Tag: reading graphs describing different situations

Questions Related to reading graphs describing different situations

The straight lines given by the equations $\displaystyle x+y=2 , x-2y=5 \ and \ \frac{x}{3}+y=0$ are?

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. concurrent

  2. intersecting to make a right triangle.

  3. intersecting to make an isosceles triangle.

  4. parallel to each other.

Show answer
Correct Option: A
Explanation:
Given lines
$x+y=2$------(1)
$x-2y=5$----(2) and $\dfrac{x}{3}+y=0$----(3)
Solving eq (1) and (2)
$x-2(2-x)=5$
$x-4+2x=5$
$x=3$ and $y=2-3=-1$
Point of intersection of line (1) and (2) is $P(3,-1)$
Solving eq (2) and (3)
$-3y-2y=5$
$-5y=5$
$y=-1$ and $x=-3y=3$
Point of intersection of line (2) and (3) is $Q(3,-1)$
Solving eq (1) and (3)
$-3y+y=2$
$-2y=2$
$y=-1$ and $x=-3y=3$
Point of intersection of line (1) and (3) is $R(3,-1)$
Here point of intersection of all line is same Hence line is concurrent

If the line ax + by + c = 0 is such that  a = 0 and b, $\displaystyle c\neq 0$ then the line is perpendicular to 

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. x-axis

  2. y-axis

  3. x + y =1

  4. x = y

Show answer
Correct Option: B
Explanation:

When $ a= 0 $ then the line equation becomes $ by + c = 0 $ or $ y = -\frac {c}{b} $

Equations of the form $ y =k $ are parallel to x-axis. This also means that they are perpendicular to $ y - $ axis as $ x-$ axis and $ y- $axis are perpendicular to each other.

Find the equation of a line passing through the point (2, -3 ) and parallel to the line 2x - 3y + 8 = 0

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. 2x - 3y =13

  2. 2x -3y = 12

  3. x - 3y =4

  4. 3x - 2y = 7

Show answer
Correct Option: A
Explanation:

Equation of the line parallel to $ 2x-3y+8 = 0 $ will be of the form $ 2x-3y + k = 0 $

Now, since it passes through $ (2,-3) $, on substituting it , we get $ 2(2) -3(-3) + k = 0  $
$ => k = -13 $

So, required eqn of parallel line is $ 2x - 3y - 13 = 0 $

$\dfrac {a}{3} + \dfrac {b}{6} = 1$
If $a$ and $b$ are positive integers in the equation above, then what is the value of $a b$?

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. $4$

  2. $2$

  3. $6$

  4. $8$

  5. $5$

Show answer
Correct Option: A
Explanation:

Given eq 

$\dfrac{a}{3}+\dfrac{b}{6}=1$
$2a+b=6$---(1)

putting $a=1$ in eq (1)
$b=4$
Hence $ab=4$

If $y$ is directly proportional to $x$ and if $y=20$  when $x=6$, what is the value of $y$ when $x=9$?

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. $\displaystyle\frac{10}{3}$

  2. $\displaystyle\frac{40}{3}$

  3. $23$

  4. $27$

  5. $30$

Show answer
Correct Option: E
Explanation:

Given that: $'y'$ is directly proportional to $'x'$,

Formally, it is expressed as
$y$ $=$ $k$$x$   where, $'k'$ is a constant
To find the value of $'k'$,
As $y$ $=$ $20$  when  $x$ $=$ $6$,
$\Rightarrow y$ $=$ $k$$x$
$\Rightarrow y$ $=$ $k$ $\times$ $x$
$\Rightarrow 20$ $=$ $k$ $\times$ $6$
$\Rightarrow k$ $=$ $\dfrac {20}{6}$
$\Rightarrow k$ $=$ $\dfrac {10}{3}$
Now, $y$ $=$ $\dfrac {10}{3}$$x$
To find $'y'$ when $x$ $=$ $9$,
$\Rightarrow y$ $=$ $\dfrac {10}{3}$$x$
$\Rightarrow y$ $=$ $\dfrac {10}{3}$ $\times$ $x$
$\Rightarrow y$ $=$ $\dfrac {10}{3}$ $\times$ $9$
$\Rightarrow y$ $=$ $30$

Therefore, the value of $'y'$ when $x$ $=$ $9$ is $'30'$.

If $y=2x+3$ and $x < 2$, which of the following represents all the possible values for $y$?

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. $y < 7$

  2. $y > 7$

  3. $y < 5$

  4. $y > 5$

  5. $5 < y < 7$

Show answer
Correct Option: A
Explanation:

Given, $y$ $=$ $2x$ $+$ $3$ and $x$ $<$ $2$

To find possible values for $y$,
$\Rightarrow 2x$ $=$ $y$ $-$ $3$
$\Rightarrow x$ $=$ $\dfrac {y \space - \space 3}{2}$
As $x$ $<$ $2$,
$\Rightarrow \dfrac {y \space - \space 3}{2}$  $<$ $2$
$\Rightarrow y$ $-$ $3$ $<$ $4$
$\Rightarrow y$ $<$ $7$
Therefore, possible values for $'y'$ are $'y$ $<$ $7'$.

The graph of equation of the form $ax + by +c=0$ where a, b are non $-$ zero numbers,
represents:

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. A triangle

  2. A ray

  3. A straight line

  4. a line segment

Show answer
Correct Option: C
Explanation:

The equation $ax+by+c=0$ represent straight line under one condition i. e, 

$|a|+|b|\neq 0$       or,     $ a\neq b\neq 0$
Also,  here
$y= \dfrac{-a}{b}x$ $  \dfrac{-c}{b}$ represent the slope interspect form where $m= (-a/b), y-m =  -c/b $ 

A right-angled triangle is formed by a straight line : $3x-4y=12$ with both the axis. Then length of perpendicular from the origin to the hypotenuse is :

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. $3.5$

  2. $2.4$

  3. $4.2$

  4. none of these

Show answer
Correct Option: B
Explanation:

Given straight line is $3x-4y=12$
$\Rightarrow\frac{x}{4}-\frac{y}{3} = 1$
this line have x intercept y & y- intercept (-3)
so the evaluation of hypotenuse s the given straight line $3x-4y=12$
so, distance of O(0, 0) form the line $3x-4y=12$ is given by
$=\frac{|3\times0-4\times0-12|}{\sqrt{(3)^2+(-4)^2}}$
$=\frac{12}{5}$
$=2.4$

The graph of the function $\displaystyle \cos x.\cos (x+2)-\cos^{2}(x+1)$ is a 

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. straight line passing through the point $\displaystyle(0,-\sin^{2}1)$ with slope $2$

  2. straight line passing through the origin

  3. parabola with vertex $\displaystyle (1,-\sin^{2}1)$

  4. straight line passing through the point $\displaystyle\left(\dfrac{\pi}2,-\sin^{2}1\right) $ and parallel to the $x-$axis

Show answer
Correct Option: D
Explanation:

Let $y=\cos { x } \cos { \left( x+2 \right)  } -\cos ^{ 2 }{ \left( x+1 \right)  } \ =\cos { \left( x+1-1 \right)  } \cos { \left( x+1+1 \right)  } -\cos ^{ 2 }{ \left( x+1 \right)  }\$
 Using $\cfrac{1}{2} \left[\cos\left(A+B+A-B\right)+\cos\left(A+B-A+B\right)\right]\
           = \cfrac{1}{2}\left[\cos 2A + \cos 2B\right]
           = \cfrac{1}{2}\left[\cos^{2}A -1 +1 -2 \sin^{2}A\right ]
           =  \cos ^{2}\left(x+1\right)-\sin^{2}1 $
$=\cos ^{ 2 }{ \left( x+1 \right)  } -\sin ^{ 2 }{ 1 } -\cos ^{ 2 }{ \left( x+1 \right)  } \ =-\sin ^{ 2 }{ 1 } $
This is a straight line which is parallel to x-axis, it passes through $\left( \cfrac { \pi  }{ 2 } ,-\sin ^{ 2 }{ 1 }  \right) $

Complete the table, to draw the graph of line $2y=3x+2$.


$x:$ $3$ $\displaystyle \frac{7}{3}$ $-2$
$y:$ $y _1$ $y _2$ $y _3$

maths banking and taxation reading graphs describing different situations using equations to plot lines basics of a straight line

  1. $y _1 = \dfrac12, y _2=\dfrac72,y _3 = -2$

  2. $y _1 = \dfrac{11}2, y _2=\dfrac92,y _3 = -2$

  3. $y _1 = \dfrac12, y _2=\dfrac72,y _3 = -3$

  4. $y _1 = \dfrac{11}2, y _2=\dfrac92,y _3 = -3$

Show answer
Correct Option: B
Explanation:

The equation of line is $2y=3x+2$

Now to complete the given table,
Substitute $x=3$ in the equation $2y=3x+2$ to fill the first column of the table
$\displaystyle 2y _1=3\times 3+2\ \Rightarrow 2y _1=9+2\ \Rightarrow 2y _!=11\ \Rightarrow y _!=\dfrac { 11 }{ 2 }$
 Substitute $x=\dfrac { 7 }{ 3 }$ in the equation $2y=3x+2$ to fill the second column of the table

$2y _2=3\times \dfrac { 7 }{ 3 } +2\\ \Rightarrow 2y _2=7+2\\ \Rightarrow 2y _2=9\\ \Rightarrow y _2=\dfrac { 9 }{ 2 }$
Finally, substitute $x=-2$ in the equation $2y=3x+2$ to fill the first column of the table
$2y _3=3\times -2+2\\ \Rightarrow 2y _3=-6+2\\ \Rightarrow 2y _3=-4\\ \Rightarrow y _3=-2$