Tag: reducing simple equations to simpler form

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

The manufacturer of a certain item can sell all he can produce at the selling price of $Rs. 60$ each. It costs him $Rs. 40$ in materials and labour to produce each item and he has overhead expenses of $Rs. 3000$ per week in order to operate the plant. The number of units he should produce and sell in order to make a profit of at least $Rs\,1000$ per week, is : 

  1. $200$
  2. $250$
  3. $300$
  4. $400$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the no of units sold be $x$ per week.

$60x$ = sales should include all the expenses and profit required to balance things out.
$60x = 1000+40x+3000$

$20x = 4000$
$x =200$

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

The value of $x$ for which $\cfrac{x-3}{4}--x< \cfrac{x-1}{2}-\cfrac{x-2}{3}$ and $2-x> 2x-8$

  1. $[-1,10/3]$
  2. $(1,10/3)$
  3. $R$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{x-3}{4}-x<\dfrac{x-1}{2}-\dfrac{x-2}{3}$


$\dfrac{-3{x}-3}{4}<\dfrac{x+1}{6}$


$\implies x>-1$

$2-x>2{x}-8\implies x<\dfrac{10}{3}$

$\implies x\in (-1,10/3)$

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

If $\sqrt { 1+\dfrac { x }{ 289 }  } =1\dfrac { 1 }{ 17 }$ then $x=$

  1. $1$
  2. $13$
  3. $35$
  4. $15$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have,

$\sqrt{1+\dfrac{x}{289}}=1\dfrac{1}{17}$

$\sqrt{\dfrac{289+x}{289}}=\dfrac{18}{17}$

 

On squaring both sides, we get

$ {{\left( \sqrt{\dfrac{289+x}{289}} \right)}^{2}}={{\left( \dfrac{18}{17} \right)}^{2}} $

$ \dfrac{289+x}{289}=\dfrac{324}{289} $

$ 289+x=324 $

$ x=324-289 $

$ x=35 $

 

Hence, this is the answer.

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

Solve for $x$:-
$\dfrac{{2x - 1}}{2}\,\,\, - \dfrac{{x + 3}}{3}\,\, = \dfrac{{x - 2}}{5}$

  1. $x=\dfrac{5}{14}$
  2. $x=\dfrac{3}{14}$
  3. $x=\dfrac{33}{14}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac{{2x - 1}}{2}\,\,\, - \dfrac{{x + 3}}{3}\,\, = \dfrac{{x - 2}}{5}\Rightarrow \dfrac{6x-3-2x-6}{6}=\dfrac{x-2}{5}\Rightarrow 20x-45=6x-12\Rightarrow 14x=33\Rightarrow x=\dfrac{33}{14}$

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

If $\sqrt{10+ \sqrt{25+ \sqrt{x+ \sqrt{154+ \sqrt{225}}}}} = 4$ find the value of $x$

  1. 110

  2. 108

  3. 100

  4. 114
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,

$\sqrt {10+\sqrt {25+\sqrt {x+\sqrt{154+\sqrt{225}}}}}=4$
$\Rightarrow \sqrt {10+\sqrt {25+\sqrt {x+\sqrt{154+15}}}}=4$
$\Rightarrow \sqrt {10+\sqrt {25+\sqrt {x+\sqrt{169}}}}=4$
$\Rightarrow \sqrt {10+\sqrt {25+\sqrt {x+13}}}=4$
On squaring both sides, we get
$10+\sqrt {25+\sqrt {x+13}}=16$
$\sqrt {25+\sqrt {x+13}}=6$
On squaring both sides, we get
$25+\sqrt {x+13}=36$
$\Rightarrow \sqrt {x+13}=11$
On squaring both sides, we get
$x+13=121$
$x=121-13=108$
Hence, $x=108$