Algebra - Class 6 Maths - Extra Questions

Solution:

$$LHS=14$$
$$RHS=27-y$$

Consider the given equation,

$$ y=a{{\left( x-a \right)}^{2}} $$

$$ y=a\left( {{x}^{2}}+{{a}^{2}}-2ax \right) $$

$$ =a{{x}^{2}}+{{a}^{3}}-2a^2x $$

Hence, this is the answer.

Type: Linear equation of one variable.

In the form of: $$ax+b=0$$   

Given: $$17=x+7 $$

$$\therefore$$   $$x - 10 = 0$$

Here $$'x'$$ is the variable.

So, This is the linear equation of one variable in the form of $$ax+b=0$$ with variable $$'x'$$ .

Hence, this is the required answer.

Standard form for linear equation in one variable is $$ax+b=0$$

$$LHS=2a-3$$
$$RHS=5$$

Equations with linear expressions in one variable only are called linear equations in one variable$$.$$

An equation is a statement of equality of two algebraic expression involving an unknown quantity. An equation involving only linear polynomials is called a linear equation.

Is not an equation, it is a numerical equation.

$$(-2x+1)+(-2x-2)$$

Consider the following question.

linear equation of one variable.

$$ax+b=0$$

& $$(t-7)>5 $$

& $$t>7+5 $$

& $$t>12 $$

These equation are linear variable of one equation so identify the variable of from the equation.

Hence, this is the required answer.

Sum of number $$a$$ and $$b=a+b$$

So,

$$\dfrac{x+y}{2}$$.

$$5$$ is taken away from $$y$$ gives the expression 

$$y-z$$.

(a) $$10 y = 80$$

Given: $$4p-3= 13$$

And age of her Grandmother $$=6x-2$$

$$10-yz$$.

$$2p < 15$$
Is not an equation $$=L.H.S \neq R.H.S$$
It has no sign of equality.

The given statement is: $$7$$ divided by $$p$$ 

$$7$$ added to $$p$$ can be expressed as $$p+7$$

$$p$$ multiplied by $$-5$$ can be expressed as $$-5p$$

$$\dfrac{pq}{4}$$.

$$x^2+y^2$$.

$$ab-(a+b)$$.

$$3mn+5$$.

$$p$$ divided by $$7$$ can be expressed as $$\dfrac{p}{7}$$

$$11$$ subtracted from $$2m$$ can be expressed as $$2m-11$$

$$11$$ added to $$2\ m$$ can be expressed as $$2m+11$$

Finding the value of a variable in a linear equation that  $$\text{satisfies}$$  the equation is called a  $$\text{root}$$  of the equation.

$$-p$$  divided by $$5$$ can be expressed as $$\dfrac{-p}{5}$$

It is given that 
Sides of a triangle are  $$2x + 3y , x + 5y$$ and $$7x - 2y$$
We know that 
Perimeter = Sum of all three sides of a triangle 
Substituting the values
$$P= 2x + 3y + x + 5y + 7x - 2y$$
By further calculation 
$$= 2x + x + 7x + 3y + 5y - 2y$$
So we get 
$$= 10x + 8y - 2x$$
$$= 10 x + 6y$$

It is given that 
Sides of a rectangle are $$6a + 9b$$ and $$8a - 4b$$
So length $$=  6 a + 9b$$ and breadth $$= 8a - 4b$$
We know that 
Perimeter $$= 2$$ (length $$+ $$ breadth)
Substituting the values 
$$ = 2 \left ( 6a + 9b + 8a - 4b \right ) $$
By further calculation
$$ = 2 \left ( 14a + 5b \right )$$ 
So we get 
$$= 28a + 10b$$

i) L.H.S $$=-(-x)$$$$=-\left (-\dfrac {11}{15}\right) = \dfrac {11}{15}$$

$$ 5 \times  t = 35 $$

$$(\dfrac{1}{7}x-\dfrac{1}{8}y)(7x^2-8y^2)$$

An equation is a statement that the two expressions are equal.

 y - 12 = 20 

$$x - \dfrac {y}{2} - 5 = 0$$

$$  {\textbf{Step -1: Expand the given series and simplify it.}} $$

                  $$  {\left( {x + 3} \right)^3} = {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3} $$

                                  $$   = {x^3} + 9{x^2} + 27x + 27 $$

                  $$  {\text{So, the co-efficient of }}x{\text{ is 27}}{\text{.}} $$

$$  {\textbf{Hence, the answer is 27}}{\textbf{.}} $$ 

If we add $$14yx,\ -3xy$$ then we'll get, $$14xy-3xy=11xy$$.

Perimeter of square $$=4S$$

Perimeter of square $$=4S$$

Perimeter $$=$$ sum of side lengths

Perimeter = sum of side lengths

We know,

Perimeter$$=$$sum of the 4 sides $$=(4+2+1+5)cm=12 cm$$

Perimeter of square $$=4S$$

Perimeter of rectangle $$=  2\times(l+b)$$

$$7$$ subtracted from $$-m$$ can be expressed as $$-m-7$$

Perimeter of rectangle

$$3x-(y-2x)=3x-y+2x=5x-y$$

Let, marks obtained by Palak$$=x.$$
According to the problem,
Abha gets twice the marks as that of Palak$$=2x$$
$$\textbf{Hence, Abha gets }  2x \ \textbf{marks}.$$

Given , sum of two number is$$ 81$$.
One is twice the other.
We know,
Small number: $$x$$
Large number: $$2 x$$ or $$81-x \ \ \textbf{[ From part a]}$$
Given, Sum of two numbers $$=81$$
$$\Rightarrow x+2 x=81$$
$$\textbf{Hence, required equation is } x+2 x=81$$

The cost of oil per litre is $$5$$ times the cost of ride per kg.

We have, $$2 x-30=0 \ \   \textbf{ [From (C)]}$$
$$\Rightarrow 2 x=30$$
$$\Rightarrow x=\frac{30}{2}$$
$$\Rightarrow x=15$$
$$\textbf{Hence, the solution of the equation is }  15$$.

$$y$$ is multiplied by $$-8$$ and then $$5$$ is added to the result can be expressed as $$-8y+5$$

Let, the first number be $$\mathrm{x}$$.
Given, the sum of two numbers is $$60$$.
Therefore,
$$x+\text { second number }=60$$
$$\Rightarrow \text { Second number }=60-x$$
Hence, the other number is $$(60-\mathrm{x})$$.

$$6$$ more than thrice a number $$x$$ can be wriiten in mathematical form as $$3x+6$$

$$3$$ added to $$5$$ times $$y$$ can be expressed as $$5y+3$$

The perimeter of an equilateral triangle $$= 3 \times side$$
Therefore, the perimeter of an equilateral triangle of side $$x$$ is $$3x$$.
The expression we obtained is $$3x$$, which has only one term, so it is a monomial.
$$\textbf{Hence, the required expression is 3x and is a monomial.}$$

$$\textbf{Step 1: Find the perimeter of one square flower bed}$$
              Given that each flower bed is a square having side $$(x+2)$$ m
              Perimeter of the square is $$=4 \times side$$
              Therefore, the perimeter  square of side $$(x+2)m$$
                           $$=4 \times x+2m$$
                           $$=4x+8$$
              There are $$5$$ square flower beds
              So, the total length of fencing $$=5 \times 4x+8=20x+40$$ metres.

$$\textbf{Step 2: Find the total cost of fencing}$$
              The cost of fencing is $$Rs. \ 50$$ per $$100$$ metres
              So, by using unitary method the cost of fencing $$1$$ metre is $$\frac{50}{100}=Rs. \  \frac{1}{2}$$
              Therefore, the cost of fencing $$20 x+40$$ metres
              $$=\frac{1}{2} \times(20 x+40)=Rs. \ 10 x+20$$

$$\textbf{Hence, the total cost of fencing the flower beds at the cost of Rs. } 50 \textbf{ per } 100 \textbf{ metres is Rs. } 10x+20$$.

$$\textbf{Step 1:Use the formula of perimeter of rectangle}$$
              The given figure is a rectangle having length $$5x-y$$ and width $$2(x+y)$$
              Perimeter of a rectangle $$=2 \times (length+width)$$
              Therefore, the perimeter of the given figure is
                                                       $$=2 \times \{{5x-y+2(x+y) \}} $$

$$\textbf{Step 2: Simplify the expression}$$
              Therefore perimeter $$=2 \times \{(5 x-y)+2(x+y)\}$$
             $$\begin{aligned}&=2 \times\{5 x-y+2 x+2 y\} \\&=2 \times\{7 x+y\} \\&=14 x+2 y\end{aligned}$$

$$\textbf{Hence, the perimeter of the given figure is } 14x+2y.$$

The perimeter of a rectangle $$=2 \times($$ length $$+$$ breadth $$)$$
Therefore, the perimeter of a rectangle with length $$p$$ and breadth $q$
$$=2 \times(p+q)=2 p+2 q$$.
The expression, we obtained is $$2 p+2 q$$ which has two unlike terms, so it is a binomial.
$$\textbf{Hence, the required expression is 2p+2q and  is a binomial.}$$

(a)
$$(3x - 4)$$ metres wire is  cut from $$(7x - 3)$$ metres long wire.
So, the remaining length of the wire
$$\begin{aligned}&=(7 x-3)-(3 x-4) \\&=7 x-3-3 x+4 \\&=(7 x-3 x)+(-3+4) \\&=4 x+1\end{aligned}$$
$$\textbf{Hence, the length of wire left is (4x+1) metres.}$$

(b)
$$\textbf{Step 1: Find the remaining length of the wire}$$
$$(3x - 4)$$ metres wire is  cut from $$(7x - 3)$$ metres long wire.
So, the remaining length of the wire
$$\begin{aligned}&=(7 x-3)-(3 x-4) \\&=7 x-3-3 x+4 \\&=(7 x-3 x)+(-3+4) \\&=4 x+1\end{aligned}$$
Hence, the length of wire left is $$(4x+1)$$ metres.

$$\textbf{Step 2:Find the length of each side of the triangle formed from the left-out wire}$$
Perimeter of an equilateral triangle $$=3 \times$$ side
Therefore, $$3 \times$$ side $$=4 x+1$$
side $$=\frac{4 x+1}{3}$$ (dividing by $$3$$ on both sides)
$$\textbf{Hence, the length of each side of the triangle so formed is } \dfrac{4x+1}{3} \textbf{ metres}$$.

(i)
Commutative property of addition says that the change in order of numbers in addition will not change the result.
Therefore, for variable x and y the commutative property of addition is $$x+y=y+x.$$

(ii)
Commutative property of multiplication says that the change in order of numbers in multiplication will not change the result.
Therefore, the commutative property of multiplication for variable x and y is $$x \times y=y \times x$$

(iii)
Associative property of addition is $$a+b+c=a+b+c$$
Therefore, the associative property of addition for variable x, y and z is $$x+y+z=x+y+z$$

(iv)
Associative property of multiplication is $$a \times b \times c=(a \times b) \times c$$.
Therefore, the associative property of multiplication for variables x, y and z is $$x \times  y \times z=(x \times y) \times z$$

(v)
Distributive property of multiplication over addition is $$a \times b+c=a \times b+a times c$$. Therefore, the distributive property of multiplication over addition multiplication for variables x, y and z is $$x \times y+z=x \times y+x \times z$$

$$\textbf{Step 1: Find the perimeter of the combined figure}$$
            Perimeter of the combined figure is length of its boundary
          $$\begin{aligned}&=E F+F C+C B+B A+A D+D E \\=& x+y+x+x+x+y \\&=4 x+2 y\end{aligned}$$

$$\textbf{Hence, the required perimeter is } 4x+2y.$$

Length of rectangle $$=L=3x-4y+6z$$

Sides of a triangle are $$5a-3b,3a+2b$$ and $$5b-2a$$
Perimeter=$$5a-3b+3a+2b+5b-2a$$
$$=8a-2a+7b-3b$$
$$=6a+4b$$

Two adjacent sides of a rectangle are $$4x+7y$$ and $$3y-x$$
Perimeter=$$2(4x+7y+3y-x)=2(3x+10y)=6x+20y$$

The perimeter of a triangle whose sides measure $$2a,b$$ and $$a+b$$ is $$2a+b+a+b$$ which is equal to $$3a+2b$$

A term of an equation can be transposed to the other side by changing its sign
For example, a linear equation is given $$x+a=0$$ where $$x=-a$$. Thus, the term $$a$$ of an equation can be transposed to the other side by changing its sign.

Let the breadth of a rectangle = x m.
Thrice its breadth=3x m
Length of a rectangle= 3x-6 m
Perimeter of  a rectangle = 2(l+b)
$$=2(3x-6+x)$$
=$$2(4x-6)$$
=$$8x-12$$
But we are given perimeter=148 m
$$8x-12=148$$
$$8x=148+12$$
$$8x= 160$$
$$x=20$$ metres
Breadth=x= 20 metres
and Length=$$ 3x- 6= 3 \times 20-6= 60-6= 54$$ metres.

It is given that
Area of a rectangular field $$25x^{2} + 20xy + 3y^{2}$$ square unit
Length $$= 5x + 3y$$ unit
We know that
Breadth = Area / Legth
So we get
$$= (25x + 20xy + 3y^{2}) / (5x + 3y)$$
$$= 5x + y$$ units
Now the perimeter of the rectangular field = 2 (length + breadth)
Substitute the values
$$= 2 95x + 3y + 5x + y)$$
So we get
$$= 2 (10x + 4y)$$
$$= 20x + 8y$$

Given, 
Perimeter of a triangle = $$7p^2 - 5p + 11$$ 
And, two of its sides are $$p^2 + 2p - 1$$ and $$3p^2 - 6p + 3$$ 
We know that, 
Perimeter of a triangle = Sum of three sides of triangle 
$$\Rightarrow \ \ 7p^2 - 5p + 11 = (p^2 + 2p - 1) + (3p^2 - 6p + 3) + (\ Third \ side \ of \ triangle) \ 7p^2 - 5p + 11 = (4p^2 - 4p + 2) + (\ Third \ side \ of \ triangle)$$ 
$$\Rightarrow  \ \ \ Third \ side \ of \ triangle = (7p^2 - 5p + 11) - (4p^2 - 4p + 2)$$ 
= $$(7p^2 - 4p^2) + (- 5p + 4p) + (11 - 2)$$ 
= $$3p^2 - p + 9$$ 
Thus, the third side of the triangle is $$3p^2 - p + 9$$. 

Let us assume the length be $$l$$, breadth be b and perimeter be $$P$$. Then the algebraic expression for the given statement is written as,
$$ P = 2(l + b) $$