MCQ Questions for Class 7 Maths Simple Equations Quiz 4

The value of

$x$ if

$23x+69=0$

0%
$-3$
0%
$0$
0%
$3$
0%
$4$

Explanation

$23x+69=0$

Subract both side by 69

$\Rightarrow$

$23x=-69$

Divide both side by 23

$\Rightarrow$

$x=\dfrac{-69}{23}$

$\Rightarrow$

$x=-3$

$a+5=7$ find

$a$

0%
$2$
0%
$57$
0%
$75$
0%
None of these

Explanation

$a+5=7$

Subract both side by 5

$a+5-5=7-5$

$a=2$

$3x-4=0$ then

$x$ is equal to

0%
$\dfrac{4}{3}$
0%
$-\dfrac{2}{3}$
0%
$-\dfrac{4}{3}$
0%
$-\dfrac{7}{3}$

Explanation

$3x-4=0$

Add both side by 4

$3x-4+4=0+4$

$\Rightarrow$

$3x=4$

Divide both side by

$3$

$\Rightarrow$

$x=\dfrac 43$

$2p=34$ then

$4p =$ ?

0%
$68$
0%
$77$
0%
$91$
0%
$97$

Explanation

$2p=34$

Divide both side by 2

$\Rightarrow$

$p=\dfrac{34}2$

$\Rightarrow$

$p=17$

$4p=4\times 17=68$

State true or false

The equation

$2x + 7 = 2(x + 5)$ has one solution.

0%
True
0%
False

The values of

$r$ which does not satisfy the equation

$r - 4 =0$ is

0%
$4$
0%
$-4$
0%
$8$
0%
$0$

Explanation

$r-4=0$

Solve for all values of

$r$

$\therefore$ for

$r=4$

$4-4=0$

$0=0$

$\therefore$ For

$r=-4$

$-4-4=-8$

$-8\neq 0$

For

$r=8$

$8-4=4$

$4\neq 0$

For

$r=0$

$0-4=-4$

$-4\neq 0$

$\therefore$ it does not satisfy for values

$-4, 8, 0$ .

$\textbf{State whether the statement is true or false}$

$x=5$ is the solution of the equation

$3x+2=20$ .

0%
True
0%
False

Explanation

To check if the given value of

$x=5$ is the solution of the given equation we substitute

$x=5$ in LHS

$LHS=3x+2$

$=3 \times 5+2$

$=17$ which is not equal to the RHS
So,

$x=5$ is not a solution of the given equation.

$\textbf{Hence, the given statement is False.}$

State whether the statement is true (T) or false (F).
An equation is true for all values of its variables.

0%
True
0%
False

One third of a number added to itself give

$8,$ can be represented as

$\dfrac{x}{3}+8=x$

0%
True
0%
False

Explanation

Let the unknown number be x,
Then one third of the number would be

$\frac{1}{3} \times x$ or

$\frac{x}{3}$
Now, one third of

$x$ added to itself would be

$\frac{x}{3}+x$
Finally, One third of a number added to itself gives 8 would be

$\frac{x}{3}+x=8$
Clearly,

$_{3}^{x}+8=x$ and

$\frac{x}{3}+x=8$ are not the same,

$\textbf{Therefore, the above statement is False}$

$\dfrac{3}{2}$ is the solution of the equation

$8x-5=7$

0%
True
0%
False

Explanation

$8x-5=7$

Add

$5$ on both the sides,

$8x-5+5=7+5$

$\Rightarrow 8x=12$

Divide

$6$ on both the sides, we get

$\dfrac {8x}{8} = \cfrac { 12 }{ 8 }$

$\Rightarrow x=\dfrac{3}{2}$

$4x-7=11$ , then

$x=4$

0%
True
0%
False

Explanation

$\Rightarrow 4x-7=11$

Add 7 on both side

$\Rightarrow 4x-7+7=11+7$

$\Rightarrow 4x=18$

Divide throughout by 4

$\Rightarrow x=\dfrac{18}{4}$

$\Rightarrow x=\dfrac{9}{2} \ne { 4}$

The solution of the equation

$3(x - 4)= 30$ is

$x =6$

0%
True
0%
False

Explanation

$x=6$ is the solution then,

$3(x-4)=30$ is satisfied by

$x=6$

LHS

$=3(6-4)=6$

RHS

$=30$

LHS

$\neq$ RHS

$\implies x=6$ is not the solution

$x = -1$ is a solution of the equation

0%
$x - 5 =6$
0%
$2x + 5 = 7$
0%
$2(x - 2) + 6 =0$
0%
$3x + 5 =4$

Explanation

${\textbf{Step -1: Considering option A.}}$

${\text{We have given,}}$ $x - 5 = 1$

${\text{Putting }}x = - 1{\text{ we get,}}$

$- 1 - 5 = 6$

$\Rightarrow - 6 = 1$ ${\text{Which is not possible}}{\text{.}}$

${\text{Here, L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S }}$

${\text{So, - 1 is not a solution of }}x - 5 = 6.$

${\textbf{Step -2: Considering option B.}}$

${\text{We have given,}}$ $2x + 5 = 7$

${\text{Putting }}x = - 1{\text{ we get,}}$

$2\left( { - 1} \right) + 5 = 7$

$\Rightarrow - 2 + 5 = 7$

$\Rightarrow 3 = 7$ ${\text{Which is not possible}}{\text{.}}$

${\text{Here, L}}{\text{.H}}{\text{.S }} \ne {\text{ R}}{\text{.H}}{\text{.S }}$

${\text{So, - 1 is not a solution of 2}}x + 5 = 7.$

${\textbf{Step -3: Considering option C.}}$

${\text{We have given,}}$ $2\left( {x - 2} \right) + 6 = 0$

${\text{Putting }}x = - 1{\text{ we get,}}$

$2\left( { - 1 - 2} \right) + 6 = 0$

$\Rightarrow 2\left( { - 3} \right) + 6 = 0$

$\Rightarrow - 6 + 6 = 0$

$\Rightarrow 0 = 0$ ${\text{Which is possible}}{\text{.}}$

${\text{Here, L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S }}$

${\text{So, - 1 is a solution of 2}}\left( {x - 2} \right) + 6 = 0.$

${\textbf{Step -4: Considering option D.}}$

${\text{We have given,}}$ $3x + 5 = 4$

${\text{Putting }}x = - 1{\text{ we get,}}$

$3\left( { - 1} \right) + 5 = 4$

$\Rightarrow - 3 + 5 = 4$

$\Rightarrow 2= 4$ ${\text{Which is not possible}}{\text{.}}$

${\text{Here, L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S }}$

${\text{So, - 1 is not a solution of }}3x + 5 = 4.$

${\textbf{ Hence, option C}}{\textbf{. 2}}\left(\mathbf {x - 2} \right)\mathbf{ + 6 = 0}{\textbf{ is correct answer.}}$

The solution of the equation

$3x - 5 =2 \, is\, x = \dfrac{7}{3}$

0%
True
0%
False

Explanation

True

$3x - 5 =2$

$\implies\, 3x = 2+ 5 =7$

$\implies\, x =\dfrac{7}{3}$

Which equation among the following don't have

$(-1)$ as a solution?

0%
$x + 1=0$
0%
$3x + 4 =1$
0%
$5x + 7= 2$
0%
$x -1 =2$

Explanation

${\textbf{Step -1: Considering option A.}}$

${\text{We have given,}}$ $x +1 = 0$

${\text{Putting }}x = - 1{\text{ we get,}}$

$- 1 +1 = 0$

$\Rightarrow 0 = 0$ ${\text{Which is possible}}{\text{.}}$

${\text{Here, L}}{\text{.H}}{\text{.S}} = {\text{R}}{\text{.H}}{\text{.S }}$

${\text{So, - 1 is a solution of }}x +1 = 0.$

${\textbf{Step -2: Considering option B.}}$

${\text{We have given,}}$ $3x + 4 = 1$

${\text{Putting }}x = - 1{\text{ we get,}}$

$3\left( { - 1} \right) + 4 = 1$

$\Rightarrow - 3 + 4 = 1$

$\Rightarrow 1 = 1$ ${\text{Which is possible}}{\text{.}}$

${\text{Here, L}}{\text{.H}}{\text{.S }} = {\text{ R}}{\text{.H}}{\text{.S }}$

${\text{So, - 1 is a solution of 3}}x + 4 = 1.$

${\textbf{Step -3: Considering option C.}}$

${\text{We have given,}}$ $5x + 7 = 2$

${\text{Putting }}x = - 1{\text{ we get,}}$

$5\left( { - 1 } \right) + 7 = 2$

$\Rightarrow -5+ 7 = 2$

$\Rightarrow 2 = 2$ ${\text{Which is possible}}{\text{.}}$

${\text{Here, L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S }}$

${\text{So, - 1 is a solution of }}5x+7 =2.$

${\textbf{Step -4: Considering option D.}}$

${\text{We have given,}}$ $x -1 = 2$

${\text{Putting }}x = - 1{\text{ we get,}}$

$-1-1= 2$

$\Rightarrow - 2 = 2$ ${\text{Which is not possible}}{\text{.}}$

${\text{Here, L}}{\text{.H}}{\text{.S}} \ne {\text{R}}{\text{.H}}{\text{.S }}$

${\text{So, - 1 is not a solution of }}x -1 = 2.$

${\textbf{ Hence, option D. }}\mathbf{ x-1 =2}{\textbf{ is correct answer.}}$

$8x-3=25+17x$ , then

$x$ is:

0%
A fraction
0%
An integer
0%
A rational number
0%
Cannot be solved

Explanation

Given,

$8x-3=25+17x$

$\Rightarrow 8x-17x=25+3$ [Transposing

$17x$ to LHS and

$-3$ to RHS]

$\Rightarrow -9x=28$

$\therefore x=\dfrac{-28}{9}$ [Dividing both side by

$-9$ ]
Hence,

$x$ is a rational number.

Therefore, option

$C$ is correct.

$\dfrac{9}{4}$ is a solution of the equation

$5x - 1= 8$ .

0%
True
0%
False

Explanation

$\dfrac{9}{8}$ is a solution then it must satisfy the given equation

so,

From LHS

$5x-1=5\times \dfrac{9}{4}-1$

$=\dfrac{45}{4}-1$

$=\dfrac{45-4}{4}$

$=\dfrac{41}{4}$

$=10.25 \neq 8$

$\implies$ LHS

$\neq$ RHS

Hence

$\dfrac{9}{8}$ is not the solution of the given equation.

Arpita's present age is thrice of Shilpa. If Shilpa's age three years ago was

$x$ . Then Arpita's present age is

0%
$3(x-3)$
0%
$3x+3$
0%
$3x-9$
0%
$3(x+3)$

Select the equations where

$x = 1$ is solution.

0%
$x+3 = 4$
0%
$x-3 = 4$
0%
$3x+3 = 6$
0%
$3x-3 = 6$

$\dfrac{x+6}{3} = 7$ , what is the value of

$x$ ? ( using trail and error method)

0%
$x= 21$
0%
$x= 20$
0%
$x= 15$
0%
$x= 14$

Explanation

We need to find

$x$ such that

$\dfrac{x+6}{3}$ is

$7$ .

We know that

$\dfrac{21}{3}$ is

$7$ .

We thus get

$7$ if

$x+6=21$

$\Longrightarrow$

$x=15$

CBSE Questions for Class 7 Maths Simple Equations Quiz 4 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Simple Equations Quiz 4 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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