180

1 + 2 = 90 .....(e 1)

TS i a e e dic la d a li e PR ch ha TSP = 5 = 90

Al , TP i e e dic la li e PQ ch ha TPQ = 90

4 + 2 = 90 .....(e 2)

O c a i g (e 1) a d (e 2) e ge ,


1 + 2 = 4 + 2

1 = 4

I RQP a d TSP

3 = 5 (each 90 )

1 = 4 (f ab e)

B AA

RQP TSP

The ef e,

ST/PQ = PS/RQ

C - l i l he

ST.RQ = PS.PQ

He ce, ed

A 17).

Gi e : SecA = 2 3

T fi d: + tanAcosA tanA

1+ sinA

B ig e ide i :

H = 2 a d B = 3

We k , H 2 = P 2 + B 2

H 2 = P 2 + B 2

2 2 = P 2 + ( ) 2 3


O l i g, P=1

We al ead ha e, H=2, B= 3

We al k , Si A = PH

S , Si A = 21

C A = BH

C A = 23

Ta A = BP

Ta A = 13

O b i i g he e al e i + tanAcosA tanA

1+ sinA

We ge , 64 + 9 3

S , + = tanAcosA tanA

1+ sinA6

4 + 9 3

[I e a , i e d he l i al ]

A 18). T e: ec 2 c 2(90 ) = c 2(90 ) + c 2 .


LHS = ec 2 c 2(90 )

= ec 2 - a 2

= 1/C 2 - i 2 /c 2

= (1 - i 2 )/ c 2

= C 2 /c 2 = 1

RHS = c 2(90- ) + c 2

Si 2 + c 2 = 1

LHS = RHS, He ce P ed

A 19).


A 20).

Gi e : M de ---> 34.5 , l = 20

T fi d:

S , acc di g he f la f de

M de = l + ( )hf1 f02f1 f0 f2

34.5 = 20 + ( )1010 820 8 x

34.5 = 20 + ( )10212 x

34.5 - 20 = ( )10212 x

14.5 = ( )2012 x

14.5 (12- ) = 20

= 12 - 2014.5

O i lif i g ---> = 12 - 2940

= 12 - 2940

= 29348 40

= 29308

= 10.62


A 21).

ha c fac 5

Le a e ha i a i al a d / ha c fac 5

= / 5

S a i g b h ide

( ) =( / ) 5

5= /

5 =

5 i a fac f

al 5 i a fac f

le = 5

5 =(5 )

5 =25

=5

S 5 i a fac f

al 5 i a fac f

He ce e aid ha 5 i a i al b acc di g c adic i 5 i i a i al.

3+ =k 5

=k-3 5

LHS i i a i al i ce i i a i al a ed 5

He ce, RHS i i a i al .


A 22). P l ial ( ) = 4 + 6 3 + 2 24 20

The gi e e e a e: 2 a d (-5)

Acc di g Fac The e

( -2) a d ( +5) a e fac a e ( )

Al ,

( -2) ( +5) = 2 + 5 - 2 - 10

2 + 3 - 10

+3 -10 i al a fac f ( )

N ,

Di i g ( ) 4 + 6 3 + 2 24 20 b +3 -10

We ge Q( ) = + 3 +2

( ) = 0

( ) = ( 2 + 3 - 10)( 2 + 3 + 2)

= ( -2)( +5)( 2 + + 2 + 2)

= ( -2)( +5)( +1)( +2)

S , he e e f ( ) a e:


-2 = 0 , = 2

+ 5 = 0 , = -5

+ 1 = 0 , = -1

+ 2 = 0 , = -2

A 23). *GRAPH*

A 24).

L =x \ T = \

S ( x+ ) km /hr

S

( x ) km /hr

i e = speeddistance


Ti e ake c e 30 k ea = 30x

Ti e ake c e 44 k d ea = 44x+

S , acc di g he e i :

+ = 1030x

44x+

Ti e ake c e 40 k ea = 40x

Ti e ake c e 55 k d ea = 55x+

S , acc di g he e i :

+ = 1340x

55x+

Le = a d = 1x

1x+

⇒ 30u+44v=10 ------> eq n1

⇒ 40u+55v=13 ------> eq n2 ⇒ (150 u+220v=50) (160u+220v=52) [ M l iplied b 5 and b ac ed bo h he e n] S , = ⅕ a d =

N , a e k = a d = 1x

1x+


S , = = ⅕ , - = 51x

A d, = , + =111x+

O b ac i g he e 2 e a i

= 8 a d = 3

H , =8 \

T =3 \

(Sh he l i al i e a )

A 25).

J i A id i f BC a D.

S ,ED = BE = ( )BC -----> e 141

I ia gle AED, AE = AD + ED -----> e 2

I ia gle ABD, AD = AB - BD -----> e 3

P i g al e f AD f (3) i (2),

AE = AB - BD + ED = AB - ( ) + ( )2BC

4BC

a BD = (1/2)BC a d ED = (1/4)BC f (1)....,,,


S e ge ....16AE = 13AB

A 26).

Gi e : ∠ABD = ∠XYD = ∠CDB = 90 . AB = a, XY = c a d CD = b,

A ABD = XYD = CDB = 90

⇒ AB XY CD

I ABD

AS AB XY

S , = BDDY c

a

⇒ DY = (BD) .-----> e (1)ca

I BDC

A CD XY

S ,


= BDBY c

b

BY = (BD) ......(2)cb

Addi g (1) a d (2), e ge

DY + BY = (BD) + (BD)ca

cb

BD = c( + )BDa1

b1

= + c1

a1

b1

= c1

aba+b

c(a+b) = ab

He ce ed.

A 27). I ABC, i g he hag a he e ,


A = 1


A 28).

c 2B(Si 2A + c 2A)/ i 2A

c 2B/ Si 2A = 2 = RHS

A 29). ec A - 1 \ ec A + 1

M l i l i g a d di idi g b SecA - 1

( ec A - 1) / ( ec A - 1)


1 ( ec A + 1 - 2 ecA) [ ec A - 1 = a A]

Di ide he e a i b Ta 2A

c A ( a A + 2 - 2 ecA) [ = c A]1tanA

1 + 2c A - 2 [ l i l i g]1cosA SinA

CosASinACosA

[ = c A = ecA]SinACosA 1

CosA

(1 + c A) + c A -2c Ac ecA

c ec A + c A - 2c ec Ac A = c ec A

(c A - c ecA)

He ce P ed!


CBSE Maths 2015

1. All questions are compulsor .2. The question paper consists of 31 questions divided into four sections A, B, C andD.3. Section A contains 4 questions of 1 mark each.4. Section B contains 6 questions of 2 marks each.5. Section C contains 10 questions of 3 marks each6. Section D contains 11 questions of 4 marks each.

Questions -

EC I N A

1. If the quadratic equation px2 2 5px + 15 = 0 has two equal roots then find the valueof p.

2. In the following figure, a tower AB is 20 m high and BC, its shadow on the ground, is20 3 m long. Find the Sun s altitude.

3. Two different dice are tossed together. Find the probabilit that the product of thetwo numbers on the top of the dice is 6.

4. In the following figure, PQ is a chord of a circle with center O and PT is a tangent. If∠QPT = 60 , find ∠PRQ.


EC I N B

5. In the following figure, two tangents RQ and RP are drawn from an external point R tothe circle with centerO, If ∠PRQ = 120 , then prove that OR = PR + RQ.

6. In the following figure, a ABC is drawn to circumscribe a circle of radius 3 cm, suchthat the segments BD and DC are respectivel of lengths 6 cm and9 cm. If the area of ABC is 54 cm2 , then find the lengths of sides AB and AC.

7. Solve the following quadratic equation for x: 4x2 + 4bx (a2 b2 ) = 0.


8. In an AP, if S5 + S7 = 167 and S10 = 235, then find the A. P., where Sn denotes the sumof its first n terms.

9. The points A (4, 7), B (P, 3) and C (7, 3) are the vertices of a right triangle, right-angledat B, Find the values of P.

10. Find the relation between x and if the points A(x, ), B ( 5, 7) and C( 4, 5) arecollinear.

EC I N C

11. The 14 th term of an A. P. is twice its 8 th term. If its 6 th term is 8, then find thesum of its first 20 terms.

12. Solve for x: 3x2 2 2x 2 3 = 0.

13. The angle of elevation of an aeroplane from point A on the ground is 60 . After flightof 15 seconds, the angle of elevation changes to 30 . If the aeroplane is fl ing at aconstant height of 1500 3m, find the speed of the plane in km/hr.

14. If the coordinates of points A and B are ( 2, 2) and (2, 4) respectivel , find thecoordinates of P such that AP = 3/7 AB, where P lies on the line segment AB.

15. The probabilit of selecting a red ball at random from a jar that contains onl red,blue and orange balls is 1/4 . The probabilit of selecting a blue ball at random from thesame jar 1/3 . If the jar contains 10 orange balls, find the total number of balls in the jar.

16. Find the area of the minor segment of a circle of radius 14 cm, when its centralangle is 60 . Also find the area of the corresponding major segment. [ Use = 22/7 ].


17. Due to sudden floods, some welfare associations jointl requested the governmentto get 100 tents fixed immediatel and offered to contribute 50% of the cost. If the lowerpart of each tent is of the form of a c linder of diameter 4.2 m and height 4 m with theconical upper part of same diameter but height 2.8 m, and the canvas to be used costsRs. 100 per sq. m, find the amount, the associations will have to pa . What values areshown b these associations? [ Use = 22/7 ].

18. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filledinto 72 c lindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquidis wasted in this transfer.

19. A cubical block of side 10 cm is surmounted b a hemisphere. What is the largestdiameter that the hemisphere can have? Find the cost of painting the total surface areaof the solid so formed, at the rate of Rs. 5 per sq. cm. [ Use = 22/7 ].

20. 504 Cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into ametallic sphere, Find the diameter of the sphere and hence find its surface area. [ Use = 22/7 ].

EC I N D

21. The diagonal of a rectangular field is 16 metres more than the shorter side. If thelonger side is 14 metres more than the shorter side, then find the lengths of the sides ofthe field.

22. Find the 60 th term of the AP 8, 10, 12, ., if it has a total of 60 terms and hencefind the sum of its last 10 terms.

23. A train travels at a certain average speed for a distance of 54 km and then travels adistance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3hours to complete the total journe , what is its first speed?


24. Prove that the lengths of the tangents drawn from an external point to a circle areequal.

25. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to thechord joining the end points of the arc.

26. Construct a ABC in which AB = 6 cm,∠A = 30 and ∠B = 60 , Construct anotherAB C similar to ABC with base AB = 8 cm.

27. At a point A, 20 metres above the level of water in a lake, the angle of elevation of acloud is 30 . The angle of depression of the reflection of the cloud in the lake, at A is60 . Find the distance of the cloud from A.

28. A card is drawn at random from a well-shuffled deck of pla ing cards. Find theprobabilit that the card drawn isi. a card of spade or an ace.j. a black king.k. neither a jack nor a kingl. either a king or a queen.

29. Find the values of k so that the area of the triangle with vertices (1, 1), ( 4, 2k) and( k, 5) is 24 sq. units.

30. In the following figure, PQRS is square lawn with side PQ = 42 metres. Two circularflower beds are there on the sides PS and QR with centre at O, the intersections of itsdiagonals. Find the total area of the two flower beds (shaded parts).


31. From each end of a solid metal c linder, metal was scooped out in hemisphericalform of the same diameter. The height of the c linder is 10 cm and its base is of radius4.2 cm. The rest of the c linder is melted and converted into a c lindrical wire of 1.4 cmthickness. Find the length of the wire. [ Use = 22/7 ]


CBSE Maths 2015Sol ions : 1. We ha e one form la if a q adratic eq ation of the form 𝑎 2 + 𝑏 + 𝑐 = 0 ha eeq al roots then 𝑏2 4𝑎𝑐 = 0.

In the gi en eq ation 𝑎 = , 𝑏 = 2 5 and 𝑐 = 15,( 2 5p)2 4 p 15 = 020 2 60 = 0

= 0 or = 30 = 0 is not possible so p = 30

2. From trigonometric e pressions clearl 𝑎 =𝐴B/𝐴𝐶 𝑎 = 20/20 3 𝑎 = 1/ 3 = 30o

3. When t o dice are tossed together there ill be 36 possibilities. Possibilities forprod ct of the n mbers is (1,6), (2,3), (3,2) and (6,1). Req ired probabilit is= 4/36 .

4. ∠𝑂𝑃𝑇 = 90o

From the fig re ∠𝑂𝑃𝑄 = ∠OPT ∠QPT = 90o 60o = 300

From the fig re ∠POQ = 2∠QPT = 2 60o = 120o

Refle angle of ∠POQ = 360o 120o = 240o

∠PRQ = 1/2 refle angle ∠POQ = 1/2 240o = 120o

5. ∠𝑂𝑃𝑇 = 900


Gi en that ∠𝑃𝑅𝑄 = 120From the propert of the circle one can get∠𝑃𝑅𝑂 = ∠𝑄𝑅𝑂 = 120/2 = 600

We kno that lengths of tangents from an e ternal points are eq al.Th s, from diagram 𝑃𝑅 = 𝑅𝑄.After 𝑂𝑃 and 𝑂𝑄.Both are the radii from the center 𝑂,𝑂𝑃 is perpendic lar to 𝑃𝑅 and 𝑂𝑄 is perpendic lar to 𝑅𝑄.Th s, 𝑂𝑃𝑅 and 𝑂𝑄𝑅 are right angled congr ent triangles.

Hence, ∠𝑃𝑂𝑅 = 900 ∠𝑃𝑅𝑂 = 900 600 = 300 .∠𝑄𝑂𝑅 = 900 ∠𝑄𝑅𝑂 = 900 600 = 30o

𝑖 ∠𝑄𝑅𝑂 = 𝑖 300 = From the diagram𝑖 300 = 𝑃𝑅/𝑂𝑅 implies 𝑃𝑅/𝑂𝑅 = 1/2𝑂𝑅 = 2𝑃𝑅𝑂𝑅 = 𝑃𝑅 + 𝑃𝑅𝑂𝑅 = 𝑃𝑅 + 𝑄𝑅.


6.

Let the gi en circle to ch the sides 𝐴𝐵 and 𝐴𝐶 of the triangle at points 𝑅 and 𝑄respecti el and let the length of line segment AR be .No , it can be obser ed that:𝐵𝑅 = 𝐵𝑃 = 6 𝑐 (tangents from point B)𝐶𝑄 = 𝐶𝑃 = 9 𝑐 (tangents from point C)𝐴𝑅 = 𝐴𝑄 = (tangents from point A)𝐴𝐵 = 𝐴𝑅 + 𝑅𝐵 = + 6𝐵𝐶 = 𝐵𝑃 + 𝑃𝐶 = 6 + 9 = 15𝐶𝐴 = 𝐶𝑄 + 𝑄𝐴 = 9 + 2 = 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴 = + 6 + 15 + 9 + = 30 + 2 = 15 +

𝑎 = 15 + 15 = 𝑏 = 15 + ( + 9) = 6 𝑐 = 15 + (6 + ) = 9

Area of the triangle 𝐴𝐵𝐶 = [ ( 𝑎)( 𝑏)( 𝑐)]54 = [(15 + ) (6)(9)]18 = [(15 + 2)(6)

2 + 15 54 = 0( + 18)( 3) = 0Distance can t be negati e then = 3𝐴𝐶 = 3 + 9 = 12𝐴𝐵 = 𝐴𝑅 + 𝑅𝐵 = 6 + = 6 + 3 = 9

7.


8. S10=10/2 (2a+9d)=235

5(2a+9d)=235

10a+45d=235

Di iding hole b 5

2a+9d=47

M ltipl ing b 6

12a+54d=282 1

S5+S7=5/2 [2a+(5-1)d]+ 7/2 [2a+(7-1)d]

5/2 (2a+4d)+ 7/2 (2a+6d)

5 (a+2d)+ 7 (a+3d)=167

5a+10d+7a+21d=167


12a+31d=167 2

S btracting 2 from 1 , e get,

23d=115

D=115/23

D=5

2a +9d = 47

2a+ 9*5=47

2a=49-47

2a=2

A=1

A.P., is 1,6,11,16


9.


10.

11. ' ' ' ,

------- (1)

,

(1)

+13 = 2( +7 )

+13 =2 +14

-2 =14 -13

- =1 = - ------------ (2)


. , +5 = -8 ------- (3)

(2) = -

(3)

- + 5 = -8 4 = -8 = -8/4 = -2

,

= -(-2) . =2 = -2

20 . =20

,

,

S 1 20 -340

12.


13.

Let 𝑄𝑅 be the height then 𝑄𝑅 = 1500 3. In 15 seconds, the aeroplane mo es from 𝑃 to𝑆. 𝑃 and 𝑆 are the points here angles of ele ations are 600 and 300 . Let 𝑄𝑃 = 𝑎 and 𝑃𝑆 =b𝑄𝑅 = 𝑎 + 𝑏In triangle 𝑅𝑄𝑃,𝑎 600 = 𝑄𝑅/𝑄𝑃

3 = 1500 3/𝑎𝑎 = 1500In triangle 𝑅𝑄𝑆,𝑎 30 0 = 𝑄𝑅/𝑄𝑆

1/ 3 = 1500 3/(𝑎 + 𝑏)𝑏 = 3000

14.


15.Gi en that the jar contains red, bl e and orange balls.For s ppose 𝑎 be the n mber of red ballsand 𝑏" be the n mber of bl e ballsN mber of orange balls is gi en as 10Total n mber of balls ill be = 𝑎 + 𝑏 + 10

Gi en that

Probabilit of selecting red ball is= 1/4 implies 𝑎/(𝑎+𝑏+10) = 4𝑎 = 𝑎 + 𝑏 + 10 implies 3𝑎 𝑏 = 10 ..(1)

Probabilit of selecting red ball is= 1/3 implies 𝑏/(𝑎+𝑏+10) = ⅓

3𝑏 = 𝑎 + 𝑏 + 10 implies 2𝑏 𝑎 = 10 ..(2)

Sol ing (1) and (2) e get,M ltipl eq. (2) be 3 then add it ith eq ation (1)5𝑏 = 40 implies 𝑏 = 8If 𝑏 = 8 then 𝑎 = 6.Total n mber of balls in the jar ill be eq al to 6 + 8 + 10 = 24.


16.Gi en radi s of the circle is 14𝑐 and center angle is 600 . We ha e the form la forfinding the area.

17.Gi en parameters areDiameter of the tent = 4.2 then radi s ill be = 2.1 ,Height of the c lindrical part is (ℎ) = 4Height of the conical part is (ℎ1) = 2.8mWe ha e the form la for slant height

= 52.8m2

18.ol me of hemisphere =2/3 r = ol me of liq id present in bo l

=2/3 22/7 (18)=2/3 22/7 18 18 18 cm

no , according to q estion 10% liq id is asted b transfer .so, rest ol me of liq id =90 of ol me of liq id present in bo l


again ,

90 of ol me of liq id present in bo l = 72 ol me of each bottle9/10 2/3 22/7 18 18 18 = 72 22/7 3 3 H

H = 5.4 cm

19.The greatest diameter a hemisphere can ha e 10 cm.

Total s rface area of the solid = S rface area of the c be + CSA of the hemisphere Area of the base of the hemisphere.

Cost of printing the block at the rate of Rs 5 per sq. cm

The cost of 678.57sq cm =678.57 5=3392.85

Hence, the amo nt is Rs 3393.

20.


21.. Let s ass me that 𝑎 be the longer length and 𝑏 be the shorter length Gi en thatdiagonal of a rectang lar field is 16 𝑒 𝑒 more than the shorter side Implies diagonal=16 + b.Longer side is 14 𝑒 𝑒 more than the shorter side implies𝑎 = 14 + 𝑏. B singP thagoras theorem(16 + 𝑏)2 = (14 + 𝑏)2 + 𝑏2

256 + 𝑏 2 + 32𝑏 = 196 + 𝑏 2 + 28𝑏 + 𝑏 2

𝑏 2 4𝑏 60 = 0(𝑏 + 6)(𝑏 10) = 0𝑏 = 6 or 𝑏 = 10Breadth ill be eq al to 10 and other length is = 24 .

22.AP=8,10,12....

=8, =10-8=2 , = +( -1) ₀ =8+(60-1)2

=8+118=126 , ₀=60/2[2 8 +(60-1)2]

=30[16+118] =4020

=60-10=50 ,


₅₀=50/2[2 8+(50-1)2] =25[16+98] =2850 S = ₀- ₅₀=4020=-2850 =1170 23.

Speed can t be negati e and hence initial speed of the train is 36 /h.

24.

In PQO and PROPO= commonQO=RO= Radi s of circle∠PQO=∠PRO=90 o

(Radi s tangent)


∴ PQO PRO (RHS)and hence PQ=PRHence pro ed (b CPCT)

25.

In the fig re, 𝐶 is the midpoint of the minor arc 𝑃𝑄,𝑂 is the center of the circle and𝐴𝐵 is tangent to the circle thro gh point 𝐶.We ha e to sho the tangent dra n at the midpoint of the 𝑎 𝑐 𝑃𝑄 of a circle is parallelto the chord joining the end points of the 𝑎 𝑐 𝑃𝑄.We ill sho 𝑃𝑄 is parallel to 𝐴𝐵.It is gi en that 𝐶 is the midpoint point of the 𝑎 𝑐 𝑃𝑄.So, 𝑎 𝑐 𝑃𝐶 = 𝑎 𝑐 𝐶𝑄Implies 𝑃𝐶 = 𝐶𝑄This sho s that the triangle 𝑃𝑄𝐶 is an isosceles triangle.Th s, the perpendic lar bisector of the side 𝑃𝑄 of triangle 𝑃𝑄𝐶 passes thro gh 𝑒 𝑒 𝐶.The perpendic lar bisector of a chord passes thro gh the center of the circle.So the perpendic lar bisector of 𝑃𝑄 passes thro gh the center 𝑂 of the circle.Th s perpendic lar bisector of 𝑃𝑄 passes thro gh the points 𝑂 and 𝐶.Implies 𝑃𝑄 is perpendic lar to 𝑂𝐶𝐴𝐵 is the tangent to the circle thro gh the point 𝐶 on the circle.Implies 𝐴𝐵 is perpendic lar to 𝑂𝐶The chord 𝑃𝑄 and the tangent 𝑃𝑄 of the circle are perpendic lar to the same line 𝑂𝐶.𝑃𝑄 is perpendic lar to 𝐴𝐵.

26.


27.

if e ass me the distance of the clo d from point A.


e.g., DE = h and CF = h - 20

no , from ADE,

tan30 = DE/AD

1/ 3 = h/AD

AD = 3h ..............(1)

from triangle ADF,

tan60 = DF/AD

3 = (FC + CD)/ 3h

3 3h = (h + 20 + 20)

3h = h + 40

h = 20m.

hence, Distance of clo d from A = 20m

28. Let s ass me that S be the sample space of dra ing a card from a ell-sh ffled deck.

(S) = 521C

i) In a deck e ha e 13 spade 4 ace card a spade can be chosen in 13 a s.An ace can be chosen in 4 a s. B t e ha e alread chosen one ace card hich isincl ded in spade.So req ired probabilit = 13+4 1/52 = 16/52 = 4/13.

ii) There are onl t o black king in a deck. A black king card is dra n in 2a s. So req ired probabilit = 2/52 = 1/26 .

iii) There are 4 jack and 4 king cards in deck. So there are 44 cards hich areneither jacks nor kings So req ired probabilit = 44/52 = 2/13 .

i ) There are 4 kings and 4 q een cards in a deck. So there are 4 + 4 = 8 cardhich are either king or q een. So req ired probabilit = 8/52 = 1/26 .

29.


Area of the triangle formed b ( 1, 1 ), ( 2, 2) and ( 3, 3) is gi en b

30.Gi en that 𝑃𝑄𝑅𝑆 is a sq are. So each side is eq al and angle bet een the adjacentsides is a right angle. Also the diagonals perpendic larl bisect each other.

Area of the shape 𝑂𝑅𝑄 = 𝐴 𝑒𝑎 𝑓 𝑒𝑐 𝑂𝑅𝑄 𝐴 𝑒𝑎 𝑓 𝑖𝑎 𝑔 𝑒 𝑅𝑂𝑄= ( 42/2 ) 2 ( /2 1)= 441(0.57) = 251.37𝑐 2

Area of the flo er bed 𝑂𝑅𝑄 = Area of the flo er bed 𝑂𝑃𝑆 = 251.37𝑐 2

31.F , .

= 10 = 4.2

V = = 176.4


V = 4/3 = 4/3 (4.2) = 98.784 98.8

T = 176.4 - 98.8 = 77.6

N F , D = 1.4 R = 1.4/2 = 0.7

T = T 77.6 = C 77.6 = (0.7)

= 77.6/0.49 = 158.36 158.4


CBSE Maths 2014

1) All q estions are comp lsor .2) The q estion paper consists of thirt q estions di ided into 4 sections A, B, Cand D. Section A comprises of ten q estions of 01 mark each, Section B comprises offi e q estions of 02 marks each, Section C comprises ten q estions of 03 marks eachand Section D comprises of fi e q estions of 06 marks each.3) All q estions in Section A are to be ans ered in one ord, one sentence or asper the e act req irement of the q estion.4) There is no o erall choice. Ho e er, internal choice has been pro ided in oneq estion of 02 marks each, three q estions of 03 marks each and t o q estions of 06marks each. Yo ha e to attempt onl one of the alternati es in all s ch q estions.5) In q estion on constr ction, dra ing sho ld be near and e actl as per the gi enmeas rements.6) Use of calc lators is not permitted.

Q estions -

1. If the height of a ertical pole is 3times the length of its shado on the gro nd, thenthe angle of ele ation of the S n at that time is ______.

2. A bag contains cards n mbered from 1 to 25. A card is dra n at random from thebag. The probabilit that the n mber on this card is di isible b both 2 and 3 is

3. T o different coins are tossed sim ltaneo sl . The probabilit of getting at least onehead is ____.

4. T o concentric circles are of radii 5 cm and 3 cm. Length of the chord of the largercircle (in cm), hich to ches the smaller circle is _______.


5. In fig re, a q adrilateral ABCD is dra n to circ mscribe a circle s ch that its sidesAB, BC, CD and AD to ch the circle at P, Q, R and S respecti el . If AB = cm, BC = 7 cm,CR = 3 cm and AS = 5 cm, find .

6. The perimeter of a triangle ith ertices (0, 4), (0, 0) and (3, 0) is _______.

7. A rectang lar sheet of paper 40 cm 22 cm, is rolled to form a hollo c linder ofheight 40 cm. The radi s of the c linder (in cm) is _______.

8. The ne t term of the A.P. 7,28,63, ... is

EC ION B

9. In fig re , XP and XQ are t o tangents to the circle ith centre O, dra n from ane ternal point X. ARB is another tangent, to ching the circle at R. Pro e that XA + AR =XB + BR.


10. Pro e that the tangents dra n at the ends of an diameter of a circle are parallel.

12. In Fig re , OABC is a q adrant of a circle of radi s 7 cm. If OD = 4 cm, find the areaof the shaded region. [Use = 22/7 ]

13. Sol e for : 3 2 22 23 = 0.

14. The s m of the first n terms of an AP is 5n n2. Find the nthterm of this A.P.

EC ION C

15. Points P, Q, R and S di ide the line segment joining the points A (1, 2) and B (6, 7) in5 eq al parts. Find the coordinates of the points P, Q and R.

16. In the fig re, from a rectang lar region ABCD ith AB = 20 cm, a right triangle AEDith AE = 9 cm and DE = 12 cm, is c t off. On the other end, taking BC as diameter, a

semicircle is added on o tside the region. Find the area of the shaded region.[Use = 3.14]


17. In the fig re, ABCD is a q adrant of a circle of radi s 28 cm and a semi circle BEC isdra n ith BC as diameter. Find the area of the shaded region. [Use = 22/7]

18. A 5 m ide cloth is sed to make a conical tent of base diameter 14 m and height24 m. Find the cost of cloth sed at the rate of Rs 25 per metre. [Use = 22/7]

19. A girl empties a c lindrical b cket, f ll of sand, of base radi s 18 cm and height 32cm, on the floor to form a conical heap of sand. If the height of this conical heap is 24cm, then find its slant height correct p to one place of decimal.

20. The s m of the first 7 terms of an A.P. is 63 and the s m of its ne t 7 terms is 161.Find the 28th term of this A.P.

21. T o ships are approaching a light-ho se from opposite directions. The angles ofdepression of the t o ships from the top of the light-ho se are 30 and 45 . If thedistance bet een the t o ships is 100 m, find the height of the light-ho se. [Use 3=1.732]

22. If 2 is a root of the q adratic eq ation 3 2+ p 8 = 0 and the q adratic eq ation4 2 2p + k = 0 has eq al roots, find the al e of k.

23. Constr ct a triangle PQR, in hich PQ = 6 cm, QR = 7 cm and PR = 8 cm. Thenconstr ct another triangle hose sides are 45 times the corresponding sides of PQR.


24. Find the al e(s) of p for hich the points (p + 1, 2p 2), (p 1, p) and (p 3, 2p 6)are collinear.

EC ION D

25. The mid-point P of the line segment joining the points A ( 10, 4) and B ( 2, 0) lies onthe line segment joining the points C ( 9, 4) and D ( 4, ). Find the ratio in hich Pdi ides CD. Also find the al e of .

26. A q adrilateral is dra n to circ mscribe a circle. Pro e that the s ms of oppositesides are eq al.

27. The angle of ele ation of the top of a chimne from the foot of a to er is 60 andthe angle of depression of the foot of the chimne from the top of the to er is 30 . If theheight of the to er is 40 m, find the height of the chimne . According to poll tioncontrol norms, the minim m height of a smoke emitting chimne sho ld be 100 m.State if the height of the abo e mentioned chimne meets the poll tion norms. Whatal e is disc ssed in this q estion?

28. A hemispherical depression is c t o t from one face of a c bical block of side 7 cm,s ch that the diameter of the hemisphere is eq al to the edge of the c be. Find thes rface area of the remaining solid. [Use = 227] .

29. If Sn Denotes the s m of the first n terms of an A.P., pro e that S30 = 3 (S20 S10).

30. A metallic b cket, open at the top, of height 24 cm is in the form of the fr st m of acone, the radii of hose lo er and pper circ lar ends are 7 cm and 14 cm respecti el .Find


(i) the ol me of ater hich can completel fill the b cket.

(ii) the area of the metal sheet sed to make the b cket.

[Use = 22/7]

31. The s m of the sq ares of t o consec ti e e en n mbers is 340. Find the n mbers.

32. Pro e that the tangent at an point of a circle is perpendic lar to the radi s thro ghthe point of contact.

33. A dice is rolled t ice. Find the probabilit that

(i) 5 ill not come p either time.

(ii) 5 ill come p e actl one time.

34.


CBSE Ma h 2014 Sol ions :- 1. A me OA a he ole and OB i i hado .

Le he leng h of he hado be .

Le be he angle of ele a ion of he o of he ole f om he g o nd.

Gi en ha he heigh of he ole (h) = 3 leng h of i hado

⇒h = 3

In OAB

an = AO OB = h = 3 = 3

⇒ an = an 60

⇒ = 60

∴ The angle of ele a ion of he S n i 60 . 2. To al n mbe of ca d = 25

To al n mbe of o ible o come = 25

A me E a he e en ha he n mbe on he ca d d a n i di i ible b bo h 2 and 3.

∴ E = 6, 12, 18, 24

∴ N mbe of o ible o come fa o able o E = 4

P(E) = N mbe of o ible o come fa o able o ETo al n mbe of o come

P(E) = 425.


3. When o coin a e o ed im l aneo l , he o ible o come a e (H, H),(H, T), (T, H), (T, T) .

∴ To al n mbe of o ible o come = 4

A me he e en of ge ing a lea one head a 'E'.

N mbe of o ible o come fa o able o E i (H, H), (H, T), (T, H) .

∴ N mbe of fa o able o come = 3

P(E) = N mbe of fa o able o come o ETo al n mbe of o ible o come

P(E) = 34.

4. In he fig re, O is he common cen re, of he gi en concen ric circles. AB is a chord of he bigger circle s ch ha i is a angen o he smaller circle P. Since, OP is he radi s of he smaller circle. ∴ OP ⊥ AB => ∠APO = 90 Also, radi s perpendic lar o a chord bisec s he chord ∴ OP bisec s AB => AP = No , in righ APO,


⠀ OA = AP + OP => 5 = AP + 3 => AP = 5 - 3 => AP = 4 => AP = 4cm => AB = 4 => AB = 2 4 = 8cm Hence, he req ired leng h of he chord AB is 8cm. 5. Tangen s dra n from an e ernal poin o a circle are al a s eq al. The q adrila eral, ABCD is dra n o circ mscribe a circle s ch ha i s sides AB, BC, CD and AD o ch he circle a P, Q, R and S respec i el . Therefore A,B,C and D are e ernal poin s from hich angen s AS=AP=5, BP=BQ, CQ=CR=3 and DR=DS are dra n BC=7. => BQ + CQ =BC=7 => BQ= 7-3=4 ∴ BP=4 AB= , => AP +BP = => 5+4 = => =9 6. A(0,4) B(0,0) C(3,0) Therefore perime er of riangle ABC is = AB+BC+AC AB= (0-0)^2+(0-4)^2 = (0)+(16) = +4,-4 BC= (3-0)^2+(0-0)^2 = 9+0 =+3,-3 AC= (3-0)^2+(0-4)^2 = 9+16 = +5,-5 Meas remen of sides canno be rejec ed herefore -4,-3,-5 is rejec ed Therefore AB=4 cm BC=3cmAC=5cm


Perime er= 12 cm 7. A ea of he ec ang la hee = l b = (40 22) cm2 = 880 cm2

A me he adi of he c linde a ' '.

Gi en ha heigh (h) of he c linde = 40 cm

A ea of he ec ang la hee i ame a he c ed face a ea of he c linde .

∴ 2 h = A ea of he ec ang la hee

⇒ 2 22/7 40 = 880

⇒ = 7/2 = 3.5

∴Radi of he c linde i 3.5 cm. 8. 7, 28, 63... = 7, (7 2 2) , (7 3 3) ... = (7 1 1) , (7 2 2), (7 3 3) ... = 7, 27, 37... Gi en ha he firs erm of he gi en AP is 7 and he common difference is (27 7) = 7 So, a = d = 7 The ne erm of he seq ence, i.e. he 4 h erm = a + 3d = 37+ 7 = 47= 7 4 4 = 112.


9.

GIVEN: XP and XQ are angen s of he circle from i h cen re O and R is a poin on he circle. TO PROVE : XA + AR = XB + BR. PROOF: XP = XQ [X is an e ernal poin ].......(1) AP = AR [A is an e ernal poin ]......(2) BQ = BR [B is an e ernal poin ]......(3) From eq 1 XP = XQ (XA + AP) = (XB + BQ) XA + AR = XB + BR [From eq 2 & 3 , AP = AR ,BQ = BR ] Hence pro ed. 10.


Le AB be a diame er of he circle. T o angen s PQ and RS are dra n a poin s A and B respec i el . Radi s dra n o hese angen s ill be perpendic lar o he angen s. Th s, OA ⊥ RS and OB ⊥ PQ ∠OAR = 90 ∠OAS = 90 ∠OBP = 90 ∠OBQ = 90 I can be obser ed ha ∠OAR = ∠OBQ (Al erna e in erior angles) ∠OAS = ∠OBP (Al erna e in erior angles) Since al erna e in erior angles are eq al, lines PQ and RS ill be parallel. 11. The n mbe ha o dice co ld ho a e 1, 2, 3, 4, 5, 6.

The o ible o come hen o dice a e olled a e:

(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)

∴ N mbe of o ible o come = 6 6 = 36.

A me 'E' a he e en ha he m of n mbe a ea ing on he o dice i 10.

The o ible o come fa o able o e en , E a e (4, 6), (5, 5), (6, 4) .

∴ N mbe of fa o able o come o e en E = 3.

P(E) = N mbe of o ible o come fa o able o ETo al n mbe of o ibleo come = 336 = 12.


12. Gi en ha , adi of he ci cle = 7 cm

A ea of he ad an OABC = 1/4 2

= 1/4 22/7 7 7 cm2= 77/2cm2 = 33.5cm2

A ea of ODC = 1/2 OD OC

= (1/2 4 7) cm2 (Since OC i he adi of he ci cle)

= 14 cm2

A ea of he haded egion = A ea of he ad an OABC A ea of ODC

A ea of he haded egion = 33.5 14 = 24.5cm2

13. Gi en he ad a ic e a ion a ,

3 2 22 23 = 0

⇒ 3 2 32 + 2 23 = 0

⇒ 3 ( 6) + 2( 6) = 0

⇒ (3 +2) ( 6) = 0

⇒ = 2/3 o = 6

14. The m of n e m of an A.P. i gi en b 5n -n2

B e kno ha Sn= n2(a + n)


he e a = fi e m Tn = n h e m

So, e ha e 5n n2= n2 ( a + n)

⇒ n(5 n) = n2( a + n)

⇒ 10 2n = a + n ...(1)

No , e ha e m of he fi e m S1= a

⇒ a = (5(1) 12) = 4

S b i e he al e of a in e a ion (1), e ge

10 2n = 4 + n

⇒ n= 6 2n .

The efo e, n h e m of he A.P. i 6 2n.

15. I is gi en ha P, Q, R and S di ide he line segmen joining he poin s A(1, 2) and B(6, 7) in 5 eq al par s.

Sec ion form la,

P di ides he line AB in 1:4.

Q di ides he line AB in 2:3.

R di ides he line AB in 3:2.


Therefore he coordina es of he poin s P, Q and R are P(2,3), Q(3,4) and R(4,5).

16. AED i an igh angle iangle,

AD2= AE2+ED2.

⇒ AD2 = (92+ 122)

= (81 + 144)

= 225 cm2

⇒ AD = 15 cm

A ea of he ec ang la egion ABCD

= AB AD

= (20 15)

= 300 cm2

A ea of AED

= 1/2 AE DE = (1/2 9 12) cm2 = 54 cm2

In a ec angle

AD = BC = 15 cm

Since, BC i he diame e of he ci cle, adi of he ci cle = 15 cm

A ea of he emi-ci cle = 1/2 2

= (1/2 3.14 15/2 15/2) = 88.3125 cm2

A ea of he haded egion = A ea of he ec angle + A ea of he emi-ci cle A ea of heiangle

= (300 + 88.3125 54) cm2


= 334.3125 cm2.

17. Gi en:

Radi ( ) of he ci cle = AB = AC = 28 cm

A ea of ad an ABPC = 1/4

= (1/4 22/7 28 28) cm

=22 28= 616 cm

A ea of ABC = 1/2 AC AB = (1/2 28 28)

= 392 cm

A ea of egmen BPC = A ea of ad an ABPC A ea of ABC

= (616 392) cm

= 224 cm .............(1)

In a igh -angled BAC

BC = BA + AC (B P hago a heo em)

BC = (28 + 28 ) cm

BC = 784 +784 cm

BC = 16 98 = 16 49 2

BC = 4 7 2= 28 2

BC= 28 2 cm

BC(Diame e ) = 28 2

Radi of emici cle= 28 2/2= 14 2 cm

A ea of emici cle BEC= 1/2


= ( 1/2 22/7 14 2 14 2) cm

= 22 2 14 2 = 22 14 2 = 44 14

= 616 cm

A ea of he haded o ion = A ea of emici cle BEC A ea of egmen BPC

= 616 224 cm = 392 cm

Hence, he a ea of he haded egion = 392 cm

18. Fo Conical en ,

b=5m

adi = diame e /2 =14/2=7m

heigh h=24 m

lan heigh l = ((h^2)+( ^2))= (576)+(49)= 625=25

C.S.A of en = l

=(22/7) 7 25

=550 m^2

A ea of clo h e i ed = C.S.A of en

l b=550

l 5=550

l=110m

Co of 1 m clo h= 25

Co of 110 m clo h= (110 25)= 2750


19.

20. Gi en :

S7 = 63 and m of i ne 7 e m i 161

B ing he fo m la ,S m of n h e m , Sn = n/2 [2a + (n 1) d]

S7 = (7/2) [ 2a + (7 - 1)d ]

63 = 7/2 [2a + 6d]

63 2/7 = [2a + 6d]

9 2 = 2a + 6d

2a + 6d = 18 ...(1)And

S m of i ne 7 e m = 161(Gi en)


S m of fi 14 e m = S m of fi 7 e m + S m of ne 7 e m .

S14 = 63 + 161 = 224

S14 = 224

B ing he fo m la ,S m of n h e m , Sn = n/2 [2a + (n 1) d]

S14 = (14/2) [ 2a + (14 - 1)d ]

224 = 7 [ 2a + 13d ]

224/7 = 2a + 13d

2a + 13d = 32 ... (2)On b ac ing e (1) f om (2)

2a + 13d = 32

2a + 6d = 18

(-) (-) (-)

-------------------

7d = 14

---------------------

d = 14/7

d = 2On ing he al e of d = 2 in e (1),

2a + 6d = 18

2a + 6(2) = 18


2a + 12 = 18

2a = 18 - 12

2a = 6

a = 6/2

a = 3Fo 28 h e m :

B ing he fo m la ,an = a + (n - 1)da28 = a + ( 28 - 1) d

a28 = 3 + (27) 2

a28 = 3 + 54

a28 = 57

Hence, 28 h e m of hi A.P i 57.

21.

AB = 100 Le AC = Le BC = 100-


Le CD = h In EAD, EA = DC ED = AC an 45 = EA/ED 1 = h/ = h

In DFB FB = CD DF = CB an 30 = FB/DF 1/ 3 = h/100- 1/ 3 = h/100-h 100-h = 3h 100-h = 1.732h 2.732h = 100 h = 100/2.732 h = 36.603m h = 36.6m

22. Gi en : 3 + - 8 = 0 .(1)4 - 2 + k = 0 (2)

On ing he al e of gi en oo i.e = 2 in e 1 .

3 + - 8 = 0

3(2) + (2) 8 = 0

3 4 + 2 - 8 = 0

12 + 2 8 = 0


4 + 2 = 0

2 = - 4

= - 4/ 2 = -2

= - 2

Hence he al e of i - 2.On ing he al e of = - 2 in e 2,

4 - 2 + k = 0

4 - 2(- 2) + k = 0

4 + 4 + k = 0

On com a ing he gi en e a ion i h a + b + c = 0

He e, a = 4, b = 4 , and c = k

D(di c iminan ) = b 4ac

Gi en : Q ad a ic e a ion ha e al oo i.e D = 0

b 4ac = 0

4 4(4)(k) = 0

16 16k = 0

16 = 16k

k = 16/16 = 1

k = 1


23. S e of con c ion:

S e 1: D a a a PX.

S e 2: PQ = 6 cm a he adi , d a an a c f om oin P in e ec ing PX a Q.

S e 3: Q a he cen e and adi e al o 7 cm, d a an a c.

S e 4: P a he cen e and adi e al o 8 cm, d a ano he a c, c ing hee io l d a n a c a R.

S e 5: Join PR and QR.

The efo e, PQR i he e i ed iangle.

S e 6: D a an a PY making an ac e angle i h PQ.

S e 7: No loca e 5 oin P1, P2, P3, P4and P5 on PY o ha PP1=P1P2= P2P3=P3P4= P4P5.

S e 8: Join P5Q and d a a line h o gh P4, hich i a allel o P5Q, o in e ec PQ aQ'.

S e 9: D a a line h o gh Q', hich i a allel o line QR, o in e ec PR a R'.

Then e ge , PQ'R' i he e i ed iangle.


24. Gi en ,

Poin = ( +1 , 2 -2 ), ( -1 , ) and ( -3 , 2 -6)

Fo he gi en oin ( ₁ , ₁ ) , ( ₂ , ₂ ) and ( ₃ , ₃) o be collinea hen

[ ₁ ( ₂ - ₃ ) + ₂( ₃ - ₁ )+ ₃( ₁- ₂ )] = 0

He e,

₁ = + 1 ₁ = 2 - 2

₂ = - 1 ₂ =

₃ = - 3 ₃ = 2 - 6

S b i ing he al e in he fo m la ,

( + 1 ) ( - ( 2 - 6 ) ) + ( - 1 ) ( 2 - 6 - ( 2 - 2 ) ) + ( - 3 ) ( 2 - 2 - ( ) ) = 0

( + 1 ) ( - 2 + 6 ) ) + ( - 1 ) ( 2 - 6 - 2 + 2 ) ) + ( - 3 ) ( 2 - 2 - ) = 0

( + 1 ) ( - + 6 ) + ( - 1 ) ( -4 ) + ( - 3 ) ( - 2 ) = 0

- - + 6 + 6 - 4 + 4 + - 3 - 2 + 6 = 0

- 4 + 16 = 0

4 = 16

Di iding bo h he ide b 4

4 / 4 = 16 / 4

= 4

Hence,

Fo he oin o be collinea , = 4


25.

26.


Gi en:- Le ABCD be he ad ila e al ci c m c ibing he ci cle i h cen e O. Thead ila e al o che he ci cle a oin P,Q,R and S.

To o e:- AB+CD+AD+BC

P oof:-

A e kno ha , leng h of angen d a n f om he e e nal oin a e e al.The efo e,

AP=AS.....(1)

BP=BQ.....(2)

CR=CQ.....(3)

DR=DS.....(4)

Adding e a ion (1),(2),(3) and (4), e ge

AP+BP+CR+DR=AS=BQ+CQ+DS

(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)

⇒AB+CD=AD+BCHence P o ed.

27.


Le PQ be he chimne

Le AB be he o er here AB = 40 m

Also, gi en ha and

Le s consider he riangle ABP

Th s, e ha e,

No , e shall consider he riangle APQ

Hence, e ha e,

Th s, he height of the chimney is 120 m

Since, he minim m heigh of a smoke emi ing chimne sho ld be 100 m and he heigh of he chimne is 120 m.

Th s, he heigh of he chimne mee s he poll ion norms.

28.


S rface area of he remaining solid = S rface area of he c be - S rface area of he circle on one of he faces of he c be + S rface area of he hemisphere.

29.

In an A.P , he m of n e m i Sn= n2[2a+ (n 1) d].

he e,

a = Fi e m

d = Common diffe ence

n = N mbe of e m in an A.P.

We ha e o o e ha , S30= 3(S20 S10).

Take R.H.S., 3(S20 S10)

= 3 [202 2a + (20 1)d 102 2a + (10 1)d ]

= 3[10 (2a + 19d) 5(2a + 9d)]


= 3[20a + 190d 10a 45d]

= 3[10a + 145d]

= 15[2a + 29d]

No ake L.H.S., e ge

S30= 302[2a + (30 1)d]S30= 15[2a + 29d]

∴ L.H.S. =R.H.S.Hence, o ed.

30. (i). ol me of f m of cone= ⅓ h ( R + R + )= ⅓ 22/7 24 ( 14 + 14 7 + 7 )= 22 8/7 ( 196 + 98 + 49 )= 22 8/7 343= 22 8 49= 8624 cm(ii). lan heigh of f m l= ( ( R - ) + h )= ( ( 14 - 7 ) + 24 )= ( 7 + 24 )= ( 49 + 576 )= 625= 25 cma ea of me allic hee = CSA of f m + a ea of ba e= l ( R + ) + = 22/7 25 ( 14 + 7 ) + 22/7 7= 550/7 21 + 22 7= 550 3 + 154= 1650 + 154= 1804 cm


31. Le he con ec i e e en n mbe be & + 2.

A/

S a e of he e n mbe i 340. ( ) + ( + 2) = 340

+ + 2 + 2 2 = 340

2 + 4 = 340 4

2 ( + 2) = 336

( + 2) = 336/2

+ 2 = 168

+ 2 168 = 0

No , B ing he middle e m li ing me hod.

+ 2 168

+ 14 12 168

( + 14) 12 ( + 14)

( + 14) ( 12)

( + 14) = 0 o , ( 12) = 0

= 14 o = 12

Rejec he nega i e al e of and ake = 12

Hence, Re i ed n mbe a e

= 12 and ( + 2) = 12+2 = 14Ve ifica ion -


12 + 14 = 340

144 + 196 = 340

340 = 340

32.Tangen i a line hich o che a ci cle e ac l a one oin . All o he line in e ec heci cle a o oin . Take a ecan PQOR. Tangen be PT.Le he adi be R. Le PQ = .

o e oin of P ci cle: = PQ * PR = ( + 2R) = + 2 R = ( + R) - R = PO - RSo o e oin of P ci cle de end onl on di ance PO and Radi .

Po e oin of P calc la ed along PT, ( ha i Q and R me ge a T): = PT * PT

Hence, e ha e : PT = PO - RIn he iangle PTO, hi mean P hago a heo em i alid. So he iangle i ighangled.

Hence PT ⊥ OT.


33. In a die ha 6 face n mbe ed 1,2,3,4,5 and 6.

N mbe of o ible o come on olling he dice ice a e

(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)

To al n mbe of o come = 36.

(i) Fa o able o come fo he e en ha 5 ill no ho ei he ime a e

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

To al n mbe of fa o able o come = 25

∴ P obabili of no ge ing 5 ei he ime = N mbe of fa o able o come To al N mbeof o come = 2536

(ii) Fa o able o come fo he e en ha 5 ill ho e ac l one ime a e

(1, 5), (2, 5), (3,5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 5)To al n mbe of fa o able o come = 10

∴ P obabili of ge ing 5 ch ha i ill come e ac l one ime

= N mbe of fa o able o come To al n mbe of o come = 1036


34.( ) ( ) 53 2x+3

3x 1 2 3x 12x+3

Le = m)( 2x+33x 1

I.e. 3m-2/m = 53m2 - 2 = 5m3m2 - 5m - 2 = 03m2 - 6m + m - 2 = 03m(m-2) + (m-2) = 0(3m+1)(m-2) = 0

m = 2 o m = -⅓

= 2)( 2x+33x 1 OR = -⅓)( 2x+3

3x 1

3 -1/2 +3 = 2 9 -3 = -2 -3 = -7 = 0

An e : = -7, 0


CBSE Maths 2013

1) All questions are compulsor .2) The question paper consists of thirt questions divided into 4 sections A, B, C

and D. Section A comprises of ten questions of 01 mark each, Section Bcomprises of five questions of 02 marks each, Section C comprises tenquestions of 03 marks each and Section D comprises of five questions of 06marks each.

3) All questions in Section A are to be answered in one word, one sentence or asper the e act requirement of the question.

4) There is no overall choice. However, internal choice has been provided in onequestion of 02 marks each, three questions of 03 marks each and two questionsof 06 marks each. You have to attempt onl one of the alternatives in all suchquestions.

5) In question on construction, drawing should be near and e actl as per the givenmeasurements.

6) Use of calculators is not permitted.


Questions -

EC ION A

1. The common difference of the A.P. 1/p, 1 p/p, 1 2p/p..................is:

2. In Fig. 1, PA and PB are two tangents drawn from an e ternal point P to a circle withcentre C and radius 4 cm. If PA ⊥ PB, then the length of each tangent is:

3. In Figure,, a circle with centre O is inscribed in a quadrilateral ABCD such that, ittouches the sides BC, AB, AD and CD at points P, Q, R and S respectivel , If AB = 29 cm,AD = 23 cm, ∠B = 90 and DS = 5 cm, then the radius of the circle (in cm.) is:

4. The angle of depression of a car, standing on the ground, from the top of a 75 m hightower, is 30 . The distance of the car from the base of the tower (in m.) is:

5. The probabilit of getting an even number, when a die is thrown once, is :


6. A bo contains 90 discs, numbered from 1 to 90. If one disc is drawn at random fromthe bo , the probabilit that it bears a prime-number less than 23, is :

7. In Fig. 3, Find the area of triangle ABC (in sq. units) is:

8. If the difference between the circumference and the radius of a circle is 37 cm, thenusing = 22/7, the circumference (in cm) of the circle is:

(A) 154

(B) 44

(C) 14

(D) 7

EC ION B

9. Solve the following quadratic equation for :

4 3 2+5 2 3= 0


10. How man three digit natural numbers are divisible b 7?

11. In Fig. 4, a circle inscribed in triangle ABC touches its sides AB, BC and AC at pointsD, E and F respectivel . If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengthsof AD, BE and CF.

12. Prove that the parallelogram circumscribing a circle is a rhombus.

13. A card is drawn at random from a well shuffled pack of 52 pla ing cards. Find theprobabilit that the drawn card is neither a king nor a queen.

14. Two circular pieces of equal radii and ma imum area, touching each other are cutout from a rectangular card board of dimensions 14 cm 7 cm. Find the area of theremaining card board. [ Use = 22/7]

EC ION C

15. For what value of k, are the roots of the quadratic equation k ( 2) + 6 = 0 equal?

16. Find the number of terms of the A.P. 18, 151/2, 13, ...., - 491/2 and find the sum of allits terms.


17. Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct anothertriangle whose sides are 2/3 times the corresponding sides of first triangle.

18. The hori ontal distance between two poles is 15 m. The angle of depression of thetop of first pole as seen from the top of second pole is 30 . If the height of the secondpole is 24 m, find the height of the first pole. [ 3= 1.732 ]

19. Prove that the points (7, 10), ( 2, 5) and (3, 4) are the vertices of an isosceles righttriangle.

20. Find the ratio in which the -a is divides the line segment joining the points ( 4, 6)and (10, 12). Also find the coordinates of the point of division.

21. In Fig.5, AB and CD are two diameters of a circle with centre O, which areperpendicular to each other. OB is the diameter of the smaller circle. If OA = 7 cm, findthe area of the shaded region.[ use = 22/7]

22. A vessel is in the form of a hemispherical bowl surmounted b a hollow c linder ofthe same diameter. The diameter of the hemispherical bowl is 14 cm and the totalheight of the vessel is 13 cm. Find the total surface area of the vessel. [ = 22/7]

23. A wooden to was made b scooping out a hemisphere of same radius from eachend of a solid c linder. If the height of the c linder is 10 cm, and its base is of radius 3.5cm, find the volume of wood in the to . [ = 22/7]


24. In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre. Find: (i) thelength of the arc (ii) area of the sector formed b the arc. [ Use = 22/7]

25. Solve the following for :

26. Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is16 cm, find the sides of the two squares.

27. If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find thesum of its first n terms.

28. Prove that the tangent at an point of a circle is perpendicular to the radius throughthe point of contact.

29. In fig, l and m are two parallel tangents to a circle with centre O, touching the circleat A and B respectivel . Another tangent at C intersects the line l at D and m at E. Provethat ∠DOE = 90


30. The angle of elevation of the top of a building from the foot of the tower is 30 andthe angle of elevation of the top of the tower from the foot of the building is 60 . If thetower is 60 m high, find the height of the building.

31. A group consists of 12 persons, of which 3 are e tremel patient, other 6 aree tremel honest and rest are e tremel kind. A person from the group is selected atrandom. Assuming that each person is equall likel to be selected, find the probabilitof selecting a person who is (i) e tremel patient (ii) e tremel kind or honest. Which ofthe above values ou prefer more?

32. The three vertices of a parallelogram ABCD are A(3, 4), B( 1, 3) and C( 6, 2). Findthe coordinates of verte D and find the area of ABCD.

33. Water is flowing through a c lindrical pipe, of internal diameter 2 cm, into ac lindrical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in thelevel of water in the tank in half an hour.

34. A bucket open at the top, and made up of a metal sheet is in the form of a frustumof a cone. The depth of the bucket is 24 cm and the diameters of its upper and lowercircular ends are 30 cm and 10 cm respectivel . Find the cost of metal sheet used in itat the rate of Rs 10 per 100cm2. [Use = 3.14]


CBSE Ma h 2013 1.The common diffe ence i - 1.

2. Gi en ha , AP PB, CA AP , CB BP.

⇒∠ACB = 90 .

And AC = CB ( adi of he ci cle)

∴APBC i a a e .

Each ide of he a e i e al o 4 cm.

The efo e, leng h of each angen i 4 cm.

3.The leng h of he angen d a n f om an e e nal oin o a ci cle a e e al.

DS = DR = 5 cm

∴ AR = AD - DR = 23 5 = 18 cm

AQ = AR = 18 cm

∴ QB = AB - AQ = 29 18 = 11 cm

QB = BP = 11 cm.

∠PBQ = 90 [Gi en]

∠OPB = 90 [Angle be een he angen and he adi a he oin of con ac i a ighangle.]


∠OQB = 90 [Angle be een he angen and he adi a he oin of con ac i a ighangle.]

∠POQ = 90 [Angle m o e of a ad ila e al.]

So, OQBP i a a e.

∴QB = BP = = 11 cm

∴ he adi of he ci cle i 11 cm.

4.in T i.ABC

an30 =AB/BC

1/ 3=75/BC

BC=75 3

5. When a die i h o n once, he am le ace S i gi en b 1, 2, 3, 4, 5, 6 .

N mbe of o ible o come = 6

Le E be he e en of ge ing an e en n mbe . E = 2, 4, 6

N mbe of fa o able o come = 3

∴ P obabili of ge ing an e en n mbe = P (E) =N mbe of fa o able o come /N mbe of o ible o come = 3/6 =


6. Gi en ha he bo con ain 90 di c .

A one di c i d a n a andom f om he bo , n mbe ed f om 1 o 90 i.e 1,2,3,.......90.

N mbe of o ible o come = 90

Le E be he e en of ge ing a ime n mbe le han 23 a e 2, 3, 5, 7, 11, 13, 17, and19.


∴ P(E) = N mbe of fa o able o come /N mbe of o ible o come = 8/90= 4/45.

7.

Gi en : BC = 5 ni and AM = 3 ni .

In ABC, BC i he ba e and AM i he heigh .

∴ A ea of iangle ABC = 1/2 ba e heigh

= 1/2 BC AM

= 1/2 5 3

= 7.5 . ni


8.

Gi en ha he diffe ence be een ci c mfe ence and adi of he ci cle i 37 cm.

A me he adi of he ci cle a ' '.

⇒2 - = 37

⇒ (2 - 1) = 37

⇒ (2 22/7 - 1) = 37

⇒ 37/7 = 37

⇒ = 7 cm

∴ Ci c mfe ence of he ci cle = 2 = 2 227 7 = 44 cm

An -> B

9.

Gi en ad a ic e a ion i

10.

Le h ee digi n mbe f om 100 o 999.


All he h ee-digi na al n mbe ha a e di i ible b 7 ill be of he fo m 7n.

∴ 100 7n 999

100/7 n 9997

142/7 n 1425/7

A n i an in ege , n mbe of 3-digi na al n mbe ha a e di i ible b 7 a e 142 14i.e. 128.

11.


12.

Since ABCD i a a allelog am ci c m c ibed in a ci cle

AB=CD........(1)

BC=AD........(2)

DR=DS (Tangen on he ci cle f om ame oin D)

CR=CQ(Tangen on he ci cle f om ame oin C)

BP=BQ (Tangen on he ci cle f om ame oin B )

AP=AS (Tangen on he ci cle f om ame oin A)

Adding all he e e a ion e ge

DR+CR+BP+AP=DS+CQ+BQ+AS

(DR+CR)+(BP+AP)=(CQ+BQ)+(DS+AS)

CD+AB=AD+BC

P ing he al e of e a ion 1 and 2 in he abo e e a ion e ge

2AB=2BC

AB=BC...........(3)

F om e a ion (1), (2) and (3) e ge


AB=BC=CD=DA

∴ABCD i a Rhomb

13.

Le E be he e en of ha he d a n ca d i nei he a king no a een.

To al n mbe of o ible o come = 52.

To al n mbe of ca d ha a e king o een in he ack of la ing ca d = 4 + 4 = 8.

∴ N mbe of ca d ha a e nei he a king no a een = 52 8 = 44.

To al n mbe of fa o able o come = 44.

∴ Re i ed obabili = P(E) = N mbe of fa o able o come /To al n mbe ofo ible o come = 44/52 = 11/13

∴ The obabili ha he d a n ca d i nei he a king no a een i 11/13.

14.

Dimen ion of he ec ang la ca d boa d = 14 cm 7 cm

T o ci c la iece of e al adii and ma im m a ea o ching each o he a e c f omhe ec ang la ca d boa d, he efo e, he diame e of each ci c la iece i 14/2 = 7 cm.

Radi of each ci c la iece = d/2=7/2 cm

∴ S m of a ea of o ci c la iece =


A ea of he emaining ca d boa d = A ea of he ca d boa d A ea of o ci c la iece .

= (14 7) (77)= 98 77 = 21 cm2.

∴The a ea of he emaining ca d boa d i 21 cm2.

15.


16.

Gi en : A.P. 18, 151/2, 13, ...., - 491/2


17.

S e 1: D a a line egmen AB = 4 cm. Con ide he oin A a cen e, d a an a c of 5cm adi .

Simila l , con ide he oin B a i cen e, d a an a c of 6 cm adi .

Bo h he a c ill in e ec each o he a C and he e i ed iangle ABC i fo med.

S e 2: D a a a AX making an ac e angle i h he line egmen AB on he o o i eide of he e e C.

S e 3: Loca e 3 oin A1, A2, A3on line AX ch ha AA1 = A1A2= A2A3.

S e 4: Join B and A3i.e. BA3and d a a line h o gh A2 a allel o BA3 o in e ec ABa oin B'.

S e 5: D a a line egmen h o gh B' a allel o BC o in e ec AC a C'.

∴ AB'C' i he e i ed iangle.


18.


19.

Gi en : Poin (7, 10), (-2, 5) and (3, -4 ) a e e ice of a iangle.

Sol ion :

Le A(7, 10), B (- 2, 5) and C (3, - 4 )

B ing di ance fo m la : ( 2 - 1) + ( 2 - 1)

Ve ice : A(7, 10), B (- 2, 5)

Leng h of ide AB = (- 2 - 7) + (5 - 10)

AB = (-9) + (-5)

AB = 81 + 25

AB = 106 ni

Ve ice : B (- 2, 5) and C (3,- 4)

Leng h of ide BC = (3 + 2) + ( - 4 - 5)

BC = (5) + (-9)

BC = 25 + 81

BC = 106 ni


Ve ice : A(7, 10),C (3, - 4 )

Leng h of ide AC = (3 - 7) + (- 4 - 10)

AC = (-4) + (-14)

AC = 16 + 196

AC = 212 ni

Since he 2 ide AB = BC = 106.

The efo e i an i o cele .

No , in ABC, b ing P hago a heo em

AC = AB + BC

( 212) = ( 106) + ( 105)

212 = 106 + 106

212 = 212

Since AC = AB + BC

Hence, he gi en e ice of a iangle i an i o cele igh iangle. 20.


21. Gi en ha AB and CD a e he diame e of a ci cle i h cen e O.

∴ OA = OB = OC = OD = 7 cm (Radi of he ci cle)

A ea of he haded egion

= A ea of he ci cle i h diame e OB + (A ea of he emi-ci cle ACDA A ea of ACD)

= 2 + ( 1/2 2 -1/2 CD OA)

= 22/7 (7/2)2+(1/2 22/7 49 1/2 14 7)

= (77/2+ 77 - 49) cm2

= 66.5 cm2

∴The a ea of he haded egion i 66.5cm2.


22.

Radi =7 cmHeigh of c lind ical o ion =13 7=6 cm

C ed face a e of c lind ical o ion can be calc la ed a follo :2 h=2 22/7 7 6=264cm2

C ed face a ea of hemi he ical o ion can be calc la ed a follo :2 2

=2 22/7 (7)2

=308cm2

Then, o al face a ea of he e el =308+264=572 cm

23.Gi en heigh of c linde =10 cmRadi of c linde =3.5 cm

The ol me of o = Vol me of c linde 2 Vol me of a hemi he e= 2h (2 2/3 3)=[ (3.5)2 10] (2 2/3 (3.5)3) = 205.251 cm3


24.

25.


-2ab = 4a + 4 + 2 b

0= 4a + 4 + 2 b + 2ab

0= 4 (a + ) + 2b( + a)

0= ( + a) (4 + 2b)

[B Ze o P od c R le]

+ a = 0 OR 4 + 2b = 0

= -a OR 4 = -2b

= -a OR = -b/2

26. A me ha he ide of he o a e be cm and cm he e > .

The a ea of he a e a e 2 and 2 and hei e ime e a e 4 and 4 .

Gi en ha , 2+ 2= 400 (1)

and 4 4 = 16

⇒ 4( ) = 16 ⇒ - = 4 (2)

⇒ = + 4

S b i ing he al e of in e n (1), e ge ( +4)2+ 2= 400

⇒ 2+16 + 8 + 2 = 400

⇒2 2+ 8 + 16 = 400


⇒ 2+ 4 - 192 = 0

⇒ 2 + 16 12 - 192 = 0

⇒ ( + 16) 12( + 16) = 0

⇒ ( + 16) ( 12) = 0

⇒ = 16 o = 12

Since he al e of canno be nega i e o he al e of = 12.

Val e of = + 4 = 12 + 4 = 16.

∴The ide of he o a e a e 16 cm and 12 cm.

27.S m of fi 7 e m =49

⇒7/2(2a+6d)=49

⇒a+3d=7

⇒S m of fi 17 e m =289

⇒17/2(2a+16d)=289

⇒a+8d=17

⇒(a+8d) (a+3d)=17 7

⇒5d=10

⇒d=2

⇒a=1

S m of fi n e m


=2n(2+2n 2)

=n2

28.

Refe ing o he fig e:

OA=OC (Radii of ci cle)

No OB=OC+BC

∴OB>OC (OC being adi and B an oin on angen )

⇒OA<OB

B i an a bi a oin on he angen .

Th , OA i ho e han an o he line egmen joining O o anoin on angen .

Sho e di ance of a oin f om a gi en line i he e endic la di ance f om haline.

Hence, he angen a an oin of ci cle i e endic la o he adi .


29.

Join OCIn iangle ODA and iangle ODCOA=OC (Radii of he ame ci cle )AD=DC (Leng h of angen d a n f om an e e nal oin o a ci cle a e e al)DO=OD (common ide)

DOA≅ ODC∴∠DOA=∠COD

DOA≅ ODC∴∠DOA=∠COD

Simila l OEB≅ OEC∴∠EOB=∠COEAOB i a diame e of he ci cle.Hence, i i a aigh line.

∴∠DOA+∠COD+∠COE+∠EOB=180⇒2∠COD+2∠COE=180⇒∠COD+∠COE=90⇒∠DOE=900

Hence o ed.


30.

Angle of ele a ion of he o of a b ilding f om he foo of a o e =300

Angle of ele a ion of he o of a he o e f om he foo of he b ilding =600

Heigh of o e = 60mLe he heigh of he b ilding be h mLe he di ance he o e and b ilding be b mIn ABC


∴ heigh of he b ilding = 20m

31. Gi en ha he g o con i of 12 e on .

∴To al n mbe of o ible o come = 12

A me E1 a he e en of elec ing e on ho a e e emel a ien .

∴ N mbe of fa o able o come o E1 i 3.

A me E2 a he e en of elec ing e on ho a e e emel kind o hone .

N mbe of eo le ho a e e emel hone i 6.

N mbe of e on ho a e e emel kind = 12 (6 + 3) = 3

∴ N mbe of fa o able o come o E2 = 6 + 3 = 9.

(i) P obabili of elec ing a e on ho i e emel a ien i P(E1).

P(E1) = N mbe of o come fa o able o E1/To al n mbe of o ible o come =3/12 = 1/4.

(ii) P obabili of elec ing a e on ho i e emel kind o hone i

P (E2) = N mbe of o come fa o able o E2 / To al n mbe of o ible o come =9/12 = 3/4.

32.



33.

Gi en:

Wa e i flo ing h o gh a c lind ical i e of in e nal diame e 2cm in o a c lind icalank of ba e adi 40cm, a a e of 0.4m/ econd.

To find:

The i e in le el of a e in he ank in half an ho .

We ha e,

In e nal diame e of c lind ical i e= 2cm

Radi of c lind ical ank,[R]= 40cm

The leng h of a e flo ing in 1 econd= 0.4m

We kno ha 1m= 100cm

So,

0.4m = (0.4 100)cm

40cm leng h of a e flo ing in 1 econd.

In e nal adi of c lind ical i e,[ ]=

In e nal adi of c lind ical i e,[ ]= 1cm.

Acco ding o he e ion:

Leng h of a e flo in half an ho ,[h]= 30min e

Leng h of a e flo in econd=(30 40 60)cm

Leng h of a e flo in econd= 72000cm

No ,

We kno ha fo m la of he ol me of c linde : h c bic ni


Vol me of a e in c lind ical ank = Vol me of a e flo f om i e

R H = h

R H= h

(40cm) H = (1cm) 72000cm

1600cm H = 72000cm

H= 45cm

34.

GIVEN :

Diame e of e ci c la end of a b cke = 30 cm

Radi of he e end of he b cke , R = 30/2 = 15 cm

Diame e of lo e ci c la end of a b cke = 10 cm

Radi of he lo e end of he b cke , R = 10/2 = 5 cm

Heigh of he b cke , h = 24 cm

Slan heigh of a b cke , l = (R - ) + h

l = (15 - 5) + 24

l = 10 + 576

l = 100 + 576

l = 676

l = 26 cm

Slan heigh of a b cke , l = 26 cm


S face a ea of he b cke = C ed face a ea of a b cke + A ea of he malleci c la ba e

= (R+ )l +

= (15 + 5) 26 + 5

= (20 26 + 25)

= (520 + 25)

= 545

= 3.14 545

= 1711.3 cm

S face a ea of he b cke = 1711.3 cm

Co of me al hee ed @ of 10 e 100 cm = 1711.3 cm 10/100

= 1711.3 0.1 = 171.13

Hence, he co of me al hee ed i 171.13. ----- ----- -----


CBSE Maths 2012

1) All questions are compulsor .2) The question paper consists of thirt questions divided into 4 sections A, B, Cand D. Section A comprises of ten questions of 01 mark each, Section B comprises offive questions of 02 marks each, Section C comprises ten questions of 03 marks eachand Section D comprises of five questions of 06 marks each.3) All questions in Section A are to be ans ered in one ord, one sentence or asper the e act requirement of the question.4) There is no overall choice. Ho ever, internal choice has been provided in onequestion of 02 marks each, three questions of 03 marks each and t o questions of 06marks each. You have to attempt onl one of the alternatives in all such questions.5) In question on construction, dra ing should be near and e actl as per the givenmeasurements.6) Use of calculators is not permitted.

Questions -

1. The length of shado of a to er on the plane ground is 3 times the height of theto er. The angle of elevation of sun is: (A) 45 (B) 30 (C) 60 (D) 90

2. If the area of a circle is equal to sum of the areas of t o circles of diameters 10 cmand 24 cm, then the diameter of the larger circle (in cm) is: (A) 34 (B) 26 (C) 17 (D) 14

3. If the radius of the base of a right circular c linder is halved, keeping the height thesame, then the ratio of the volume of the c linder thus obtained to the volume of originalc linder is: (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1

4. T o dice are thro n together. The probabilit of getting the same number on bothdice is: (A) 1/2 (B) 1/3 (C) 1/6 (D) 1/12

5. The coordinates of the point P dividing the line segment joining the points A(1,3) andB(4,6) in the ratio 2 : 1 are: (A) (2,4) (B) (3,5) (C) (4,2) (D) 5,3)


6. If the coordinates of the one end of a diameter of a circle are (2,3) and thecoordinates of its centre are (-2,5), then the coordinates of the other end of the diameterare: (A) (-6,7) (B) ( 6,-7) (C) (6,7) (D) (-6,-7)

7. The sum of first 20 odd natural number is : (A) 100 (B) 210 (C) 400 (D) 420

8. If 1 is a root of the equations a 2 + a + 3 = 0 and 2 + + b = 0, then ab equals:

(A) 3 (B) -7/2 (C) 6 (D) -3

9. In Fig., the sides AB, BC and CA of a triangle ABC, touch a circle at P, Q and Rrespectivel . If PA = 4 cm, BP = 3 cm and AC = 11 cm, then the length of BC (in cm) is:(A) 11 (B) 10 (C) 14 (D) 15

10. In Fig., a circle touches the side DF of EDF at H and touches ED and EF produced atK and M respectivel . If EK = 9 cm, then the perimeter of EDF (in cm) is: (A) 18 (B) 13.5(C) 12 (D) 9


SECTION B

11. If a point A(0,2) is equidistant from the points B(3,p) and C(p,5) then find the value ofp.

12. A number is selected at random from the first 50 natural numbers. Find theprobabilit that it is a multiple of 3 and 4.

13. The volume of a hemisphere is 24251/2 cm3 . Find its curved surface area.

[Use = 22/7]π

14.Tangents PA and PB are dra n from an e ternal point P to t o concentric circle ithcentre O and radii 8 cm and 5 cm respectivel , as sho n in Fig., If AP = 15 cm, then findthe length of BP.

15.In fig., an isosceles triangle ABC, ith AB = AC, circumscribes a circle. Prove that thepoint of contact P bisects the base BC.

OR

In fig., the chord AB of the larger of the t o concentric circles, ith centre O, touchesthe smaller circle at C. Prove that AC = CB


16. In fig., OABC is a square of side 7 cm. If OAPC is a quadrant of a circle ith centre O,then find the area of the shaded region.

[Use = 22/7]π

17.Find the sum of all three digit natural numbers, hich are multiples of 7.

18.Find the values (s) of k so that the quadratic equation 3 2 2k + 12 = 0 has equalroots.

SECTION C

19.A point P divides the line segment joining the points A(3,-5) and B(-4,8) such thatAP/PB = K/1 . If P lies on the line + = 0, then find the value of K.

20.If the vertices of a triangle are (1,-3), (4,p) and (-9,7) and its area is 15 sq. units, findthe value (s) of p.


21.Prove that the parallelogram circumscribing a circle is a rhombus.OR

Prove that opposite sides of a quadrilateral circumscribing a circle subtendsupplementar angles at the centre of the circle.

22. From a solid c linder of height 7 cm and base diameter 12 cm, a conical cavit ofsame height and same base diameter is hollo ed out. Find the total surface area of theremaining solid.

[Use = 22/7]π

OR

A c lindrical bucket, 32 cm high and ith radius of base 18 cm, is filled ith sand. Thisbucket is emptied on the ground and a conical heap of sand is formed. If the height ofthe conical heap is 24 cm, then find the radius and slant height of the heap.

23. In fig., PQ and AB are respectivel the arcs of t o concentric circles of radii 7 cmand 3.5 cm and centre O. If ㄥPOQ = 30o, then the area of the shaded region is?

24. Solve for : 4 2 4a + (a2 b2 ) = 0

25. A kite is fl ing at a height of 45 m above the ground. The string attached to the kiteis temporaril tied to a point on the ground. The inclination of the string ith the groundis 60o. Find the length of the string assuming that there is slack in the string.

26. Dra a triangle ABC ith side BC = 6 cm, ㄥC = 30o and ㄥA = 105o. Then constructanother triangle hose sides are 2/3 times the corresponding sides of triangle ABC.

27. The 16th term of an AP is 1 more than t ice its 8th term. If the 12th term of the APis 47, then find its nth term.

28.A card is dra n from a ell shuffled deck of 52 cards. Find the probabilit of getting(i) a king of red colour (ii) a face card (iii) the queen of diamonds.


SECTION D29. A bucket is in the form of a frustum of a cone and its can hold 28.49 litres of ater.If the radii of its circular ends are 28 cm and 21 cm, find the height of the bucket.

[Use = 22/7]π

30.The angle of elevation of the top of a hill at the foot of a to er is 60 and the angle ofdepression from the top of the to er of the foot of the hill is 30. If the to er is 50 mhigh, find the height of the hill.


32.A shopkeeper bu s some books for 80. If he had bought 4 more books for the sameamount, each book ould have cost 1 less. Find the number of books he bought.

33.Sum of the first 20 terms of an AP is -240, and its first term is 7. Find its 24th term.

34.A solid is in the shape of a cone standing on a hemisphere ith both their radii beingequal to 7 cm and the height of the cone is equal to its diameter. Find the volume of thesolid.

[Use = 22/7]π


CBSE Ma h 2012S -

1. B

2. B

3. C

4. C

5. B

6. A

7. C

8. A

9. B

10. A

11. I A (0, 2) B(3, ) C( , 5).

S , AB = AC, AB2 = AC2

U ing di ance fo m la, e ha e:

12. The o al n mbe of o come i 50. Fa o able o come = 12, 24, 36, 48

Re i ed p obabili = N mbe of fa o able o come = 4 = 2 T 50 25


13. G : 2425 1/2 3 = 4851/2 3 N ,

= 2/3 3

= 2/3 3 = 4851/2= ⅔ 22/7 3 = 4851/2R3 = 4851/2 3/2 7/22 = (21/2)3

The efo e = 21/2 cmSo, C ed face a ea of he hemi phe e = 2 2

693 .cm

14. Gi en: Tangen PA and PB a e d a n f om an e e nal poin P o oconcen ic ci cle i h cen e O and adii OA = 8 cm, OB = 5 cme pec i el . Al o, AP = 15 cm

To find: Leng h of BPCon c ion: We join he poin O and P.

Sol ion: OA AP ; OB BP[U ing he p ope ha adi i pe pendic la o he angen a he poin of con ac ofa ci cle]In igh angled iangle OAP,OP2 = OA2 + AP2 [U ing P hago a Theo em]= (8)2 + (15)2 = 64 + 225 = 289∴ OP = 17 cmIn igh angled iangle OBP,OP2 = OB2 + BP2

BP2 = OP2 - OB2

= (17)2 (5)2 = 289 25 = 264∴ BP = . cm√264 = 2√66


15. Gi en: ABC i an i o cele iangle, he e AB = AC, ci c m c ibing a ci cle.To p o e: The poin of con ac P bi ec he ba e BC.i.e. BP = PCP oof: I can be ob e ed haBP and BR ; CP and CQ; AR and AQ a e pai of angen d a n o he ci clef om he e e nal poin B , C and A e pec i el .So, appl ing he e l ha he angen d a n f om an e e nal poin o a ci cle, e geBP = BR --- (i)CP = CQ --- (ii)AR = AQ --- (iii)Gi en ha AB = AC⇒ AR + BR = AQ + CQ⇒ BR = CQ [f om (iii)]⇒ BP = CP [f om (i) and (ii)]∴ P bi ec BC.Hence p o ed.

OR

Gi en: The cho d AB of he la ge of he o concen ic ci cle , i h cen e O, o chehe malle ci cle a

C.To p o e: AC = CBCon c ion: Le join OC.

P oof: In he malle ci cle, AB i a angen o he ci cle a he poin of con ac C.∴ OC AB ------ (i)(U ing he p ope ha he adi of a ci cle i pe pendic la o he angen a he poinof con ac )Fo he la ge ci cle, AB i a cho d and f om (i) e ha e OC AB∴ OC bi ec AB


(U ing he p ope ha he pe pendic la d a n f om he cen e o a cho d of a ci clebi ec he cho d)∴ AC = CB

16. Gi en, OABC i a a e of ide 7 cmi.e. OA = AB = BC = OC = 7cm∴ A ea of a e OABC = ( ide)2 = 72 = 49 .cmGi en, OAPC i a ad an of a ci cle i h cen e O.∴ Radi of he ec o = OA = OC = 7 cm.Sec o angle = 90o

∴ A ea of ad an OAPC 90o/360o 2

17. Fi h ee- digi n mbe ha i di i ible b 7 = 105Ne n mbe = 105 + 7 = 112The efo e he e ie i 105, 112, 119,The ma im m po ible h ee digi n mbe i 999.When e di ide b 7, he emainde ill be 5.Clea l , 999 5 = 994 i he ma im m po ible h ee digi n mbe di i ible b 7.The e ie i a follo :105, 112, 119, ., 994He e a = 105, d = 7Le 994 be he n h e m of hi A.Pan = a + (n-1)d994 = 105 + (n-1)7(n-1)7 = 889(n-1) = 127n = 128So, he e a e 128 e m in he A.P. S m = n/2 fi e m + la e m= 128/2 a1 + a128

= 64 105+994 = (64)(1099) = 70336

18. Gi en ad a ic e a ion i 3 2 2k + 12 = 0He e a = 3, b = -2k and c = 12


The ad a ic e a ion ill ha e e al oo if = 0b2 - 4ac = 0P ing he al e of a,b and c e ge

Con ide ing a e oo on bo h ide ,k = ±√36 = 6The efo e, he e i ed al e of k a e 6 and -6.

19.

Le he co-o dina e of poin P be ( , )Then ing he ec ion fo m la co-o dina e of P a e.

= K + 14K + 3 = K + 1

8K 5

Since P lie on + =0

+ = 0K + 14K + 3

K + 18K 5

4K - 2 = 0K = 2/4K =


20.

The a ea of a iangle , ho e e ice a e ( 1, 1), ( 2, 2) and ( 3, 3) i

S b i ing he gi en coo dina e

A ea of iangle = 1(p-7) + 4(7+3) + (-9)(-3-p)On ol ing -p = -3 o -9An hence he al e of p=-3 o -9

21. Le ABCD be a pa allelog am ch ha i ide o ching a ci cle i h cen e O.We kno ha he angen o a ci cle f om an e e io poin a e e al in leng h.

AP = AS [F om A] (i)BP = BQ [F om B] (ii)CR = CQ [F om C] (iii)and, DR = DS [F om D] (i )Adding (i), (ii), (iii) and (i ), e ge


AP+BP+CR+DR=AS+BQ+CQ+DS(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)AB+CD = AD+BC2 AB = 2 BC [ ABCD i a pa allelog am he efo e AB=CD and BC = AD]AB=BCTh , AB=BC=CD=ADHence, ABCD i a homb .

OR

A ci cle i h cen e O o che he ide AB, BC, CD, and DA of a ad ila e al ABCD ahe poin P, Q, R and S e pec i el .

TO PROVE : ㄥAOB + ㄥCOD = 180 and, ㄥAOD + ㄥBOC = 180

CONSTRUCTIONJoin OP, OQ, OR and OS.

PROOFSince he o angen d a n f om a e e nal poin o a ci cle b end e al angle ahe cen e.

The efo e ㄥ1=ㄥ2, ㄥ3=ㄥ4,ㄥ5=ㄥ6, ㄥ7=ㄥ8No ㄥ1+ㄥ2+ㄥ3+ㄥ4+ㄥ5+ㄥ6+ㄥ7+ㄥ8 = 360o

[S m of all he angle b ended a a poin i 360o]2( ㄥ2 + ㄥ 3 + ㄥ 6 + ㄥ 7) = 360 and 2( ㄥ 1 + ㄥ 8 + ㄥ4 + ㄥ 5) = 360o

[ ㄥ2+ㄥ3 = ㄥAOB, ㄥ6+ㄥ7 = ㄥCOD ][ ㄥ1+ㄥ8 = ㄥAOD and ㄥ4+ㄥ5 = ㄥBOC ]

ㄥAOB + ㄥCOD = 180o


ㄥAOD + ㄥBOC = 180o

Hence P o ed

22.

Gi en: adi of c l= adi of cone= =6cmHeigh of he c linde =heigh of he cone=h=7cmSlan heigh of he cone= l√7.7 .6+ 6= √85To al face a ea of he emaining olid= c ed face a ea of he c linde + a ea of he ba e of he c linde + c ed facea ea of he cone(2 h + 2 + l)= 2 22/7 6 7 + 22/7 6.6 + 22/7 6 √85= 377.1 + 132/7 cm2√85

ORVol me of he conical heap= ol me of he and emp ied f om he b cke .Vol me of he conical heap=⅓ 2h = 1/3 2 24cm2 (heigh of he conei 24)----------(1)

Vol me of he and in he b cke = 2h= (18)232 cm2 -------(2)

E a ing 1 and 2⅓ 2 24 = (18)232

2 = (18)232 324

= 36cm


23. [U e = 22/7]

A ea of he haded egion=A ea of ec o POQ-A ea of ec o AOB

R2 - 2360 360

A ea of Shaded egion = 30/360 22/7 (72 - 3.52)= 77/8 cm2

24.

[(2 - a) - b) = 0 o [(2 -a) + b] = 0

= (a+b)/2 ; = (a-b)/2

25. Gi en: Po i ion of ki e i B.Heigh of ki e abo e g o nd= 45 mAngle of inclina ion = 60


Re i ed leng h of ing = AB

In igh angled iangle AOB,

Hence, he leng h of he ing i m03 √3

26.


27. Le a and d e pec i el be he fi e m and he common diffe ence of he AP.We kno ha he n h e m of an AP i gi en b an = a + (n 1)dAcco ding o he gi en info ma ion,A16 = 1 + 2 a8

a + (16 1)d = 1 + 2[a + (8 1)d]a + 15d = 1 + 2a + 14d

a + d = 1 (1)Al o, i i gi en ha , a12 = 47a + (12 1)d = 47a + 11d = 47 (2)Adding (1) and (2), e ha e:12d = 48d = 4F om (1),

a + 4 = 1a= 3Hence, an = a + (n 1)d = 3 + (n 1)(4) = 3 + 4n 4 = 4n 1Hence, he n h e m of he AP i 4n 1.

28. To al n mbe of o come =52(i) P obabili of ge ing a ed kingHe e he n mbe of fa o able o come =2P obabili = no.of fa o able o come

o al n mbe of o come= 12/52= 3/13

(iii)P obabili of een of diamondn mbe of een of diamond=1,hence

P obabili = no.of fa o able o comeo al n mbe of o come

1/52


29.

He e, R = 28 cm and = 21 cm, e need o find h.Vol me of f m = 28.49 L = 28.49 1000 cm3 = 28490 cm3

No , Vol me of f m = (R2+R + 2)h/3On p ing al e -h = 28490 21 = 15 22 1813

Heigh of b cke i 15 cm.

30. Gi en :-

AB i he hill ha ing h me e heighDC he o e ha ing 50 me e heigh .Angle ACB i 60 and Angle DBC i 30 .The heigh of hill H = 50 + m

Taking he iangle DBC .

He e Pe pendic la i 50m ( DC ) and Ba e i m(CB)


Angle DCB i 90 and angle DBC i 30 .

No ba e of iangle DBC i e i alen o he ba e of iangle ABC.

No aking iangle ABC

He e Pe pendic la i AB( H me e ) and ba e i CB(50 3 m)

Angle ABC i 90 and Angle ACB i 60

31. Gi en: AB i a angen o a ci cle i h cen e O.To p o e: OP i pe pendic la o AB.


Con c ion: Take a poin Q on AB and join OQ.

P oof: Since Q i a poin on he angen AB, o he han he poin of con ac P, o Q illbe o ide he ci cle.Le OQ in e ec he ci cle a R.No OQ = OR + RQ

OQ > OR OQ > OP [a OR = OP]

OP < OQTh OP i ho e han an o he egmen among all and he ho e leng h i hepe pendic la f om Oon AB. OP AB. Hence p o ed.

32. L

T = R . I ,

T = R .


OR

OR

OR A , , , -20 . T = 16 33. G : S20 = -240 = 7 S = /2 2 +( -1) S20 = 20/2 2 7 +19 = -240 O , 10(14 + 19 ) = -240 19 = -38

= -2 24 = + 23 = 7 + 23(-2) 24 = -39

34.

R - = 7 R = 7 H = = 14

= + H -


= 2h + ⅔ 3 = 2(h+2 )= ⅓ 22/7 49 (14+14)= ⅓ 22/7 49 28= 4312/2 cm3


CBSE Maths 2011

1) All questions are compulsor .

2)

The question paper consists of 34 questions divided into 4 sections A, B, C and D.Section A comprises of ten questions of 01 mark each, Section B comprises of eightquestions of 02 marks each, Section C comprises ten questions of 03 marks eachand Section D comprises of si questions of 06 marks each.

3) All questions in Section A are to be ans ered in one ord, one sentence or as perthe e act requirement of the question.

4)

There is no overall choice. Ho ever, internal choice has been provided in onequestion of 02 marks each, three questions of 03 marks each and t o questions of06 marks each. You have to attempt onl one of the alternatives in all suchquestions.

5) In question on construction, dra ing should be near and e actl as per the givenmeasurements.

6) Use of calculators is not permitted.

Questions -

EC I N A

1. The roots of the equation 2 3 m(m+3)=0, here m is a constant, are:

2. If the common differences of an A.P. is 3, then a20 a15 is:


3. In figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ= 70 , then ∠TPQ is equal to ________.

4. In Figure,, AB and AC are tangents to the circle ith centre O such that ∠BAC = 40 .Then ∠BOC is equal to ________.

5. The perimeter (in cm) of a square circumscribing a circle of radius a cm, is _________.

6. The radius (in cm) of the largest right circular cone that can be cut out from a cube ofedge 4.2 cm is _________.

7. A to er stands verticall on the ground. From a point on the ground hich is 25 ma a from the foot of the to er, the angle of elevation of the top of the to er is foundto be 45 . Then the height (in meters) of the to er is _________ m.

8. If P(a2, 4) is the mid-point of the line-segment joining the points A ( 6, 5) and B( 2, 3),then the value of a is


9. If A and B are the points ( 6, 7) and ( 1, 5) respectivel , then the distance. 2AB isequal to __________.

10. A card is dra n from a ell-shuffled deck of 52 pla ing cards. The probabilit thatthe card ill not be an ace is ________.

EC I N B

11. Find the value of m so that the quadratic equation m ( 7) + 49 = 0 has t o equalroots.

12. Find ho man t o-digit numbers are divisible b 6.

13. In Figure 3, a circle touches all the four sides of a quadrilateral ABCD hose sidesare AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of the side AD.

14. Dra a line segment AB of length 7 cm. Using ruler and compasses, find a point Pon AB such that AP/AB= 3/5.


15. Find the perimeter of the shaded region in Figure 4, if ABCD is a square of side 14cm and APB and CPD are semicircles. [Use = 22/7 ]

16. T o cubes each of volume 27 cm3 are joined end to end to form a solid. Find thesurface area of the resulting cuboid.

(EITHER Q.15 OR Q.16)

17. Find the value of for hich the distance bet een the points A(3, 1) and B(11, ) is10 units.

18. A ticket is dra n at random from a bag containing tickets numbered from 1 to 40.Find the probabilit that the selected ticket has a number hich is a multiple of 5.

EC I N C

19. Find the roots of the follo ing quadratic equation: 2 35 +10=0

20. Find the A.P. hose fourth term is 9 and the sum of its si th term and thirteenthterm is 40.

21. In Figure 5, a triangle PQR is dra n to circumscribe a circle of radius 6 cm such thatthe segments QT and TR into hich QR is divided b the point of contact T, are oflengths 12 cm and 9 cm respectivel . If the area of PQR = 189 cm 2 , then find thelengths of sides PQ and PR.


22. Dra a pair of tangents to a circle of radius 3 cm, hich are inclined to each other atan angle of 60 .

23. A chord of a circle of radius 14 cm subtends an angle of 120 at the centre. Find thearea of the corresponding minor segment of the circle. Use = 22/7 and 3= 1.73.

24. An open metal bucket is in the shape of a frustum of a cone of height 21 cm ithradii of its lo er and upper ends as 10 cm and 20 cm respectivel . Find the cost of milk

hich can completel fill the bucket at Rs. 30 per litre. [Use = 22/7]

25. Point P( , 4) lies on the line segment joining the points A( 5, 8) and B(4, 10). Findthe ratio in hich point P divides the line segment AB. Also find the value of .

26. Find the area of the quadrilateral ABCD, hose vertices are A( 3, 1), B ( 2, 4),C(4, 1) and D (3, 4).

27. From the top of a vertical to er, the angles of depression of t o cars, in the samestraight line ith the base of the to er, at an instant are found to be 45 and 60 . If thecars are 100 m apart and are on the same side of the to er, find the height of the to er.[Use 3= 1.73]


28. T o dice are rolled once. Find the probabilit of getting such numbers on the t odice, hose product is 12.

EC I N D


30. The first and the last terms of an A.P. are 8 and 350 respectivel . If its commondifference is 9, ho man terms are there and hat is their sum?

31. A train travels 180 km at a uniform speed. If the speed had been 9 km/hour more, itould have taken 1 hour less for the same journe . Find the speed of the train.

32. In the given figure, three circles each of radius 3.5 cm are dra n in such a a thateach of them touches the other t o. Find the area enclosed bet een these three circles(shaded region). [Use = 22/7 ]


33. Water is flo ing at the rate of 15 km/hour through a pipe of diameter 14 cm into acuboidal pond hich is 50 m long and 44 m ide. In hat time ill the level of ater inthe pond rise b 21 cm?

34. The angle of elevation of the top of a vertical to er from a point on the ground is60 . From another point 10 m verticall above the first, its angle of elevation is 30 . Findthe height of the to er.


CBSE Ma h 2011 Sol ions:

1.

Gi en ad a ic e a ion: 2 3 m(m+3)=0.

I can be i en a

2 (m+3) m m(m+3)=0

2 (m+3) +m m(m+3)=0

- (m + 3) + m - (m + 3) = 0

( + m) - (m + 3) =0

The efo e, = - m and = (m + 3) a e he e oe of he e a ion.

2.

Le a be he fi e m of he A.P. and d be he common diffe ence.

n h e m of an A.P. = an=a+(n 1)d

a20 a15= [a + (20 1)d] [a + (15 1)d]

= 19d 14d

= 5d

= 5 3

= 15.


3.

Con ide he iangle in he ci cle. i.e., OPQ:

OP = OQ (a bo h a e adii of he ci cle)

⇒ ∠OQP = ∠OPQ (E al ide of a iangle ha e e al angle o o i e o hem)

∠POQ + ∠OPQ + ∠OQP = 180 (ba ed on he angle m o e of iangle )

⇒ 70 + 2∠OPQ = 180

⇒ 2∠OPQ = 180 70 = 110

⇒ ∠OPQ = 55

The angen i e endic la o he adi h o gh he oin of con ac .

∴ ∠OPT = 90

⇒ ∠OPQ + ∠TPQ = 90

⇒ 55 + ∠TPQ = 90

⇒ ∠TPQ = 90 55 = 35 .

4.

The angen o a ci cle i e endic la o he adi h o gh he oin of con ac .

∴ ∠ABO = ∠ACO = 90

In ad ila e al ABOC, acco ding o he angle m o e :

∠ABO + ∠BOC + ∠ACO + ∠BAC = 360

⇒ 90 + ∠BOC + 90 + 40 = 360

⇒ ∠BOC + 220 = 360

⇒ ∠BOC = 360 220 = 140 .


5.

Le he adi of he ci cle be a cm

The diame e of he ci cle become o ime he adi

2 a cm = 2a cm.

The efo e, he ide of he ci c m c ibing a e = Diame e of he ci cle = 2a cm.

∴ Pe ime e of he ci c m c ibing a e = 4 2a cm = 8a cm.

6.

In a c be, he la ge igh ci c la cone ha a diame e ha i e al o he ide of ac be.

Simila l , i heigh i e al o he ide of he c be.

So, if he adi of he cone i cm,

∴ 2 = 4.2 cm.

⇒ = 4.22 cm = 2.1 cm.

7.

In he fig e abo e AB i he o e and C i he oin on he g o nd 25 m a a f om hefoo of he o e in ch a a ha ∠ACB = 45 .

The efo e, in ABC,

an 45 = AB AC

⇒1 = AB 25

⇒AB = 25 m

The efo e, he heigh of he o e AB i 25 m.


8.

P(a2, 4) i he mid- oin of he line egmen joining oin A ( 6, 5) and B ( 2, 3).

∴ Coo dina e of he oin P can be fo nd b ing he mid oin fo m la.

P = ( 6+( 2)2, 5+32)= ( 82, 82) = (-4, 4)

Gi en ha P(a2,4) i he mid- oin , he efo e

(-4, 4) = (a2,4)

∴ a2 = - 4

⇒a = 8

So, he al e of a i 8.

9.

Gi en ha he coo dina e of he oin A and B a e ( 6, 7) and ( 1, 5) e ec i el .

AB = ( 6 ( 1))2+(7 ( 5))2

= ( 6+1)2+(7+5)2

= ( 5)2+(12)2

= 25+144

= 169

= 13

∴ 2AB = 2 13 = 26.

Hence, he di ance 2AB i 26 ni .


10.

N mbe of ace ca d in a deck of 52 la ing ca d i 4.

∴ N mbe of non-ace ca d in he ack = 52 4 = 48

A me E a he e en of d a ing a non-ace ca d.

To al n mbe of o ible o come = 52


P(E) = N mbe of fa o able o come To al n mbe of o come = 4852= 1213.

11.e ha e o find he al e of m o ha he ad a ic e a ion m ( -7)+49=0 ha o

e al oo .

Com a e abo e e a ion i h he gene al ad a ic e a ion , e ge

a=m, b=-7m, c=49

A e kno if he ad a ic e a ion ha e al oo hen he di c iminan ill bee al o 0 i.e

⇒

⇒

im lie 4, 0


12.

The o digi n mbe ha a e di i ible b 6 a e 12, 18, 24, , 96

He e, Fi e m = a = 12

Common diffe ence d = 18 12 = 6,

La e m = an = 96

Since, n h e m i gi en b :

an = a + (n - 1)d

96 = 12 + (n 1)6

96 12 = (n 1)6

84 = 6n 6

84 + 6 = 6n

90 = 6n

n = 15

The e a e 15 o-digi n mbe di i ible b 6.

13.

He e, ABCD i a ad ila e al.

Le a me ha he ad ila e al ABCD o che he ci cle a oin P, Q, R and S.

Th , AB, BC, CD and AD a e angen o he ci cle

P o e of angen a e ha "Tangen d a n f om an e e nal poin o a ci cle a eeq al in leng h".

Th , e ge

AP = AS

Le AP = AS =

Al o, BP = BQ, CQ = CR and DR = DS


Since AP =

Th , BP = AB - AP =

No , BQ = BP =

Al o, CQ = CB - BQ =

No , CR = CQ =

Al o, DR = CD - CR =

No , DS = DR =

Finall ,

AD = AS + DS

AD = + 5 -

AD = 5 cm

Th , leng h of ide AD i 5cm

14.

Con c ion gi en belo :


15.

ABCD i a a e of ide 14 cm. ∴

∴ AB = BC = CD = AD = 14 cm

The e a e o emici cle in he a e of adi = 14/2 = 7 cm

Pe ime e of he haded egion = BC + AD + Pe ime e of emi-ci cle APB + Pe ime e ofemi-ci cle CPD

Pe ime e of emici cle APB = 1/2 2 adi = 1/2 2 7 = 22/7 7 = 22 cm.

Pe ime e of emici cle CPD = 1/2 2 adi = 1/2 2 7 = 22/7 7 = 22 cm.

∴ Hence, he e ime e of he haded egion = 22 + 22 + 14 + 14 = 72 cm.


16.

17.

Coo dina e of he gi en oin a e A (3, 1) and B (11, ).

Di ance be een he oin A and B = AB = 10 ni

∴(11 3)2+[ ( 1)]2 = 10

⇒(64)+( +1)2=10

⇒ 64 + ( +1)2=100

⇒( +1)2 = 100 64

⇒( +1)2= 36

⇒( + 1) = 6

⇒( + 1) = 6 o ( + 1) = - 6

⇒ = 6 1 = 5 o 6 1 = 7

Hence, he al e of = 5 o 7.


18.

A me he e en of d a ing a m l i le of 5 a E.

Since he e a e 40 icke , he o al n mbe of o come = 40

The o come of e en E a e 5, 10, 15, 20, 25, 30, 35 and 40.

So, he n mbe of fa o able o come of e en E = 8

P obabili ha he elec ed icke ha a n mbe ha i a m l i le of 5:

P(E) = N mbe of fa o able o come To al n mbe of o come

= 840

= 15

Hence, he obabili of d a ing a icke hich i a m l i le of 5 = 15.

19.

Com a e he gi en ad a ic e a ion i h he anda d fo m of ad a ic e a ion i.e.,a 2 + b + c = 0.

a = 1, b = 35, c = 10

D =b2 4ac

= ( 35)2 4 1 10

= 45 40

= 5

∴ = ( b D)/2a

= 35 5/2 1

= 35+ 5/2 OR 35- 5/2


20.

A me he fi e m of he A.P. a 'a' and he common diffe ence a 'd'.

4 h e m of he A.P. = 4= 9.

The m of he i h and hi een h e m i 40 ⇒ 6+ 13= 40.

If 4= 9

⇒a + (4 1)d = 9 [ n = a + (n 1)d]

⇒a + 3d = 9 ---------- (1)

6+ 13= 40

⇒ a + (6 - 1)d + a + (13 - 1)d = 40

⇒ a + 5d + a + 12d = 40

⇒2a + 17d = 40 ---------- (2)

Sol ing he linea e a ion (1) and (2)

F om (1):

a = 9 3d

S b i ing he al e of a in (2):

2 (9 3d) + 17d = 40

⇒ 18 + 11d = 40

⇒ 11d = 22

⇒ d = 2

∴ a = 9 3 2 = 3

Hence, he gi en A.P. = a, a + d, a + 2d , he e a = 3 and d = 2.

So, he A.P. i 3, 5, 7, 9


21.

Since he iangle i ci c m c ibing he ci cle, he ci cle i he inci cle of he iangle. Allhe ide of he iangle a e angen o he ci cle.

Gi en RT= 9cm and QT = 12cm

Since RT and RU a e angen o a ci cle f om a oin R, he a e e al

Th RT = RU = 9cm

imila l , QT = QS = 12cm

Le PS = PU = cm

adi of ci cle, = 6cm

ide of iangle a e:

PQ = (12+ ) cm

QR = 12+9 = 21cm

PR = (9+ ) cm

Gi en ha a ea = 189cm


h 21+ = 31.5

⇒ = 31.5 - 21

⇒ = 10.5

Side of iangle a e:

PQ = (12+ ) cm = 12+10.5 = 22.5cm

QR = 12+9 = 21cm

PR = (9+ ) cm = 9+10.5 = 19.5cm

22.


23.The adi of he ci cle i 14 cm. The b end angle i 120 deg ee a he cen e.

The a ea of mino ec o i

A ea of iangle i

The fo m la o find he a ea of mino egmen i


The efo e he a ea of mino egmen of he ci cle i 120.46 cm .

24.

Heigh of f m of cone, h = 21 cm

Radi of lo e end, = 10 cm

Radi of e end, R = 20 cm

Vol me of f m of cone = 1/3 h(R2 + 2 + R )

Vol me of milk e i ed o fill he b cke

= 1/3 22/7 21 [(20)2+(10)2+20 10]cm3

= 22 (400 + 100 + 200)

= 22 700

= 15400 cm3

= 15400/1000 li e

= 15.4 li e

Co of milk = R 30 e li e

Hence, he co of milk fo filling he b cke i R (30 15.4) i.e. R 462.

25.



26.

27.


AB i he o e

The angle of de e ion of o ca in he ame aigh line i h he ba e of he o ea an in an a e fo nd o be 45 deg ee and 60 deg ee i.e. ∠ACB = 60 and ∠ADB = 45

The ca a e 100 m a a i.e. CD = 100 m

Le CB be

So, DB = DC+CB = 100+

In ABC

---1

In ABD

------2

Wi h 1 and 2

So,


Hence he heigh of o e i 236.602 m

28.

Li o he am le ace of he gi en e e imen

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)

The efo e, n (S) = 36

A me he e en of ge ing he n mbe ho e od c i 12 a E.

E = (2, 6), (3, 4), (4, 3), (6, 2)

The efo e, n (E) = 4

Hence, he e i ed obabili = n(E)n(S)= 436= 19.

29.

Gi en : A ci cle C (0, ) and a angen l a oin A.

To o e : OA ⊥ l

Con c ion : Take a oin B, o he han A, on he angen l. Join OB. S o e OB meehe ci cle in C.

P oof: We kno ha , among all line egmen joining he oin O o a oin on l, hee endic la i ho e o l.

OA = OC (Radi of he ame ci cle)


No , OB = OC + BC.

∴ OB > OC

⇒ OB > OA

⇒ OA < OB

B i an a bi a oin on he angen l. Th , OA i ho e han an o he line egmenjoining O o an oin on l.

He e, OA ⊥ l

30.

n h e m of AP = a+(n-1)d

he e a i he fi e m , d i he common diffe ence.

350= 8+(n-1)9

342=(n-1)9

n-1 = 38

n = 39

N mbe of e m = 39.

S m = (39/2)(8+350)

= (39/2)(358)

= 39*179

= 6981


31.


32.

Con c ion:- D a a line connec ing he cen e of each ai of adjacen ci clec ea ing a e ila e al iangle of ide leng h 7 cm.

No ,

A ea of iangle =

The e a e fo a ea in ide of hi e ila e al iangle.

[Th ee 60 ci cle ec o .]

A ea of ec o =

A ea of h ee ec o =

Th , a ea of Shaded o ion = a ea of e ila e al iangle - a ea of h ee ec o


33.adi =7cm=7/100

h=21cm=21/100m

ol me of i e = ol me of ond

2h=l b h

22/7 7/100 7/100 h=50 44 21/100

22/100 7/100 h=50 44 21/100

h=50 44 21 100 100

22 7 100

h=30000m

h=30km

heigh of i e = di ance a elled b a e

ime=di ance/ eed

ime=30/15

ime=2 ho

34.


Le he heigh of he o e be h cm.

No , In

And, In

Since, AB = CD, SO, e a ion (2) become ,

E a ing e a ion (1) and (3), e ge ,

Hence, he heigh of he o e ill be 15 cm.


Documents

&%6( 35(9,286