Section C

Solution.13


In this case LCM of 520 and 468 is,

520 = 2³ x 5 x 13

468 = 2² x 3² x 13

= 2² x 3² x 5 x 13

= 4680

4680 is exactly divisible by 520 and 468, reduce this number by 21

4680 - 21 = 4659

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Solution.14


a²b²y² + b²y - (a²y + 1) = 0

b²y(a²y + 1) - (a²y + 1) = 0

(b²y - 1)(a²y + 1) = 0

b² - 1 = 0

y = 1/b²

a²y + 1 = 0

y = -(1/a²)

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Solution.15
Let the digit at unit place be x and the digit at ten's place by y. Then

y + x = 12 ---- ---------1

Number = 10y + x

Number formed by reversing the digits = 10x + y

10y + x - (10x + y) = 18

9(y - x) = 18

y - x = 2 -----------------2

On solving equations

y = 7 and x = 5

Hence number = 10y + x = 10 x 7 + 5 = 75

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Solution.16
If three points are collinear then area bounded by these points will be zero

area of triangle =

1/2

lx₁(y₂ - y₃) + x₂(y₃ - y₁)+ x₃(y₁ - y₂)l

On solving ab = ay + bx , dividing both sides by ab

x/a

+

y/b

= 1

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Solution.17
a cosθ - b sinθ = c

Squaring both sides

(a cosθ - b sinθ)² = c²

a² cos² θ + b² sin² θ - 2ab sinθ cosθ = c²

a² (1 - sin²θ) + b² (1 - cos²θ) - 2ab sinθ . cosθ = c²

a² - a² sin² θ + b² - b² cos² θ - 2ab sinθ . cosθ = c²

- a² sin² θ - b² cos² θ - 2ab sinθ . cosθ = c² - a² - b²

a² sin² θ + b² cos² θ + 2ab sinθ . cosθ = a² + b² - c²

(a sinθ + b cosθ)² = a² + b² - c²

a sinθ + b cosθ = ±√(a² + b² - c²)

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Solution.18
Let 0 be the center of the circle and P be the mid-point of BC.Since triangle of ABC is isosceles So,point of P is the mid-point of BC.

Let AP = y and PB = PC = x .

Applying Pythagoras theorem in triangles of APB and OPB, OB² = OP² + BP² and AB² = BP² + AP²



36 = y² + x² ------------- 1

81 = (9 - y)² + x²--------- 2

81 = 81 + y² - 18y + x²

x² + y² - 18y = 0

On solving we get

36 = 18y

y = 2 cm

Put y = 2 in equation 1

36 = 4 + x²

x² = 32

x = 4√2 cm

Hence, Area of triangle ABC =

1/2

x Base x Height

1/2

x 8√2 x 2

= 8√2 cm²

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Solution.19
Given AB = AR + RB = 27cm

AC = AS + SC = 45cm

AR/AB

=

9/27

=

1/3

AS/AC

=

15/45

=

1/3

AR/AB

=

AS/AC

Thus, in triangles ARS and ABC, we have

AR/AB

=

AS/AC

and angle A = angle A (common)

Therefore, by SAS criterion of similarity triangle, we have

triangle ARS and triangle ABC are congruent triangles.

AR/AB

=

AS/AC

=

RS/BC

AS/AC

=

RS/BC

15/45

=

RS/BC

1/3

=

RS/BC

BC = 3RS

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Solution.20
∆OAB is right triangle
AB = r tanθ
Area of shaded portion = Area of ∆OAB - Area of sector
=

1/2

x r x rtanθ -

θ/360^o

x π r²
=

r²/2

(tanθ -

θπ/180^o

)

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Solution.21
Let radius of the original cone be R and radius of the smaller cone be r . h is the height of the smaller cone. H is the height of original

volume of original cone =

1/3

πR²H

volume of smaller cone =

1/3

πr²h

1/3

πr²h =

1/8

x

1/3

πR²H

h =

1/8

(

R/r

)² x 40

Now, find

R/r

. in the terms of h with the help of congruency property of triangle



∆ABC ∽ ∆ADE

h/H

=

r/R

h/40

=

r/R

R/r

=

40/h

h = (

1/8

) (

40/h

)² x 40

h =

1/8

x

1600 x 40/h²

h³ =

64000/8

= 8000

h = 20 cm

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Solution.22
this is ungrouped type data,so we can find mean using formula x̄ =

∑fx/∑f

∑ f = 41 + a

∑ fx = 303 + 9a

8 =

303 + 9a/41 + a

328 + 8a = 303 + 9a

a = 25

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