vedanta institute institute sco 209 second floor sector 36-d chandigarh ph. 8054369515 a = 1/2(180-...

VEDANTA INSTITUTE

SCO 209 SECOND FLOOR SECTOR 36-D CHANDIGARH PH. 8054369515www.vedantainstitute.in

ELEMENTARY MATHEMATICS - ANSWER KEY (DEMO)

1. Solution: I. diagonal of the rhombus are not equal.

II statement is true

2. Solution: ΔBCA and ΔACD are similar

cmCD

CDCD

AD

CA

AB

4.6

10

86

8

6

3. Solution:

7.16

144

100

24

2

2

ABCar

ABCar

EF

BC

DEFar

ABCar

4. Solution:

OM2= OA2－AM

2 = 100-36= 64

OM= 8cm

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MF= 10-8 = 2cm

5. Solution: Perpendicular bisector of a chord always passes through the centre. (property)

II. AB and CD are two chords intersecting at a point E

Therefore AE×EB= ED×EC

EB

EC

ED

AE

is true

Both the statements are true

6. Solution: I.

In Δ ABC

2 2 + 3= 2 1

3= 2 1 - 2 2

In ΔDBC

4+ 2 = 1

4 = 1 - 2

4 = 1/2 ( 3)

I is true

II .

In ΔABD

a-d+b = 90 →d= b+a-90 …………..1

2a+b+c = 180

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a = 1/2(180- b - c)

Putting the value of a in 1, we get

D= b + 90- 1/2 b - 1/2 c - 90

D= 1/2 (b-c)

2 statement is false

7. Solution: AB2+BC

2 = AC

2

4x2+x

4+1 -2x

2 = AC

2

AC = x2+1

Since AD is an internal bisector, it will bisects BC in the ration AB:AC

xx

x

BC

BD

x

x

DC

BD

21

2

1

222

=

2)1(

2

x

x

8. Solution: Median of equilateral triangle is equal to circumradius + inradius

= 32

2

323

aaaa

= 2

3a

9.Solution:

BCAC

AC

BC o

2

60cos

Ar ΔABC= 2

1

BC×AB= 4

ABAC

I is true

II. BD is median→AD=DC

Also circumcentre of a right angle triangle is at point D

Therefore AD=DC=BD

Both the statements are true

10. Solution:

It is a property that if external bisectors of 2 angles meets at a certain point then the angle between the bisectors is

equal to the 90- half of the third angle.

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11. Solution:

ΔBCD and ΔABD are similar triangles

AD

BD

BD

CD

BD2= AD.CD

BD2 = 16×9= 144

BD= 12

12. Solution: since all the perpendiculars are equal implies that interior point is incentre

Length of inradius = 30/3 = 10cm

1032

a

a= 20 3

13. Solution: Sum of exterior angles of every polygon is same is equal to 360

II. Sum of interior angles = )2( n

Sum of interior angles increases as n increases

Hence both the statements are true.

14. Solution: A rhombus is formed on joining the mid points of rectangle.

Area of rhombus = 1/2 (d1 ×d2)

= 1/2 (10×12)

15.Solution: Interior angle + exterior angle = 180

Exterior angle= 180-135=

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Exterior angle = 45

Number of sides = 360/ exterior angle

N= 8

Number of diagonals =

202

)3(

nn

16. Solution:

ΔBDC and ΔFEC are similar

2

2

CE

CD

FECar

BDCar

arΔBDC=

1254

2025

arΔBDC = ar ΔADC

As both the triangles have same base and are between the same parallel lines

Ar ΔADC= 125 cm2

17.Solution:

As OA=OB →A=B= 120/2 = 60o

→triangle is equilateral →AB= radius = half of diameter

18. Solution: I.

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Both angles are in the same segment , hence equal

II. PA×PB =PC×PD

PB

PD

PC

PA

Both the statements are true.

19. Solution: for the triangle the circle is circumcircle

Radius of the circumcircle = 3

a

=20 →a=20 3

Radius of incircle =

1032

320

32

a

20.Solution: I. it is a property of circle that if perpendicular through the centre always bisects the chord.

II. Perpendicular bisectors always passes through the centre.

21. Solution: total expenditure = 30 lakhs = 360o

2 lakhs =

24230

360

22. Solution: Rent and Hiring = 5 lakhs

Total expenditure = 30 lakhs

Required % =

%3

216100

30

5

23-24. Solution:

Marks of student No of students cf

5-10 8 8

10-15 10 18

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15-20 12 30

20-25 x 30+x

25-30 28 58+x

30-35 y 58+x+y

35-40 4 62+x+y

40-45 20 82+x+y

Total 100

82+x+y = 100

x+y= 18

Median = 27.5 →median interval = 25-30

L= 25, N/2= 50, cf= 30+x, H=5

Median=

hf

cfn

l

2

27.5= 25 +

528

3050

x

x=6

→y= 12

25. Solution:

Marks of student No of students cf

5-10 8 8

10-15 10 18

15-20 12 30

20-25 6 36

25-30 28 64

30-35 12 76

35-40 4 80

40-45 20 100

Total 100

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Mode =

hfff

ffl

201

01

2

Modal class = 25-30 as it has the largest frequency 28

L= 25, f1=28, f0= 6, f2 = 12, h =5

Putting the values in the formula , we get mode = 25.57

26. only 1 is correct

27. Solution: Since , both the hour hand and minute hand have interchanged their positions.

Sum of the angles formed by hour hand and minute hand= 3600

Let he was out for x minutes

Then angle formed by the minute hand = 6x

And the angle formed by the hour hand = 0.5x

6x+ 0.5x= 360

x= 55.38 min

28. Solution: To form a triangle we need 3 lines

Among 10 lines 3 lines can be chosen as 10

C3 =

12023!7

!78910

!3!7

!10

29. Solution: a=x (alternate angles)

y= b (alternate angles)

x+ y + 180 - a+180 - b = 360

x+y = a+b

30. Solution: Let the length, breadth and height of the parallelopiped = x, 6x , 3x

Surface area of cube = surface area of parallelopiped

6a2 = 2(6x

2+ 18x

2 + 3x

2 )

a= 3x

Volume of cube /volume of parallelopiped = a3/x×6x×3x =3:2

31. Solution: Increase in the height of the water level

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= πr2h= 1540

15407

7722

h

h= 10 cm

32. Solution: Curved surface area of cylinder/ Curved surface area of cone =

5:86436

822

rl

rh

33. Solution: I. Radius gets double

Let originally radius = r , new radius = 2r

Volume = πr2h

New volum2 = π(2r)2h = 4πr

2h

Volume becomes 4 times

II. CSA = 2πrh

New CSA = 2π2rh = 4πrh

CSA becomes twice the original CSA

Only 2 statement follows

34. Solution: length of the longest rod = a 3 = 12 3 cm

35. Solution: 100

1444100

1001007

77500022

x

x= 8 hours

36. Solution: Volume of cone = Volume of cylinder - volume of the remaining portion

44002110107

221010

7

22

3

1 h

h=21 cm

37. Solution: Canvas required = CSA of cylinder +CSA of cone

1138563322

105

7

22

2

rlrh

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38. Solution: Total surface area = 2(lb+bh+hl) = 22

2(2+2h+h) = 22

h=3

Diagonal= 14941222 hbl

39. Solution: Let radius = r

Perimeter of the semicircular path = πr+2r = 36

r= 7

Area =

772

77

7

22

2

2

r

40. Solution: Volume of sphere = Volume of hemisphere

33

3

2

3

4rR

rR 31

2

Ratio of CSA =

1:2

2

2

2

4 31

232

2

2

2

R

R

r

R

41. Solution: Semi-perimeter of triangle =

152

11109

Area = 23045615))()(( csbsass

42. Solution: Area of the bounder region = Area of square - area of 4 quadrants

6×6 - 4

34

2

= 36-9π = 9(4-π)

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43. Solution: Perimeter = 4a = 40

a= 10

352

103

60sin 0

OB

AB

OB

52

101

60cos 0

OB

AB

OA

BD = 2OB= 10 3

AC= 2OA = 10

Area =

350103102

1

44. Solution: Radius of circumscribed circle= diagonal/2 = 1/2 (a 2 ) =7 2

Radius of inscribe circle = a/2 = 7

Ratio= 1: 2

45. Solution:

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72

3

1

108

3108

30tan108

xx

Base

x

46.Solution: Let height of chimney = y metre

hxhy

xyxh

xyh

h

yhBB

yh o

cot

tantan

tan

45tan

47. Solution: 4sinθ+4cosθ= 5sinθ-5cosθ

Sinθ=9cosθ

Tanθ=9

1tan

1tan2

2

=

80

82

40

41

48. Solution: For x= π , secπ = -1 and tanπ = 0

→secx< tanx for x=π

→statement 1 is incorrect

Sec2x= 1 +tan

2x is identically true for every x

II statement is true

49. Solution: m

AecpnA

A 2

2

2

cossin

cos

mcot2A+n = p(1 + cot

2A)

Cot2A(m-p)= p-n

Cot2A=

pm

np

50.Solution: ap-bq = a2sinA+abcosA-bacosA - b

2sinA

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= (a2-b

2)sinA

aq-bp = a2cosA +absinA - absinA - b

2cosA

=(a2-b

2)cosA

(ap-bq)2 +(aq- bp)

2 = (a

2-b

2)2 (sin

2A+cos

2A) = (a

2-b

2)2

51. Solution: cosecA - cotA = m ……..(1)

Cose2A - cot2A = 1

(cosecA- cotA )(cosecA +cotA ) = 1

cosecA+ cotA = 1/m …….(2)

Adding 1 and 2

2cosecA = m

m 12

SinA = 1

22 m

m

52.Solution: x= sinAsecB and y = cosAcosecB

22

21

yx

y

=

B

BABA

BAB

BB

BABA

B

AB

BecABA

BecA

2

2222

222

22

2222

2

22

2222

22

cos

)sin1(cossin)cos1(

coscossin

sincos

coscossinsin

sin

cossin

coscossecsin

coscos1

53. Solution: cot(A+B ) = 1/tan(A+B) =

BABA

BA

BA

BA

BA

BABA

BA

cotcot

1

tantan

1

tan

1

tan

1

1

tantan

1

tantan

tantan

tantan

1

tantan

tantan1

= ba

11

54. Solution: tan200+ tan40

o + tan60

0+ ………+ tan180

0

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=tan200+ tan40

o + tan60

0+ ………+tan(180

0-40

0) + tan(180

o- 20

0) + tan180

0

=tan200+ tan40

o + tan60

0+ ………- tan40

o - tan20

0 + tan180

0 = 0

55. Solution: m(1-2sin2A) = 3(4- 4 sin

2A )

m-2msin2A = 12 - 12 sin

2A

(12-2m) sin2A = 12 - m

Sin2A = m

m

212

12

≥0

12-m ≥0 and 12-2m≥0 or 12-m≤0 ans 12-2m ≤0

m≤6 or m≥12

56. Solution: 4cos2A - 3cosA - 10 = 0

4cos2A - 8cosA+5cosA - 10 = 0

(cosA- 2)(4cosA +5) = 0

cosA = 2 , 4

5

I statement is true

II. For A = 0, 4-3 -10 ≠0

II statement in incorrect

57. Solution: MΠPΠC = 18

Maths and physics alone = MΠP = 30 -18 = 12

Maths and chemistry only = 28 - 18 = 10

Mathematics only = 100 -10-12-18 = 60

58. Solution: remainders = 6 and 5

Remainder on dividing ab by 8 is same as the remainder on dividing 6×5 = 30 by 8

Ans is 6

59. Solution: 6619305

Sum of digits at odd places = 15

Sum of digits at even places = 15

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15-15 = 0 →divisible by 11

Hence the number is divisible by 77

60. Solution: if n is even then unit unit digit of 32n+1

is 3 and unit digit of 22n+1

is 2

If n is odd then unit digit of 32n+1

is 7 and unit digit of 22n+1

is 8

In both the cases sum is 5

61. Solution: unit digit of 567453

is given by 753

on dividing 53 by 4 remainder = 1 , 71 = 7

Unit digit of 349569

is given by 969

, on dividing 69 by 4 remainder is 1, 91 = 9

7+9 = 16 =→6

62. Solution: See which option satisfies x = 1

Only option C,

63. Solution: put x = 2

8+ 4 - 10 +a = 0

→a = -2

64. Solution: x3-1 = (x-1)(x

2+x+1)

X4+x

2+1 = (x

2+x+1)(x

2-x+1)

X4-5x

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

LCM =(x2-1)(x

2-4)(x

2+x+1)(x

2+1-x)

65. Solution: 5612.0 =

9900

1244

9900

121256

2475

311

66. Solution: put x= a =b= c= 1

bac

a

cacb

c

bcba

b

a

x

x

x

x

x

x

1

67. Solution: 3 = k1/x

, 5 = k1/y

, 75 = k1/z

52×3 = 75

K2/y

×k1/x

= k1/z

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zxy

112

Z = yx

xy

2

68. Solution: 911,79,35,57

= 911

2,

79

2,

35

2,

57

2

Greater the Denominator, less will be the fraction

= 35

2

is greatest

69. Solution: let distance between A and C = 2D

AB = BC = D

2

13

vu

D

vu

D

………….1

92

vu

D

Putting the value in 1

2

4

2

9

2

13

vu

D

→

42

vu

D

70. Solution: Train A takes 1 hour and train B takes 3/2 hours to cover the same distance.

Let distance = 30 km

Speed of A = 30km/h

Speed of B = 20 km/h

Time =

h5

3

2030

30

= 36 mins

Trains will meet each other at 4:36 pm

71. Solution: time = distance/ speed

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5

56

18100

x

Speed of train = X = 78 km/h

Speed of car = y

6

578

18100

y

Y = 18

72. Solution: let distance = D

60

6

5040

DD

D = 20 km

Correct time = 20/40 - 11

30 - 11 mins = 19 mins

73. Solution: Distance covered by thief from 9 to 9:30 am = 10 km ( as 20×1/2)

Time in police will catch thief =

min302

1

20

10

At 10 am. Police will catch thief

74. Solution:

140

23

4045

xx

8x + 9x + 207 = 360

17x = 153

X = 9 days

75. Solution: A can cultivate 2/5 in 6 days

A can cultivate whole land in 15 days

B can cultivate whole land in 30 days

Together A and B = 30

3

30

1

15

1

In 10 days

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76. Solution: 25× 20 = (16+9) x

X = 20 mins

77. Solution: let his income is 6000

Expenditure = 5000 and savings = 1000

Income increased by 20% that is 1200, new income = 7200

Expenditure by 8% that is 400, new expenditure = 5400

New Savings= 1800

Savings increased by 80%

78. Solution: milk is 60% in 30 litres , that is 18 litres

18 litres = 20%

100% = 90

90-30 = 60 litres water must be added to make 20% milk solution

79. Solution: if time is same ratio of principal and amount remains same in the case of compound interest

For 4 years , ratio is 6000/4800 = 5/4

After 12 years 60004

5

4

5 =9375

80. Solution: difference = principal100100

RR

Principal = 40×10×10 = 4000

81. Solution: one revolution in 1 sec , 5/3 sec and 5/2 sec

LCM (1,5/3, 5/2) = 5sec

82. Solution: HCF ( 184, 230, 276) = 46

Numbers of bags required = 184/46 + 230/46 + 276/46 = 4+5+6 = 15 bags

83. Solution: ratio of earnings of man woman boy and girl are

M:W = 7:6

W:B = 3:2

B:G = 5:4

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M:W:B:G = 35:30:20:16

16 p= 16

1p = 1

35 p = 35 rs

84. Solution:

CBA5

2

4

1

6

1

10A= 15B=24C

A:B:C = 12: 8: 5

25 P = 2250

1P = 90

C’s share = 90×5 = 450

85. Solution: A:B = 3:2 and B:C = 3:2

A:B:C = 9:6:4

19P = 342

1P = 18

Runs scored by B = 6×18 =108

86. Solution: His age = 19 +3 + 2 = 24

24 +1(after 1 year job) = 25

25+3 = 28 years

87. Solution: total profit = 100%

After 9% charity , profit left 91%

B’s sahre = 4/7 of 91% = 1196

1% = 23

100% = 2300

88. Solution: average of whole class = 6.1150

12301120

89. Solution: difference of age = 55 - 39 =16

Average age diminished by 16/8 = 2 years

New average = 38 years

90. Solution:

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5P = 3000

2P = 1200

91. Solution: 21010

22

1010 )10log10(log)10log5(log = 1

92. Solution: 4x2 + 12 x + k = (2x)

2 + 2×2x×3 + k

→k = 32 = 9

93. Solution: System of equations have no solution when

3

3

2

2

1

1

b

a

b

a

b

a

For k = 6 , the system has no solution

94. Solution: by putting options

95. Solution: 2x+y = 5

X+2y = 4

B options satisfies both the equations

96. Solution: distance walked by him is 2 + 1 + 1/2 + 1/4 + 1/8……

1st term = 2 and ratio = 1/2

Sum = r

a

1 = 4 kms

97. Solution: angle = uteshours min2

1130

210 - 110 = 100

98. Solution: a3+b

3 = (a+b) (a

2-ab +b

2)

=1/132

99. Solution: x3+y

3+z

3 - 3xyz = (x+y+z)(x

2+y

2+z

2 +{x

2+y

2+z

2 - (x+y+z)

2}/2)

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= 9×(35 + (35-81)/2) = 9 ×12 = 108

100. Solution: x = 0.0064

Y = 156.25

Z = -0.1536

vedanta institute institute sco 209 second floor sector 36-d chandigarh ph. 8054369515 a = 1/2(180-...

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