Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

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:

  1. $3$
  2. $-3$
  3. $1$
  4. $\dfrac12$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\cfrac{1}{2}$

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

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

  1. $80$
  2. $100$
  3. $120$
  4. $135$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$40=120$

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

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

  1. $1000$
  2. $1100$
  3. $1200$
  4. $1400$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $=1200$

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

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

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

  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. $a^5 + b^5 + c^5 + 5abc = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $a^2 + b^2 + c^2 + 2abc = 1$

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

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

  1. $x=12/5 , y=6/5$
  2. $x=2/5 , y=1/5$
  3. $x=9 , y=9/2$
  4. $x=2 , y=1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

$\therefore x=\dfrac{12}{5}\,\,,y=\dfrac{6}{5}$
Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

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

  1. Only one solution

  2. infinite solution

  3. no solution

  4. two solution
Reveal answer Fill a bubble to check yourself
A Correct answer
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.