Permutations and Combinations Test 52

Result

Permutations and Combinations Test 52

Score
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Rank
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TIME Taken - -

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Weekly Quiz Competition

Question 1
1 / -0

If $$A = (-3,4) , B =(-1,-2) , C=(5,6) D= (x,-4) $$ are the vertices of a quadrilateral such that area triangle $$ABD= 2 \times$$ (area of a triangle $$ACD$$), then $$x =$$

A
6

B
9

C
69

D
96

Solution
Question 2
1 / -0

$$\dfrac { 7 } { 11 } : \dfrac { 336 } { 110 } : ? \quad : \quad \dfrac { 720 } { 272 }$$

A
$$\dfrac { 9 } { 17 }$$

B
$$\dfrac { 9 } { 13 }$$

C
$$\dfrac { 11 } { 13 }$$

D
$$\dfrac { 11 } { 17 }$$

Solution
Question 3
1 / -0

A line L passes through the points $$ (1,1) $$and $$ (2,0) $$ and another line $$ L^{\prime} $$ passes through $$ \left(\frac{1}{2}, 0\right) $$ and perpendicular to L.Then the area of the triangle formed by the lines $$ L, L^{\prime} $$ and $$ y- $$ axis, is

A
$$15 / 8 $$

B
$$25 / 4 $$

C
$$25 / 8 $$

D
$$25 / 16 $$

Solution

$$\textbf{Hint: Product of slopes of two perpendicular lines is -1.}$$

$$\textbf{Step 1: Find the equation of two lines.}$$

$$\text{Using two points form, equation of line L is given by,}$$

$$\Rightarrow y-0=\dfrac{0-1}{2-1}(x-2)$$

$$\Rightarrow y=-(x-2)$$

$$\Rightarrow x+y=2.......(1)$$

$$\text{Slope of L = - 1}$$

$$\text{L and }$$ $$\text{L' are perpendicular to each other}$$

$$\text{Slope of L' =}$$ $$\dfrac{-1}{-1}=1$$

$$\text{Using point slope form, equation of line L' is given by,}$$

$$\Rightarrow y-0=1(x-\dfrac{1}{2})$$

$$\Rightarrow 2x-2y=1 ......(2)$$

$$\textbf{Step 2: Find the required area.}$$

$$\text{Let, L and L' intersect each other at B.}$$

$$\text{Solving eq(1) and eq(2) , we get,}$$

$$\Rightarrow A(x,y)=\left(\dfrac{5}{4},\dfrac{3}{4}\right)$$

$$\text{The intersection points of lines L and L' with y-axis are B(0,2) and C}$$$$\left(0,-\dfrac{1}{2}\right)$$ $$\text{respectively.}$$

$$\text{Using distance formula,}$$
$$AB=\sqrt{\left(0-\dfrac{5}{4}\right)^2+\left(2-\dfrac{3}{4}\right)^2}=\dfrac{5}{4}\sqrt2$$

$$\text{And,}$$
$$AC=\sqrt{\left(\dfrac{5}{4}-0\right)^2+\left(\dfrac{3}{4}+\dfrac{1}{2}\right)^2}=\dfrac{5}{4}\sqrt2$$

$$\text{Area of the triangle}$$ $$=\dfrac{1}{2}\times AB\times AC=\dfrac{1}{2}\times\dfrac{5}{4}\sqrt2\times\dfrac{5}{4}\sqrt2\ sq.unit $$

$$=\dfrac{25}{16}\ sq.unit$$

$$\textbf{Hence, the correct option is D.}$$
Question 4
1 / -0

The area of the triangle formed by the lines $$x=0;y=0$$ and $$x\sin { { 18 }^{ 0 } } +y\cos { { 36 }^{ 0 } } +1=0$$ is

A
1

B
2

C
3

D
4

Solution
Question 5
1 / -0

If $$P , Q$$ are two points on the line $$3 x + 4 y + 15 = 0$$ such that $$O P = O Q = 9$$ then the area of $$\Delta O P Q$$ is

A
6$$\sqrt { 2 }$$

B
9$$\sqrt { 2 }$$

C
12$$\sqrt { 2 }$$

D
18$$\sqrt { 2 }$$

Solution
Question 6
1 / -0

$${\log _5}2,\,\,{\log _6}\,2,\,\,{\log _{12}}\,\,2\,\,$$ are in

A
A.P

B
G.P

C
H.P

D
none of these

Solution
Question 7
1 / -0

Area of a triangle whose vertices are $$(a\cos \theta ,b\sin \theta ),(-a\sin \theta ,b\cos \theta)$$ and $$(-a\cos \theta ,-b\sin \theta )$$ is-

A
$$ab\sin \theta \cos \theta $$

B
$$a\cos \theta \sin \theta$$

C
$$\frac 12ab$$

D
ab

Solution
Question 8
1 / -0

Area of the triangle formed by the tangents at the points $$\left( {4,6} \right),\left( {10,8} \right)$$ and $$\left( {2,4} \right)$$ on the parabola $${y^2} - 2x = 8y - 20,$$is (in sq. units)

A
4

B
2

C
1

D
8

Solution
Question 9
1 / -0

The area of the triangle inscribed in the parabola $$y^2$$ = $$4x$$ , the ordinates of whose vertices are 1 , 2 and 4 is :

A
7/2

B
5/2

C
3/2

D
3/4
Question 10
1 / -0

Let A(-4, 0) & B(4,0). Then the number of points C=(x,y) on the circle $${x^2} + {y^2} = 16$$ lying in first quadrant $$(x,y \geqslant 0)$$ such that the area of the triangle whose vertices are A,B,C is a integer is

A
14

B
15

C
16

D
None of these

Solution