Tag: business maths

Questions Related to business maths

Who has coined the term Regression?

  1. Francis Galton

  2. R.A. Fisher

  3. Daniel

  4. Karl Pearson


Correct Option: A
Explanation:

Regression is a technique for determining the statistical relationship between two or more variables where a change in a dependable variable is associated with and depends on a change in one or more independant variables. The term regression has been coined by Francis Galton.

The pair of lines represented by $\displaystyle :3ax^{2}+5xy+\left ( a^{2}-2 \right )y^{2}= 0$ and at right angles to each other, then value $ \left ( s \right )$ of $a$ is/are:

  1. $\displaystyle :\frac{-3+\sqrt{17}}{2}$

  2. $\displaystyle :\frac{-3-\sqrt{17}}{2}$

  3. $\displaystyle :\frac{3+\sqrt{17}}{2}$

  4. $\displaystyle :\frac{3-\sqrt{17}}{2}$


Correct Option: A,B
Explanation:

The pair of lines represented by $\displaystyle :3ax^{2}+5xy+\left ( a^{2}-2 \right )y^{2}= 0$ and at right angles to each other.
pair of lines given by $ax^2+2hxy+by^2=0$ are at right angles if $a+b=0$
$\Rightarrow 3a+a^2-2=0$
$\Rightarrow a^2+3a-2=0$
$\therefore a=\displaystyle\frac{-3\pm \sqrt{17}}{2}$
Hence, options A and B.

If one of the line given by the equation $a _{1}x^{2}+2h _{1}xy+b _{1}y^{2}=0$ coincides with one of the lines given by $a _{2}x^{2}+2h _{2}xy+b _{2}y^{2}=0$ and the other lines represented by them be perpendicular then $\dfrac {h _{1}a _{2}b _{2}}{b^{2}-a _{2}}\dfrac {h _{2}a _{1}b _{1}}{b _{1}-a _{1}}=\dfrac {1}{2}\sqrt {-a _{1}a _{2}b _{1}b _{2}}$.

  1. True

  2. False


Correct Option: A

The triangle formed by the lines whose combined equation is  $\displaystyle (y^{2}-4xy-x^{2}) ( x+y-1  )=0$ is

  1. equilateral.

  2. right angled.

  3. isosceles.

  4. obtuse angled.


Correct Option: B
Explanation:

Given equation of the pair of straight lines is  $x^{2}-4xy-y^{2}=0$

Since, coefficient of $x^2$ + coefficient of $y^2$ $= 0$ 

The two lines are perpendicular.

Therefore, triangle formed is right angled.

Find the equation of the line perpendicular to $x-7y+5=0$ and having x-intercept 3.

  1. $x+y-21=0$

  2. $x-7y+21=0$

  3. $7x+y-21=0$

  4. None of these


Correct Option: C
Explanation:

The equation of a line perpendicular to $x-7y+5=0$ is $7x+y+\lambda =0$.


Its x-intercept is 3. This means that the line cuts x-axis at a distance of 3 units from the origin. Consequently, it passes through the point (3,0) on x-axis.


Therefore, $21+0+\lambda =0$

$\lambda = -21$

Thus, the equation of the required line is $7x+y-21=0$.

If $2x^{2}+3xy+my^{2}=0$ represents two real and mutually perpendicular lines then $m$ is

  1. any negative real number

  2. any positive real number

  3. $-2$

  4. none of these


Correct Option: C
Explanation:

From above formula
$\tan90^{\circ}=\left | \dfrac{2\sqrt{(\frac{3}{2})^{2}+2m}}{2+m} \right |=\dfrac{1}{0}$
$m=-2$

The product of the perpendiculars from origin to the pair of lines $ a x ^ { 2 } + 2 h x y + b y ^ { 2 } + 2 g x + 2 f y + c = 0 $ is

  1. $

    \frac { | c | } { \sqrt { ( a + b ) ^ { 2 } + 4 h ^ { 2 } } }

    $

  2. $

    \frac { | c | } { \sqrt { ( a + b ) ^ { 2 } - 4 h ^ { 2 } } }

    $

  3. $

    \frac { | c | } { \sqrt { ( a - b ) ^ { 2 } + 4 h ^ { 2 } } }

    $

  4. $

    \frac { | c | } { \sqrt { ( a - b ) ^ { 2 } - 4 h ^ { 2 } } }

    $


Correct Option: A

The equation $\displaystyle ax^{3}-9yx^{2}-y^{2}x+4y^{3}=0 $ represents three straight lines. If two of the lines are perpendicular to each other, then the value of $a$ is:

  1. 5

  2. -5

  3. 4

  4. -4


Correct Option: A,D
Explanation:

The given equation is $ax^3 - 9yx^2 -y^2x + 4y^3 = 0$, which represents three straight lines.


We can see that all the lines passes from origin $(0,0)$

Let's assume the lines are given by equation $y = mx$

Putting $y = mx$ in the given equation of three straight lines, we get,

$\Rightarrow ax^3- 9(mx)(x^2) - (mx)^2x + 4(mx)^3 = 0$

$\Rightarrow (a - 9m - m^2 + 4m^3) x^3 = 0$

$\Rightarrow 4m^3 -m^2 -9m +a = 0$ ....$(1)$

This equation in $m$ has three roots, $m _1$, $m _2$ and $m _3$, which are three slopes of three lines respectively.

If the two lines are perpendicular then let's assume $m _1.m _2 = -1$,

Product of roots in equation $(1)$ is $m _1.m _2.m _3 = \dfrac{-a}{4}$

Hence $m _3 = \dfrac{a}{4}$

also from equation $(1)$, $m _1 + m _2 + m _3 = \dfrac{1}{4}$

$\Rightarrow m _1m _2 + m _3(m _1 + m _2) = \dfrac{-9}{4}$

$\Rightarrow m _1 + m _2 = \dfrac{1-a}{4}$

$\Rightarrow a(1-a) = -20$

By Solving the above equation we get $a = 5, -4$

Equation $\displaystyle ax^{3}-9yx^{2}-y^{2}x+4y^{3}=0$ represents three straight lines. If two of the lines are perpendicular to each other then the value of a is

  1. 5

  2. -5

  3. 4

  4. -4


Correct Option: A,D
Explanation:

The given equation is $ax^3 - 9yx^2 -y^2x + 4y^3 = 0$, which represents three straight lines.


We can see that all the lines passes from origin $(0,0)$

Let's assume the lines are given by equation $y = mx$

Putting $y = mx$ in the given equation of three straight lines, we get,

$\Rightarrow ax^3- 9(mx)(x^2) - (mx)^2x + 4(mx)^3 = 0$

$\Rightarrow (a - 9m - m^2 + 4m^3) x^3 = 0$

$\Rightarrow 4m^3 -m^2 -9m +a = 0$ ....$(1)$

This equation in $m$ has three roots, $m _1$, $m _2$ and $m _3$, which are three slopes of three lines respectively.

If the two lines are perpendicular then let's assume $m _1.m _2 = -1$,

Product of roots in equation $(1)$ is $m _1.m _2.m _3 = \dfrac{-a}{4}$

Hence $m _3 = \dfrac{a}{4}$

also from equation $(1)$, $m _1 + m _2 + m _3 = \dfrac{1}{4}$

$\Rightarrow m _1m _2 + m _3(m _1 + m _2) = \dfrac{-9}{4}$

$\Rightarrow m _1 + m _2 = \dfrac{1-a}{4}$

$\Rightarrow a(1-a) = -20$

By Solving the above equation we get $a = 5, -4$

The pair of lines represented by $3ax^{2}+5xy+\left ( a^{2}-2 \right )y^{2}= 0$ and $\perp $ to each other for

  1. two values of $a$

  2. for all $a$

  3. for one value of $a$

  4. for no values of $a$


Correct Option: A
Explanation:

 Using fact: Pair of lines $\displaystyle Ax^{2}+2hxy+By^{2}=0$ are 


$\displaystyle \perp $ to each other if $\displaystyle A+B=0$ 

$\displaystyle \Rightarrow 3a+a^{2}-2=0 $ $\displaystyle \Rightarrow a^{2}+3a-2=0 $ $\displaystyle\Rightarrow $ There exist two value of a as $\displaystyle D> 0$