Unit 12: Algebraic Expression

## Exercise 1: (Multiple Choice Questions and Answers 1-16)

In each of the questions 1 to 16, out of the four options, only one is correct. Write the correct answer.

# Question: 1

An algebraic expression containing three terms is called a:

a.    Monomial

b.   Binomial

c.    Trinomial

d.   All of these

## Solution

(c)

An algebraic expression containing three terms is called trinomial.

# Question: 2

Number of terms in the expression $3{x}^{2}y-2{y}^{2}z-{z}^{2}x+5$ is

a.    $2$

b.   $3$

c.    $4$

d.   $5$

## Solution

(c)

Total number of terms in the expression are $4$. They are  $3{x}^{2}y,-2{y}^{2}z,-{z}^{2}x$ and $5$.

# Question: 3

The terms of expression $4{x}^{2}-3xy$ are:

a.    $4{x}^{2}\text{\hspace{0.17em}}\text{and}-3xy$

b.

c.

d.

## Solution

(a)

Terms in the expression $4{x}^{2}-3xy$ are $4{x}^{2}\text{\hspace{0.17em}}\text{and}-3xy.$

# Question: 4

Factors of $-5{x}^{2}{y}^{2}z$ are

a.    $-\text{\hspace{0.17em}}\text{\hspace{0.17em}}5×x×y×z$

b.   $-\text{\hspace{0.17em}}\text{\hspace{0.17em}}5×{x}^{2}×y×z$

c.    $-\text{\hspace{0.17em}}\text{\hspace{0.17em}}5×x×x×y×y×z$

d.

## Solution

(c)

$-5{x}^{2}{y}^{2}z$ Can be factorized as $-5×x×x×y×y×z$.

# Question: 5

Coefficient of $x$ in  is

a.    $9yz$

b.   $-9yz$

c.    $9{y}^{2}z$

d.   $-9{y}^{2}z$

## Solution

(d)

Coefficient of $x$ in  is $-9{y}^{2}z$.

# Question: 6

Which of the following is a pair of like terms?

a.    $-\text{\hspace{0.17em}}7x{y}^{2}z,\text{\hspace{0.17em}}\text{\hspace{0.17em}}-7{x}^{2}yz$

b.   $-10ay{z}^{2}$, $3ay{z}^{2}$

c.

d.

## Solution

(b)

Like terms are those terms, having same algebraic factor.

Hence, $-10ay{z}^{2}$ and $3ay{z}^{2}$ are like terms.

# Question: 7

Identify the binomial out of the following:

a.    $3x{y}^{2}+5y-{x}^{2}y$

b.   ${x}^{2}y-5y-{x}^{2}y$

c.    $xy+yz+zx$

d.   $3x{y}^{2}+5y-x{y}^{2}$

## Solution

(d)

Taking option (d),

$\begin{array}{l}3x{y}^{2}+5y-x{y}^{2}\\ =2{x}^{2}y+5y\end{array}$

As it contains only two terms, hence it is binomial.

# Question: 8

The sum of ${x}^{4}-xy+2{y}^{2}$ and $-{x}^{4}+xy+2{y}^{2}$ is

a.    Monomial and polynomial in $y$

b.   Binomial and Polynomial

c.    Trinomial and polynomial

d.   Monomial and polynomial in $x$

## Solution

(a)

Required sum $=\left({x}^{4}-xy+2{y}^{2}\right)+\left(-{x}^{4}+xy+2{y}^{2}\right)$

$=\text{\hspace{0.17em}}\left[\left({x}^{4}+\left(-{x}^{4}\right)\right]+\left(-xy+xy\right)+\left(2{y}^{2}+2{y}^{2}\right)$

$=0+0+4{y}^{2}=4{y}^{2}$

$4{y}^{2}$ is a monomial and polynomial in $y$.

# Question: 9

The subtraction of 5 times of $y$ from $x$ is

a.    $5x-y$

b.   $y-5x$

c.    $x-5y$

d.   $5y-x$

## Solution

(c)

$5$ times of $y$ is $5y$. Now, subtraction of $5$ times of $y$ from $x$ is $x-5y$

# Question: 10

is equal to

a.    $-1×b$

b.   $1-b-0$

c.    $0-\left(-1\right)×b$

d.   $-55$

## Solution

(a)

is equal to $-1×b$

# Question: 11

The length of the top of square table is $x$. The expression for perimeter is:

a.    $4+x$

b.   $2x$

c.    $4x$

d.   $8x$

## Solution

(c)

Given, length of a square table $=x$

$\therefore$ Perimeter of a square

$\begin{array}{l}=4×\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Side}\text{\hspace{0.17em}}\\ =4×x\\ =4x.\end{array}$

# Question: 12

The number of scarfs of length half meter that can be made from $y$ meters of cloth is:

a.    $2y$

b.   $\frac{2y}{2}$

c.    $y+2$

d.   $y+\frac{1}{2}$

## Solution

(a)

We have, length of 1 $\text{scarf}=1/2\text{\hspace{0.17em}}m$

So, number of scarfs which can be made from $y$ metres

$\begin{array}{l}=\frac{y}{1/2}\\ =2y.\end{array}$

# Question: 13

$123{x}^{2}y-138{x}^{2}y$ is a like term of:

a. $10xy$

b.   $-15xy$

c. $-15x{y}^{2}$

d.   $10{x}^{2}y$

## Solution

(d)

We have, $123{x}^{2}y-138{x}^{2}y=-15{x}^{2}y$

Hence, it is like term of $10{x}^{2}y$ as both contain ${x}^{2}y$.

# Question: 14

The value of $3{x}^{2}-5x+3$ when $x=1$ is

a.    $1$

b.   $0$

c.    $-1$

d.   $11$

## Solution

(a)

Put $x=1$ in given equation, we get

$\begin{array}{l}3{x}^{2}-5x+3\\ =3{\left(}^{1}-5\left(1\right)+3\\ =3-5+3\\ =1\end{array}$

# Question: 15

The expression for the number of diagonals that we can make from one vertex of a $n$ sided polygon is:

a.    $2n+1$

b.   $n-2$

c.    $5n+2$

d.   $n-3$

## Solution

(d)

Since, vertex is formed by joining two sides. Diagonal is line segment joining the two opposite vertex. So, number of diagonals formed by one vertex $=n-3$

# Question: 16

The length of a side of square is given as $2x+3$. Which expression represents the perimeter of the square?

a.    $2x+16$

b.   $6x+9$

c.    $8x+3$

d.   $8x+12$

## Solution

(d)

Side of the square $=\left(2x+3\right)$          [Given]

$\therefore$ Perimeter of square $\begin{array}{l}=4×\left(\text{Side}\right)\\ =4×\left(2x+3\right)\\ =8x+12\end{array}$

In questions 17 to 32, fill in the blanks to make the statements true.

# Question: 17

Sum or difference of two like terms is ________.

## Solution

Sum or difference of two like terms is a like terms.

# Question: 18

In the formula, area of circle $=\pi {r}^{2}$, the numerical constant of the expression $\pi {r}^{2}$ is ________.

## Solution

In $\pi {r}^{2}$, r is variable, so the numerical constant is $\pi$.

# Question: 19

are ________ terms.

## Solution

$3{a}^{2}b$ and $-7b{a}^{2}$ are like terms as both have same variable factor ${a}^{2}b$.

# Question: 20

are ________ terms.

## Solution

$-5{a}^{2}b$ and $-5{b}^{2}a$ are unlike terms.

# Question: 21

In the expression $2\pi r$, the algebraic variable is ________.

## Solution

$r$ is algebraic variable in the expression $2\pi r$.

# Question: 22

Number of terms in a monomial is ________.

## Solution

Number of terms in a monomial is one.

# Question: 23

Like terms in the expression $n\text{(}n+1\text{)}+6\text{\hspace{0.17em}}\text{(}n\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1\text{)}$ are ___________and________.

## Solution

We have, $n\left(n+1\right)+6\left(n-1\right)={n}^{2}+n+6n-6$

Hence, like terms in the expression $n\left(n+1\right)+6\left(n-1\right)$ are $n$ and $6n$.

# Question: 24

The expression $13+90$ is a ________.

## Solution

$13+90=103$ is a constant term.

# Question: 25

The speed of car is $55$ km/hrs. The distance covered in $y$ hours is________.

## Solution

Given, speed of car $=55\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{km/hrs}$

We know that, Distance $=$ Speed $×$ Time

$\therefore$ Distance covered in $y$ hours $=55xy$ km

# Question: 26

$x+y+z$ is an expression which is neither monomial nor ________.

## Solution

$x+y+z$ contains three terms, so it is trinomial.

Hence, $x+y+z$ is an expression which is neither monomial nor binomial.

# Question: 27

If $\left({x}^{2}y+{y}^{2}+3\right)$ is subtracted from $\left(3{x}^{2}y+2{y}^{2}+5\right)$, then coefficient of $y$ in the result is ________.

## Solution

We have, $\left(3{x}^{2}y+2{y}^{2}+5\right)-\left({x}^{2}y+{y}^{2}+3\right)$

$\begin{array}{l}=3{x}^{2}y+2{y}^{2}+5-{x}^{2}y-{y}^{2}-3\\ =2{x}^{2}y+{y}^{2}+2\end{array}$

$\therefore$  Coefficient of $y=2{x}^{2}$

# Question: 28

is same as  ( ________ ).

## Solution

$-\text{\hspace{0.17em}}a-b-c=-a-\left(b+c\right)$

So, $-a-b-c$ is same as $-a-\left(b+c\right)$.

# Question: 29

The unlike terms in perimeters of following figures are ________ and ________.

Fig. (i)

Fig. (ii)

## Solution

In above fig. (i), Perimeter = Sum of all sides

$=2x+y+2x+y=4x+2y$

In above fig. (ii), Perimeter = Sum of all sides

$\begin{array}{l}=x+{y}^{2}+x+{y}^{2}\\ =2x+2{y}^{2}\end{array}$

Unlike terms in perimeters are $2y$ and $2{y}^{2}$

# Question: 30

On adding a monomial _____________ to , the resulting expression becomes a binomial.

## Solution

We can add $2x$ to the expression to make it binomial.

# Question: 31

$3x+23{x}^{2}+6{y}^{2}+2x+{y}^{2}+$ ____________ $=5x+7{y}^{2}.$

## Solution

Assume

$\left(3x+23{x}^{2}+6{y}^{2}+2x+{y}^{2}\right)+\text{N}=5x+7{y}^{2}$

$⇒\text{N}=5x+7{y}^{2}-\left(3x+23{x}^{2}+6{y}^{2}+2x+{y}^{2}\right)$

$⇒\text{N}=5x+7{y}^{2}-3x-23{x}^{2}-6{y}^{2}-2x-{y}^{2}$

$⇒\text{N}=-23{x}^{2}$

# Question: 32

If Rohit has $5xy$ toffees and Shantanu has $20xy$ toffees, then Shantanu has ___________ more toffees.

## Solution

We have, Rohit’s toffees $=5xy$

Shantanu’s toffees $=20xy$

Difference:

$\begin{array}{l}=20xy-5xy\\ =15xy\end{array}$

Hence, Shantanu had $15xy$ more toffees.

--

In questions 33 to 52, state whether the statements given are True or False.

# Question: 33

$1+\frac{x}{2}+{x}^{3}$ is a polynomial.

## Solution

True

Expression with three or more than three terms is called a polynomial.

# Question: 34

is a binomial.

## Solution

False

We have,

$=3a-b+3-a-b$

$=3a-a-b-b+3$

$=2a-2b+3$

The expression has three terms, it is a trinomial.

# Question: 35

A trinomial can be a polynomial.

## Solution

True

Trinomial is a polynomial.

# Question: 36

A polynomial with more than two terms is a trinomial.

## Solution

False

A trinomial have exact three terms.

# Question: 37

Sum of $x$ and $y$ is $x+y$.

## Solution

True

Sum of $x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x+y$

# Question: 38

Sum of $2$ and $p$ is $2p$.

## Solution

False

Sum of $2$ and $p$ is $2+p$

# Question: 39

A binomial has more than two terms.

## Solution

False

Binomial has exactly two terms.

# Question: 40

A trinomial has exactly three terms.

## Solution

True

A trinomial has exactly three unlike terms.

# Question: 41

In like terms, variables and their powers are the same.

## Solution

True

In like terms, variables and their powers are the same.

# Question: 42

The expression $x+y+5x$ is a trinomial.

## Solution

False

$\therefore x+y+5x=6x+y$ it is a binomial.

# Question: 43

$4p$ is the numerical coefficient of .

## Solution

False

Numerical coefficient of ${q}^{2}$ in $-4p{q}^{2}=-4$.

# Question: 44

$5a$ and $5b$ are unlike terms.

## Solution

True

Both the terms have different algebraic factors.

# Question: 45

Sum of ${x}^{2}+x$ and .

## Solution

False

Sum $=\left({x}^{2}+x\right)+\left(y+{y}^{2}\right)={x}^{2}+x+y+{y}^{2}$

$={x}^{2}+{y}^{2}+x+y$

# Question: 46

Subtracting a term from a given expression is the same as adding its additive inverse to the given expression.

## Solution

True

Additive inverse is the negation of a number or expression.

# Question: 47

The total number of planets of Sun can be denoted by the variable $n$.

## Solution

False

As, Sun has infinite planets around it.

# Question: 48

In like terms, the numerical coefficients should also be the same.

## Solution

False

e.g. $-3{x}^{2}y$ and $4{x}^{2}y$ are like terms as they have same algebraic factor ${x}^{2}y$ but have different numerical coefficients.

# Question: 49

If we add a monomial and binomial, then answer can never be a monomial.

## Solution

False

If we add a monomial and a binomial, then answer can be a monomial, e.g.

Add ${x}^{2}&-{x}^{2}+{y}^{2}$

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

$\begin{array}{l}={x}^{2}-{x}^{2}+{y}^{2}\\ ={y}^{2}\end{array}$

# Question: 50

If we subtract a monomial from a binomial, then answer is at least a binomial.

## Solution

False

If we subtract a monomial from a binomial, then answer is at least a monomial.

# Question: 51

When we subtract a monomial from a trinomial, then answer can be a polynomial.

## Solution

True

When we subtract a monomial from a trinomial, then answer can be binomial or polynomial.

# Question: 52

When we add a monomial and a trinomial, then answer can be a monomial.

## Solution

False

When we add a monomial and a trinomial, then it can be binomial or trinomial.

# Question: 53

Write the following statements in the form of algebraic expression and write whether it is monomial, binomial or trinomial.

a.    $x$ is multiplied by itself and then added to the product of $x$ and $y$.

b.   Three times of $5$ and two times of $q$ are multiplied and then subtracted from $r$.

c.    Product of $p$, twice of $q$ and thrice of $r$.

d.   Sum of the products of $a$ and $b$, $b$ and $c$ and $c$ and $a$.

e.    Perimeter of an equilateral triangle of
side $x$.

f.Perimeter of a rectangle with length $p$ and breadth $q$.

g.   Area of a triangle with base m and
height $n$.

h.   Area of a square with side $x$.

i.  Cube of $s$ subtracted from cube of $t$.

j.  Quotient of $x$ and 15 multiplied by $x$.

k.   The sum of square of $x$ and cube of $z$.

l.  Two times q subtracted from cube of $q$.

## Solution

a.    ${x}^{2}+xy$

b.   $\begin{array}{l}r-\left(3p×2q\right)\\ =r-6pq\end{array}$

c.    $\begin{array}{l}p×2q×3r\\ =6pqr\end{array}$

d.   $ab+bc+ca$

e.    $3x$

f.$\begin{array}{l}2\left(p+q\right)\\ =2p+2q\end{array}$

g.   $\frac{1}{2}mn$

h.   ${x}^{2}$

i.   ${t}^{3}-{s}^{3}$

j.  $\begin{array}{l}\frac{x}{15}×x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{or}\text{\hspace{0.17em}}\frac{{x}^{3}}{15}\end{array}$

k.   ${x}^{2}+{z}^{3}$

l.  ${q}^{3}-2q$

# Question: 54

Write the coefficient of ${x}^{2}$ in the following:

(i)             ${x}^{2}-x+4$

(ii)         ${x}^{3}-2{x}^{2}+3x+1$

(iii)      $1+2x+3{x}^{2}+4{x}^{3}$

(iv)       $y+{y}^{2}x+{y}^{3}{x}^{2}+{y}^{4}{x}^{3}$

## Solution

(i)             $1$

(ii)         $-2$

(iii)      $3$

(iv)       ${y}^{3}$

# Question: 55

Find the numerical coefficient of each of the terms:

(i)             ${x}^{3}{y}^{3}z,\text{\hspace{0.17em}}\text{\hspace{0.17em}}x{y}^{2}{z}^{2},-3x{y}^{2}{z}^{3},\text{\hspace{0.17em}}5{x}^{3}{y}^{3}z,-7{x}^{2}{y}^{2}{z}^{2}$

(ii)         $10xyz,-7x{y}^{2}z,-9xyz,2x{y}^{2}z,2{x}^{2}{y}^{2}{z}^{2}$

## Solution

(i)             Coefficient of:

${x}^{3}{y}^{2}z=1$

(ii)         Coefficient of:

$10\text{\hspace{0.17em}}xyz=10$

$-7x{y}^{2}z=-7$

$-9xyz=-9$

$2x{y}^{2}z=2$

$2{x}^{2}{y}^{2}{z}^{2}=2$

# Question: 56

Simplify the following by combining the like terms and then write whether the expression is a monomial, a binomial or a trinomial.

a.    $3{x}^{2}y{z}^{2}-3x{y}^{2}z+{x}^{2}y{z}^{2}+7x{y}^{2}z$

b.

c.    ${p}^{3}{q}^{2}r+p{q}^{2}{r}^{2}+3{p}^{2}q{r}^{2}-9{p}^{2}q{r}^{2}$

d.

e.

## Solution

a.    $3{x}^{2}y{z}^{2}-3x{y}^{2}z+{x}^{2}y{z}^{2}+7x{y}^{2}z$

By combining the like terms

$\begin{array}{l}=3{x}^{2}y{z}^{2}+{x}^{2}y{z}^{2}-3x{y}^{2}z+7x{y}^{2}z\\ =4{x}^{2}y{z}^{2}+4x{y}^{2}z\end{array}$

The expression contains $3$ terms.
So, it is trinomial.

b.   ${x}^{4}+3{x}^{3}y+3{x}^{2}{y}^{2}-3{x}^{3}y-3x{y}^{3}$

$+{y}^{4}-3{x}^{2}{y}^{2}$

By combining the like terms

$\begin{array}{l}={x}^{4}+3{x}^{3}y-3{x}^{3}y+3{x}^{2}{y}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3{x}^{2}{y}^{2}-3{x}^{3}y+{y}^{3}\\ ={x}^{4}-0+0-3{x}^{3}y+{y}^{4}\\ ={x}^{4}+{y}^{4}-3{x}^{3}y\end{array}$

The expression contains $3$ terms.
So, it is trinomial.

c.    ${p}^{3}{q}^{2}r+p{q}^{2}{r}^{3}+3{p}^{2}q{r}^{2}-9{p}^{2}q{r}^{3}$

By combining the like terms

$\begin{array}{l}={p}^{3}{q}^{2}r+p{q}^{2}{r}^{3}+3{p}^{2}q{r}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-9{p}^{2}q{r}^{2}\end{array}$

$={p}^{3}{q}^{2}r+p{q}^{2}{r}^{3}-6{p}^{2}q{r}^{2}$

The expression contains terms.
So, it is trinomial.

d.   $2a+2b+2c-2a-2b-2c$

$-2b+2c+2a$

By combining the like terms

$\begin{array}{l}=2a-2a+2a+2b-2b-2b\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+2c-2c+2c\\ =2a-2b+2c\end{array}$

The expression contains $3$ terms.
So, it is trinomial.

e.    $50{x}^{3}-21x+107+41{x}^{3}-x+1-93$

$+\text{\hspace{0.17em}}71x-31{x}^{3}$

By combining the like terms

$\begin{array}{l}=50{x}^{3}+41{x}^{3}-31{x}^{3}-21x-x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\text{\hspace{0.17em}}71x+107+1-93\\ =60{x}^{3}+49x+15\end{array}$

The expression contains $3$ terms.
So, it is trinomial.

# Question: 57

a.

b.

c.

d.

e.

f.

g.

h.    ${c}^{2}+3ca-ab$

i.  $\frac{5}{8}{p}^{4}+2{p}^{2}+\frac{5}{8};\frac{1}{8}-17p+\frac{9}{8}{p}^{2}\text{\hspace{0.17em}}\mathrm{and}$ ${p}^{5}-{p}^{3}+7a$

j.

## Solution

a.    ${p}^{2}-7pq-{q}^{2}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}$

$-3{p}^{2}-2pq+7{q}^{2}$

$={p}^{2}-7pq-{q}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3{p}^{2}-2pq+7{q}^{2}$

By combining the like terms, we get

$\begin{array}{l}={p}^{2}-3{p}^{2}-7pq-2pq-{q}^{2}+7{q}^{2}\\ =-2{p}^{2}-9pq+6{q}^{2}\end{array}$

b.   ${x}^{3}-{x}^{2}y-x{y}^{2}-{y}^{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}$

${x}^{3}-2{x}^{2}y+3x{y}^{2}+4y$

By combining the like terms, we get

$={x}^{3}+{x}^{3}-{x}^{2}y-2{x}^{2}y-x{y}^{2}+3x{y}^{2}-{y}^{3}+4y$

$=2{x}^{3}-3{x}^{2}y+2x{y}^{2}-{y}^{3}+4y$

c.    $ab+bc+ca\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}-bc-ca-ab$

$=ab+ac+bc-bc-ca-ab$

By combining the like terms,

$\begin{array}{l}=ab-ab-bc+bc+ca-ca\\ =0+0+0=0\end{array}$

d.   ${p}^{2}-q+r,{q}^{2}-r+p\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}-p+q$

$={p}^{2}-q+r+{q}^{2}-r+p+{r}^{2}-p+q$

By combining the like terms,

$\begin{array}{l}={p}^{2}+{q}^{2}+{r}^{2}-q+q+r-r+p-p\\ ={p}^{2}+{q}^{2}+{r}^{2}+0+0+0={p}^{2}+{q}^{2}+{r}^{2}\end{array}$

e.    ${x}^{3}{y}^{2}+{x}^{2}{y}^{3}+3{y}^{4}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}{x}^{4}+3{x}^{2}{y}^{3}+4{y}^{4}$

$={x}^{3}{y}^{2}+{x}^{2}{y}^{3}+3{y}^{4}+{x}^{4}+3{x}^{2}{y}^{3}+4{y}^{4}$

By combining the like terms,

$\begin{array}{l}={x}^{3}{y}^{2}+{x}^{2}{y}^{3}+3{y}^{4}+{x}^{4}+3{x}^{2}{y}^{3}+4{y}^{4}\\ ={x}^{4}+7{y}^{4}+{x}^{3}{y}^{2}+4{x}^{2}{y}^{3}\end{array}$

f.     ${p}^{2}qr+p{q}^{2}r+pq{r}^{2}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}-3p{q}^{\text{2}}r-2pq{r}^{2}$

$={p}^{2}qr+p{q}^{2}r+pq{r}^{2}\text{\hspace{0.17em}}-3p{q}^{2}r-2pq{r}^{2}$

By combing the like terms,

$\begin{array}{l}={p}^{2}qr+p{q}^{2}r-3p{q}^{2}r+pq{r}^{2}-2pq{r}^{2}\\ ={p}^{2}qr-2p{q}^{2}r-pq{r}^{2}\end{array}$

g.   $\begin{array}{l}uv-vw,vw-wu\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}wu-uv\end{array}$

$=uv-vw+vw-wu+wu-uv$

By combining like terms,

= 0

h.      $\begin{array}{l}{a}^{2}+3ab-bc,\text{\hspace{0.17em}}{b}^{2}+3bc-ca\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{c}^{2}+3ca-ab\end{array}$

$\left({a}^{2}+3ab-bc\right)+\left({b}^{2}+3bc-ca\right)+\left({c}^{2}+3ca-ab\right)$

${a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$

i.        $\begin{array}{l}\frac{5}{8}{p}^{4}+2{p}^{2}+\frac{5}{8};\frac{1}{8}-17p+\frac{9}{8}{p}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{p}^{5}-{p}^{3}+7\end{array}$

$=\frac{5}{8}{p}^{4}+2{p}^{2}+\frac{5}{8}+\frac{1}{8}-17p+\frac{9}{8}{p}^{2}+{p}^{5}-{p}^{3}+7$

By combining like terms,

$={p}^{5}+\frac{5}{8}{p}^{4}-{p}^{3}+\left(2+\frac{9}{8}\right){p}^{2}-17p$

$+\left(\frac{5}{8}+\frac{1}{8}+7\right)$

$={p}^{5}+\frac{5}{8}{p}^{4}-{p}^{3}+\left(\frac{16+9}{8}\right){p}^{2}-17p$

$+\left(\frac{5+1+56}{8}\right)$

$={p}^{5}+\frac{5}{8}{p}^{4}-{p}^{3}+\frac{25}{8}{p}^{2}-17p+\frac{62}{8}$

$={p}^{5}+\frac{5}{8}{p}^{4}-{p}^{3}+\frac{25}{8}{p}^{2}-17p+\frac{31}{4}$

j.        $t-{t}^{2}-{t}^{3}-14;\text{\hspace{0.17em}}\text{\hspace{0.17em}}15{t}^{3}+13+9t-8{t}^{2};$

$12{t}^{2}-19-24\mathrm{t}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4t-9{t}^{2}+19{t}^{3}$

$=t-{t}^{2}-{t}^{3}-14+15{t}^{3}+13+9t-8{t}^{2}+12{t}^{2}$

$-19-24t+4t-9{t}^{2}+19{t}^{3}$

By combining like terms,

$=t+9t-24t+4t-{t}^{2}-8{t}^{2}+12{t}^{2}$

$-9{t}^{2}-{t}^{3}+15{t}^{3}+19{t}^{3}-14+13-19$

$\begin{array}{l}=-10t-6{t}^{2}+33{t}^{3}-20\\ =33{t}^{3}-6{t}^{2}-10t-20\end{array}$

# Question: 58

Subtract

a.

b.

c.

d.

e.

f.

g.

h.

i.

j.

## Solution

a.    We have,

$\begin{array}{l}-3{p}^{2}qr-\left(-7{p}^{2}qr\right)\\ =-3{p}^{2}qr+7{p}^{2}qr\\ ={p}^{2}qr\left(-3+7\right)\\ =4{p}^{2}qr\end{array}$

b.   We have,

$\begin{array}{l}{b}^{2}+ab-\left(-{a}^{2}-ab\right)\\ ={b}^{2}+ab+{a}^{2}+ab\end{array}$

By combining like terms,

c.    We have,

${x}^{3}+3x{y}^{2}-{x}^{2}y-\left(-4{x}^{2}y-{y}^{3}\right)$

$={x}^{3}+3x{y}^{2}-{x}^{2}y+4{x}^{2}y+{y}^{3}$

$={x}^{3}+{y}^{3}+3{x}^{2}y+3x{y}^{2}$

d.   We have,

$\begin{array}{l}2{x}^{4}-{x}^{3}{y}^{3}+7{y}^{4}-\left({x}^{4}+3{x}^{3}{y}^{3}+5{y}^{4}\right)\\ =2{x}^{4}-{x}^{3}{y}^{3}+7{y}^{4}-{x}^{4}-3{x}^{3}{y}^{3}-5{y}^{4}\end{array}$

By combining like terms,

$\begin{array}{l}=2{x}^{4}-{x}^{4}-{x}^{3}{y}^{3}-3{x}^{3}{y}^{3}+7{y}^{4}-5{y}^{4}\\ ={x}^{4}-4{x}^{3}{y}^{3}+2{y}^{4}\end{array}$

e.    We have,

$-ab+bc+ca-\left(ab-bc-ca\right)$

$=-ab+bc+ca-ab+bc+ca$

By combining like terms,

$\begin{array}{l}=-ab-ab+bc+bc+ca+ca\\ =-2ab+2bc+2ca\end{array}$

f.We have,

$\begin{array}{l}\left(-{a}^{2}-{b}^{2}+2ab\right)-\left(-2{a}^{2}-2{b}^{2}\right)\\ =-{a}^{2}-{b}^{2}+2ab+2{a}^{2}+2{b}^{2}\end{array}$

By combining like terms,

$\begin{array}{l}=-{a}^{2}+2{a}^{2}-{b}^{2}+2{a}^{2}+2ab\\ ={a}^{2}-{b}^{2}+2ab\end{array}$

g.   We have,

${x}^{4}+{y}^{4}+3{x}^{2}{y}^{2}-x{y}^{3}-\left({x}^{3}{y}^{2}+3{x}^{2}{y}^{2}-7x{y}^{3}\right)$

$={x}^{4}+{y}^{4}+3{x}^{2}{y}^{2}-x{y}^{3}-{x}^{3}{y}^{2}-3{x}^{2}{y}^{2}+7x{y}^{3}$

By combining the like terms,

$\begin{array}{l}={x}^{4}+{y}^{4}+3{x}^{2}{y}^{2}-3{x}^{2}{y}^{2}-x{y}^{3}+7x{y}^{3}-{x}^{3}{y}^{2}\\ ={x}^{4}+{y}^{4}+6x{y}^{3}-{x}^{3}{y}^{2}\end{array}$

h.   We have,

$\begin{array}{l}-ab-bc-ca-2\left(ab+bc+ca\right)\\ =-ab-bc-ca-2ab-2bc-2ca\end{array}$

By combining the like terms,

$\begin{array}{l}=-ab-2ab-bc-2bc-ca-2ca\\ =-3ab-3bc-3ca\end{array}$

i.      We have,

$\begin{array}{l}5{x}^{4}-32{x}^{2}-7.3x-\left(4.5{x}^{5}-3.4{x}^{2}+5.7\right)\\ =5{x}^{4}-3.2{x}^{2}-7.3x-4.5{x}^{5}+3.4{x}^{2}-5.7\end{array}$

By combining the like terms,

$\begin{array}{l}=-4.5{x}^{5}+5{x}^{4}-3.2{x}^{2}+3.4{x}^{2}-7.3x-5.7\\ =-4.5{x}^{5}+5{x}^{4}+0.2{x}^{2}-7.3x-5.7\end{array}$

j.      We have,

$\begin{array}{l}{y}^{3}-15{y}^{2}-y-11-\left(11-15{y}^{2}\right)\\ ={y}^{3}-15{y}^{2}-y-11-11+15{y}^{2}\end{array}$

By combining the like terms,

$\begin{array}{l}={y}^{3}-15{y}^{2}+15{y}^{2}-y-11-11\\ ={y}^{3}-y-22\end{array}$

# Question: 59

a.    What should be added to

b.   What should be added to

$3pq+5{p}^{2}{q}^{2}+{p}^{3}$ to get ${p}^{3}+2{p}^{2}{q}^{2}+4pq$

## Solution

a.    Subtract: ${x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}$ from ${x}^{3}+{y}^{3}$

Required expression is

$\begin{array}{l}{x}^{3}+{y}^{3}-\left({x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\right)\\ ={x}^{3}+{y}^{3}-{x}^{3}-3{x}^{2}y-3x{y}^{2}-{y}^{3}\end{array}$

By combining like terms,

$\begin{array}{l}={x}^{3}-{x}^{3}+{y}^{3}-{y}^{3}-3{x}^{2}y-3x{y}^{2}\\ =-3{x}^{2}y-3x{y}^{2}\end{array}$

So, if we add $-3{x}^{2}y-3x{y}^{2}$ in

${x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}$ we get ${x}^{3}+{y}^{3}$

b.   Subtract: $3pq+5{p}^{2}{q}^{2}+{p}^{3}$ from ${p}^{3}+2{p}^{2}{q}^{2}+4pq$

Required expression is,

$\begin{array}{l}{p}^{3}+2{p}^{2}{q}^{2}+4pq-\left(3pq+5{p}^{2}{q}^{2}+{p}^{3}\right)\\ ={p}^{3}+2{p}^{2}{q}^{2}+4pq-3pq-5{p}^{2}{q}^{2}-{p}^{3}\end{array}$

By combining like terms,

$\begin{array}{l}={p}^{3}-{p}^{3}+2{p}^{2}{q}^{2}-5{p}^{2}{q}^{2}+4pq-3pq\\ =\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3{p}^{2}{q}^{2}+pq\end{array}$

So, if we add $-3{p}^{2}{q}^{2}+pq$ in $3p{q}^{2}+5{p}^{2}{q}^{2}+{p}^{3}$,

we get ${p}^{3}+2{p}^{2}{q}^{2}+4pq$

# Question: 60

a.    What should be subtracted from

$2{x}^{3}-3{x}^{2}y+2x{y}^{2}+3{y}^{3}$ to get

b.   What should be subtracted from

$-7mn+2{m}^{2}+3{n}^{2}$ to get

## Solution

a.    Subtract: ${x}^{3}-2{x}^{2}y+3x{y}^{2}+4{y}^{3}$ from

$2{x}^{3}-3{x}^{2}y+2x{y}^{2}+3{y}^{3}$

Required expression is

$\begin{array}{l}=2{x}^{3}-3{x}^{2}y+2x{y}^{2}+3{y}^{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{x}^{3}+2{x}^{2}y-3x{y}^{2}-4{y}^{3}\end{array}$

By combining like terms,

$\begin{array}{l}=2{x}^{3}-{x}^{3}-3{x}^{2}y+2{x}^{2}y+2x{y}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-3x{y}^{2}+3{y}^{3}-4{y}^{3}\end{array}$

$={x}^{3}-{x}^{2}y-x{y}^{2}-{y}^{3}$

So, if we subtract ${x}^{3}-{x}^{2}y-x{y}^{2}-{y}^{3}$ from

$2{x}^{3}-3{x}^{2}y+2x{y}^{2}+3{y}^{3}$, then we get

${x}^{3}-2{x}^{2}y+3x{y}^{2}+4{y}^{3}$

b.   Subtract: ${m}^{2}+2mn+{n}^{2}$ from $-7mn+2{m}^{2}+3{n}^{2}$

Required expression is,

$-7mn+2{m}^{2}+3{n}^{2}-\left({m}^{2}+2mn+{n}^{2}\right)$

$=-7mn+2{m}^{2}+3{n}^{2}-{m}^{2}-2mn-{n}^{2}$

By combining like terms,

$=-7mn-2mn+2{m}^{2}-{m}^{2}+3{n}^{2}-{n}^{2}$

$=-9mn+{m}^{2}+2{n}^{2}$

So, if we subtract ${m}^{2}+2{n}^{2}-9mn$ from

$-7mn+2{m}^{2}+3{n}^{2}$, then we get

${m}^{2}+2mn+{n}^{2}$

# Question: 61

How much is $21{a}^{3}-17{a}^{2}$ less than $89{a}^{3}-64{a}^{2}+6a+16?$

## Solution

Required expression is

$89{a}^{3}-64{a}^{2}+6a+16-\left(21{a}^{3}-17{a}^{2}\right)$

$=89{a}^{3}-64{a}^{2}+6a+16-21{a}^{3}+17{a}^{2}$

By combining the like terms,

$\begin{array}{l}=89{a}^{3}-21{a}^{3}-64{a}^{2}+17{a}^{2}+6a+16\\ =68{a}^{3}-47{a}^{2}+6a+16\end{array}$

So, $21{a}^{3}-17{a}^{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{\hspace{0.17em}}68{a}^{3}-47{a}^{2}+6a+16$ less than

$89{a}^{3}-64{a}^{2}+6a+16$.

# Question: 62

How much is ${y}^{4}-12{y}^{2}+y+14$ greater than $17{y}^{3}+34{y}^{2}-51y+68?$

## Solution

Required expression is

${y}^{4}-12{y}^{2}+y+14-\left(17{y}^{3}+34{y}^{2}-51y+68\right)$

$={y}^{4}-12{y}^{2}+y+14-17{y}^{3}-34{y}^{2}+51y-68$

By combining like terms,

$={y}^{4}-12{y}^{2}-34{y}^{2}+y+51y+14-68-17{y}^{3}$

$={y}^{4}-46{y}^{2}+52y-17{y}^{3}-54$

$={y}^{4}-17{y}^{3}-46{y}^{2}+52y-54$

So, ${y}^{4}-12{y}^{2}+y+14$ is ${y}^{4}-17{y}^{3}-46{y}^{2}+52y-54$ greater than $17{y}^{3}+34{y}^{2}-51y+68$

# Question: 63

How much does $93{p}^{2}-55p+4$ exceed $13{p}^{3}-5{p}^{2}+17p-\text{\hspace{0.17em}}\text{\hspace{0.17em}}90?$

## Solution

Required expression is

$\begin{array}{l}93{p}^{2}-55p+4-\left(13{p}^{3}-5{p}^{2}+17p-90\right)\\ =93{p}^{2}-55p+4-13{p}^{3}+5{p}^{2}-17p+90\end{array}$

By combining the like terms,

$=93{p}^{2}+5{p}^{2}-55p-17p+4+90-13{p}^{3}$

$\begin{array}{l}=98{p}^{2}-72p+94-13{p}^{3}\\ =-13{p}^{3}+98{p}^{2}-72p+94\end{array}$

So, $93{p}^{2}-55p+4$ is $-13{p}^{3}+9{p}^{2}-72p+94$

more from $13{p}^{3}-5{p}^{2}+17p-90.$

# Question: 64

To what expression must $99{x}^{3}-33{x}^{2}-13x-41$ be added to make the sum zero?

## Solution

In order to find the solution, we will subtract $99{x}^{3}-33{x}^{2}-13x-41$ from 0.

Required expression is

$\begin{array}{l}0-\left(99{x}^{3}-33{x}^{2}-13x-41\right)\\ =0-99{x}^{3}+33{x}^{2}+13x+41\end{array}$

So, if we add $-99{x}^{3}+33{x}^{2}+13x+41$ to $99{x}^{3}-33{x}^{2}-13x-41$, then the sum is zero.

# Question: 65

Subtract $9{a}^{2}-15a+3$ from unity.

## Solution

In order to find the solution, we will subtract $9{a}^{2}-15a+3$ from unity, i.e. $1$.

Required expression is $1-\left(9{a}^{2}-15a+3\right)$

$\begin{array}{l}=1-9{a}^{2}+15a-3\\ =-9{a}^{2}+15a-2\end{array}$

# Question: 66

Find the values of the following polynomials at  and $b=3$:

a.    ${a}^{2}+2ab+{b}^{2}$

b.   ${a}^{2}-2ab+{b}^{2}$

c.    ${a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$

d.   ${a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}$

e.    $\frac{{a}^{2}+{b}^{2}}{3}$

f.$\frac{{a}^{2}-{b}^{2}}{3}$

g.   $\frac{a}{b}+\frac{b}{a}$

h.   ${a}^{2}+{b}^{2}-ab-{b}^{2}-{a}^{2}$

## Solution

Given, $a=-2$ and $b=3$

Put $a=-2$ and $b=3$ in the given expressions, we get

a.    ${a}^{2}+2ab+{b}^{2}$

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

$\begin{array}{l}=4-12+9\\ =1\end{array}$

b.   ${a}^{2}-2ab+{b}^{2}$

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

$\begin{array}{l}=4+12+9\\ =25\end{array}$

c.    ${a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$

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

d.   ${a}^{3}-3{a}^{2}b+3a{b}^{2}-{b}^{3}$

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

$\begin{array}{l}=-8-36-54-27\\ =-125\end{array}$

e.    $\frac{{a}^{2}+{b}^{2}}{3}$

$=\frac{{\left(-2\right)}^{2}+{\left(3\right)}^{2}}{3}$

$\begin{array}{l}=\frac{4+9}{3}\\ =\frac{13}{3}\end{array}$

f.$\frac{{a}^{2}-{b}^{2}}{3}$

$=\frac{{\left(-2\right)}^{2}-\left(3\right)}{3}$

$\begin{array}{l}=\frac{4-9}{3}\\ =-\frac{5}{3}\end{array}$

g.   $\frac{a}{b}+\frac{b}{a}$

$=-\frac{2}{3}+\frac{3}{-2}$

$\begin{array}{l}=-\frac{2}{3}-\frac{3}{2}\\ =\frac{-4-9}{6}\end{array}$

$=-\frac{13}{6}$

h.   ${a}^{2}+{b}^{2}-ab-{b}^{2}-{a}^{2}$

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

$\begin{array}{l}=4+9+6-9-4\\ =6\end{array}$

# Question: 67

Find the values of following polynomials at  and $p=2$:

a.    $m+n+p$

b.   ${m}^{2}+{n}^{2}+{p}^{2}$

c.    ${m}^{3}+{n}^{3}+{p}^{3}$

d.   $mn+np+pm$

e.    ${m}^{3}+{n}^{3}+{p}^{3}-3mnp$

f.${m}^{2}{n}^{2}+{n}^{2}{p}^{2}+{p}^{2}{m}^{2}$

## Solution

Given, $m=1,\text{\hspace{0.17em}}n=-1$

and $p=2$

Put $m=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}n=-1$

and $p=2$ in the expression, we get

a.    $\begin{array}{l}m+n+p\\ =1-1+2\\ =2\end{array}$

b.   ${m}^{2}+{n}^{2}+{p}^{2}$

$={\left(1\right)}^{2}+{\left(-1\right)}^{2}+\left(2\right)$

$=1+1+4$

c.    ${m}^{3}+{n}^{3}+{p}^{3}$

$={\left(1\right)}^{3}+{\left(-1\right)}^{3}+{\left(2\right)}^{3}$

$=1-1+8=8$

d.   $mn+np+pm$

$=\left(1\right)\left(-1\right)+\left(-1\right)\left(2\right)+\left(2\right)\left(1\right)$

$\begin{array}{l}=-1-2+2\\ =-1\end{array}$

e.    ${m}^{3}+{n}^{3}+{p}^{3}-3mnp$

$={\left(1\right)}^{3}+{\left(-1\right)}^{3}+{\left(2\right)}^{3}-3\left(1\right)\left(-1\right)\left(2\right)$

$\begin{array}{l}=1-1+8+6\\ =14\end{array}$

f.${m}^{2}{n}^{2}+{n}^{2}{p}^{2}+{p}^{2}{n}^{2}$

$={\left(1\right)}^{2}{\left(-1\right)}^{2}+{\left(-1\right)}^{2}{\left(2\right)}^{2}+{\left(2\right)}^{2}{\left(1\right)}^{2}$

$\begin{array}{l}=1+4+4\\ =9\end{array}$

# Question: 68

If

and $\text{C}=4{x}^{2}-7x+3,$ then find:

1.   (A + B) - C

2.   B + C - A

3.   A + B + C

## Solution

Given, $\text{A}=3{x}^{2}–4x+1,$

1.   $\text{(A}+\text{B)}-\text{C}$

$=\left(3{x}^{2}-4x+1+5{x}^{2}+3x-8\right)-\left(4{x}^{2}-7x+3\right)$

By combining the like terms,

$=\left(3{x}^{2}+5{x}^{2}–4x+3x+1–8\right)–\left(4{x}^{2}–7x+3\right)$

$=8{x}^{2}-x-7-4{x}^{2}+7x-3$

$=4{x}^{2}+6x-10$

2.   $\text{B}+\text{C}-\text{A}$

$=5{x}^{2}+3x-8+4{x}^{2}-7x+3-\left(3{x}^{2}-4x+1\right)$

By combining the like terms,

$=\left(5{x}^{2}+4{x}^{2}+3x-7x-8+3\right)-\left(3{x}^{2}–4x+1\right)$

$=\left(9{x}^{2}-4x-5\right)-\left(3{x}^{2}–4x+1\right)$

$=9{x}^{2}-4x-5-3{x}^{2}+4x-1$

$\begin{array}{l}=9{x}^{2}–3{x}^{2}–4x+4x–5-1\\ =6{x}^{2}-6\end{array}$

3.   $\text{A}+\text{B}+\text{C}$

$=3{x}^{2}-4x+1+5{x}^{2}+3x-8+4{x}^{2}-7x+3$

By combining the like terms,

$=3{x}^{2}+5{x}^{2}+4{x}^{2}-4x+3x-7x+1–8+3$

$=12{x}^{2}-8x-4$

# Question: 69

If  and

, find $a$, when $\text{P}+\text{Q}+\text{R}=ax.$

## Solution

Given,

Also given, $\text{P}+\text{Q}+\text{R}=ax$

Put the values of P, Q and R on LHS, we get

$-\left(x-2\right)+\left[-2\left(y+1\right)\right]+\left(-x+2y\right)=ax$

$-x+2+\left(-2y-2\right)-x+2y=ax$

By combining the like terms,

$\begin{array}{l}-x–x–2y+2y+2–2=ax\\ -2x=ax\end{array}$

By comparing LHS and RHS, we get $a=-2$

# Question: 70

From the sum of  and $1-{x}^{2}-{y}^{2}$ subtract .

## Solution

Sum of ${x}^{2}-{y}^{2}-1,{y}^{2}-{x}^{2}-1$ and  $\begin{array}{l}1-{x}^{2}-{y}^{2}\\ ={x}^{2}-{y}^{2}-1+{y}^{2}-{x}^{2}-1+1-{x}^{2}-{y}^{2}\end{array}$

By combining like terms,

$\begin{array}{l}={x}^{2}–{x}^{2}–{x}^{2}–{y}^{2}+{y}^{2}–{y}^{2}–1–1+1\\ =-{x}^{2}–{y}^{2}–1\end{array}$

Now, subtract $–\left(1+{y}^{2}\right)$ from $–{x}^{2}–{y}^{2}-1$

$=-{x}^{2}–{y}^{2}-1–\left[-\left(1+{y}^{2}\right)\right]$

$\begin{array}{l}=–{x}^{2}–{y}^{2}–1+1+{y}^{2}\\ =-{x}^{2}–{y}^{2}+{y}^{2}–1+1\\ =-{x}^{2}\end{array}$

# Question: 71

Subtract the sum of $12ab-10{b}^{2}-18{a}^{2}$ and $9ab+12{b}^{2}+14{a}^{2}$ from the sum of $ab+2{b}^{2}$ and $3{b}^{2}-{a}^{2}$.

## Solution

Sum of

$9ab+12{b}^{2}+14{a}^{2}$

$=12ab–10{b}^{2}–18{a}^{2}+9ab+12{b}^{2}+14{a}^{2}$

By combining the like terms,

$\begin{array}{l}=12ab+9ab–10{b}^{2}+12{b}^{2}–18{a}^{2}+14{a}^{2}\\ =21ab+2{b}^{2}–4{a}^{2}\end{array}$

Sum of

$\begin{array}{l}=ab+2{b}^{2}+3{b}^{2}–{a}^{2}\\ =ab+5{b}^{2}–{a}^{2}\end{array}$

Now, subtracting
we get,

$=ab+5{b}^{2}–{a}^{2}–21ab–2{b}^{2}+4{a}^{2}$

By combining the like terms,

$=ab–21ab+5{b}^{2}–2{b}^{2}–{a}^{2}+4{a}^{2}$

$\begin{array}{l}=–20ab+3{b}^{2}+3{a}^{2}\\ =3{a}^{2}+3{b}^{2}–20ab\end{array}$

# Question: 72

Each symbol given below represents an algebraic expression:

$=2{x}^{2}+3y,$

$=5{x}^{2}+3x,$

$=8{y}^{2}-3{x}^{2}+2x+3y$

The symbols are then represented in the expression:

$+$                   $-$

Find the expression which is represented by the above symbols.

## Solution

Given,

$=2{x}^{2}+3y,$

$=5{x}^{2}+3x,$

$=8{y}^{2}-3{x}^{2}+2x+3y$

$\therefore$          $+$               $-$

$=\left(2{x}^{2}+3y\right)+\left(5{x}^{2}+3x\right)-\left(8{y}^{2}-3{x}^{2}+2x+3y\right)$

$=2{x}^{2}+3y+5{x}^{2}+3x-8{y}^{2}+3{x}^{2}-2x-3y$

By combining the like terms,

$=2{x}^{2}+5{x}^{2}+3{x}^{2}+3y-3y+3x-2x-8{y}^{2}$

# Question: 73

Observe the following nutritional chart carefully:

 Food Item (Per Unit $=100$ g) Carbohydrates Rajma $60$ g Cabbage $5$ g Potato $22$ g Carrot $11$ g Tomato $4$ g Apples $14$ g

Write an algebraic expression for the amount of carbohydrates in ‘g’ for

a. $y$ units of potatoes and 2 units of rajma

b. $2x$ units tomatoes and y units apples.

## Solution

(a)

By unitary method,

$\therefore$ $1$ unit of potatoes contain carbohydrates

$=22\text{\hspace{0.17em}}\text{g}$

of potatoes contain carbohydrates

Similarly,

$\therefore$ $1$ unit of rajma contain carbohydrates

$\therefore$ $2$ units of rajma contain carbohydrates

Hence, required expression is $22y+120.$

(b)

By unitary method,

$\therefore$ $1$ unit of tomatoes contain carbohydrates

$\therefore$ $2x$ units of tomatoes contain carbohydrates

Similarly,

$\therefore$ $1$ unit apples contain carbohydrates

$y$ units apples contain carbohydrates

Hence, the required expression is $8x+14y.$

# Question: 74

Arjun bought a rectangular plot with length $x$ and breadth $y$ and then sold a triangular part of it whose base is $y$ and height is $z$. Find the area of the remaining part of the plot.

## Solution

Given,

Arjun bought a rectangular plot with length $x$ and breadth $y$

$\therefore$ area of rectangular plot

$\begin{array}{l}=l×b\\ =x×y\\ =xy\end{array}$

Also, given triangular part of it whose base is $y$ and height is $z$ so, area of triangular part

$\begin{array}{l}=\frac{\text{1}}{2}×y×z\\ =\frac{\text{1}}{2}×yz\end{array}$

Area of remaining part of the plot

$=$ Area of rectangular plot $–$ Area of triangular plot

$\begin{array}{l}=xy-\frac{\text{1}}{2}yz\\ =y\left(x-\frac{\text{1}}{2}z\right)\end{array}$

# Question: 75

Amisha has a square plot of side m and another triangular plot with base and height each equal to m. What is the total area of both plots?

## Solution

Given,
side of square plot $=m$ and
height & base of triangular plot $=m$

Area of square plot, ${\left(\text{side}\right)}^{2}={m}^{2}$

Area of triangular plot,

$\begin{array}{l}\left(\frac{1}{2}×h×b\right)\\ =\frac{1}{2}×m×m\\ =\frac{{m}^{2}}{2}\end{array}$

Total area of both plots

$=$ Area of square plot $+$ Area of triangular plot

$\begin{array}{l}={m}^{2}+\frac{{m}^{2}}{2}\\ =\frac{2{m}^{2}+{m}^{2}}{2}\\ =\frac{3{m}^{2}}{2}\end{array}$

[taking LCM of $1&2$ is $2$ ]

# Question: 76

A taxi service charges $Rs$ 8 per km and levies a fixed charge of $Rs$ 50.Write an algebraic expression for the above situation, if the taxi is hired for $x$ km.

## Solution

As per the given information, taxi service charged  per km and fixed charged of 50. If taxi is hired for $x$ km. Then, algebraic expression for the situation $=8×x+50=8x+50$

Hence, the required expression is $8x+50.$

# Question: 77

Shiv works in a mall and gets paid $\text{Rs}$ 50 per hour. Last week he worked for 7 hours and this week he will work for $x$ hours. Write an algebraic expression for the money paid to him for both the weeks.

## Solution

Given, money paid to shiv

$\therefore$ Money paid last week $\begin{array}{l}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}50×7\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}350\end{array}$

So, money paid this week $\begin{array}{l}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}50×x\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}50x\end{array}$

Total money paid to shiv $\begin{array}{l}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\left(350+50x\right)\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}50\left(x+7\right)\end{array}$

# Question: 78

Sonu and Raj have to collect different kinds of leaves for science project. They go to a park where Sonu collects $12$ leaves and Raj collects $x$ leaves.

After some time Sonu loses $3$ leaves and Raj collects $2x$ leaves. Write an algebraic expression to find the total number of leaves collected by both of them.

## Solution

According to the question,

Sonu collected leaves $\begin{array}{l}=12-3\\ =9\end{array}$

Raj collected leaves $\begin{array}{l}=x+2x\\ =3x\end{array}$

$\therefore$ Total leaves collected $=9+3x$

Hence, the required expression is $9+3x.$

# Question: 79

A school has a rectangular play ground with length $x$ and breadth $y$ and a square lawn with side $x$ as shown in the figure given below.

What is the total perimeter of both of them combined together?

## Solution

Given, Length of rectangular playground, $\text{AB}=x$

& breadth of rectangular playground, $\text{BC}=y$

FCDE is a square, i.e., $\text{FC}=\text{CD}=\text{EF}=\text{DE}=x$

ABCF is a rectangle, i.e., $\text{AB}=\text{FC}=x\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{BC}=\text{AF}=y$

Now, perimeter of combined (playground $+$ lawn)
$=$ Sum of all sides

$=\text{AB}+\text{BC}+\text{CD}+\text{DE}+\text{EF}+\text{FA}$

$=x+y+x+x+x+y=4x+2y$

# Question: 80

The rate of planting the grass is $\text{Rs}$$x$ per square meter. Find the cost of planting the grass on a triangular lawn whose base is $y$ meters and height is $z$ meters.

## Solution

Given,
base of triangular lawn is $y$ meters and height $z$ meters.

Area of triangular lawn
$\begin{array}{l}=\frac{1}{2}×y×z\\ =\frac{1}{2}yz\text{\hspace{0.17em}}{\text{m}}^{2}\end{array}$

Cost of planting the grass on lawn
$\begin{array}{l}=\frac{1}{2}yz×x\\ =\text{\hspace{0.17em}}\text{Rs}\frac{1}{2}xyz\end{array}$

# Question: 81

Find the perimeter of the figure given below:

## Solution

We know that, perimeter is the sum of all sides. Perimeter of the given figure
$=\text{AB}+\text{BC}+\text{CD}+\text{DA}$

$=2\left(x+y\right)+\left(5x-y\right)+2\left(x+y\right)+\left(5x-y\right)$

$=2x+2y+5x-y+2x+2y+5x-y$

On combining the like terms,

$\begin{array}{l}=5x+2x+5x+2x-y+2y-y+2y\\ =14x+2y\end{array}$

# Question: 82

In a rectangular plot, $5$ square flower beds of side $\left(x+2\right)$ meters each have been laid (see figure given below). Find the total cost of fencing the flower beds at the cost of $\text{Rs}\text{\hspace{0.17em}}50$ per $100$ meters.

## Solution

Given,
side of one square flower bed

Perimeter of one square flower bed

Now, total perimeter of such square flower beds
$=5x$ perimeter of $1$ square

$=5×4\left(x+2\right)$

Cost of fencing of

Cost of

Cost of

$\begin{array}{l}=\frac{50}{100}×20\left(x+2\right)\\ =10\left(x+2\right)\end{array}$

$=Rs\text{\hspace{0.17em}}\left(10x+20\right)$

# Question: 83

A wire is $\left(7x-3\right)$ meters long. A length of $\left(3x-4\right)$ meters is cut for use. Now, answer the following questions:

a.    How much wire is left?

b.   If this left out wire is used for making an equilateral triangle. What is the length of each side of the triangle so formed?

## Solution

Given,
length of wire  and wire cut for use has length

a. Left wire $=\left(7x-3\right)-\left(3x-4\right)$

$\begin{array}{l}=7x-3-3x+4\\ =7x-3x-3+4\end{array}$

b. $\therefore$ Left wire

$\therefore$ Perimeter of equilateral triangle $=$ Length of wire left

$⇒3×\left(\text{side}\right)=4x+1$

# Question: 84

Rohan's mother gave him $\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{\hspace{0.17em}}x{y}^{2}$ and his father gave him $\text{Rs}$$5\left($