Tag: solution of linear equations in one variable

Questions Related to solution of linear equations in one variable

If $\left(\displaystyle\frac{2}{3}\right)^{rd}$ of a number is $20$ less than the original number, then the number is ___________.

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. $60$

  2. $40$

  3. $80$

  4. $120$

Show answer
Correct Option: A
Explanation:

Let the number is $x$

Therefore, $x-\dfrac{2}{3}x =20$
$\Rightarrow \dfrac{1}{3}x=20$
$\Rightarrow x=60$
Therefore, the  number is $60$.

A lady reaches her office $20$ minutes late by traveling at a speed of $20$ km/h and reaches $15$ minutes early by traveling at $30$ km/h. By how much time will she be early or late if she travels at $25$ km/h?

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. $1$ minute early

  2. $5$ minutes early

  3. $1$ minute late

  4. $5$ minutes late

Show answer
Correct Option: A
Explanation:

Let the distance of the office be $x$ km


Time taken at $20 km/hr = \dfrac{x}{20}hrs$

Given that she reaches $\dfrac{20}{60} = \dfrac{1}{3} hrs$ late

Time taken at $30 km/hr = \dfrac{x}{30}hrs$

Given that she reaches $\dfrac{15}{60} = \dfrac{1}{4} hrs$ early

$\therefore \dfrac{x}{20} - \dfrac{x}{30} = \dfrac{1}{3} + \dfrac{1}{4}$

$\Rightarrow x = 35km$

Time taken at $30km/hr = \dfrac{35}{30}\times 60 = 70$ minutes

Time taken at $25 km/hr = \dfrac{35}{25}\times60 = 84$ minutes

So travelling at $25 km/hr$ she reaches $70+15-84 = 1$ minute early

The average age of $3$ sisters is $15$. If the ages of $2$ sisters are $12$ years and $15$ years, the age of the third sister is-

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. 21 years

  2. 17 years

  3. 18 years

  4. 16 years

Show answer
Correct Option: C
Explanation:

Let the ages of sisters be $x,y,z$


Given that

$\Rightarrow \dfrac{x+y+z}{3}=15$

$\Rightarrow x+y+z=45$

Given that $x,y=12,15$

$\Rightarrow 12+15+z=45$

$\Rightarrow z=18$

Therefore, age of third sister is $18 years$

The difference of two numbers is $72$ and the quotient obtained by dividing one by the other is $3$. Find the numbers.

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. $36$ $and$ $108$

  2. $16$ $and$ $88$

  3. $63$ $and$ $135$

  4. $\text{none}$

Show answer
Correct Option: A
Explanation:

$Let\>the\>numbers\>be\>x\>and\>y,\>then\>x-y=72\\and\>(\frac{x}{y})=3\\or\>x\>=\>3y\\\therefore\>3y-y=72\\2y=72\\\therefore\>y=36,\>then\>x\>=\>108$

In an orchard, $\dfrac{1}{5}$ are orange trees, $\dfrac{3}{13}$ are mango trees and the rest are banana trees.  If the banana trees are $148$ in number, find the total number of trees in the orchard.

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. $252$

  2. $360$

  3. $260$

  4. $352$

Show answer
Correct Option: C
Explanation:

$let \>total \>number\> of \>trees=x \\then \>banana\> trees =148\\x-(\frac{x}{5})-(\frac{3x}{13})=148\\(\frac{65x-13x-15x}{65})= 148\\\therefore x= (\frac{148\times 65}{37})=260$

At present anil is $1.5$ times of purvis age. $8\ yr$ later, the respective ratio between Anil and Purvis ages will be $25:18$. What is Purvis present age?

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. $50\ yr$

  2. $28\ yr$

  3. $42\ yr$

  4. $36\ yr$

Show answer
Correct Option: B
Explanation:
Let Present age of Purvis is $x$ then, Age of Anil will be $1.5x$
 After $8 yr$,
                   Age of anil $=1.5x+8$ And Age of Purvis $=x+8$
             $\dfrac{25}{18}=\dfrac{1.5x+8}{x+8}$
                        $x=28$
 So Age of Purvis $=28 yr$

Solve for $x : \dfrac { x + 2 } { 6 } - \left[ \dfrac { 11 - x } { 3 } - \dfrac { 1 } { 4 } \right] = \dfrac { 3 x - 4 } { 12 }$

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. $\dfrac { 6 } { 11 }$

  2. 10

  3. 14

  4. 11

Show answer
Correct Option: D
Explanation:
$\dfrac{x+2}{6}-\left[\dfrac{11-x}{3}-\dfrac{1}{4}\right]=\dfrac{3x-4}{12}$

$\Rightarrow \dfrac{2(x+2)}{12}-\left[\dfrac{4(11-x)}{12}-\dfrac{3}{12}\right]=\dfrac{3x-4}{12}$

$\Rightarrow 2x+4-[44-4x-3]=3x-4$

$\Rightarrow 2x+4-44+3+4x=3x-4$

$\Rightarrow 6x-37=3x-4$

$\Rightarrow 3x=33$

$\Rightarrow x=11$.

Seven times a two digit number is equal to four times the number obtained by reversing the order of digits. Find the number, if the difference between its digits is $3$. 

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. $14$

  2. $25$

  3. $36$

  4. $47$

Show answer
Correct Option: C
Explanation:

Let one's digit be $x$ and the tens be $x-3$


Number = $10(x-3) +x$ 

Reversed no. = $10x +x-3$ 

$ 7(10(x-3) +x) = 4(x-3 +10x)\ 70x -210 + 7x = 4x -12 +40x\ 33x = 198\ x = 6$ 

Number = $36$

Solve: $\displaystyle \frac{2x\, +\,1}{10}\, -\, \frac{3\, -\, 2x}{15}\, =\, \frac{x\, -\, 2}{6}$.


Hence, find y, if $\displaystyle \frac{1}{x}\, +\, \frac{1}{y}\, +\, 1\, = 0$.

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. $\displaystyle x\, =\, -\frac{7}{5}; \, y\, =\, -\frac{7}{2}$

  2. $\displaystyle x\, =\, -\frac{2}{5}; \, y\, =\, \frac{7}{2}$

  3. $\displaystyle x\, =\, -\frac{6}{5}; \, y\, =\, -\frac{7}{2}$

  4. $\displaystyle x\, =\, -\frac{12}{5}; \, y\, =\, \frac{7}{2}$

Show answer
Correct Option: A
Explanation:

$ \dfrac {2x + 1}{10} - \dfrac {3-2x}{15} = \dfrac {x-2}{6} $

On taking LCM and simplifying, we get

$ \dfrac {6x + 3 - 6 + 4x}{30} = \dfrac {x - 2}{6} $

$ => \dfrac {10x - 3}{30} = x - 2 $


$ => 10x - 3 = 5x - 10 $

$ 5x = -7 $

$ x = -\dfrac {7}{5} $

Now, substituting x in $ \dfrac {1}{x} + \dfrac {1}{y} + 1 = 0 $, we get

$ - \dfrac {5}{7} + \dfrac {1}{y} + 1 = 0 $

$ => \dfrac {1}{y} = - 1 + \dfrac {5}{7} =  - \dfrac {2}{7} $

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

An altitude of a triangle is five-third the length of its corresponding base. If the altitude is increased by $4 cm$ and the base is decreased by $2 cm$, the area of the triangle remains same. Find the base and the altitude of the triangle.

maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

  1. The base of the triangle is $12 cm$ and altitude is $20 cm$.

  2. The base of the triangle is $4 cm$ and altitude is $34 cm$.

  3. The base of the triangle is $16 cm$ and altitude is $12 cm$.

  4. The base of the triangle is $8 cm$ and altitude is $32 cm$.

Show answer
Correct Option: A
Explanation:
Let the base of the triangle be $x$ cm. 
Then, the altitude of the triangle $=\cfrac { 5x }{ 3 } $
So, area of the triangle $=\cfrac { 1 }{ 2 } \times base\times altitude$
$=\cfrac { 1 }{ 2 } \times x\times \cfrac { 5 }{ 3 } x\\ =\cfrac { 5 }{ 6 } { x }^{ 2 }$        ...(1)
On increasing the altitude by $4 cm$ and the decreasing base by $2 cm$, the area remains the same.
Therefore, $\cfrac { 1 }{ 2 } \times (x-2)\times \left( \cfrac { 5x }{ 3 } +4 \right) =\cfrac { 5 }{ 6 } { x }^{ 2 }$       ...[using (1)]
$\Longrightarrow \cfrac { 1 }{ 2 } \times (x-2)(\cfrac { 5x+12 }{ 3 } )=\cfrac { 5 }{ 6 } { x }^{ 2 }$
$ \Longrightarrow (x-2)(5x+12)=5{ x }^{ 2 }$
$ \Longrightarrow 5{ x }^{ 2 }-10x+12x-24=5{ x }^{ 2 }$
$ \Longrightarrow 2x-24=0$
$ \Longrightarrow 2x=24$ or $x = 12$.
 Now, altitude of the triangle $=\cfrac { 5x }{ 3 } =\cfrac { 5\times 12 }{ 3 } =20 cm$
Hence, the base of the triangle is $12 cm$ and altitude is $20 cm$.