Linear Equations In One Variable - Class 8 Maths - Extra Questions

Let  page numbers of open pages be  $$x$$ and $$x+1$$

Then according to the question,

$$x+x+1=373$$

$$2x=372$$

$$x=186$$

$$2\sqrt { \dfrac { x }{ a }  } +3\sqrt { \dfrac { a }{ x }  } =\dfrac { b }{ a } +\dfrac { 6a }{ b }$$

Let $$\sqrt{ \dfrac { x }{ a }  }=t$$ then $$\sqrt{ \dfrac { a }{ x }  } =\dfrac { 1 }{ t }$$

$$2t+\dfrac { 3 }{ t } =\dfrac { b }{ a } +\dfrac { 6a }{ b } \\ \dfrac { 2{ t }^{ 2 }+3 }{ t } =\dfrac { { b }^{ 2 }+{ 6a }^{ 2 } }{ ab } \\ 2ab{ t }^{ 2 }+3ab={ b }^{ 2 }t+{ 6a }^{ 2 }t\\ 2ab{ t }^{ 2 }-{ b }^{ 2 }t-{ 6a }^{ 2 }t+3ab=0\\ bt(2at-b)-3a(2at-b)=0\\ (bt-3a)(2at-b)=0\\ t=\dfrac { 3a }{ b } ,\dfrac { b }{ 2a } \\ \sqrt { \dfrac { x }{ a }  } =t\quad \\ \Rightarrow x=a{ t }^{ 2 }$$

For $$t=\dfrac { 3a }{ b }$$

$$x=a{ \left( \dfrac { 3a }{ b }  \right)  }^{ 2 }=\dfrac { 9{ a }^{ 3 } }{ { b }^{ 2 } }$$

For $$t=\dfrac { b }{ 2a }$$

$$x=a{ \left( \dfrac { b }{ 2a }  \right)  }^{ 2 }=\dfrac { { b }^{ 2 } }{ 4a }$$

So the values of $$x$$ are $$\dfrac { 9{ a }^{ 3 } }{ { b }^{ 2 } }$$ and $$\dfrac { { b }^{ 2 } }{ 4a }$$

$${\textbf{Step-1: Solving linear equation in Variable p}}$$

                 $$p – 8 =23$$

                  $${\text{Therefore,}}$$ $$p = 23 + 8$$

                  $$=31$$

$${\textbf{Hence, value of}}$$ $$p = 31$$

Let the three consecutive number is $$x,x+1$$ and $$x+2$$

Then, According to the given question.

$$x+\left( x+1 \right)+\left( x+2 \right)=18$$

$$3x+3=18$$

$$3x=15$$

$$x=5$$

Then,

$$x=5$$

$$x+1=5+1=6$$

$$x+2=5+2=7$$

We have,

$$-x+3+\left( -19 \right)+15+\left( -10 \right)=0$$

$$-x+3-19+15-10=0$$

$$-x-16+5=0$$

$$-x-11=0$$

$$x=-11$$

Hence, the value of $$x$$ is $$-11$$ .

We have,

$$\dfrac{1}{2}\left( x+5 \right)-\dfrac{1}{3}\left( x-2 \right)=4$$

$$\dfrac{3\left( x+5 \right)-2\left( x-2 \right)}{6}=4$$

$$3x+15-2x+4=24$$

$$x+19=24$$

$$x=5$$

Hence, this is the answer.

Let the three integers be x-1 , x and x +1 consecutively. We have to find x. Just add x-1 , x and x+1 and equate to 39.

x-1+x+x+1= 39

3x = 39. 

Divide both sides by 3 and we get

x = 13

first number=x-1=13-1=12

second number=x=13

third number=x+1=13+1=14

Let the number be x.

Writing the statement of problem in equation form

=70 - 5 x = 30

= 70 - 30 = 5 x

=5 x = 40

Dividing both sides by 5, we get

x=8

So,number = 8

$$\textbf{Hint: Use the method of solving linear equation in one variable.}$$

Solution:

$$\textbf{Step 1: Find the L.C.M of the denominator of both the terms in L.H.S}$$

$$\quad\quad\quad$$ We have given: $$\dfrac{{3t - 2}}{4} - \dfrac{{2t + 3}}{3} = \dfrac{2}{3} - t$$             ….(1)

$$\quad\quad\quad$$ LCM of $$3$$ and $$4$$ is $$3\times 4=12$$

$$\textbf{Step 2: Multiply the given equation by the L.C.M obtained}$$

$$\quad\quad\quad$$ Multiplying both sides of eq(1) by 12, we get

$$\quad\quad\quad$$ $$\therefore {\rm{\;}}12\left( {\dfrac{{3t - 2}}{4} - \dfrac{{2t + 3}}{3}} \right) = 12\left( {\dfrac{2}{3} - t} \right)$$

$$\quad\quad\quad$$ $$\Rightarrow 12\left( {\dfrac{{3t - 2}}{4}} \right) - 12\left( {\dfrac{{2t + 3}}{3}} \right) = 12\left( {\dfrac{2}{3} - t} \right)$$

$$\quad\quad\quad$$ $$\Rightarrow 3\left( {3t - 2} \right) - 4\left( {2t + 3} \right) = 8 - 12t$$

$$\quad\quad\quad$$ $$\Rightarrow 9{\rm{t}} - 6 - 8{\rm{t}} - 12 = 8 - 12t$$

$$\textbf{Step 3: Bring the like terms together}$$

$$\quad\quad\quad$$ $$\Rightarrow \left( {9t + 12t - 8{\rm{t}}} \right) = 8 + 18$$

$$\quad\quad\quad$$ $$\Rightarrow 21t - 8t = 8 + 18$$

$$\quad\quad\quad$$ $$\Rightarrow 13t = 26$$

$$\quad\quad\quad$$ $$\Rightarrow t = \dfrac{{26}}{{13}}$$

$$\quad\quad\quad$$ $$\Rightarrow t = 2$$

$$\textbf{Hence, the solution of the given linear equation is}$$   $$t = 2$$ .

As given,the three consecutive multiples of 7 will be of the form:

$$x,x+7$$ and $$x+14$$

Then according to the question,

$$x+x+7+x+14=777$$

$$3x+21=777$$

$$3x=756$$

$$x=\dfrac{756}{3}=252$$

Let the length of the park be $$x$$

Then according to the question,

Breadth= $$x+26$$ and $$2(x+x+26)=400$$ (perimeter is 400m)

$$2x+26=200$$                                            

$$2x=174$$

$$x=87m$$

Let the number be $$x$$

Then according to the question,

$$\dfrac{2}{5}x=\dfrac{x}{7}+36$$

$$\dfrac{2}{5}x-\dfrac{x}{7}=36$$

$$\dfrac{14x-5x}{35}=36$$

$$\dfrac{9x}{35}=36$$

$$\textbf{Step-1: Assume the third side to be equal to some variable & then apply all information's given.}$$

                  $$\text{Let the third side be x m}$$

                  $$\text{Then according to the question, other two sides are (2x-5) meters each.}$$

                  $$\text{Then,}$$ $$2x-5+2x-5+x=55$$               $$[\because$$ $$\text{Perimeter = sum of all sides]}$$

$$\textbf{Step-2: Simplify the above results to get the required unknown.}$$

                  $$\Rightarrow$$ $$5x-10=55$$

                  $$\Rightarrow$$ $$5x=65$$

                  $$\Rightarrow$$ $$x=13m$$

Let the number of 5 rupee notes be $$x$$ ,then

The number of 10 rupee notes = $$90-x$$

Then ,according to the question

$$5x+10(90-x)=500$$

$$5x+900-10x=500$$

$$5x=400$$

$$x=80$$

$${\textbf{Step - 1: Assume one integer and form equation with condition}}{\text{.}}$$

                    $${\text{Suppose the first integer is x}}{\text{.}}$$

                    $${\text{Given that their difference is 36}}{\text{.}}$$

                    $${\text{So , another integer will be x}} - 36$$   $$\quad \quad \text{.......eqn(i)}$$

                    $${\text{We have also given that, when one integer be divided by other the quotient will be 4}}{\text{.}}$$

                    $$\Rightarrow \dfrac{x}{{x - 36}} = 4$$

                    $$\Rightarrow x = 4\left( {x - 36} \right)$$

$${\textbf{Step - 2: Solve the above equation to find the two required integers}}{\text{.}}$$

                    $${\text{So , }}x = 4\left( {x - 36} \right)$$

                    $$\Rightarrow x = 4x - 144.$$

                    $$\Rightarrow 144 = 3x$$

                    $$\Rightarrow x = \dfrac{{144}}{3}$$

                    $$\Rightarrow x = 48$$

                    $${\text{So one is 48, another will be 48}} - 36 = 12$$   $$\quad \quad \textbf{[From eqn (i)]}$$

$${\textbf{Thus, the integers are 48 and 12}}{\text{.}}$$

$${\textbf{Step -1: Assume the number to be }}\mathbf{x}$$

                $${\text{Also, it is given that one number is 16 times the other number}}{\text{.}}$$

                $${\text{Assume that, first number is }}$$ $$x.$$

                $${\text{Then the second number will be}}$$ $$16x.$$

                $${\text{Now, product of these two number is 1296}}{\text{.}}$$

                $$\Rightarrow x \times \left( {16x} \right) = 1296 \ldots \left( 1 \right)$$

$${\textbf{Step -2: Solve equation }}\left(\mathbf 1 \right)$$

                $$\Rightarrow x \times \left( {16x} \right) = 1296$$

                $$\Rightarrow 16{x^2} = 1296$$

                $$\Rightarrow {x^2} = \dfrac{{1296}}{{16}}$$

                $$\Rightarrow {x^2} = \dfrac{{81 \times 16}}{{16}}$$

                $$\Rightarrow {x^2} = 81$$

                $$\Rightarrow x =  \pm 9$$

$${\textbf{Step -3: Find the other number if }}$$ $$\mathbf{x = 9.}$$

                $${\text{Second number}}$$ $$= 9 \times 16$$

                $$\Rightarrow {\text{Second number}}$$ $$= 144$$

$${\textbf{Step -4: Find the other number if }}$$ $$\mathbf {x = 9.}$$

                $${\text{Second number}}$$ $$=  - 9 \times 16$$

                $$\Rightarrow {\text{Second number}}$$ $$=  - 144$$

                $${\text{Thus, two numbers are either 9 and 144 or  - 9 and  - 144}}{\text{.}}$$

$${\textbf{Final Answer: Hence, if first number is 9 then the second number will be 144 and}}$$  

                                $${\textbf{If first number is  - 9 then the second number will be  - 144.}}$$

$${\textbf{Step  - 1: Writing mathematical equation according to question}}$$

                    $${\text{Let Laxmi's age be x,}}$$

                     $${\text{Then according to question, }}$$

                    $$\text{Laxmi's father is 49 years old. He is 4 years older than three time Laxmi's age}$$

                     $${\text{49  =  4  +  3x}}$$

$${\textbf{Step  - 2: Finding Laxmi's age}}$$

                     $$\Rightarrow {\text{ 49  =  4  +  3x}}$$

                     $$\Rightarrow {\text{ 45  =  3x}}$$

                     $$\Rightarrow {\text{ x  =  15}}$$

$${\textbf{Hence, Laxmi's age is 15 years}}{\text{.}}$$

The earning of a labourer in seven days is $${\rm{Rs}}\;672$$ .

The earning of a labourer in one day will be $$\cfrac{{672}}{7} = {\rm{Rs}}\;96$$ .

The earning of a labourer in 18 days will be $$18 \times 96 = {\rm{Rs}}\;1728$$ .

Let $$x$$ be the missing number.

$$296 + x =  - 148$$

$$x =  - 148 - 296$$

$$x =  - 444$$

The cost of $$8\;{\rm{m}}$$ cloth is $${\rm{Rs}}\;250$$ .

The cost of $$1\;{\rm{m}}$$ cloth will be $${\rm{Rs}}\;\cfrac{{250}}{8}$$ .

The cost of $$5.8\;{\rm{m}}$$ cloth will be $$5.8 \times \cfrac{{250}}{8} = {\rm{Rs}}\;181.25$$ .

Let the dfraction is $$\dfrac{x}{y}$$

Then, according to the given question,

  $$\dfrac{x+1}{y+1}=\dfrac{4}{5}$$

  $$\Rightarrow 5x+5=4y+4$$

  $$\Rightarrow 5x-4y+1=0\,\,\,........\,\,\left( 1 \right)$$

Again, according to the given question,

  $$\dfrac{x-1}{y-1}=\dfrac{1}{2}$$

  $$\Rightarrow 2x-2=y-1$$

  $$\Rightarrow 2x=y+1$$

  $$\Rightarrow x=\dfrac{y+1}{2}\,\,\,\,\,.......\,\,\,\left( 2 \right)$$

Put the value of x by equation (2)  in to (1), we get,

  $$5x-4y+1=0$$

  $$\Rightarrow 5\left( \dfrac{y+1}{2} \right)-4y+1=0$$

  $$\Rightarrow 5y+5-8y+2=0$$

  $$\Rightarrow -3y+3=0$$

  $$\Rightarrow y=1$$

Put $$y=1$$ in equation (2) and we get,

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

Hence, fraction is  $$\dfrac{x}{y}=\dfrac{1}{1}$$

$$\\(\dfrac{x-5}{5})+(\dfrac{x-3}{2})=1\\(\dfrac{2(x-5)+5(x-3)}{10})=1\\2x-10+5x-15=10\\7x=35\\\therefore\>x=5\\now(\dfrac{2}{x})+(\dfrac{3p}{45})=1\\(\dfrac{2}{5})+(\dfrac{3p}{45})=1\\(\dfrac{2}{5})+(\dfrac{p}{15})=1\\(\dfrac{6+p}{15})=1\\6+p=15\therefore\>p=9$$

We have,

$$\dfrac{x}{5} \to$$ Defective

Remaining $$= x - \dfrac{x}{5} = \dfrac{{4x}}{5}$$

Damaged no. of dolls $$= \dfrac{{25}}{{100}} \times \dfrac{{4x}}{5} = \dfrac{x}{5}$$

No. of dolls left over were $$= x - \dfrac{x}{5} - \dfrac{{4x}}{5}$$

$$= \dfrac{{3x}}{5}$$

$$\dfrac{{3x}}{5} = 4800$$

$$x=8000$$

Hence,

Original number of dolls is $$8000$$ .

$$let\>the\>numbers\>be\>n,n+2,n+4,n+6,n+8\\then\>$$

$$\\sum=n+(n+2)+(n+4)+(n+6)+(n+8)$$

$$85=5n+20$$

$$13=n+4$$

$$\\\therefore\>n=13-4=9$$

So the numbers are  $$9, 11, 13, 15, 17$$

We have,

Let the total number of flower $$=x$$

At reach Ist temple $$=x-\dfrac{x}{2}=\dfrac{x}{2}$$

At reach second temple

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

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

At reach third temple

$$=\dfrac{x}{4}-\dfrac{x}{4}\times \dfrac{1}{2}$$

$$=\dfrac{x}{4}-\dfrac{x}{8}$$

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

According to given question,

$$\dfrac{x}{8}=3$$

$$\Rightarrow x=24$$

Hence, this is the answer.