Free Linear Equations in One Variable 01 Practice Test - 8th Grade 

An algebraic expression consists of operands (variables and numbers) and operators $(+,-,\times ,÷)$.

When an algebraic expression is equated to another algebraic expression (which may be even 0), we get an algebraic equation.

So, 3x - 5y + 6z is not an equation, but an expression.

By substituting n = 10 in the given equation, we can find out the ${10}^{th}$ term i.e.,
$2.5x-10$ = (2.5 $\times$ 10) - 10
                 = 25 - 10
                 = 15

Perimeter of swimming pool $=64\left(Given\right)$

Perimeter of  rectangle $=2(length+breadth)$

Since length and breadth are given to be in the ratio $5:3$, we can assume them to be $5x$ and $3x$ respectively.

$\Rightarrow$ Length = $5x$ and breadth = $3x$

Substituting these in the formula of perimeter, we get
$64=2(5x+3x)\Rightarrow 8x=32\Rightarrow x=4$

$\therefore$ Length of the swimming pool
 $=5x=5\times 4=20m$

$\frac{3}{7}+x=\frac{17}{7}$

$x=\left(\frac{17}{7}\right)–\left(\frac{3}{7}\right)=\frac{14}{7}=2$

Hence, the solution of the equation $\frac{3}{7}+x=\frac{17}{7}$ is 2.

Hence the given statement is true.

$\text{Let, the number which Anita}\text{initially thought of be x.}$

$\text{She subtracts}\frac{3}{2}\text{from x and then}\text{multiplies it by 8.}$ 
 $\Rightarrow (x–\frac{3}{2})\times 8$

$\text{According to the question,}(x–\frac{3}{2})\times 8=5\times x8x–12=5x8x–5x=123x=12x=\frac{12}{3}x=4$

Thus, the number that Anita initially thought of is 4.

Let the digit in units place be $x$.
Then, the digit in tens place will be $2x$.

 So, the number will be
$(10\times 2x)+(1\times x)=20x+x=21x$

When we interchange the digits, the number will be
$(10\times x)+(1\times 2x)=10x+2x=12x$

It is given that,
$21x+12x=66$
$\Rightarrow 33x=66$
$\Rightarrow x=\frac{66}{33}$
$\Rightarrow x=2$

$\therefore$The digit in units place $=2$
The digit in tens place $=4$
The original number is $42$.

Verification:
Original number $=42$
Interchanging the digits of $42$, we get $24$.
Sum of the original number and the interchanged number is $42+24=66$.

$\text{Let Vinay’s present age be}x\text{Then, his mother’s present age}=3x$

$\text{After 5 years,}\text{Vinay’s age =}x+5\text{Vinay's mother’s age =}3x+5$

Sum of Vinay and Vinay's Mother's age after 5 years = 70

$\Rightarrow x+5+3x+5=70\Rightarrow 4x+10=70\Rightarrow 4x=60\Rightarrow x=15\Rightarrow 3x=45$

$\therefore \text{Vinay’s age is 15 years and his}$$\text{mother’s age is 45 years.}$
 

Let Arjun be x years old now

After 20 years, his age will be x + 20
According to question,
x + 20 = 3x or 3x = x + 20
3x - x = 20
2x = 20
x = 10
Therefore, Arjun's present age is 10 yrs.

Let the total number of students be $x$

According to the question,
Number of students dancing $=\frac{x}{2}$

Number of students remaining $=x-\frac{x}{2}$
$=\frac{x}{2}$

Students playing cricket 
$=\left(\frac{3}{4}\right)\left(\frac{x}{2}\right)$
$=\frac{3x}{8}$

Remaining students $=9$

$\Rightarrow (x-\frac{x}{2})-\frac{3x}{8}=9\Rightarrow \frac{x}{8}=9\Rightarrow x=8\times 9\Rightarrow x=72$ 

Hence, the total number of students in the class = 72

$\text{Let the base of the triangle be}xcm.$

$\text{Then, the corresponding altitude}\text{of the triangle would be}\frac{3x}{5}cm.$

$\text{If altitude is increased by 4 cm,}\text{the new altitude}=\frac{3x}{5}+4cm$

$\text{New base}=x-2cm$

$\text{Area of triangle}=\frac{1}{2}\times \text{base}\times \text{height}$

$\text{Since the area remains the same,}\frac{1}{2}\times x\times \frac{3x}{5}=\frac{1}{2}\times (\frac{3x}{5}+4)\times (x-2)$
$\Rightarrow \frac{3x^2}{5}=\frac{3x^2}{5}+4x-\frac{6x}{5}-8$
$\Rightarrow 4x-\frac{6x}{5}=8\Rightarrow \frac{20x-6x}{5}=8$
$\Rightarrow 14x=40$
$\Rightarrow x=\frac{40}{14}=\frac{20}{7}\Rightarrow x=2\frac{6}{7}\text{cm}$ 
$\text{Hence, base of the triangle is}2\frac{6}{7}cm\text{long.}$