Tag: maths

Questions Related to maths

Solve the following equations: $\cfrac{8x-3}{3x}=2$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $x = \cfrac{5}{2}$

  2. $x = \cfrac{5}{6}$

  3. $x = \cfrac{9}{8}$

  4. $x = \cfrac{3}{2}$

Show answer
Correct Option: D
Explanation:

Given, $\dfrac{8x-3}{3x}=2$

Multiply $3x$ on both the sides, we get
$8x-3=6x$
$8x-6x=3$
$2x=3$
$\therefore x=\dfrac{3}{2}$

Solve the following equations: $\cfrac{z}{z+15}=\cfrac{4}{9}$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $2$

  2. $9$

  3. $12$

  4. $16$

Show answer
Correct Option: C
Explanation:

Given, $\dfrac{z}{z+15}=\dfrac{4}{9}$

$9z=4(z+15)$
$9z=4z+60$
Subtract $4z$ from both the sides, we get
$9z-4z=4z+60-4z$
$5z=60$
$\therefore z=12$

In the expression $\cfrac { x+1 }{ x-1 } $ each $x$ is replaced by $\cfrac { x+1 }{ x-1 } $. The resulting expression, evaluated for $x=\cfrac { 1 }{ 2 } $ equals:

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $3$

  2. $-3$

  3. $1$

  4. $\dfrac12$

Show answer
Correct Option: D
Explanation:

Expression is $\cfrac{x+1}{x-1}=y$

If $x$ is replaced by $\cfrac{x+1}{x-1}$
$\implies \cfrac{\cfrac{x+1}{x-1}+1}{\cfrac{x+1}{x-1}-1}$
$\implies \cfrac{x+1+x-1}{x+1-x+1}=\cfrac{2x}{2}$
The resultant expression is $x$.
When $x=\cfrac{1}{2}$
The value is $\cfrac{1}{2}$

A bag contains Rs. $90$ in coins. If coins of $50$ paise, $25$ paise, and $10$ paise are in the ratio $2 : 3: 5$, the number of $25$ paise coins in the bag is

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $80$

  2. $100$

  3. $120$

  4. $135$

Show answer
Correct Option: C
Explanation:

Let  the number coins are $ 2x,3x$, and $5x$.

then 
rupees from 50 paise coins= $2x$ $\times$$ \dfrac{1}{2}=x$

rupees from 25 paise coins= $3x$ $\times$ $\dfrac{1}{4}=\dfrac{3}{4}x$

rupees from 10 paise coins=$5x$ $\times$ $\dfrac{1}{10}=\dfrac{x}{2}$
Now given $x+\dfrac{3}{4}x+\dfrac{x}{2}=90$

or, $4x+3x+2x=90$$\times$$4$
or, $9x=360$
or, $x=40$
so  number of 25 paise coins = $3x$=$3$x$40=120$

A candidate should score $45\%$ marks of the total marks to pass the examination. He gets $520$ marks and fails by $20$ marks. The total marks in the examination are

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $1000$

  2. $1100$

  3. $1200$

  4. $1400$

Show answer
Correct Option: C
Explanation:

Given candidate should get $45\%$ of total marks to pass.

Let the total marks be $x$.
Given he got $520$ marks and fails by $20$ marks
Then to just pass
$540=\cfrac{45}{100}\times x$
$x=1200$
So, total marks $=1200$

Find the Solution  : $x - cy - bz = 0 $

                                 $ cx - y +az = 0 $
                                 $bx+ ay -z = 0 $

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $a^3 + b^3 + c^3 + 3abc = 1$

  2. $a^2 + b^2 + c^2 + 2abc = 1$

  3. $a^4 + b^4 + c^4 + 4abc = 1$

  4. $a^5 + b^5 + c^5 + 5abc = 1$

Show answer
Correct Option: B
Explanation:

The given equations are 

$x-cy-bz = 0$
$cx - y + az = 0$
$bx + ay - z = 0$
Since x, y, z are not all zero, the system will have non-trivial solution if
$\triangle = \begin{vmatrix} 1 & -c & -b \ c & -1 & a \ b & a & -1\end{vmatrix} = 0$
or $1(1 - a^2) + c (-c -ab) -b (ac + b) = 0$
or $1 - a^2 - c^2 -abc - abc - b^2 = 0$
or $a^2 + b^2 + c^2 + 2abc = 1$

 Solve:$\dfrac{{7x - 2y}}{{xy}} = 5$ and $x=2y$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $x=12/5 , y=6/5$

  2. $x=2/5 ,  y=1/5$

  3. $x=9 , y=9/2$

  4. $x=2 , y=1$

Show answer
Correct Option: A
Explanation:
$\dfrac{7x-2y}{xy}=5$

$\Rightarrow 7x-2y=5xy$

$\Rightarrow \dfrac{7x}{xy}-\dfrac{2y}{xy}=5$

$\Rightarrow \dfrac{7}{y}-\dfrac{2}{x}=5$

Substitute $x=2y$ in the above equation, we get

$\dfrac{7}{y}-\dfrac{2}{2y}=5$

$\Rightarrow \dfrac{14-2}{2y}=5$

$\Rightarrow \dfrac{12}{2y}=5y$

$\Rightarrow 5y=6$

$\therefore y=\dfrac{6}{5}$

Put $y=\dfrac{6}{5}$ in $x=2y$ we get

$x=2\times \dfrac{6}{5}=\dfrac{12}{5}$

$\therefore x=\dfrac{12}{5}\,\,,y=\dfrac{6}{5}$

$\dfrac{2x-3}{2}-\dfrac{(x+1)}{3}=\dfrac{3x-8}{4}$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $x=4$

  2. $x=-2$

  3. $x=2$

  4. $x=-4$

Show answer
Correct Option: C
Explanation:

$\cfrac { 2x-3 }{ 2 } -\cfrac { x+1 }{ 3 } =\cfrac { 3x-8 }{ 4 } \ =>(6x-9-2x-2)\times 4=(3x-8)\times 6\ =>16x-44=18x-48\ =>2x=4\ =>x=2$

Solve: 

$3(x+1)=6$

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. $x=-1$

  2. $x=\dfrac{1}{3}$

  3. $x=\dfrac{-1}{3}$

  4. $x={1}$

Show answer
Correct Option: D
Explanation:
$3(x+1)=6$

$ \Rightarrow 3x+3=6$

$ \Rightarrow 3x=3$

$ \Rightarrow x=1$

The equation $x-\dfrac {8}{|x-3|}=3--\dfrac {8}{|x-3|}$ has

maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

  1. Only one solution

  2. infinite solution

  3. no solution

  4. two solution

Show answer
Correct Option: A
Explanation:

Consider the following equation.

$ x-\dfrac{8}{|x-3|}=3-\dfrac{8}{|x-3|} $

$ x-\dfrac{8}{\pm \left( x-3 \right)}=3-\dfrac{8}{\pm \left( x-3 \right)} $

$ x-\dfrac{8}{\left( x-3 \right)}=3-\dfrac{8}{\left( x-3 \right)} $

$ x=3 $

 $ x+\dfrac{8}{\left( x-3 \right)}=3+\dfrac{8}{\left( x-3 \right)} $

$ x=3 $

Hence, this is the correct answer.