Tag: solution of a linear equation in one variable

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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

If $\displaystyle \frac{x^2\, -\, (x\, +\, 1)(x\, +\, 2)}{5x\, +\, 1}\, =\, 6$, then $x$ is equal to

  1. $\displaystyle \frac{8}{33}$
  2. $\displaystyle \frac{8}{3}$
  3. $\displaystyle \frac{-8}{33}$
  4. $\displaystyle \frac{-6}{33}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $\displaystyle \frac{x^2\, -\, (x\, +\, 1)(x\, +\, 2)}{5x\, +\, 1}\, =\, 6$
$\Rightarrow x^2\, -\, (x^2\, +\, 3x\, +\, 2)\, =\, 6(5x\, +\, 1)$, .....(on cross multiplying )
$\Rightarrow x^2\, -\, x^2\, -\, 3x\, -\, 2\, =\, 30x\, +\, 6$
$\Rightarrow -3x - 2 = 30x + 6$
$\Rightarrow -3x - 30x = 6 + 2 \Rightarrow -33x = 8$
$\Rightarrow x\, =\, \displaystyle \frac{-8}{33}$
Hence, the solution is, $x=-\cfrac{8}{33}$.

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

After receiving two successive raises Hrash's salary became $\dfrac {15}{8}$ times of his initial salary. By how much percent was the salary raised the first time if the second raise was twice as much as high (in percent) as the first ?

  1. $15 \%$
  2. $20 \%$
  3. $25 \%$
  4. $30 \%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let initial salary was Rs. $100$
After two raise it become $\dfrac {15}{8}$ i.e. $\dfrac {(15 \times 100) }{ 8} =$ Rs. $187.5$
Raise $= 187.5 - 100 = 87.5$
Using formula,
[( first raise  + second raise) + (first raise * 2nd raise) / 100]  = 87.5
$x + 2x +\dfrac { (2x ^2)}{100} = 87.53$
$300x + 2x ^2 = 8750$
$x ^2 + 150x = 4375$
$x ^2 + 150x - 4375 = 0$
$x ^2 + 175x - 25x - 4375 = 0$
$x = -175, 25$ .....(Negative value is not possible)
So, $x= 25\%$
Second raise was $2x = 2 \times 25 = 50\%$.

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

A brand new car costs $ \$35,000$. For the first $50,000$ miles, it will depreciate approximately $\$0.15$ per mile driven. For every mile after that, it will depreciate by $\$0.10$ per mile driven until the car reaches its scrap value. Find the net worth of the car after it is driven $92,000$ miles.

  1. $\$11,300$
  2. $\$13,800$
  3. $\$17,000$
  4. $\$23,300$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given the cost of car $=\$35000$

First $50000$ miles it will depreciated app $\$ 0.15$ per mile 
And After that it will depreciate by $\$0.10$ per miles 
Then  depreciate after $50000$ miles $=$  $50000\times $0.15=$7500$
And depreciate after $92000-50000=42000$ miles $= $ $42000\times $0.10=$4200$
Total  depreciate after $92000$ mole $=7500+4200=\$11700$
Then worth after it is driven $92000=35000-11700=\$23300$

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

If  $\sqrt{x+16} = x-4$, then the value of extraneous solution of the above equation is:

  1. $0$
  2. $4$
  3. $5$

  4. There are no extraneous solutions
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $\sqrt{x+16}=x-4$
Squaring both sides, we get
$x+16=(x-4)^{2}$
$\Rightarrow x+16=x^{2}-8x+16$
$\Rightarrow x^{2}-8x-x=16-16$
$\Rightarrow x^{2}-9x=0$
$\Rightarrow x(x-9)=0$
Then $x=0$ or $x=9$
Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

A neighborhood recreation program serves a total $280$ children who are either $11$ years old or $12$ years old. The sum of the children's ages is $3,238$ years. How many $11$ year old children does the recreation program serve?

  1. $54$
  2. $122$
  3. $132$
  4. $158$
  5. $208$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the number of $11$ years children is $x$ and $12$ year children is $y$.

$\therefore  11x+12y=3238$.....(1)
$\Rightarrow x+y=280$
$\Rightarrow y=280-x$
Sustitute the value of $y$ in (1)
$\Rightarrow 11x+12(280-x)=3238$
$\Rightarrow 11x+3360-12x=3238$
$\Rightarrow 11x-12x=3238-3360$
$\Rightarrow x=122$
Hence, number of $11$ year children are $122$.