Sequences and Series Test 47

Result

Sequences and Series Test 47

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Question 1
1 / -0

Find the number .

A
$$473$$

B
$$623$$

C
$$389$$

D
$$584$$

Solution

$$ \begin{array}{l} 4^{3}+7^{3}+3^{3}=434 \\ \text { similarly, } \\ 6^{3}+5^{3}+2^{3}=349 \\ \Rightarrow \text { ? }=8^{3}+4^{3}+2^{3} \\ \Rightarrow \text { ? }=584 \\ \text { option } D \text { is correct. } \end{array} $$
Question 2
1 / -0

There is a specific relationship between the numbers that are given in the following figures. On the basis of the relationship choose the correct alternative to replace the question mark.

A
$$210$$

B
$$266$$

C
$$288$$

D
$$318$$
Question 3
1 / -0

If three lines $$x-3y=p,ax+2y=q$$ and $$ax+y=r$$ form a right angled triangle, then

A
$${ a }^{ 2 }-9a+18=0$$

B
$${ a }^{ 2 }-6a-12=0$$

C
$${ a }^{ 2 }-6a+9=0$$

D
$${ a }^{ 2 }-9a+12=0$$

Solution
Question 4
1 / -0

If in a progression $${a_1},{a_2},{a_3},...,etc.,\left( {{a_r} - {a_{r + 1}}} \right)$$ bears a constant ratio with $${a_r}.{a_{r + 1}}$$ then the term of the progression are in

A
AP

B
GP

C
HP

D
none of there

Solution
Question 5
1 / -0

If the tangent at $$\theta =\frac { \pi }{ 4 } $$ to the curve $$x=a\cos ^{ 3 }{ \theta } ,y=a\sin ^{ 3 }{ \theta } $$ meets the x and y axes in A and B then the area of the triangle OAB is

A
$$\frac { { a }^{ 2 } }{ 4 } sq.units$$

B
$$\frac { { a }^{ 2 } }{ 2 } sq.units$$

C
$$\frac { { 3a }^{ 2 } }{ 4 } sq.units$$

D
$$\frac { { 5a }^{ 2 } }{ 4 } sq.units$$

Solution
Question 6
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 7
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 8
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 9
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
a b sin $$ \theta \cos \theta $$

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

C
$$ \frac{1}{2} a b $$

D
$$ a b $$

Solution
Question 10
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