MCQ Questions for Class 7 Maths Simple Equations Quiz 3

$20 = 6 + 2x$ , then the value of

$x$ is:

0%
$6$
0%
$9$
0%
$8$
0%
$7$

Explanation

Given,

$20=6+2x$

Subtract

$6$ on both the sides,

$20-6=6+2x-6$

$14=2x$

Divide

$2$ on both the sides, we get

$\dfrac {14}{2} = \dfrac { 2x}{ 2}$

$\Rightarrow x=7$

Hence the answer is option

$D$ .

If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation:

0%
changes
0%
remains the same
0%
changes in case of multiplication only
0%
changes in case of division only

Explanation

Let us take an equation

$x+2y=5$ .....(1)
Put

$x=1$ in eqn (1)

$\Rightarrow y=2$
So,

$x=1,y=2$ is a solution of the equation
Multiplying both sides of equation 1() by

$3$ , we get

$3x+6y=15$
Put

$x=1, y=2$ in LHS
LHS

$=3\times 1+6\times 2=15=$ RHS
Hence, on multiplying an equation by a non-zero number , the solution remains same.

Let us divide both sides of equation (1) by

$2$

$\Rightarrow \displaystyle \frac{x}{2}+y=\frac{5}{2}$
Put

$x=1,y=2$ in above equation
LHS

$=\displaystyle \frac{1}{2}+2=\frac{5}{2}=$ RHS
Hence, on dividing or multiplying an equation by a non-zero number, the solution remains the same.

If the mean of observations

$x, x + 2, x + 4, x + 6$ and

$x + 8$ is

$11$ , find the value of

$x$ :

0%
$8$
0%
$5$
0%
$6$
0%
$7$

Explanation

Mean of observations

$x, x + 2, x + 4, x + 6, x + 8$ is

$11$
Mean

$= \cfrac{\text{Sum}}{\text{Number of observation}}$
Mean

$= \cfrac{x + x +2 + x + 4 +x + 6 + x + 8}{5} = 11$

$\cfrac{5x + 20}{5} = 11$

$5x + 20 = 55$

$5x = 55 - 20$

$5x = 35$

$x = 7$

Hence the answer is option

$D$

$x = 5, y = 2$ is a solution of the linear equation

0%
$x + 2 y = 7$
0%
$5x + 2y = 7$
0%
$x + y = 7$
0%
$5 x + y = 7$

Explanation

$\textbf{Step 1: Check which of the equation passes through (5,2).}$

$\text{The solution should satisfy the appropriate equation.}$

$\text{So, let us investigate each option by substituting }x=5$ &

$y=2$

$(A)$

$x+2y=7$ .

$\text{L.H.S. }=5+2\times 2=9\neq$

$\text{R.H.S.}$

$\therefore$

$\text{It is not the appropriate equation.}$

$(B)$

$5x+2y=7$

$\text{L.H.S. }=5\times 5+2\times 2=29\neq$

$\text{R.H.S.}$

$\therefore$

$\text{It is not the appropriate equation.}$

$(C)$

$x+y=7$

$\text{L.H.S. }=5+2=7=$

$\text{R.H.S}$

$\therefore$

$\text{It is the appropriate equation.}$

$(D)$

$5x+y=7$

$\text{L.H.S. }=5\times 5+2=27\neq$

$\text{R.H.S.}$

$\therefore$

$\text{It is not the appropriate equation.}$

$\textbf{Hence, option C is correct.}$

State true or false:

If we multiply both sides of an equation by zero, then we can create an equivalent equation.

0%
True
0%
False

Explanation

Let the equation be

$y=7$

By multiplying zero to both sides, we will get

$y \times 0 = 7 \times 0$

$0=0$
Hence, this is a true statement.

The average of 11, 12, 13, 14, and x isThe value of x is

0%
$17$
0%
$21$
0%
$15$
0%
None

Explanation

$\displaystyle \frac{11+12+13+14+x}{5}=13$

$\displaystyle \frac{50+x}{5}=13$

$50+x = 65$

$x = 65-50$

$x = 15$

Hence, the correct option is

$C$

State true or false:

The statement "

$x$ multiplied by

$5$ and added to

$10$ " can be represented as

$10+5x$

0%
True
0%
False

Explanation

Here, the given statement is

$x$ multiplied by

$5$ and added to

$10$ .

The word 'multiplied by' suggests multiplication that is

$x$ multiplied by

$5$ means

$x\times 5=5x$ and the word 'added to' suggests addition. So, the verbal expression

$x$ multiplied by

$5$ and added to

$10$ can be represented by the algebraic expression

$10+5x$ .

Hence, the statement

$x$ multiplied by

$5$ and added to

$10$ can be rewritten as

$10+5x$ .

So,

$TRUE$

The average of

$11, 12, 13, 14$ and x is

$13$ . The value of x is _____________.

0%
$17$
0%
$21$
0%
$15$
0%
$20$

Explanation

The average of

$11, 12, 13, 14$ and

$x$ is

$13$

$\therefore$

$\dfrac{11+12+13+14+x}{5}=13$

$\therefore$

$50+x=65$

$x=15$

Hence, correct option is

$C$ .

The average of four consecutive even numbers is

$15$ . The

$2$ nd highest number is

0%
$12$
0%
$18$
0%
$14$
0%
$16$

Explanation

Let the numbers be

$x - 2,x,x + 2,x + 4$

According to the question,

$\dfrac{x-2 + x + x + 2 + x + 4}{4} = 15$

$4x + 4 = 60$

$4x = 56$

$\Rightarrow x = 14$

Hence, the second highest number i.e.

$x + 2 = 14 + 2 = 16$

If the mean of

$x, x+2, x+4, x+6, x+8$ is

$20$ then

$x$ is

0%
$32$
0%
$16$
0%
$8$
0%
$4$

Explanation

mean=

$\dfrac{x+x+2+x+4+x+6+x+8}{5}=20$

$5x=80$

$x=16$

Convert this statement in mathematical form.
The number is

$8$ greater than

$80.$

0%
$x+8=80$
0%
$x-8=80$
0%
$8x=80$
0%
$x=80\times8$

Explanation

Let the number be

$x$ . Here, the given statement is the number is

$8$ greater than

$80$ .

The word 'greater than' suggests addition. So, the verbal expression the number is

$8$ greater than

$80$ can be represented by the algebraic expression:

$x=80+8\\ \Rightarrow x-8=80$

Hence, the statement the number is

$8$ greater than

$80$ can be rewritten as

$x-8=80$ .

$6$ more than twice the number is

$22$ . Express the statement in mathematical form.

0%
$2x+6=22$
0%
$2x-6=22$
0%
$2x+22=6$
0%
$2x-22=6$

Explanation

Let the number be

$x.$

Twice the number

$=2x$

Six more than

$2x$ will be

$6+2x.$

This is equal to

$22.$

Hence, the answer is

$6+2x=22.$

The weight of Rahul is

$5\ kg$ more than Sachin .The weight of Samir is

$12\ kg$ less than double the weight of Sachin .If the total weight of all the three is

$93$ form the equation

0%
$4x-7=93$
0%
x+5=0
0%

$x+ (x+5)+ (2x-12)=93$
0%
2x-12=21

Explanation

Let weight of sachin

$=x \ kg$

Acc. to question.

Weight of Rahul

$=(x+5)\ kg$

Weight of Samir

$=(2x-12)\ kg$

Total weight of all

$=93\ kg$

That is ,

$x+ (x+5)+ (2x-12)=93$

$4x-7=93$

$6$ ,

$p$ ,

$12$ ,

$8$ and

$9$ mean of the data is

$9$ then

$p=$ ?

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

Arithmetic mean,

$A=\dfrac { S }{ N } =\dfrac { 6+p+12+8+9 }{ 5 } =9$

$\dfrac { 35+p }{ 5 } =9$

$35+p=45$

$p=45-35=10$

$\textbf{State whether the statement is True or False.}$

$5$ is the solution of the equation

$3x+2=17$ .

0%
True
0%
False

Explanation

Consider the given equation,

$3 x+2=17$
On transposing

$2$ to RHS.
We get,

$\Rightarrow 3 x=17-2$

$\Rightarrow 3 x=15$

$\Rightarrow x=\frac{15}{3}$

$\Rightarrow x=5$

$\therefore 5 \textbf{ is the solution of the equation } 3 x+2=17$ .

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

State true or false:

$4x-5=7$ does not have an integer as its solution.

0%
True
0%
False

Explanation

Given linear equation is

$4x-5=7$

Adding

$5$ on both sides, we get:

$4x-5+5=7+5$

$\Rightarrow 4x=12$

Dividing the complete equation by

$4$ , we get:

$x=\dfrac{12}{4}$

$\therefore \ x=3$

Hence, the solution of this equation is

$3$ , which is an integer.

Therefore, the given statement is false.

If the mean of

$2, 4, 6, 8, x, y$ is

$5$ , then find the value of

$x + y$ ?

0%
$6$
0%
$8$
0%
$10$
0%
$12$

Explanation

Mean

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

$\Rightarrow \cfrac{2+4+6+8+x+y}{6}=5$

$\cfrac{20+x+y}{6}=5$

$20+x+y=5(6)$

$x+y=30-20=10$

State true or false.

$x=5$ is the solution of the equation

$3x-2=13$ .

0%
True
0%
False

Explanation

Given,

$13=3x-2$

Add

$2$ on both the sides we get,

$13+2=3x-2+2$

$15=3x$

Divide both sides by

$3$ we get,

$\dfrac {15}{3} = \cfrac { 3x }{ 3 }$

$\therefore x =5$

Hence, the given statement is True.

$16t-4=0$ then find

$3t$

0%
$\dfrac{3}{4}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{9}{4}$
0%
$1$

Explanation

$16t-4=0$

Add

$4$ both the sides

$\Rightarrow$

$16t=4$

Multiply both the sides by

$16$

$\Rightarrow$

$t=\dfrac 4{16}$

$\Rightarrow$

$t=\dfrac 14$

$\Rightarrow$

$3t=\dfrac 34$

Find

$x$ , if

$6-\left[ 5-\left\{ x-\left( 2-\dfrac { 3 }{ 2 } \right) \right\} \right] =3$

0%
$\dfrac {5}{2}$
0%
$\dfrac {3}{2}$
0%
$\dfrac {7}{2}$
0%
$\dfrac {9}{2}$

Explanation

Given,

$6-\left[ 5-\left\{ x-\left( 2-\dfrac { 3 }{ 2 } \right) \right\} \right] =3$

or,

$6-\left[ 5-\left\{ x-\dfrac { 1 }{ 2 } \right\} \right] =3$

or,

$6-\left[ 5- x+\dfrac { 1 }{ 2 } \right] =3$

or,

$6-5+ x-\dfrac { 1 }{ 2 } =3$

or,

$1+ x-\dfrac { 1 }{ 2 } =3$

or,

$x+\dfrac { 1 }{ 2 } =3$

or,

$x =3-\dfrac{1}{2}$

or,

$x=\dfrac{5}{2}$ .

CBSE Questions for Class 7 Maths Simple Equations Quiz 3 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Simple Equations Quiz 3 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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