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CBSE Questions for Class 7 Maths Simple Equations Quiz 3 - MCQExams.com
CBSE
Class 7 Maths
Simple Equations
Quiz 3
If $$20 = 6 + 2x$$, then the value of $$x$$ is:
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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:
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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 $$x$$:
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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
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$$x + 2 y = 7$$
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.
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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
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$$17$$
0%
$$21$$
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$$15$$
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
$$10+5x$$
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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 _____________.
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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
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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
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$$32$$
0%
$$16$$
0%
$$8$$
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$$4$$
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 $$80.$$
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$$x+8=80$$
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.
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$$2x+6=22$$
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
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$$4x-7=93$$
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x+5=0
0%
$$x+ (x+5)+ (2x-12)=93$$
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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 ,
$$x+ (x+5)+ (2x-12)=93$$
$$4x-7=93$$
If $$6$$, $$p$$, $$12$$, $$8$$ and $$9$$ mean of the data is $$9$$ then $$p=$$ ?
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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 $$3x+2=17$$.
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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.
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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 $$x + y $$?
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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 $$3x-2=13$$.
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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.
If $$16t-4=0$$ then find $$3t$$
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$$\dfrac{3}{4}$$
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 $$6-\left[ 5-\left\{ x-\left( 2-\dfrac { 3 }{ 2 } \right) \right\} \right] =3$$
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$$\dfrac {5}{2}$$
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, $$x=\dfrac{5}{2}$$.
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