Tag: solving linear equations with variable on both sides

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Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides 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 ___________.

  1. $60$
  2. $40$
  3. $80$
  4. $120$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $60$.

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

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?

  1. $1$ minute early
  2. $5$ minutes early
  3. $1$ minute late
  4. $5$ minutes late
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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-

  1. 21 years

  2. 17 years

  3. 18 years

  4. 16 years
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $18 years$

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

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

  1. $36$ $and$ $108$
  2. $16$ $and$ $88$
  3. $63$ $and$ $135$
  4. $\text{none}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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.

  1. $252$
  2. $360$
  3. $260$
    <span class="MathJax"><span class="math">
  4. $352$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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?

  1. $50\ yr$
  2. $28\ yr$
  3. $42\ yr$
  4. $36\ yr$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $=28 yr$
Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

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

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

  2. 10

  3. 14

  4. 11
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\dfrac{x+2}{6}-\left[\dfrac{11-x}{3}-\dfrac{1}{4}\right]=\dfrac{3x-4}{12}$

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

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

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

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

$\Rightarrow 3x=33$

$\Rightarrow x=11$.
Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

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 $3$. 

  1. $14$
  2. $25$
  3. $36$
  4. $47$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 = $36$

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

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


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

  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. $\displaystyle x\, =\, -\frac{12}{5}; \, y\, =\, \frac{7}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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} $

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

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.

  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 $32 cm$.
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $20 cm$.