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CBSE Questions for Class 7 Maths Simple Equations Quiz 1 - MCQExams.com
CBSE
Class 7 Maths
Simple Equations
Quiz 1
Solve for $$x$$: $$3x+9=33$$
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$$8$$
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$$3$$
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$$2$$
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$$5$$
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:
$$y+3 = 10$$
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$$3$$
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$$7$$
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$$10$$
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$$13$$
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.
If $$\cfrac{t}{5} = 10$$, then $$t$$ is equal to
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$$25$$
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$$10$$
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$$50$$
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$$100$$
Explanation
Given, $$ \dfrac { t }{ 5 } =10$$
To eliminate $$5$$ from the denominator, we multiply both sides by $$5$$
So, $$5 \times \dfrac {t}{5}=10\times 5$$
$$\Rightarrow t=50$$
I
f $$\dfrac{2x}{3}= 18$$, then $$x $$ is equal to
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$$36$$
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$$54$$
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$$32$$
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$$27$$
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:
$$x-2 = 7$$
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$$7$$
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$$2$$
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$$9$$
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$$11$$
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.
$$\text{If } 6x=12, \text{ then } x \text{ is}$$
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$$12$$
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$$2$$
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$$72$$
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$$6$$
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?
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$$60$$
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$$30$$
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$$15$$
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$$10$$
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:
$$6 = z+2$$
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$$4$$
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$$2$$
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$$6$$
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$$8$$
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
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$$4$$
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$$3$$
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$$0$$
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$$2$$
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 $$x=3$$.
If $$7x-9 = 16$$, then $$x$$ is equal to:
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$$x = \dfrac{16}{2}$$
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$$x = \dfrac{7}{7}$$
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$$x = \dfrac{25}{7}$$
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$$x = \dfrac{2}{16}$$
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 $$x$$: $$x-7=-8$$
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$$1$$
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$$-1$$
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$$0$$
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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
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$$x+6$$
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$$x+3$$
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$$x+9$$
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$$x+12$$
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
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$$3$$
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$$4$$
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$$5$$
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$$6$$
Explanation
A
rithmetic
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 $$39$$.
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$$\displaystyle 3x-16=39$$
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$$\displaystyle 3x+16=-39$$
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$$\displaystyle -3x+16=39$$
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$$\displaystyle 3x+16=39$$
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
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Rrror method
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Trial and error method
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Testing method
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Checking method
Explanation
The required method is called "Trial and error method"
Sunita's mother is 36 years old. She is 3 years older than 3 times sunita's age. What is sunita's age?
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6
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7
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8
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11
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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 $$89$$.
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$$14y+9=89$$
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$$14+9y=89$$
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$$14-9y=-89$$
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$$14+y=89$$
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 $$40$$
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$$6a - 40$$
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$$40a - 6$$
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$$40 - 6a$$
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$$0$$
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?
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$$=$$
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$$\displaystyle \neq $$
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$$\displaystyle >$$
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$$\displaystyle <$$
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.
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$$2y + 119 =373$$
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$$3y - 119=373$$
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$$2y - 119=373$$
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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.
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