Tag: equation

Questions Related to equation

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

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: B
Explanation:
Given $\cfrac { 3t-2 }{ 4 } -\cfrac { (2t+3) }{ 3 }  = \cfrac { 2 }{ 3 }  - t$
L.C.M. of $4,3,3$ is $ 12$
$\Rightarrow  3(3t-2) - 4(2t+3) = 8 - 12t$
$\Rightarrow 9t - 6 -(8t +12) = 8 -12t$
$\Rightarrow t -18  = 8 -12t$
$\Rightarrow 13 t = 26$
$\Rightarrow t = 2$

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

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

  1. $m = \cfrac{7}{5}$

  2. $m = \cfrac{2}{3}$

  3. $m = \cfrac{10}{3}$

  4. $m = \cfrac{3}{8}$

Show answer
Correct Option: A
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 }$

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

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

  1. $\displaystyle y = -\frac{4}{5}$

  2. $\displaystyle y = \frac{4}{5}$

  3. $\displaystyle y = -\frac{5}{4}$

  4. $\displaystyle y = \frac{5}{4}$

Show answer
Correct Option: A
Explanation:

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

Solve the following equations: $\cfrac{3y+4}{2-6y}=\cfrac{-2}{5}$

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

  1. $2$

  2. $-4$

  3. $4$

  4. $-8$

Show answer
Correct Option: D
Explanation:

Given, $\dfrac{3y+4}{2-6y}=\dfrac{-2}{5}$

On cross multiplying, we get
$5(3y+4)=-2(2-6y)$
$\Rightarrow 15y+20=-4+12y$
$\Rightarrow 3y=-24$
$\Rightarrow y=-8$

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

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

  1. $x = \cfrac{25}{72}$

  2. $x = \cfrac{35}{33}$

  3. $x = \cfrac{45}{39}$

  4. $x = \cfrac{22}{45}$

Show answer
Correct Option: B
Explanation:

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

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

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 $=1200$