Tag: reducing simple equations to simpler form

Questions Related to reducing simple equations to simpler form

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

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

  1. $\displaystyle \frac{8}{33}$

  2. $\displaystyle \frac{8}{3}$

  3. $\displaystyle \frac{-8}{33}$

  4. $\displaystyle \frac{-6}{33}$

Show answer
Correct Option: C
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, $x=-\cfrac{8}{33}$.

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 ?

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

  1. $15 \%$

  2. $20 \%$

  3. $25 \%$

  4. $30 \%$

Show answer
Correct Option: C
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 $2x = 2 \times 25 = 50\%$.

Reduce the following linear equation: $6t - 1 = t - 11$

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

  1. $t=-1$

  2. $t=-2$

  3. $t=-3$

  4. $t=-4$

Show answer
Correct Option: B
Explanation:

$6t - 1 = t - 11$
On transposing $t$ to the L.H.S.,  we obtain
$6t - t = -11 + 1$
$5t = -10$
$t = -2$

Solve the equation: $\dfrac{a+4}{6-3a}=\dfrac{1}{3}$

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

  1. $a=1$

  2. $a=-1$

  3. $a=2$

  4. $a=-2$

Show answer
Correct Option: B
Explanation:

Given, $\dfrac{a+4}{6-3a}=\dfrac{1}{3}$


On multiplying $6 - 3a$ on both sides, we get
$a+4=\dfrac{6-3a}{3}$
$3a + 12 = 6 - 3a$
$6a = 6 - 12$
$6a = -6$
$a = -1$

Solve the linear equation: $5x - 12 = 2x + 18$

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

  1. $x=8$

  2. $x=9$

  3. $x=10$

  4. $x=11$

Show answer
Correct Option: C
Explanation:

$5x - 12 = 2x + 18$
On transposing $2x$ to the L.H.S and $12$ to RHS, we obtain
$5x - 2x = 18 + 12$
$3x = 30$
$x = 10$

Reduce the linear equation: $x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$

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

  1. $x=12$

  2. $x=10$

  3. $x=8$

  4. $x=-6$

Show answer
Correct Option: D
Explanation:

Given, $x + 3-\dfrac{2x}{3}+\dfrac{x}{6}=0$
L.C.M of the denominator $3$ and $6$ is $6$.
Multiplying both the sides by $6$, we get
$6x + 18 - 4x + x = 0$
$7x - 4x + 18 = 0$
$3x = -18$
$x = -6$

Reduce the linear equation: $\dfrac{x}{2}+\dfrac{2x}{4}= 10$

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

  1. $x=6$

  2. $x=7$

  3. $x=8$

  4. $x=10$

Show answer
Correct Option: D
Explanation:

Given, $\dfrac{x}{2}+\dfrac{2x}{4}= 10$
L.C.M of the denominator $2$ and $4$ is $4$.
Multiplying both the sides by $4$, we get
$2x + 2x = 40$
$4x = 40$
$x = 10$

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.

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

  1. $\$11,300$

  2. $\$13,800$

  3. $\$17,000$

  4. $\$23,300$

Show answer
Correct Option: D
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 $92000=35000-11700=\$23300$

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

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

  1. $0$

  2. $4$

  3. $5$

  4. There are no extraneous solutions

Show answer
Correct Option: A
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 $x=9$

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?

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

  1. $54$

  2. $122$

  3. $132$

  4. $158$

  5. $208$

Show answer
Correct Option: B
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 $122$.