Unit 4: Simple Equations

## Exercise: 1 (Multiple Choice Questions and Answers 1-18)

In the Questions 1 to 18, there are four options, out of which only one is correct. Write the correct one.

# Question: 1

The solution of the equation $ax+b=0$ is

a.    $\frac{a}{b}$

b.   $-b$

c.    $-\frac{b}{a}$

d.   $\frac{b}{a}$

## Solution

(c)

Given equation $ax+b=0$

# Question: 2

If $a$ and $b$ are positive integers, then the solution of the equation $ax=b$ will always be $a$

a.    positive number

b.   negative number

c.    $1$

d.   $0$

## Solution

(a)

Given equation $ax=b$

On dividing the equation by a, we get $x=\frac{b}{a}$

Now, if $a$ and $b$ are positive integers, then the solution of the equation is also positive number as division of two positive integers is also a positive number.

# Question: 3

Which of the following is not allowed in a given equation?

a.    Adding the same number to both sides of the equation.

b.   Subtracting the same number from both sides of the equation.

c.    Multiplying both sides of the equation by the same non-zero number.

d.   Dividing both sides of the equation by the same number.

## Solution

(d)

Dividing both sides of the equation by the same non-zero number is allowed in $a$ given equation, division of any number by zero is not allowed as the division of number by zero is not defined.

Note: If we add or subtract the same number to both sides of the equation while adding or subtracting, then there will be no change in the given equation.

# Question: 4

The solution of which of the following equations is neither a positive fraction nor an integer?

a.    $2x+6=0$

b.   $3x-5=0$

c.    $5x-8=x+4$

d.   $4x+7=x+2$

## Solution

(d)

Let us solve the equations:

a.    Given equation is $2x+6=0$

b.   Given equation is $3x-5=0$

c.    Given equation is $5x-8=x+4$

d.   Given equation is $4x+7=x+2$

Which is neither a positive fraction nor an integer.

# Question: 5

The equation which cannot be solved in integers is

a.    $5y-3=-18$

b.   $3x-9=0$

c.    $3z+8=3+z$

d.   $9y+8=4y-7$

## Solution

(c)

Let us solve the equations

a.    Given equation is $5y-3=-18$

$\begin{array}{l}⇒5y=-18+3\\ ⇒5y=-15\\ ⇒y=-3\text{}\text{(integer)}\end{array}$

b.   Given equation is $3x-9=0$

$\begin{array}{l}⇒3x=9\\ ⇒x=3\end{array}$

c.    Given equation is $3z+8=3+z$

On transposing  to LHS & RHS respectively, we get

d.   Given equation is $9y+8=4y-7$

On transposing  to LHS & RHS respectively, we get

# Question: 6

If $7x+4=25\text{,}$ then $x$ is equal to

a.    $\frac{29}{7}$

b.   $\frac{100}{7}$

c.    $2$

d.   $3$

## Solution

(d)

Given equation is $7x+4=25$

On dividing the above equation by $7\text{,}$ we get

$⇒x=3$

Hence, the solution of the given eq. is $3.$

# Question: 7

The solution of the equation $3x+7=-20$ is

a.    $\frac{17}{7}$

b.   $-9$

c.    $9$

d.   $\frac{13}{3}$

## Solution

(b)

Given equation is $3x+7=-20$

$\begin{array}{l}⇒3x=-20-7\\ ⇒3x=-27\\ ⇒x=\frac{-27}{3}=-9\end{array}$

Hence, the solution of eq. is $-9.$

# Question: 8

The value of $y$ for which the expressions $\left(y-15\right)$ and $\left(2y+1\right)$ become equal is

a.    $0$

b.   $16$

c.    $8$

d.   $-16$

## Solution

(d)

It is given that both the equation are equal. So, the equation is

$\begin{array}{l}⇒y-15=2y+1\\ ⇒y-2y=1+15\\ ⇒-y=16\end{array}$

Multiplying both sides by $-1\text{,}$ we get

$⇒y=-16$

# Question: 9

If $k+7=16\text{,}$ then the value of $8k-72$ is

a.    $0$

b.   $1$

c.    $112$

d.   $56$

## Solution

(a)

Given equation is $k+7=16$

On transposing $7$ to RHS, we get

$⇒k=16-7=9$

Put the value of $k$ in the equation $\left(8k-72\right)\text{,}$ we get

$⇒8\left(9\right)-72=72-72=0$

# Question: 10

If  then the value of m is

a.    $0.002$

b.   $0.02$

c.    $0.2$

d.   $2$

## Solution

(a)

Given equation is

On dividing the given equation by $43\text{,}$ we get

$⇒m=\frac{0.086}{43}$

If we remove the decimal, we get $1000$ in denominator

$⇒m=\frac{86}{43}×\frac{1}{1000}=\frac{2}{1000}=0.002$

# Question: 11

$x$ exceeds $3$ by $7\text{,}$ can be represented as

a.    $x+3=2$

b.   $x+7=3$

c.    $x-3=7$

d.   $x-7=3$

## Solution

The given statement means $x$ is $7$ more than $3.$

So, the equation is $x-7=3$

We can also write it as $x-3=7.$

# Question: 12

The equation having $5$ as a solution is:

a.    $4x+1=2$

b.   $3-x=8$

c.    $x-5=3$

d.   $3+x=8$

## Solution

(d)

Let us solve the equations:

a.    Given equation is $4x+1=2$

$\begin{array}{l}⇒4x=2-1\\ ⇒4x=1\\ ⇒x=\frac{1}{4}\end{array}$

b.   Given equation is $3-x=8$

$\begin{array}{l}⇒-x=8-3\\ ⇒-x=5\\ ⇒x=-5\end{array}$

c.    Given equation is $x-5=3$

$\begin{array}{l}⇒x=3+5\\ ⇒x=8\end{array}$

d.   Given equation is $3+x=8$

$\begin{array}{l}⇒x=8-3\\ ⇒x=5\end{array}$

# Question: 13

The equation having $-3$ as a solution is:

a.    $x+3=1$

b.   $8+2x=3$

c.    $10+3x=1$

d.   $2x+1=3$

## Solution

(c)

Let us solve the equations:

a.    Given equation is $x+3=1$

$\begin{array}{l}⇒x=1-3\\ ⇒x=-2\end{array}$

b.   Given equation is $8+2x=3$

$\begin{array}{l}⇒2x=3-8\\ ⇒2x=-5\\ ⇒x=-\frac{5}{2}\end{array}$

c.    Given equation is $10+3x=1$

$\begin{array}{l}⇒3x=1-10\\ ⇒3x=-9\\ ⇒x=-3\end{array}$

Further, there is no need to solve for option d as we have obtained correct option.

# Question: 14

Which of the following equations can be formed starting with $x=0\text{?}$

a.    $2x+1=-1$

b.   $\frac{x}{2}+5=7$

c.    $3x-1=-1$

d.   $3x-1=1$

## Solution

(c)

We have $x=0$

On multiplying both the sides by $3\text{,}$ we get

$\begin{array}{l}⇒3×x=3×0\\ ⇒3x=0\end{array}$

On adding $\left(-1\right)$ to both sides, we get

$\begin{array}{l}⇒3x+\left(-1\right)=0+\left(-1\right)\\ ⇒3x-1=-1\end{array}$

# Question: 15

Which of the following equations cannot be formed using the equation $x=7\text{?}$

a.    $2x+1=15$

b.   $7x-1=50$

c.    $x-3=4$

d.   $\frac{x}{7}-1=0$

## Solution

(b)

We have $x=7$

On multiplying both the sides by $7\text{,}$ we get

$\begin{array}{l}⇒7×x=7×7\\ ⇒7x=49\end{array}$

On adding $\left(-1\right)$ both sides, we get

$\begin{array}{l}⇒7x+\left(-1\right)=49+\left(-1\right)\\ ⇒7x-1=49-1\\ ⇒7x-1=48\end{array}$

# Question: 16

If $\frac{x}{2}=3\text{,}$ then the value of $3x+2$ is

a.    $20$

b.   $11$

c.    $\frac{13}{2}$

d.   $8$

## Solution

(a)

Given $\frac{x}{2}=3$

On multiplying both the sides by $2\text{,}$ we get $\frac{x}{2}×2=3×2$

$⇒x=3×2=6$

Put $x=6$ in equation $3x+2\text{,}$ we get

$⇒3\left(6\right)+2=18+2=20$

# Question: 17

Which of the following numbers satisfy the equation $-6+x=-12\text{?}$

a.    $2$

b.   $6$

c.    $-6$

d.   $-2$

## Solution

(c)

Let us put the values given in the options in equation $-6+x=-12$

a.    Put $x=2$

$\begin{array}{l}⇒-6+2=-12\\ ⇒-4=-12\end{array}$

LHS $\ne$ RHS

b.   Put $x=6$

$\begin{array}{l}⇒-6+\left(6\right)=-12\\ ⇒0=-12\end{array}$

LHS $\ne$ RHS

c.    Put $x=-6$

$\begin{array}{l}⇒-6+\left(-6\right)=-12\\ ⇒-12=-12\end{array}$

LHS $=$ RHS    (satisfied)

Now, there is no need to check the next option.

Hence, $x=-6$ satisfied the given equation.

# Question: 18

Shifting one term from one side of an equation to another side with a change of sign is known as

a.    commutativity

b.   transposition

c.    distributivity

d.   associativity

## Solution

(b)

Transposition means shifting one term from one side of an equation to another side with a change of sign.

In Questions 19 to 48, fill in the blanks to make the statements true.

# Question: 19

The sum of two numbers is $60$ and their difference is $30.$

a.    If smaller number is $x\text{,}$ the other number is ________.(use sum)

b.   The difference of numbers in term of $x$ is ________.

c.    The equation formed is ________.

d.   The solution of the equation is ________.

e.    The numbers are ________ and ________.

## Solution

a.    If the smaller number is $x\text{,}$ then the other number is $\left(60-x\right)\text{,}$ since the sum of both numbers is $60.$

b.   Given, one number

Then, other number $=\left(60-x\right)$

$\therefore$ Difference $=\left(60-x\right)-x=60-2x$

c.    Given, difference between two no. is $30$

So, the equation is $60-2x=30$

d.   Let us solve the equation for $x$

$60-2x=30$

$\begin{array}{l}⇒-2x=30-60\\ ⇒-2x=-30\\ ⇒2x=30\end{array}$

$⇒2x=30$

On dividing the equation by $2\text{,}$ we get

$\begin{array}{l}⇒\frac{2x}{2}=\frac{30}{2}\\ ⇒x=15\end{array}$

Hence, the solution of the equation is $15$

e.    The numbers are

Now, put the value of $x\text{,}$ we get

First number $=15$

Second number $=60-15=45$

# Question: 20

Sum of two numbers is $81.$ One is twice the other.

a.    If smaller number is $x\text{,}$ the other number is ________.

b.   The equation formed is ________.

c.    The solution of the equation is ________.

d.   The numbers are ________ and ________.

## Solution

a.    We are given that one number is twice the other. If smaller number is $x\text{,}$ then the other number is $2x\text{.}$

b.   We are given that sum of two numbers is $81.$ So, the equation will be $x+2x=81$

c.    Now, solve the equation for $x\text{,}$

$\begin{array}{l}⇒x+2x=81\\ ⇒3x=81\\ ⇒x=\frac{81}{3}\\ ⇒x=27\end{array}$

Hence, the solution of the equation is $27.$

d.   The two numbers are

# Question: 21

In a test Abha gets twice the marks as that of Palak. Two times Abha's marks and three times Palak's marks make $280.$

a.    If Palak gets $x$ marks, Abha gets ________ marks.

b.   The equation formed is ________.

c.    The solution of the equation is ________.

d.   Marks obtained by Abha are ________.

## Solution

a.    If Palak gets $x$ marks, Abha gets twice the marks as that of Palak, i.e., $2x$

b.   Two times of Abha’s marks $=2\left(2x\right)=4x$ and three times the Palak marks $=3\left(x\right)=3x$

Now, two times Abha’s marks and three times Palak’s marks make $280.$ So, the equation formed is $4x+3x=280.$

c. Solve the equation for $x\text{,}$

$\begin{array}{l}⇒4x+3x=280\\ ⇒7x=280\\ ⇒x=\frac{280}{7}\\ ⇒x=40\end{array}$

Hence, the solution of the equation is $40$

d. Marks obtained by Abha are $2x$ i.e., $2×40=80.$

# Question: 22

The length of a rectangle is two times its breadth. Its perimeter is

a.    If the breadth of rectangle is  the length of the rectangle is ________.

b.   Perimeter in terms of $x$ is ________.

c.    The equation formed is ________.

d.   The solution of the equation is ________.

## Solution

a.    It is given that the length of the rectangle is two times its breadth.

$\therefore$ Length

b.   Perimeter of rectangle $=2\left(\text{Length}+\text{Breadth}\right)=2\left(2x+x\right)$

c.    As we are given that perimeter of rectangle is

So, the equation formed is $2\left(2x+x\right)=60$

d.   On solving,

$\begin{array}{l}⇒2\left(3x\right)=60\\ ⇒6x=60\end{array}$

On dividing the equation by $6\text{,}$ we get

$\begin{array}{l}⇒\frac{6x}{6}=\frac{60}{6}\\ ⇒x=10\end{array}$

Hence, the solution of the equation is $10.$

# Question: 23

In a bag there are $5$ and $2$ rupee coins. If they are equal in number and their worth is $Rs\text{\hspace{0.17em}}70\text{,}$ then

a.    The worth of $x$ coins of $Rs\text{\hspace{0.17em}}5$ each ________.

b.   The worth of $x$ coins of $Rs\text{\hspace{0.17em}}2$ each ________.

c.    The equation formed is ________.

d.   There are ________ $5$ rupee coins and ________ $2$ rupee coins.

## Solution

Let number of coins of $Rs\text{\hspace{0.17em}}5=x$

Then, number of coins of $Rs\text{\hspace{0.17em}}2=x$

a.    Number of coins of $Rs5=x$

So, the worth of $Rs5$ of $x$ coins $=\text{\hspace{0.17em}}Rs5×x=\text{\hspace{0.17em}}Rs5x$

b.   Similarly, the worth of $12$ of x coins $=\text{\hspace{0.17em}}Rs2x$

c.    As we are given that they are equal in number & the worth is $70$

So, the equation is

$⇒\text{\hspace{0.17em}}5x+2x=70$

d.    On solving,

$\begin{array}{l}⇒\text{\hspace{0.17em}}5x+2x=70\\ ⇒7x=70\\ ⇒\frac{7x}{7}=\frac{70}{7}\\ ⇒x=10\end{array}$

Therefore, there are 10 coins of both 5 rupee and 2 rupee.

# Question: 24

In a Mathematics quiz, $30$ prizes consisting of $1\text{st}$ and $2\text{nd}$ prizes only are to be given. $1\text{st}$ and $2\text{nd}$ prizes are worth $Rs2000$ and $Rs1000\text{,}$ respectively. If the total prize money is $Rs52\text{,}000$ then show that:

a.    If $1\text{st}$ prizes are $x$ in number the number of $2\text{nd}$ prizes are ________.

b.   The total value of prizes in terms of $x$ are ________.

c.    The equation formed is ________.

d.   The solution of the equation is ________.

e.    The number of $1\text{st}$ prizes are ________ and the number of $2\text{nd}$ prizes are ________.

## Solution

Given, number of prizes $=30$

Total prize money $=\text{\hspace{0.17em}}Rs52\text{,}000\text{,}$

$1\text{st}$ and $2\text{nd}$ prizes are worth $Rs2000$ and $Rs1000\text{,}$ respectively.

a.    $1\text{st}$ prizes are $x$ in number, the number of $2\text{nd}$ prizes are $\left(30-x\right)\text{,}$ because total number of prizes are $30.$

b.   Total values of prizes in terms of $x$ are $2000x+1000\left(30-x\right)\text{.}$

c.    The equation formed is

$2000x+1000\left(30-x\right)=52000$

d.    On solving,

$\begin{array}{l}⇒2000x+30000-1000x=52000\\ ⇒1000x+30000=52000\end{array}$

$\begin{array}{l}⇒1000x=22000\\ ⇒x=\frac{22000}{1000}\end{array}$

The solution of the equation is $22.$

e.     So, Number of ${2}^{\text{nd}}$ prizes $=30-22=8$.

The number of ${1}^{\text{st}}$ prizes are $22$ & the number of ${2}^{\text{nd}}$ prizes are $8.$

# Question: 25

If $z+3=5\text{,}$ then $z=$ ________.

## Solution

On solving the given equation, we get

$\begin{array}{l}⇒z+3=5\\ ⇒z=5-3\\ ⇒z=2\end{array}$

# Question: 26

_________ is the solution of the equation $3x-2=7.$

## Solution

On solving the equation for $x\text{,}$

$\begin{array}{l}⇒3x-2=7\\ ⇒3x=7+2\\ ⇒3x=9\\ ⇒x=\frac{9}{3}=3\end{array}$

# Question: 27

__________ is the solution of  $3x+10=7.$

## Solution

Solving the equation for $x\text{,}$

# Question: 28

If $2x+3=5\text{,}$ then value of $3x+2$ is __________.

## Solution

Solving the equation for $x\text{,}$

$\begin{array}{l}2x+3=5\\ ⇒2x=5-3\\ ⇒2x=2\\ ⇒x=\frac{2}{2}=1\end{array}$

Put the value of $x\text{,}$ in $3x+2\text{,}$ we get

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

# Question: 29

In integers, $4x-1=8$ has __________ solution.

## Solution

Solving the equation for $x\text{,}$

$\begin{array}{l}4x-1=8\\ ⇒4x=8+1\\ ⇒4x=9\\ ⇒x=\frac{9}{4}\end{array}$

Since the solution of the equation is not an integer, hence the equation has no solution.

# Question: 30

In natural numbers, $4x+5=-7$ has ______ solution.

## Solution

Solve the equation for $x\text{,}$

$\begin{array}{l}4x+5=-7\\ ⇒4x=-7-5\\ ⇒4x=-12\\ ⇒x=-\frac{12}{4}=-3\end{array}$

Since, the value of $x$ is not a natural number, hence equation has no solution in natural numbers.

# Question: 31

In natural numbers, $x-5=-5$ has ______ solution.

## Solution

Solving the equation for $x\text{,}$

$\begin{array}{l}x-5=-5\\ x=-5+5\\ x=0\end{array}$

Since, natural numbers do not contain zero, hence the equation has no solution.

# Question: 32

In whole numbers, $x+8=12-4$ has ______ solution.

## Solution

Solving the equation for $x\text{,}$

$\begin{array}{l}x+8=12-4\\ x+8=8\\ x=8-8\\ x=0\end{array}$

Since, zero is in the range of whole numbers, hence the equation has one solution.

# Question: 33

If $5$ is added to three times a number, it becomes the same as $7$ is subtracted from four times the same number. This fact can be represented as __________.

## Solution

Let the number be $x\text{.}$

Now, $5$ is added to $3$ times the number $5+3x\text{.}$

It is same as $7$ is subtracted from $4$ times the number, i.e. $4x-7.$

So, the equation formed is $5+3x=4x-7.$

# Question: 34

$x+7=10$ has the solution _________.

## Solution

Solving the equation for $x\text{,}$

$\begin{array}{l}x+7=10\\ ⇒x=10-7\\ ⇒x=3\end{array}$

# Question: 35

when $3x=12.$

## Solution

Given, $3x=12$

$\begin{array}{l}⇒\frac{3x}{3}=\frac{12}{3}\\ ⇒x=4\\ ⇒x-0=4-0=4\end{array}$

# Question: 36

when $2x=2.$

## Solution

Given, $2x=2$

$\begin{array}{l}⇒\frac{2x}{2}=\frac{2}{2}\\ ⇒x=1\\ ⇒x-1=1-1=0\end{array}$

# Question: 37

$x-\text{__________}=15\text{;}$ when $\frac{x}{2}=6.$

## Solution

Given, $\frac{x}{2}=6$

$\begin{array}{l}⇒x=12\\ ⇒12-?=15\\ ⇒-?=15-12\\ ⇒-?=3\\ ⇒?=-3\end{array}$

Hence, $x-\left(-3\right)=15$

# Question: 38

The solution of the equation $x+15=19$ is ________.

## Solution

Solving the equation for $x\text{,}$

$\begin{array}{l}x+15=19\\ ⇒x=19-15\\ ⇒x=4\end{array}$

Hence, the solution of the given equation is $4.$

# Question: 39

Finding the value of a variable in a linear equation that _______ the equation is called a _______ of the equation.

## Solution

Finding the value of a variable in a linear equation that satisfies the equation is called a root of the equation.

# Question: 40

Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the _______ of the term.

## Solution

Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the sign of the term.

# Question: 41

If $\frac{9}{5}x=\frac{18}{5}\text{,}$ then $x=\text{_______}\text{.}$

## Solution

Given, $\frac{9}{5}x=\frac{18}{5}$

$\begin{array}{l}⇒\frac{9}{5}x÷\frac{9}{5}=\frac{18}{5}÷\frac{9}{5}\\ ⇒x=\frac{18}{5}×\frac{5}{9}=2\end{array}$

# Question: 42

If $3-x=-4\text{,}$ then $x=\text{_______}\text{.}$

## Solution

Solving the equation for $x\text{,}$

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

# Question: 43

If $x-\frac{1}{2}=-\frac{1}{2}\text{,}$ then $x=\text{_______}\text{.}$

## Solution

Given, $x-\frac{1}{2}=-\frac{1}{2}$

$\begin{array}{l}⇒x=-\frac{1}{2}+\frac{1}{2}\\ ⇒x=0\end{array}$

# Question: 44

If $\frac{1}{6}-x=\frac{1}{6}\text{,}$ then $x=\text{_______}\text{.}$

## Solution

Given, $\frac{1}{6}-x=\frac{1}{6}$

$\begin{array}{l}⇒-x=\frac{1}{6}-\frac{1}{6}\\ ⇒-x=0\\ ⇒x=0\end{array}$

# Question: 45

If $10$ less than a number is $65\text{,}$ then the number is _______.

## Solution

Let the number be $x\text{.}$

Then, the equation will be $x-10=65$

Now, solving the equation for $x\text{.}$

$\begin{array}{l}⇒x=65+10\\ ⇒x=75\end{array}$

Hence, the number is $75.$

# Question: 46

If a number is increased by $20\text{,}$ it becomes $45.$ Then the number is _______.

## Solution

Let the number be $x\text{.}$

If it is increased by $20\text{,}$ it becomes $\left(x+20\right)\text{,}$

So, the equation formed is $x+20=45$

$\begin{array}{l}⇒x=45-20\\ ⇒x=25\end{array}$

Hence, the number is $25.$

# Question: 47

If $84$ exceeds another number by $12\text{,}$ then the other number is _______.

## Solution

Equation formed: $84-x=12$

Solving the equation for $x\text{,}$

$\begin{array}{l}84-x=12\\ ⇒-x=12-84\\ ⇒-x=-72\\ ⇒x=72\end{array}$

Hence, the other number is $72.$

# Question: 48

If $x-\frac{7}{8}=\frac{7}{8}\text{,}$ then $x=\text{_______}\text{.}$

## Solution

Given, $x-\frac{7}{8}=\frac{7}{8}$

$\begin{array}{l}⇒x=\frac{7}{8}+\frac{7}{8}\\ ⇒x=\frac{7+7}{8}\\ ⇒x=\frac{14}{8}\\ ⇒x=\frac{7}{4}\end{array}$

In Questions 49 to 55, state whether the statements are True or False.

# Question: 49

$5$ is the solution of the equation $3x+2=17.$

## Solution

True

Solving the equation for $x\text{,}$

$\begin{array}{l}⇒3x+2=17\\ ⇒3x=17-2\\ ⇒3x=15\\ ⇒x=\frac{15}{3}=5\end{array}$

# Question: 50

$\frac{9}{5}$ is the solution of the equation $4x-1=8.$

## Solution

False

Solving the equation for $x\text{,}$

$\begin{array}{l}⇒4x-1=8\\ ⇒4x=8+1\\ ⇒4x=9\\ ⇒x=\frac{9}{4}\end{array}$

# Question: 51

$4x-5=7$ does not have an integer as its solution.

## Solution

False

Given equation $4x-5=7$

# Question: 52

One third of a number added to itself gives $10\text{,}$ can be represented as $\frac{x}{3}+10=x\text{.}$

## Solution

False

Let the number be $x\text{.}$

Then, the equation formed is $\frac{x}{3}+x=10$

# Question: 53

$\frac{3}{2}$ is the solution of the equation $8x-5=7.$

## Solution

True

Solving the equation for $x\text{,}$

$\begin{array}{l}8x-5=7\\ 8x=7+5\\ 8x=12\\ x=\frac{12}{8}\\ x=\frac{3}{2}\end{array}$

# Question: 54

If $4x-7=11\text{,}$ then $x=4.$

## Solution

False

Solving the equation for $x\text{,}$

$\begin{array}{l}4x-7=11\\ 4x=11+7\\ 4x=18\\ x=\frac{18}{4}=\frac{9}{2}\end{array}$

# Question: 55

If $9$ is the solution of variable $x$ in the equation $\frac{5x-7}{2}=y\text{,}$ then the value of $y$ is $28.$

## Solution

False

Given, $x=9$

Put the value of $x$ in the equation, we get

$\begin{array}{l}\frac{5\left(9\right)-7}{2}=y\\ \frac{45-7}{2}=y\\ \frac{38}{2}=y\\ y=19\end{array}$

# Question: 56

Match each of the entries in Column I with the appropriate entries in Column II.

 Column I Column II i. $x+5=9$ (a)      $-\frac{5}{3}$ ii.   $x-7=4$ (b)     $\frac{5}{3}$ iii.             $\frac{x}{12}=-5$ (c)      $4$ iv.              $5x=30$ (d)     $6$ v.   The value of $y$ which satisfies $3y=5$ (e)      $11$ vi.              If $p=2\text{,}$ then the value of $\frac{1}{3}\left(1-3p\right)$ (f)       $-60$ (g)     $3$

## Solution

i. (c)

Given, $x+5=9$

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

ii.   (e)

Given, $x-7=4$

$\begin{array}{l}x=4+7\\ x=11\end{array}$

iii.             (f)

Given, $\frac{x}{12}=-5$

$\begin{array}{l}12×\frac{x}{12}=-5×12\\ x=-60\end{array}$

iv.              (d)

Given, $5x=30$

$x=\frac{30}{5}=6$

v.   (b)

Given, $3y=5$

$y=\frac{5}{3}$

vi.              (a)

Given, $p=2$

Put the value of $p$ in the equation $=\frac{1}{3}×\left(1-3p\right)\text{,}$ we get

$\begin{array}{l}=\frac{1}{3}×\left(1-3×2\right)\\ =\frac{1}{3}×\left(1-6\right)\\ =\frac{1}{3}×\left(-5\right)\\ =-\frac{5}{3}\end{array}$

In Questions 57 to 67, express each of the given statements as an equation.

# Question: 57

$13$ subtracted from twice of a number gives $3.$

## Solution

Let the number be $x\text{.}$

$13$ is subtracted from twice of a number i.e, $2x-13$ and it results into $3.$

So, the equation formed is $2x-13=3$

# Question: 58

One-fifth of a number is $5$ less than that number.

## Solution

Let the number be $x$

Then $\frac{1}{5}\text{th}$ of the number $=\frac{x}{5}$

Now, $\frac{x}{5}$ is $5$ less than $x$

So, the equation formed is $\frac{x}{5}=x-5.$

# Question: 59

A number is $7$ more than one-third of itself.

## Solution

Let the number be $x$

Then, $\frac{1}{3}\text{rd}$ of the number $=\frac{x}{3}$

So, the equation formed is $x=7+\frac{x}{3}$

# Question: 60

Six times a number is $10$ more than the number.

## Solution

Let the number be $x$

Then, $6$ times of the number $=6x$

So, the equation formed is $6x=10+x$

# Question: 61

If $10$ is subtracted from half of a number, the result is $4.$

## Solution

Let the number be $x$

Then, $10$ is subtracted from $\frac{x}{2}$ i.e, $\frac{x}{2}-10$ & its result into $4.$

So, the equation formed is $\frac{x}{2}-10=4$

# Question: 62

Subtracting $5$ from $p\text{,}$ the result is $2.$

## Solution

Subtract $5$ from $p\text{,}$ i.e., $p-5$ & result is $2.$

Hence, the equation formed is $p-5=2$

# Question: 63

Five times a number increased by $7$ is $27.$

## Solution

Let the number be $x\text{.}$ Then, five times of number be $5x\text{.}$

Since it is increased by $7$ i.e, $5x+7$ & it gives result as $27.$

Hence, the equation formed is $5x+7=27$

# Question: 64

Mohan is $3$ years older than Sohan. The sum of their ages is $43$ years.

## Solution

Let age of Sohan be

Then, the age of Mohan is

$\therefore$ Sum of their ages $=43$

So, the equation formed is $x+\left\{x+3\right\}=43$

# Question: 65

If $1$ is subtracted from a number and the difference is multiplied by $\frac{1}{2}\text{,}$ the result is $7.$

## Solution

Let the number be $x$

Then, $1$ is subtracted from a number & the difference multiplied by $\frac{1}{2}$ i.e., $\frac{1}{2}\left(x-1\right)$ it gives result as $7$

So, the equation formed is $\frac{1}{2}\left(x-1\right)=7$

# Question: 66

A number divided by $2$ and then increased by $5$ is $9.$

## Solution

Let the number be $x$

Then, $x$ is divided by $2$ & increased by $5\text{,}$ i.e., $\frac{x}{2}+5$ & gives result as $9.$

So, the equation formed is $\frac{x}{2}+5=9$

# Question: 67

The sum of twice a number and $4$ is $18.$

## Solution

Let the number be $x$.

Then, sum of twice of a number and $4$ gives result $18.$

Hence, $2x+4=18$ is the equation.

# Question: 68

The age of Sohan Lal is four times that of his son Amit. If the difference of their ages is $27$ years, find the age of Amit.

## Solution

Let  be the age of Amit.

Then, age of Sohan Lal

According to question,

$\begin{array}{l}4x-x=27\\ 3x=27\\ x=\frac{27}{3}=9\end{array}$

Hence, the age of Amit is .

# Question: 69

A number exceeds the other number by $12.$ If their sum is $72\text{,}$ find the numbers.

## Solution

Let $x$ be a number, then another number will be $x+12$.

According to question, $x+x+12=72$

$\begin{array}{l}2x=72-12\\ 2x=60\\ x=\frac{60}{2}=30\end{array}$

Hence, the number are $30$ & $\left(30+12\right)$ i.e.,

# Question: 70

Seven times a number is $12$ less than thirteen times the same number. Find the number.

## Solution

Let the number be $x$.

The, $7$ times of this number $=7x$

times of this number $=13x$

According to question,

$\begin{array}{l}13x-7x=12\\ 6x=12\\ x=\frac{12}{6}=2\end{array}$

Hence, the required number is $2.$

# Question: 71

The interest received by Karim is $Rs30$ more than that of Ramesh. If the total interest received by them is $Rs70\text{,}$ find the interest received by Ramesh.

## Solution

Let the interest received by Karim is $Rsx\text{,}$ Then interest received by Ramesh will be $Rs\left(x-30\right)\text{.}$ So, the interest received by both will be $Rs\left(x+x-30\right)\text{.}$

According to question,

$\begin{array}{l}x+x-30=70\\ 2x=70+30\\ 2x=100\\ x=\frac{100}{2}=50\end{array}$

So, the interest received by Ramesh $=\text{\hspace{0.17em}}Rs\left(x-30\right)$

$=\text{\hspace{0.17em}}Rs\left(50-30\right)=\text{\hspace{0.17em}}Rs20$

# Question: 72

Subramaniam and Naidu donate some money in a Relief Fund. The amount paid by Naidu is $Rs125$ more than that of Subramaniam. If the total money paid by them is $Rs975\text{,}$ find the amount of money donated by Subramaniam.

## Solution

Let $Rsx$ be the amount donated in a relief fund by Subramaniam. Then, the amount donated by Naidu will be $Rsx+125$

According to question,

$\begin{array}{l}x+x+125=975\\ 2x+125=975\\ 2x=975-125\\ 2x=850\\ x=\frac{850}{2}=\text{\hspace{0.17em}}Rs425\end{array}$

Hence, the amount of money donated by Subramaniam is $Rs425.$

# Question: 73

In a school, the number of girls is $50$ more than the number of boys. The total number of students is $1070.$ Find the number of girls.

## Solution

Let $x$ be the number of boys in the school. Then the number of girls in the school will be $x+50.$

According to question,

$\begin{array}{l}x+\left(x+50\right)=1070\\ 2x+50=1070\\ 2x=1070-50\\ x=\frac{1020}{2}=510\end{array}$

So, the number of boys in the school $=510$.

Then number of girls in the school $=510+50=560$.

# Question: 74

Two times a number increased by $5$ equals $9.$ Find the number.

## Solution

Let the number be $x$.

It is given that $2$ times this number increased by $5$ equals $9.$

$\begin{array}{l}2x+5=9\\ 2x=9-5\\ 2x=4\\ x=\frac{4}{2}=2\end{array}$

Hence, the required number is $2.$

# Question: 75

$9$ added to twice a number gives $13.$ Find the number.

## Solution

Let the number be $x$.

It is given that $9$ added to twice this number gives $13.$

$\begin{array}{l}2x+9=13\\ 2x=13-9\\ 2x=4\\ x=\frac{4}{2}=2\end{array}$

Hence, the required number is $2.$

# Question: 76

$1$ subtracted from one-third of a number gives $1.$ Find the number.

## Solution

Let the number be $x$.

Then, one-third of the number $=\frac{1}{3}x$

According to question,

$\begin{array}{l}\frac{1}{3}x-1=1\\ \frac{1}{3}x=1+1\\ \frac{1}{3}x=2\\ x=3×2\\ x=6\end{array}$

Hence, the required number is $6.$

# Question: 77

After $25$ years, Rama will be $5$ times as old as he is now. Find his present age.

## Solution

Let Rama’s present age be

Then, Rama’s age after

It is given that after  Rama’s age will be $5$ times his present age.

Therefore, the equation is

$\begin{array}{l}x+25=5x\\ 25=5x-x\\ 25=4x\\ \frac{25}{4}=x\\ 6\frac{1}{4}=x\end{array}$

Hence, the present age of Rama is

# Question: 78

After $20$ years, Manoj will be $5$ times as old as he is now. Find his present age.

## Solution

Let the present age of Manoj be

Then, Manoj’s age after

It is given that after  Manoj’s age will be $5$ times his present age.

Therefore, the equation is

$\begin{array}{l}x+20=5x\\ 20=5x-x\\ 20=4x\\ x=\frac{20}{4}=5\end{array}$

Hence, the present age of Manoj is

# Question: 79

My younger sister's age today is $3$ times, what it will be $3$ years from now minus $3$ times what her age was $3$ years ago. Find her present age.

## Solution

Let age of my younger sister be

Then, her age after

Also, her age

It is given that her present age is $3$ times her age after  minus $3$ times her age  age.

Therefore, we obtain the following equation

Hence, her present age is

# Question: 80

If $45$ is added to half a number, the result is triple the number. Find the number.

## Solution

Let $x$ be the number.

Then, half of the number is $\frac{x}{2}$

According to question,

$\begin{array}{l}\frac{x}{2}+45=3x\\ \frac{x+90}{2}=3x\\ x+90=6x\\ x-6x=-90\\ -5x=-90\\ x=\frac{-90}{-5}=18\end{array}$

Hence, the number is $18.$

# Question: 81

In a family, the consumption of wheat is $4$ times that of rice. The total consumption of the two cereals is  Find the quantities of rice and wheat consumed in the family.

## Solution

Given, total consumption of two cereals

Let $x$ be the consumption of rice.

Then, consumption of wheat $=4x$

According to question,

Consumption of wheat

Hence, the consumption of rice & wheat are  respectively.

# Question: 82

In a bag, the number of one rupee coins is three times the number of two rupees coins. If the worth of the coins is $Rs120\text{,}$ find the number of $1$ rupee coins.

## Solution

Let the number of two rupee coins be $y\text{.}$ Then, the number of one rupee coins is $3y$

Total money by two rupees coins $=2y$

Total money by one rupee coins $=3y$

Total worth of coins $=\text{\hspace{0.17em}}Rs120$

The equation formed is,

$\begin{array}{l}2y+3y=120\\ 5y=120\\ y=\frac{120}{5}=24\end{array}$

Number of two rupees coins $=y=24$ & number of one rupee coins $=3y=3×24=72$

# Question: 83

Anamika thought of a number. She multiplied it by $2\text{,}$ added $5$ to the product and obtained $17$ as the result. What is the number she had thought of?

## Solution

Let $x$ be the number thought by Anamika.

If she multiplied it by $2\text{,}$ then the number would be $2x\text{.}$

Added $5$ to it & obtained $17$ as result.

$\begin{array}{l}2x+5=17\\ 2x=17-5\\ 2x=12\\ x=\frac{12}{2}=6\end{array}$

Hence, the number is $6$ thought by Anamika.

# Question: 84

One of the two numbers is twice the other. The sum of the numbers is $12.$ Find the numbers.

## Solution

Let $x$ be the number. Then, other number is twice the first one $=2x$

According to question,

$\begin{array}{l}x+2x=12\\ 3x=12\\ x=\frac{12}{3}=4\end{array}$

Hence, the number are .

# Question: 85

The sum of three consecutive integers is $5$ more than the smallest of the integers. Find the integers.

## Solution

Let one number be $x\text{.}$ Then, the next two consecutive numbers will be

Sum of these three numbers $=x+\left(x+1\right)+\left(x+2\right)=3x+3$

According to question,

$\begin{array}{l}3x+3=x+5\\ 3x-x=5-3\\ 2x=2\\ x=\frac{2}{2}=1\end{array}$

Hence, the number are

# Question: 86

A number when divided by $6$ gives the quotient $6.$ What is the number?

## Solution

Let the required no. be $x\text{.}$ Then, $x$ divided by $6=\frac{x}{6}$

Given, when $x$ is divided by $6\text{,}$ gives quotient as $6.$

So, we obtain the following equation,

$\begin{array}{l}\frac{x}{6}=6\\ \frac{x}{6}×6=6×6\\ x=36\end{array}$

Hence, the required number is $36.$

# Question: 87

The perimeter of a rectangle is  The length of the rectangle is  less than $5$ times its breadth. Find the length of the rectangle.

## Solution

Given, the perimeter of rectangle is

Let $x$ be the breadth of rectangle.

Then, length of rectangle $=5x-4$

According to question,

$\begin{array}{l}2\left[x+\left(5x-4\right)\right]=40\\ 2x+2\left(5x-4\right)=40\\ 2x+10x-8=40\\ 12x=48\\ x=\frac{48}{12}=4\end{array}$

Hence, the length of the rectangle $=5x-4$

# Question: 88

Each of the $2$ equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is  find the length of each side of the triangle.

## Solution

Let third side of an isosceles triangle be $x$

Then, two other equal sides are twice.

So, the both equal sides are

Perimeter of triangle is sum of all sides.

According to question,

Third side

So, the other equal sides are  and .

# Question: 89

The sum of two consecutive multiples of  Find the numbers.

## Solution

Let two consecutive multiplies of $2$ be .

According to question,

$\begin{array}{l}2x+2x+2=18\\ 4x+2=18\\ 4x=18-2\\ 4x=16\\ x=\frac{16}{4}=4\end{array}$

Hence, the required numbers are $2x=2×4=8$ and $2x+2=2×4+2=10.$

# Question: 90

Two complementary angles differ by $20°\text{.}$ Find the angles.

## Solution

Let one of the angle be x, then other will be $x-20$

According to question,

$\begin{array}{l}x+\left(x-20\right)=90°\\ x+x-20=90°\\ 2x-20=90°\\ 2x=90+20\\ x=\frac{110}{2}=55°\end{array}$

Hence, the required angles are $55°$ and $\left(55-20\right)°$ i.e., $55°$ and $35°$.

# Question: 91

$150$ has been divided into two parts such that twice the first part is equal to the second part. Find the parts.

## Solution

Let one part be $x$ & other part be $2x$

Since, $150$ has been divided into above two parts.

According to question,

$\begin{array}{l}x+2x=150\\ 3x=150\\ x=\frac{150}{3}=50\end{array}$

Hence, first part is $50$ & the second part is $2×50=100$

# Question: 92

In a class of $60$ students, the number of girls is one third the number of boys. Find the number of girls and boys in the class.

## Solution

Given, the total number of students in the class $=60.$

Let $x$ be the number of boys in the class. Then, the number of girls in the class $=\frac{x}{3}$

According to question,

$x+\frac{x}{3}=60$

$\frac{3x+x}{3}=60$

$\frac{4x}{3}=60$

$4x=60×3$

$4x=180$

$x=\frac{180}{4}=45$

Hence, the number of boys in the class is $45$ & the number of girls in the class is $\frac{45}{3}=15$.

# Question: 93

Two-third of a number is greater than one-third of the number by $3.$ Find the number.

## Solution

Let the no. be $x$.

Two third of the number $=\frac{2}{3}x$ and one third of this number $=\frac{1}{3}x$

According to question,

$\frac{2}{3}x=\frac{1}{3}x+3$

$\frac{2}{3}x-\frac{1}{3}x=3$

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

$\frac{x}{3}=3$

$\frac{x}{3}×3=3×3$

$x=9$

Hence, the required number is $\text{9}\text{.}$

# Question: 94

A number is as much greater than $27$ as it is less than $73.$ Find the number.

## Solution

Let number be ( $x$ greater than $27$ ) and ( $x$ less than $73$ ).

Now according to question,

$\begin{array}{l}27+x=73-x\\ x+x=73-27\\ 2x=46\\ x=\frac{46}{2}=23\end{array}$

So, Required number is $27+x=27+23=50.$

Alternatively,

The required number would be the average or mean of $27$ and $73.$ So Required number $=\frac{27+73}{2}=50.$

Hence, the number is $50.$

# Question: 95

A man travelled two fifth of his journey by train, one-third by bus, one-fourth by car and the remaining  on foot. What is the length of his total journey?

## Solution

Let his total journey length be $x$.

Then, travelled by train $=\frac{2}{5}x$

Travelled by bus $=\frac{1}{3}x$

and travelled by car $=\frac{1}{4}x$

Total journey $=\frac{2}{5}x+\frac{1}{3}x+\frac{1}{4}x$

$\begin{array}{l}=\frac{12×2x+20×x+15×x}{60}\\ =\frac{24x+20x+15x}{60}\\ =\frac{59x}{60}\end{array}$

Remaining journey $=\frac{x}{1}-\frac{59x}{60}=\frac{60x-59x}{60}=\frac{x}{60}$

According to question, remaining journey is

Hence, the total journey is of

# Question: 96

Twice a number added to half of itself equals $24.$ Find the number.

## Solution

Let the no. be $x\text{.}$

Twice of this number $=2x$

& half of this number $=\frac{1}{2}x$

According to question, $2x+\frac{1}{2}x=24$

Multiplying both sides by $2$

$\begin{array}{l}4x+x=48\\ 5x=48\\ x=\frac{48}{5}=9.6\end{array}$

Hence, the required number is $9.6$

# Question: 97

Thrice a number decreased by $5$ exceeds twice the number by $1.$ Find the number.

## Solution

Let the number be $x\text{.}$

Then thrice of this number $=3x$

and twice of this number $=2x$

If we decrease thrice of $x$ by $5\text{,}$ we get $\left(3x-5\right)$

According to question,

$\begin{array}{l}\left(3x-5\right)-\left(2x\right)=1\\ 3x-5-2x=1\\ x-5=1\\ x=1+5=6\end{array}$

Hence, the required number is $6.$

# Question: 98

A girl is  younger than her father. The sum of their ages is  Find the ages of the girl and her father.

## Solution

Let $x$ years be the age of girl.

Then, age of her father

According to question,

$\begin{array}{l}x+\left(x+28\right)=50\\ 2x+28=50\\ 2x=50-28\\ 2x=22\\ x=\frac{22}{2}=11\end{array}$

Hence, age of the girl is

& her father’s age is

# Question: 99

The length of a rectangle is two times its width. The perimeter of the rectangle is  Find the dimensions of the rectangle.

## Solution

Let x be the width of rectangle. Then, length of the rectangle will be $2x$

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

According to question,

Hence, width of the rectangle is  & length of the rectangle is .

# Question: 100

Look at this riddle?

If she answers the riddle correctly however will she pay for the pencils?

## Solution

Let the cost of $1$ pencil be $Rsx$

Now, cost of such $7$ pencils will be $Rs7x$ and $5$ pencils will be $Rs5x$.

Given, cost of $7$ pencils is $Rs6$ more than cost of $5$ pencils. Therefore, we get

$\begin{array}{l}7x-5x=6\\ 2x=6\\ x=\frac{6}{2}=3\end{array}$

Since, cost of $1$ pencil $=\text{\hspace{0.17em}}Rs3$

Cost of $10$ pencils $=\text{\hspace{0.17em}}Rs30$

Thus, she have to pay $Rs30$ for $10$ pencils.

# Question: 101

In a certain examination, a total of $3768$ students secured first division in the years $2006$ and $2007.$ The number of first division in $2007$ exceeded those in  How many students got first division in $2006\text{?}$

## Solution

Let the number of students who got first division in year $2006$ be $x\text{.}$

Since, the number of first division in year $2007$ exceeded those in year $2006$ by $34\text{,}$ therefore the number of students who got first division in year $2007$ will be $\left(x+34\right)\text{.}$

It is given that total number of students who got first division in years $2006$ and $2007$ is $3768.$

According to question,

Hence, $1867$ students got first division in year $2006.$

# Question: 102

Radha got $Rs17\text{,}480$ as her monthly salary and over-time. Her salary exceeds the over-time by $Rs10\text{,}000.$ What is her monthly salary?

## Solution

Radha’s monthly salary & over-time $=\text{\hspace{0.17em}}Rs17\text{,}480$

Let $x$ be the her monthly salary.

Then, overtime $\left(x-10000\right)$

$\begin{array}{l}17480-x=x-10000\\ -x-x=-10000-17480\\ -2x=-27480\\ 2x=27480\\ x=13740\end{array}$

Hence, her monthly salary is $Rs13740.$

# Question: 103

If one side of a square is represented by $18x-20$ and the adjacent side is represented by $42-13x\text{,}$ find the length of the side of the square.

## Solution

Given, one side of a square is $18x-20$ and adjacent side is $42-13x\text{,}$

We know that, all the sides of a square are always equal

Hence, side of the square is

# Question: 104

Follow the directions and correct the given incorrect equation, written in Roman numerals:

a.    Remove two of these matchsticks to make a valid equation:

b.   Move one matchstick to make the equation valid. Find two different solutions.

## Solution

a.    Given, $\text{IX}-\text{VI}=\text{V}$

according to question, we have to remove two matchsticks to make a valid equation.

Hence, $\text{X}-\text{V}=\text{V}$

$10-5=5$

b.   Given, $\text{VI}-\text{IV}=\text{XI}$

according to question, we have to move one matchstick to make a valid equation.

i.      $\text{VI}+\text{IV}=\text{X}$

$6+4=10$

ii.   $\text{VI+V}=\text{XI}$

$6+5=11$

# Question: 105

What does a duck do when it flies upside down? The answer to this riddle is hidden in the equation given below:

For riddle answer : substitute the number for the letter it equals

## Solution

We have, $i+69=70$

We have, $4a=-5a+45$

We have, $-5t-60=-70$

We have, $\frac{5}{3}p+9=24$

$\begin{array}{l}\frac{5}{3}p=24-9\\ \frac{5}{3}p=15\\ p=15×\frac{3}{5}=9\end{array}$

We have, $3c=c+12$

$\begin{array}{l}3c-c=12\\ 2c=12\\ c=\frac{12}{2}=6\end{array}$

We have, $3\left(k+1\right)=24$

$\begin{array}{l}\frac{3\left(k+1\right)}{3}=\frac{24}{3}\\ k+1=8\\ k=8-1\\ k=7\end{array}$

By substituting the number for the letter it equals, we get

# Question: 106

The three scales below are perfectly balanced if $•=3.$ What are the values of $�$ and  $\text{*?}$

a.

b.

c.

## Solution

Let the value of $\Delta$  & $\text{*}$ be $x$ and $y$ & it is given that $•=3.$

From (a), $y+y+y+y+y=x+x+3+3$

From (b), $x+x=y+y+3+3$

From (c), $y+y+y+3+3+3=x+x+x$

From eq.

$x=y+3$

On putting $x=y+3$ in eq. (i), we get

$\begin{array}{l}2\left(y+3\right)-5y=-6\\ 2y+6-5y=-6\\ -3y+6=-6\\ -3y=-6-6=-12\\ y=\frac{12}{3}=4\end{array}$

On putting $y=4$ in eq. (ii), we get

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

Value of $\Delta$  $=x=7$ & value of $\text{*}=y=4.$

# Question: 107

The given figure represents a weighing balance. The weights of some objects in the balance are given. Find the weight of each square and the circle.

## Solution

Weight on LHS $=20+20=40$ kg

Weight on RHS $=14+4=18$ kg

Weight should be equal,

Therefore, circle weight $=40-18=22$ kg