MCQ Questions for Class 7 Maths Simple Equations Quiz 1

Solve for

$x$ :

$3x+9=33$

0%
$8$
0%
$3$
0%
$2$
0%
$5$

Explanation

Given,

$3x+9=33$

Subract 9 on both side

$\Rightarrow 3x+9-9=33-9$

$\Rightarrow \displaystyle 3x=24$

On dividing throughout by

$3$ , we get

$\Rightarrow x=\dfrac{24}{3}=8$
Hence, required solution is

$x=8$ .

Solve the following equation:

$y+3 = 10$

0%
$3$
0%
$7$
0%
$10$
0%
$13$

Explanation

Given,

$y+3=10$

Subtracting

$3$ on both the sides, we get:

$y+3-3=10-3$

$\Rightarrow y=7$

Hence, option B is correct.

$\cfrac{t}{5} = 10$ , then

$t$ is equal to

0%
$25$
0%
$10$
0%
$50$
0%
$100$

Explanation

Given,

$\dfrac { t }{ 5 } =10$

To eliminate

$5$ from the denominator, we multiply both sides by

$5$

So,

$5 \times \dfrac {t}{5}=10\times 5$

$\Rightarrow t=50$

$\dfrac{2x}{3}= 18$ , then

$x$ is equal to

0%
$36$
0%
$54$
0%
$32$
0%
$27$

Explanation

Given,

$\cfrac { 2x }{ 3 } =18$

Multiply

$3$ on both the sides, we get

$2x = 18\times 3$

$\Rightarrow 2x= 54$

Divide both side by 2

$\Rightarrow x =27$

Solve the following equation:

$x-2 = 7$

0%
$7$
0%
$2$
0%
$9$
0%
$11$

Explanation

Given,

$x-2=7$

Add

$2$ on both the sides, we get

$x-2+2=7+2$

$\Rightarrow x=9$

Hence, option C is correct.

$\text{If } 6x=12, \text{ then } x \text{ is}$

0%
$12$
0%
$2$
0%
$72$
0%
$6$

Explanation

Given,

$6x =12$

Dividing both sides by

$6$ , we get:

$\dfrac {6x}{6} = \cfrac { 12 }{ 6 }$

$\therefore \ x =2$

Hence, option B is correct.

When

$6$ is subtracted from four times a number, the result is

$54$ . What is the number?

0%
$60$
0%
$30$
0%
$15$
0%
$10$

Explanation

Let the number be

$x$ .

Then,

$4x-6=54$

Add 6 on both side

$\Rightarrow \displaystyle 4x-6+6=54+6=>4x=60$

Divide throughout by 4

$\Rightarrow x=\dfrac{60}{4}=15$
Hence, the required number is

$15$ .

Solve the following equation:

$6 = z+2$

0%
$4$
0%
$2$
0%
$6$
0%
$8$

Explanation

Given,

$6=z+2$

Subtract

$2$ on both the sides,

$6-2=z+2-2$

$\Rightarrow z=4$

Hence, option A is correct.

The solution of the equation

$7+3(x+5)=31$ is

0%
$4$
0%
$3$
0%
$0$
0%
$2$

Explanation

Given,

$7+3(x+5)=31$

$\Rightarrow 7+3x+15=31$

$\Rightarrow 22+3x=31$

$\Rightarrow \displaystyle 3x=31-22=9$

$\Rightarrow x=\dfrac{9}{3}=3$ .
Hence, required solution is

$x=3$ .

$7x-9 = 16$ , then

$x$ is equal to:

0%
$x = \dfrac{16}{2}$
0%
$x = \dfrac{7}{7}$
0%
$x = \dfrac{25}{7}$
0%
$x = \dfrac{2}{16}$

Explanation

Given,

$7x - 9 = 16$

Add

$9$ on both the sides, we get

$7x-9+9=16+9$

$\therefore 7x =25$

Divide

$7$ on both the sides,

$\therefore x = \dfrac { 25 }{ 7 }$

Therefore, C is the correct answer.

Find the value of

$x$ :

$x-7=-8$

0%
$1$
0%
$-1$
0%
$0$
0%
None of the above

Explanation

$x-7=-8$

Add

$7$ to both sides,

$\Longrightarrow x-7+7=-8+7$

$\Longrightarrow x=-8+7$

$\Longrightarrow x=-1$

Hence, answer is option

$B$ .

The mean of

$x, x+3, x+ 6, x+9$ and

$x+12$ is

0%
$x+6$
0%
$x+3$
0%
$x+9$
0%
$x+12$

Explanation

By definition,
Average

$\:\displaystyle =\frac{x+\left ( x+3 \right )+\left ( x+6 \right )+\left ( x+9 \right )+\left ( x+12 \right )}{5}$

$\displaystyle =\frac{5x+30}{5}=x+6$

The arithmetic mean of 6,10, x and 12 is 8 The value of x is

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

Arithmetic Mean

$=\cfrac{{Sum \; of\; the\; terms}}{{Total\; number\; of\; terms}}$

$\Rightarrow 8=\dfrac{6+10+x+12}{4}$

$\Rightarrow 32=6+10+x+12$

$\Rightarrow x=32-28$

$\Rightarrow x=4$

Frame the statement into an equation:

Adding

$16$ to

$3$ times

$x$ is

$39$ .

0%
$\displaystyle 3x-16=39$
0%
$\displaystyle 3x+16=-39$
0%
$\displaystyle -3x+16=39$
0%
$\displaystyle 3x+16=39$

Explanation

$3 \times x=3x$

As per the given condition,

Adding

$16$ to

$3x$ results in

$16+3x=39$

So, option

$D$ is correct.

The method of finding solution by trying out various values for the variable is called

0%
Rrror method
0%
Trial and error method
0%
Testing method
0%
Checking method

Sunita's mother is 36 years old. She is 3 years older than 3 times sunita's age. What is sunita's age?

0%
6
0%
7
0%
8
0%
11
0%
14

Explanation

We do not know sunita’s age.

Let us take it to be

$y$ years. .

Sunita’s mother’s age is 3years older than

$3y$ ;

It is also given that Sunita’s mother is 36 years old.

Therefore,

$3y + 3 = 36$

Subract

$3$ from both side

$3y+3-3=36-3$

$3y=33$

Divide both side by 3

$y=11$

Convert the statement into an equation : Adding

$14$ to

$9$ times

$y$ is

$89$ .

0%
$14y+9=89$
0%
$14+9y=89$
0%
$14-9y=-89$
0%
$14+y=89$

Explanation

According to the question,

$9$ times

$y$ means

$9 \times y = 9y$ .

Adding

$14$ to it means

$9y+14$ .

$\therefore$ the equation is

$9y+14=89$ .

Algebraic expression for the statement:

$6$ times

$a$ taken away from

$40$

0%
$6a - 40$
0%
$40a - 6$
0%
$40 - 6a$
0%
$0$

Explanation

Given,

$6$ times

$a$ takes away from

$40$ .

Required algebraic equation is given by,

$40-6\times a=40-6a$

Which of the sign should always be there in an equation?

0%
$=$
0%
$\displaystyle \neq$
0%
$\displaystyle >$
0%
$\displaystyle <$

Explanation

An equation is a mathematical statement that two things are equal. It consists of two expressions, one on each side of an 'equals' sign. For example

$x=y$ is an equation where two expressions

$x$ and

$y$ are equal.

Hence,

$=$ sign should always be there in an equation.

Frame the statement "Twice a number decreased by

$119$ equals

$373$ " into an equation.

0%
$2y + 119 =373$
0%
$3y - 119=373$
0%
$2y - 119=373$
0%
None of the above.

Explanation

Let

$y$ be the number.

According to the question,
Twice a number

$y$ , decreased by

$119$ means

$2y-119=373$ .

So, option

$C$ is correct.

CBSE Questions for Class 7 Maths Simple Equations Quiz 1 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Simple Equations Quiz 1 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

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