Linear Equations in One Variable Test

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Linear Equations in One Variable Test - 40

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Question 1
1 / -0

If $$\cfrac{7}{m-\sqrt{3}} = \cfrac{\sqrt{3}}{m} + \cfrac{4}{2m}$$, calculate the value of $$m$$.

A

$$m=\dfrac { -4\sqrt { 2 } -6 }{ 10-2\sqrt { 3 } }$$

B

$$m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { 3 } }$$

C
$$m=\dfrac { -4\sqrt { 3 } +6 }{ 10-2\sqrt { 3 } }$$

D
$$m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { -3 } }$$

Solution

Given $$\dfrac { 7 }{ m-\sqrt { 3 } } =\dfrac { \sqrt { 3 } }{ m } +\dfrac { 4 }{ 2m } =\dfrac { 4+2\sqrt { 3 } }{ 2m } $$
$$\Rightarrow 14m=(4+2\sqrt { 3 } )m-\sqrt { 3 } (4+2\sqrt { 3 } )$$
$$ \Rightarrow (10-2\sqrt { 3 } )m=-4\sqrt { 3 } -6$$
$$\Rightarrow m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { 3 } }$$
Question 2
1 / -0

If $$3(x + 5) - (x + 2) = 2x - 3x + 4$$, then find the value of $$x$$.

A
$$-3$$

B
$$-1$$

C
$$0$$

D
$$1$$

E
$$3$$

Solution

Given, $$3(x+5)-(x+2)=2x-3x+4$$

$$\Rightarrow 3x+15-x-2=2x-3x+4$$ ...[By Distribution Law]

$$\Rightarrow 2x+13=-x+4$$ ...[On simplifying].

Transposing $$x$$ terms to one side, we get,
$$\Rightarrow 2x+x=4-13$$

$$\Rightarrow 3x=-9$$

$$\Rightarrow x=-3$$.

Hence, option $$A$$ is correct.
Question 3
1 / -0

Find the value of $$x: \dfrac {1}{x} + \dfrac {4}{5x} = \dfrac {2}{x + 5}$$

A
$$0.71$$

B
$$3.57$$

C
$$5.8$$

D
$$45$$

Solution

Given $$\dfrac { 2 }{ x+5 } =\dfrac { 1 }{ x } +\dfrac { 4 }{ 5x } $$
Taking RHS:

$$\dfrac { 1 }{ x } +\dfrac { 4 }{ 5x } $$

LCM is $$5x$$
$$\Rightarrow \dfrac { 5 }{ 5x } +\dfrac { 4 }{ 5x } \\ \Rightarrow \dfrac { 9 }{ 5x } $$
Now taking LHS:
$$\dfrac { 2 }{ x+5 } $$
LHS $$=$$ RHS
$$\dfrac { 2 }{ x+5 } =\dfrac { 9 }{ 5x } $$
$$\Rightarrow 5x\times 2=9(x+5)\\ \Rightarrow 10x=9x+45\\ \Rightarrow x=45$$
Question 4
1 / -0

Find the value of $$\dfrac {4}{y} + 4$$ given that $$\dfrac {4}{y} + 4 = \dfrac {20}{y} + 20$$

A
$$-1$$

B
$$0$$

C
$$1$$

D
$$4$$

Solution

The value of $$\frac{20}{y}+20=5\left ( \frac{4}{y}+4 \right )$$
Given $$\frac{4}{y}+4=5\left ( \frac{4}{y}+4 \right )$$
Given one is possible only when the value of $$\frac{4}{y}+4=0$$
Question 5
1 / -0

If $$\dfrac{3}{9}=\dfrac{3}{x+2}$$, what is the value of $$x$$?

A
$$-\dfrac{5}{9}$$

B
$$\dfrac{7}{3}$$

C
$$3$$

D
$$7$$

E
None of the above

Solution

Given, $$\dfrac {3}{9}=\dfrac {3}{x+2}$$
On cross multiplying, we get
$$\Rightarrow 3(x+2)=9(3)$$
$$\Rightarrow (x+2)=(3)$$
$$\Rightarrow x=9-2$$
$$\Rightarrow x=7$$
Question 6
1 / -0

If $$\dfrac {2}{3x + 12} = \dfrac {2}{3}$$, then the value of $$x + 4 $$ is

A
$$\dfrac {1}{2}$$

B
$$1$$

C
$$\dfrac {3}{2}$$

D
$$2$$

Solution

Given $$\dfrac { 2 }{ 3x+12 } =\dfrac { 2 }{ 3 } $$
$$\Rightarrow 3(2)=2(3x+12)$$
$$\Rightarrow 6=6x+24$$
$$\Rightarrow 6x=-18$$
$$\Rightarrow x=-3$$
Therefore $$x+4=-3+4=1$$
Question 7
1 / -0

If $$13$$ is added to one-half of a certain number, the result is $$37$$. What is the original number?

A
$$24$$

B
$$40$$

C
$$48$$

D
$$61$$

Solution

Let the original number be $$'x'$$.
Given that $$13$$ is added to one half of $$'x'$$ and the result is $$37$$.
In the Mathematical Equation form, we can write the above statement as
$$\dfrac {x}{2}$$ $$+$$ $$13$$ $$=$$ $$37$$
$$\Rightarrow \dfrac {x}{2}$$ $$=$$ $$37$$ $$-$$ $$13$$

$$\Rightarrow \dfrac {x}{2}$$ $$=$$ $$24$$
$$\Rightarrow x$$ $$=$$ $$24$$ $$\times$$ $$2$$
$$\Rightarrow x$$ $$=$$ $$48$$
Therefore, the original number is $$'48'$$.
Question 8
1 / -0

If $$10 + x$$ is $$5$$ more than $$10$$, what is the value of $$2x$$ ?

A
$$-5$$

B
$$5$$

C
$$10$$

D
$$25$$

Solution

Given that
$$10$$ $$+$$ $$x$$ is $$5$$ more than $$10$$
Formally, we can express this
$$10$$ $$+$$ $$x$$ $$=$$ $$10$$ $$+$$ $$5$$
$$\Rightarrow x$$ $$=$$ $$15$$ $$-$$ $$10$$
$$\Rightarrow x$$ $$=$$ $$5$$
To find $$2x$$,
$$2x$$ $$=$$ $$2$$ $$\times$$ $$x$$
$$=$$ $$2$$ $$\times$$ $$5$$
$$=$$ $$10$$
Therefore, the value of $$'2x'$$ is $$'10'$$.
Question 9
1 / -0

A triangle has a perimeter of $$13$$ and one side of length $$3$$. If the lengths of the other two sides are equal, what is the length of each of them?

A
$$4$$

B
$$5$$

C
$$6$$

D
$$7$$

E
$$8$$

Solution

Let $$'x'$$ be the length of one of the equal sides.
The perimeter of the triangle with sides a$$, b$$ and $$c$$ is $$P=a+b+c$$.
Substitute the values pf $$P=13$$ and $$a=3, b=x , c=x$$ in $$P=a+b+c$$ as shown below:
$$\Rightarrow 13=3+x+x$$
$$\Rightarrow 13=3+2x$$
$$\rightarrow 2x=13-3$$
$$\Rightarrow 2x=10$$
$$\Rightarrow x=5$$
Therefore, the length of each of the side is $$5$$.
Hence, option B is correct.
Question 10
1 / -0

A number is doubled and $$9$$ is added. If the result is tripled it becomes $$75$$. What is that number?

A
$$3.5$$

B
$$6$$

C
$$8$$

D
$$7$$

Solution

Let $$x =$$ the number
So according to given conditions,
$$3(2x+9)=75$$
$$\Rightarrow 6x+27=75$$
$$\Rightarrow 6x=75-27$$
$$\Rightarrow 6x=48$$
$$\Rightarrow x= \dfrac{48}{6}$$
$$\Rightarrow x= 8$$

Hence. the number is $$8$$.

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Linear Equations in One Variable Test - 40

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Linear Equations in One Variable Test - 40

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Question 1

$$m=\dfrac { -4\sqrt { 2 } -6 }{ 10-2\sqrt { 3 } }$$

$$m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { 3 } }$$

$$m=\dfrac { -4\sqrt { 3 } +6 }{ 10-2\sqrt { 3 } }$$

$$m=\dfrac { -4\sqrt { 3 } -6 }{ 10-2\sqrt { -3 } }$$

Solution

Question 2

$$-3$$

$$-1$$

$$0$$

$$1$$

$$3$$

Solution

Question 3

$$0.71$$

$$3.57$$

$$5.8$$

$$45$$

Solution

Question 4

$$-1$$

$$0$$

$$1$$

$$4$$

Solution

Question 5

$$-\dfrac{5}{9}$$

$$\dfrac{7}{3}$$

$$3$$

$$7$$

None of the above

Solution

Question 6

$$\dfrac {1}{2}$$

$$1$$

$$\dfrac {3}{2}$$

$$2$$

Solution

Question 7

$$24$$

$$40$$

$$48$$

$$61$$

Solution

Question 8

$$-5$$

$$5$$

$$10$$

$$25$$

Solution

Question 9

$$4$$

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 10

$$3.5$$

$$6$$

$$8$$

$$7$$

Solution

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