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Linear Equations in One Variable Test - 23

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Linear Equations in One Variable Test - 23
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  • Question 1
    1 / -0
    Solve the equation: $$\dfrac{2z}{1-z}=6$$
    Solution
    Given, $$\dfrac{2z}{1-z}=6$$
    On multiplying both sides by $$1 - z$$, we get
    $$2z = 6(1 - z)$$
    $$2z = 6 - 6z$$
    $$8z = 6$$
    $$z = \dfrac{3}{4}$$
  • Question 2
    1 / -0
    If $$\dfrac {3n}{2} = \dfrac {4n + 3}{3}$$, then $$n =$$?
    Solution
    Given, $$\dfrac { 3n }{ 2 } =\dfrac { 4n+3 }{ 3 } $$

     $$\Rightarrow 3(3n)=2(4n+3)$$  …[Cross-multiplying the denominators]

     $$\Rightarrow 9n=8n+6$$

    Transposing $$n$$ terms to one side, we get,
     $$\Rightarrow 9n-8n=6$$

     $$\Rightarrow n=6$$.

    Hence, option $$A$$ is correct.
  • Question 3
    1 / -0
    Solve the equation: $$-(x-5)=x-2$$.
    Solution
    Given, $$-\left( x-5 \right) =x-2$$
    $$\implies$$ $$-x+5=x-2$$  ...[By Distribution Law].

    Transposing $$x$$ terms to one side, we get,
    $$\implies$$ $$x+x=5+2$$
    $$\implies$$ $$2x=7$$
    $$\implies$$ $$x=\dfrac { 7 }{ 2 } $$.

    Hence, option $$A$$ is correct.
  • Question 4
    1 / -0
    Solve the equation: $$-2x+6=4x-2$$.
    Solution
    Given, $$-2x+6=4x-2$$.

    Transposing $$x$$ terms to one side, we get,
    $$4x+2x=6+2$$

    $$\Rightarrow 6x=8$$

    $$\Rightarrow x=\dfrac { 8 }{ 6 } $$

    $$\Rightarrow x=\dfrac { 4 }{ 3 } $$.

    Hence, option $$E$$ is correct.
  • Question 5
    1 / -0
    Three more than twice a integer is equal to $$4$$. What is the integer? 
    Solution
    Let the number be $$'n'$$
    Twice the number gives $$'2n'$$
    Three more than twice the number is $$2n+3$$
    Given that three more than twice the number $$=$$ $$4$$
    $$\Rightarrow 2n$$ $$+$$ $$3$$ $$=$$ $$4$$
    $$\Rightarrow 2$$ $$\times$$ $$n$$ $$=$$ $$4$$ $$-$$ $$3$$
    $$\Rightarrow n$$ $$=$$ $$\dfrac {1}{2}$$
    $$\Rightarrow n$$ $$=$$ $$0.5$$

    Therefore, the number is $$0.5$$ 
  • Question 6
    1 / -0
     If $$3$$ times a number is equal to $$\dfrac{3}{2}$$, what is the number? 
    Solution
    Let the number be $$'n'$$
    Given that $$3$$ times the number $$'n'$$ is equal to $$\dfrac {3}{2}$$
    $$\Rightarrow 3$$ $$\times$$ $$n$$ $$=$$ $$\dfrac {3}{2}$$
    $$\Rightarrow n$$ $$=$$ $$\dfrac{\frac {3}{2}}{3}$$

    $$\Rightarrow n$$ $$=$$ $$\dfrac {3}{6}$$

    $$\Rightarrow n$$ $$=$$ $$\dfrac {1}{2}$$
    Therefore, the number is $$'$$$$\dfrac {1}{2}$$$$'$$.
  • Question 7
    1 / -0
    If $$\dfrac { 1 }{ 7 } +\dfrac { x }{ 7 } =3$$, calculate the value of $$x$$.
    Solution
    Given :- $$\frac{1}{7} + \frac{x}{7} = 3$$
     multiplying $$7$$ to both sides
     $$\Rightarrow 1+ x = 21$$
    $$\Rightarrow x = 21-1$$
    $$\Rightarrow x = 20 $$
  • Question 8
    1 / -0
    If $$\displaystyle \frac{a-b}{b}=\frac{3}{7}$$, which of the following must also be true?
    Solution
    Given: $$\displaystyle \frac {a-b}{b}=\frac 37$$
    Separating denominators,
    $$\Rightarrow \displaystyle \frac ab -1=\frac 37$$
    $$\Rightarrow \displaystyle \frac ab=\frac 37+1=\frac {10}{7}$$
    $$\Rightarrow \dfrac {a}{b}=\dfrac {10}{7}$$
    Therefore, option B is correct.
  • Question 9
    1 / -0
    If $$a\neq 0$$ and $$\dfrac{5}{x}=\dfrac{5+a}{x+a}$$, what is the value of $$x$$?
    Solution
    Given:
    $$a$$ $$\neq$$ $$0$$ and $$\dfrac {5}{x}$$ $$=$$ $$\dfrac {5 \space + \space a}{x \space + \space a}$$
    To find the value of $$x$$,
    $$\Rightarrow \dfrac {5}{x}$$ $$=$$ $$\dfrac {5 \space + \space a}{x \space + \space a}$$
    $$\Rightarrow 5$$ $$\times$$ $$(x$$ $$+$$ $$a)$$ $$=$$ $$x$$ $$\times$$ $$(5$$ $$+$$ $$a)$$
    $$\Rightarrow 5x$$ $$+$$ $$5a$$ $$=$$ $$5x$$ $$+$$ $$xa$$
    Get the co-efficients of $$'x'$$ on one-side,
    $$\Rightarrow 5x$$ $$+$$ $$xa$$ $$-$$ $$5x$$ $$=$$ $$5a$$
    $$\Rightarrow xa$$ $$=$$ $$5a$$
    As $$(a$$ $$\neq$$ $$0)$$,    $$x$$ $$=$$ $$5$$
    Therefore, the value of $$'x'$$ is $$'5'$$.
  • Question 10
    1 / -0
    Which of the following equations does not have a solution in integers?
    Solution
    A. $$x+1 = 1$$
    $$\implies x= 1 -1 =0$$, which is an integer

    B. $$x-1 = 3$$
    $$\implies x= 3+1 =4$$, which is an integer

    C.$$2x+1 = 6$$
    $$\implies 2x= 6-1 =5$$
    $$\implies x= \dfrac{5}{2}$$, which is not an integer

    D.$$1-x = 5$$
    $$\implies -x= 5 -1 =4$$, 
    $$\implies x=-4$$,which is an integer
    $$\therefore$$ Eqn $$2x+1=6$$ does not have solution as an integer
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