Maths paper with Solution

7/29/2019 Maths paper with Solution

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SOLUTIONS

SECTION-A

Ans 1.(d)

2

2 2

tan = cot

1tan =

tan

tan =1.

Now,since 1+tan Sec 1 1 2

Sec 2

θ θ

θθ

⇒ θ

θ = θ = + =

⇒ θ =

(1)

Ans 2.(d)

2 2 2

2 2

2 2 2

3Given:cotA = andAC = 10 cm.

4

AC 10 3

BC x 4

10 4BC

3

Now,by pythagoras theorem in right traingleABC,wehave

AB AC BC

10 4 16 25 10 510 10 1 103 9 9 3

10 5 50AB cm cm

3 3

⇒ = =

×⇒ =

= +

× × = + = + = =

× ⇒ = =

(1)

Ans 3.(c)

7

Decimalexpansionof 0.056,125which terminates after 3 places.

= (1)

Ans 4.(a)



1 1

2 2

1 1

2 2

1 1

2 2

Thegiven pairof linesare :

5x + 4y = 20 and 10x + 8y = 16

a b5 1 4 1Thus,we get ,

a 10 2 b 8 2

a b 1 c1 20 5,but

a b 2 c2 16 4

a b c1.

a b c2

Thus, there is nosolution.

= = = =

⇒ = = = =

⇒ = ≠

(1)

Ans 5.(b)

(1)

Ans 6.(a) (1)

The medianof thedataisgivenby theabsicca

of the point of int er sec tionof the less thanogive

andmore than ogive 28.5

−

=

Ans 7. (b)

5secA = cosecB =

3

3 2CosA SinB CosB SinA

5 5

Thus,Cos(A B) CosA.CosB SinA.SinB

3 2 2 3. . 0

5 5 5 5

Cos(A B) 0

(A B) 90

⇒ = = ⇒ = =

+ = +

⇒ − =

⇒ + =

⇒ + = °



Sin3 Cos( 26 )

Cos(90 3 ) Cos( 26 )

(90 3 ) ( 26 )

90 26 4

1264

29

θ = θ − °

⇒ ° − θ = θ − °

⇒ ° − θ = θ − °

⇒ ° + ° = θ

°⇒ = θ

⇒ θ = °

(1)

Ans 8.(a)

Euclid's division algorithm enablesus to find HCF of a andb. (1)

Ans 9.(c)

Thenumber of zereosofp(x) no.of timesthegraphinter sec ts thex axis.

3

= −

=(1)

Ans 10.(a)We know,by thm: If a line is drawn parallel to one side of a triangle to

intersect the other sides in distinct points, then the line drawn, divides the two

sides in the same ratio.Thus, AD DE

AB BC3 x 3 14

x 6cm7 14 7

=

×⇒ = ⇒ = =

(1)

SECTION-B

Ans 11.

NO (½)

Degree of remainder = Degree of divisor (1)

Division is incomplete. One more step of division should be there. (½)

Ans 12.

Yes,the given number is composite by fundamental theorem of arithmetic since it

can be expressed as the product of 2 primes,and this factorization is unique. (1½)



Ans 13.

Writing equations

x + y = 30 ; x – y = 14 because opposite sides of a rectangle are equal (½)

Solving and getting the values of x and y (1)

Writing solution as x = 22, y = 8 (½)

Ans 14.

Here,themaximum class frequencyis8,

andtheclasscorrespondingto thisfrequencyis3 5.

So,themodalclassis3 5.

Now,modalclass 3 5.

Lower limit(l)of themodalclass 3 (1

−

−

= −

=

1

0

2

1 0

1 0

)

classsize(h) 3

Frequency(f )of themodalclass 8

frequency(f )of theclasspreceedingthemodalclass 7

frequency(f )ofclasssucceedingthemodalclass 2

Now,letussubstitute thevaluesin theformula

f f mode = l

2f f f

=

=

=

=

−+

− − 2

h (1)

8 7 23 2 3 3.286 (1)

2 8 7 2 7

×

− = = × = + = × − −

Ans 15.

3Given sin(A + B) = cos(A - B) =2

and A, B (A > B) are acute angles.

thus we get sin(A + B) =sin60 (1)

and cos(A - B) =cos30

(A + B) =60 and(A - B

°

°

⇒ ° ) =30 (1)

thus,solvingweget A=45 andB=15 (1)

°

° °

(8 7 6 5 4 3 4) 4(8 7 6 5 3 1) (½)× × × × × + ⇒ × × × × +



OR

5Giventan

12

cos +sin 1+tanThen, = (1)cos -sin 1-tan

(dividingnm.anddm.bycos )

51+

cos +sin 12 5 1712=5cos -sin 12 5 7

1-12

θ =

θ θ θ

θ θ θ

θ

θ θ +⇒ = =

θ θ −(1 1)+

Ans 16. Cumulative Frequency Distribution of the less than type. (2)

Marks(Out of 50) No. of students. Marks less than Cumulative

frequency.

0-10 5 10 5

10-20 13 20 5+13=18

20-30 12 30 18+12=30

30-40 20 40 30+20=50

40-50 10 50 50+10=60

TOTAL 60

Ans 17.



SincePQ||CB,

AP AQ(1)

PC QB

[By Basic proportionality theorem] (1)

SincePR ||CD,AP AR

(2)PC RD

AQ ARFrom(1)and(2),weget

QB RD

= − − −

= − − −

=

Takingreciprocal,weget

QB RD

AQ AR

Adding1onbothsides,gives :

QB AQ RD AR(1)

AQ AR

AB AD

AQ AR

Takingreciprocal,we

=

+ +=

⇒ =

get

AR AQ

AD AB=

Ans 18.

2 2 2

2 2 2

2 2

2 2 2

2

Wehaveinright PQR,

PQ QR PR (1)

Inright PMR,PR PM MR (2)

6 8 36 64 100 (1)

PR 10cm

From(1),PQ (26) (10) 676 100 576

PQ

∆

= − − − −

∆= + − − −

= + = + =

⇒ =

= − = − =

⇒ = 576

PQ 24cm (1)⇒ =

SECTION-C

Ans 19.

Let3

2 5is a rational number (½)

3 a

b2 5=

Where a and b are coprime integers (½)



3b5

2a= (½)

Now, a, b, 2 and 3 are integers

Therefore,3b

2a

is a rational number. (½)

⇒ 5 is a rational number. (½)

Which is a contradiction.

Therefore3

2 5is an irrational number. (½)

OR

Let 3 + 5 is a rational number (½)

Therefore, 3 + 5 =a

b

where a and b are coprime integers. (½)

Now,a

5 3b

= − (½)

As a and b are integers

Therefore,a

3b

−

is a rational number. (½)

5⇒ is a rational number

But 5 is an irrational number. (½)

Therefore, our assumption is wrong

3 5⇒ + is an irrational number. (½)

Ans 20. Let the fixed charges = Rs x

and the subsequent charge = Rs y (½)

Writing equations

x + 4y = 27

and x + 2y = 21 (1)

Solving for x and y by subtracting, (1)

Thus, the solution is x = Rs 15 and y = Rs 3 (½)

OR

x + y = 5 (½)

10x + y = (10y + x) – 9 (1)

Solving for x and y the equations:



x+y-5=0 and 9x-9y+9=0 ,we get x=2 and y=3

(1)

Writing answer as 23. (½)

Ans 21. x2 – 4x + 3

Zeroes of polynomial = 1 and 3 (1)

3α = 3 and 3β = 9 (½)

Quadratic polynomial

= x2 – (α + β)x + αβ (½)

= x2 – 12x + 27 (1)

Ans 22.

(½)

In right angled ∆ACB

AB2 =BC2 + AC2 (Pythagores Theorem) (1)

= BC2 + AC2 (As BC = AC) (1)

= 2BC2 (½)

Ans 23. Let ‘a’ is any odd the integer and b = 4 using euclid division lemma

a = 4q + r where o < r < 4 (1)

⇒ a = 4q or 4q + 1 or 4q + 2 or 4q + 3 o r 4< < Q (1)

⇒ a = 4q + 1 pr 4q + 3 a is odd int egersa 4q or 4q 2

∴ ≠ + Q (1)

∴ any odd integer is of the form 4q + 1 or 4q + 3 (1)



Ans 24. L.H.S ( )2

cos ec cotθ − θ

2cos1

sin sin

θ= −

θ θ

=2

1 cos

sin

− θ

θ (1)

=( )

2

2

1 cos

sin

− θ

θ

=( )

2

2

1 cos

1 cos

− θ

− θ(1)

=( )

( ) ( )

21 cos

1 cos 1 cos

− θ

− θ + θ(1)

=1 cos

1 cos

− θ

+ θ= RHS

Ans 25.

In AOB and DOC∆ ∆

( )AOB COD Vertically opp.∠ = ∠ (1)

OAB OCD Alternate angles as. AB || CD∠ = ∠ (1)

AOB COD∴ ∆ ∆ (AA similarity) (1)

OA OB

OC OD⇒ =

Ans 26.



p2 = m2 sin2θ + n2 cos2 θ + 2mn sin θ cos θ (1)

q2 = m2 cos2 θ + n2 – 2mn sin θ cos θ (1)

p2 + q2 = m2(sin2θ + cos2 θ ) + n2 (cos2

θ + sin2 θ ) (1)

= m2 + n2 (sin2θ + cos2 θ =1)

Ans 27. Making correct C.F table (1)

Class interval Frequency Cumulative Frequency

500-600 40 40

600-700 28 68

700-800 35 103

800-900 22 125

900-1000 25 150

Total 150

Here n=150 , =n

75

2

(½)

Median class= 700-800

Median =

ncf

2h

f

−

+ ×l (½)

= = = =

− −

= + × = + × =

l 700,h 100,cf 68, f 35

ncf

75 682median l h 700 ( ) 100 720f 35

(1)



Ans 28. . Making correct table (1)

Class fi Xi di=xi – aui =

xi a

h

−

fi ui

100-150 4 125 -100 -2 -8

150-200 5 175 -50 -1 -5

200-250 12 225 = a 0 0 0

250-300 2 275 50 1 2

300-350 2 325 100 2 4

fi∑ = 225 and fiui∑ = -7 (½+½)

X = a + hfiui

fi

∑∑

= 225 + 502 × 7

25

− = 225 – 14 = 211 (1)

OR

Making correct table (1)

Classes xi fi Xi fi

0-10

10-20

20-30

30-40

40-50

5

15

25

35

45

5

18

15

P

6

25

270

275

35p

270



fi 44 p= +∑ fi xi 940 35p= +∑

x =fi xi

fi

∑∑

(1)

⇒ 25 =940 35p

44 p

+

+

⇒ 1100 + 25p = 940 + 35 p

⇒ 160 = 10p ⇒ p = 16 (1)

SECTION-D

Ans 29. Given pair of linear equations:-

3x + y – 5 = 0 – (1)

2x – y – 5 = 0 – (2)

The corresponding to equation (1) & (2) are :- (1+1)

Tables of values

For (1)

X 2 0 3

y -1 5 -4

For (2),

X 0 2 2.5

y -5 -1 0

The graph is given below :- (1)



The intersection point of (1) & (2) is (2,-1) (1)

Ans 30. p(x) = x4 + x3 – 9x2 – 3x + 18

x = ( )3, 3 are zeros of p x−

( ) ( ) ( )x 3 x 3 is a factor of p x∴ − +

( )

2x 3 is a factor of p x⇒ − (1)

Dividing p(x) by x2 = 3 and getting the quotient as x2 x – 61

12

∴ p(x) = (x2 – 3) (x2 + x – 6)

Other zeros of p(x) are given by x2 + x – 6 = 0 ⇒ (x + 3) (x – 2) = 0



⇒ x = -3, + 2 (1)

∴All the zeros of p(x) are 3, 3, 3, 2− − + (1/2)

Ans 31. LHS = sec4θ – sec2θ

= sec2θ(sec2θ – 1) (1)

= (1 + tan2θ)(– 1) since 1 + tan2θ=sec2θ (2)

= tan2θ + tan4θ = RHS (1)

Ans 32. LHS =2 2sin (1 cos )

sin (1 cos )

θ + + θ

θ + θ(½)

2 2sin 1 2cos cossin (1 cos )

θ + + θ + θ=θ + θ

(1)

2 2sin cos 1 2cos

sin (1 cos )

θ + θ + + θ=

θ + θ(½)

2 2cos

sin (1 cos )

+ θ=

θ + θ(½)

2(1 cos )

sin (1 cos )

+ θ=

θ + θ(1)

2cosec= θ (½)

= RHS

OR

o o o o o

o o o o o o o o

cos(90 20 ) cos55 cosec(90 55 )

sin20 [tan5 tan25 tan 45 tan(90 25 ) tan(90 5 )

− −+

− −(1)

o o o

o o o o o o

sin20 cos55 sec55

sin20 tan5 tan25 tan 45 cot 25 cot 5= + (1)

o

11

45= + (1)

= 1 + 1= 2 (1)

Ans 33. Cumulative frequency Distribution of more than type (1)



Profit (in lakh

Rs.)

No of shops

(Frequency)

Profit more than or

equal to

Cumulative

Frequency

0 – 5

5 – 10

10 – 15

15 – 20

20 – 25

3

14

5

6

2

0

5

10

15

20

30

27

13

8

2

Total 30

Now mark the tower class limits on x-axis and cumulative frequencies along y-axis

on suitable scales.

Thus, we plot the pts (0,30) (5,27), (10,13), (15,8) ,(15,8)& (20,2) (1)

By joining these points by a free smooth hand given, we obtain an ogive by more

than method as shown below : - (1+1)



Ans 34. Statement If a line is drawn parallel of one side of a triangle to intersect

the other two sides in distinct points, the other two sides are divided in the same

ratio. (1)

Given:A triangle ABC in which a line parallel to side BC intersects other two

sides AB and AC at D and E respectively (see fig.)

To prove thatAD AE

.BD EC

=

Construction:Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.

Proof :Now, area of 1 1

ADE base height AD EN.2 2

∆ = × = ×

(1)

Note that ∆BDE and DEC are on the same base DE and between the same

parallels BC and DE.

( )

( )

( )

( ) ( )

( )

( )

( )

( )

Letusdenotethe area of ADE is denoted as are ADE .1

So, ar ADE AD EN2

1Similarly, ar BDE DB EN.

21 1

ar ADE AE DM and ar DEC EC DM.2 2

1AD ENar ADE AD2Therefore,

1ar BDE DBDB EN

21

AE DMar ADE AE2

and 1ar DEG ECEC DM2

∆

= ×

= ×

= × = ×

×= =

×

×

= =×



So, ar(BDE) = ar(DEG)

Therefore, from (1), (2) and (3), we have :

AD AE

DB EC=

(2)

OR

Pythagoras Theorem : Statement:In a right angled triangle,the square of the

hypotenuse is equal to the sum of squares of the other two sides.

Given: A right triangle ABC right angled at B. (1/2)

To prove: that AC2 = AB2 + BC2 (1/2)

Construction:Let us draw BD ⊥ AC (See fig.) (1/2)

(1/2)

Proof : (2)

Now, ∆ ADB ∼ ∆ ABC (Using Theorem:If a perpendicular is drawn from the

vertex of the right angle of a right triangle to the hypotenuse ,then triangles on both

sides of the perpendicular are similar to the whole triangle and to each other)

So,AD AB

AB AC= (Sides are proportional)

Or, AD.AC = AB2 (1)

Also, ∆ BDC ∼ ∆ ABC (Theorem)

So,CD BC

BC AC=

Or, CD. AC = BC2



Adding (1) and (2),

AD. AC + CD. AC = AB2 + BC2

OR, AC (AD + CD) = AB2 + BC2

OR, AC.AC = AB2 + BC2

OR AC2 = AB2 + BC2

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Maths paper with Solution