Tag: equation

Quizzes (2) Questions (111)

Questions Related to equation

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

Solve the following linear equations. If $\cfrac{3t-2}{4}-\cfrac{2t+3}{3} = \cfrac{2}{3}-t$, then $t  $ is equal to

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given $\cfrac { 3t-2 }{ 4 } -\cfrac { (2t+3) }{ 3 }  = \cfrac { 2 }{ 3 }  - t$
L.C.M. of $4,3,3$ is $ 12$
$\Rightarrow  3(3t-2) - 4(2t+3) = 8 - 12t$
$\Rightarrow 9t - 6 -(8t +12) = 8 -12t$
$\Rightarrow t -18  = 8 -12t$
$\Rightarrow 13 t = 26$
$\Rightarrow t = 2$
Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

Solve the following linear equations: $m-\cfrac{m-1}{2} = 1-\cfrac{m-2}{3}$

  1. $m = \cfrac{7}{5}$
  2. $m = \cfrac{2}{3}$
  3. $m = \cfrac{10}{3}$
  4. $m = \cfrac{3}{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $m - \cfrac { (m-1) }{ 2 }  = 1-\cfrac { (m-2) }{ 3 }$ 
L.C.M. of $2$ and $3$ is $6$
$6m -3(m-1) = 6 - 2(m-2)$
$ 6m - 3m +3 = 6 - 2m +4$
$5m = 7$
$ m = \cfrac { 7 }{ 5 }$
Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

Which of the following is the solution of the equation $\displaystyle \frac{7y+4}{y+2}=\frac{-4}{3}$ ?

  1. $\displaystyle y = -\frac{4}{5}$
  2. $\displaystyle y = \frac{4}{5}$
  3. $\displaystyle y = -\frac{5}{4}$
  4. $\displaystyle y = \frac{5}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{7y+4}{y+2}=\dfrac{-4}{3}$
$21y+12=-4y-8$
$25y=-20$
$y=\dfrac{-20}{25} = \dfrac{-4}{5}$

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

Solve the following equations: $\cfrac{9x}{7-6x}=15$

  1. $x = \cfrac{25}{72}$
  2. $x = \cfrac{35}{33}$
  3. $x = \cfrac{45}{39}$
  4. $x = \cfrac{22}{45}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\dfrac{9x}{7-6x}=15$
$9x=105-90x$

Add $90x$ on both the sides, we get
$9x+90x=105-90x+90x $
$99x=105$
$\therefore x=\dfrac{105}{99}=\dfrac{35}{33}$

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 $=1200$