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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)
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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
⇒ = = ⇒ = =
+ = +
⇒ − =
⇒ + =
⇒ + = °
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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)
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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)× × × × × + ⇒ × × × × +
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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.
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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)
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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:
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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)
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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.
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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)
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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
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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)
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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
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⇒ 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)
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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)
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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
∆
= ×
= ×
= × = ×
×= =
×
×
= =×
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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
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