Tag: maths

Questions Related to maths

Consider the line: y= -x +4

Which of the following is correct.

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

  1. The line passes through (0,4) and m=1.

  2. The line passes through (0,4) and m=-1.

  3. The line passes through (0,0) and m=-1.

  4. The line passes through (4,0) and m=-1.

Show answer
Correct Option: B
Explanation:
Given line 
$y=-x+4$
on comparing above eq with $y=mx+c$
$slope(m)=-1$
y-intercept$=4$
Hence it passes through (0,4) with $m=-1$

Draw the graph for each linear equation:
$\displaystyle y=\frac{3}{2}x+\frac{2}{3}$

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

  1. The line passes through $(4/9,0)$ and $m=-\dfrac32$

  2. The line passes through $(-0.4/9,0)$ and $m=\dfrac32$

  3. The line passes through $(-4/9,0)$ and $m=\dfrac32$

  4. The line passes through $(-0.9/4,0)$ and $m=-\dfrac32$

Show answer
Correct Option: C
Explanation:

The given line equation is $\frac{3}{2}x-y+\frac{2}{3}$

The slope of given line is $-(\frac { \frac { 3 }{ 2 }  }{ -1 } )=\frac { 3 }{ 2 } $
If we put $y=0$ , then the value of $x= -\frac{4}{9}$
Therefore line passes through $(-\frac{4}{9},0)$ and slope is $\frac{3}{2}$
So the correct option is $C$

Consider the equation of the line $\displaystyle x-3=\frac{2}{5}\left ( y-1 \right )$. Which of the following is correct?

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

  1. The line passes through $(6,5)$  and $m=-2/5$.

  2. The line passes through $(5,6)$  and $m=-5/2$.

  3. The line passes through $(6,5)$  and $m=2/5$.

  4. The line passes through $(5,6)$  and $m=5/2$.

Show answer
Correct Option: D
Explanation:
Given line 
$x-3=\dfrac{2}{5}(y-1)$
$5(x-3)=2(y-1)$
$5x-15=2y-2$
$2y=5x-13$
$y=\dfrac{5x}{2}-\dfrac{13}{2}$
on comparing above eq with $y=mx+c$
$slope(m)=\dfrac{5}{2}$

when $x=5$
$2y=25-13$
$2y=12$
$y=6$
Hence it passes through (5,6) with $m=\dfrac{5}{2}$

For the pair of linear equations given below, draw graph and then state, whether the lines drawn are,
$\displaystyle y=3x-1$
$\displaystyle \frac{x}{2}+\frac{y}{3}=1$

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

  1. Perpendicular

  2. Parallel

  3. Intersecting but not at right angles

  4. Options B & C

Show answer
Correct Option: C

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are 
$\displaystyle 3x+4y=24$
$\displaystyle \frac{x}{4}+\frac{y}{3}=1$

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

  1. intersecting but not at right anglesl

  2. Options B & D

  3. perpendicular

  4. parallel

Show answer
Correct Option: D

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 $'30'$.