Unit 4: Simple Equations
In the Questions 1 to 18, there are four options, out of which only one is correct. Write the correct one.
The solution of the equation $ax+b=0$ is
a. $\frac{a}{b}$
b. $b$
c. $\frac{b}{a}$
d. $\frac{b}{a}$
(c)
Given equation $ax+b=0$
$\begin{array}{l}ax=b\text{[transposing}b\text{toRHS]}\\ x=\frac{b}{a}\text{[ondividingbothsidesby}a\text{]}\end{array}$
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. $0$
(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.
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 nonzero number.
d. Dividing both sides of the equation by the same number.
(d)
Dividing both sides of the equation by the same nonzero 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.
The solution of which of the following equations is neither a positive fraction nor an integer?
a. $2x+6=0$
b. $3x5=0$
c. $5x8=x+4$
d. $4x+7=x+2$
(d)
Let us solve the equations:
a. Given equation is $2x+6=0$
$\begin{array}{l}\Rightarrow 2x=6\text{}\text{[transposing}6\text{toRHS]}\\ \Rightarrow x=\frac{6}{2}\\ \Rightarrow x=3\text{}\text{(integer)}\end{array}$
b. Given equation is $3x5=0$
$\begin{array}{l}\Rightarrow 3x=5\text{}\text{[transposing}6\text{toRHS]}\\ \Rightarrow x=\frac{5}{3}\text{}\text{(fraction)[dividingbothsidesby}3\text{]}\end{array}$
c. Given equation is $5x8=x+4$
$\begin{array}{l}\Rightarrow 5x=x+4+8\text{}\text{[transposing}8\text{toRHS]}\\ \Rightarrow 5x=x+12\\ \Rightarrow 5xx=12\text{}\text{[transposing}x\text{toLHS]}\\ \Rightarrow 4x=12\\ \Rightarrow x=\frac{12}{4}=3\text{}\text{(integer)}\end{array}$
d. Given equation is $4x+7=x+2$
$\begin{array}{l}\Rightarrow 4x+7x=2\text{}\text{[transposing}x\text{toLHS]}\\ \Rightarrow 3x=27\text{}\text{[transposing}7\text{toRHS]}\\ \Rightarrow 3x=5\\ \Rightarrow x=\frac{5}{3}\text{}\text{[dividingbothsidesby}3\text{]}\end{array}$
Which is neither a positive fraction nor an integer.
The equation which cannot be solved in integers is
a. $5y3=18$
b. $3x9=0$
c. $3z+8=3+z$
d. $9y+8=4y7$
(c)
Let us solve the equations
a. Given equation is $5y3=18$
$\begin{array}{l}\Rightarrow 5y=18+3\\ \Rightarrow 5y=15\\ \Rightarrow y=3\text{}\text{(integer)}\end{array}$
b. Given equation is $3x9=0$
$\begin{array}{l}\Rightarrow 3x=9\\ \Rightarrow x=3\end{array}$
c. Given equation is $3z+8=3+z$
On transposing $z\text{and}8$ to LHS & RHS respectively, we get
$\begin{array}{l}\Rightarrow 3zz=38\\ \Rightarrow 2z=5\\ \Rightarrow z=\frac{5}{2}\text{}\text{[dividingbothsidesby}2\text{]}\end{array}$
d. Given equation is $9y+8=4y7$
On transposing $4y\text{}8$ to LHS & RHS respectively, we get
$\begin{array}{l}\Rightarrow 9y4y=78\\ \Rightarrow 5y=15\\ \Rightarrow \frac{5y}{5}=\frac{15}{5}\text{}\text{[dividingbothsidesby}5\text{]}\\ \Rightarrow y=3\text{}\text{(integer)}\end{array}$
If $7x+4=25\text{,}$ then $x$ is equal to
a. $\frac{29}{7}$
b. $\frac{100}{7}$
c. $2$
d. $3$
(d)
Given equation is $7x+4=25$
$\begin{array}{l}\Rightarrow 7x=254\text{}\text{[transposing}4\text{toRHS]}\\ \Rightarrow 7x=21\end{array}$
On dividing the above equation by $7\text{,}$ we get
$\Rightarrow x=3$
Hence, the solution of the given eq. is $3.$
The solution of the equation $3x+7=20$ is
a. $\frac{17}{7}$
b. $9$
c. $9$
d. $\frac{13}{3}$
(b)
Given equation is $3x+7=20$
$\begin{array}{l}\Rightarrow 3x=207\\ \Rightarrow 3x=27\\ \Rightarrow x=\frac{27}{3}=9\end{array}$
Hence, the solution of eq. is $9.$
The value of $y$ for which the expressions $\left(y15\right)$ and $\left(2y+1\right)$ become equal is
a. $0$
b. $16$
c. $8$
d. $16$
(d)
It is given that both the equation are equal. So, the equation is
$\begin{array}{l}\Rightarrow y15=2y+1\\ \Rightarrow y2y=1+15\\ \Rightarrow y=16\end{array}$
Multiplying both sides by $1\text{,}$ we get
$\Rightarrow y=16$
If $k+7=16\text{,}$ then the value of $8k72$ is
a. $0$
b. $1$
c. $112$
d. $56$
(a)
Given equation is $k+7=16$
On transposing $7$ to RHS, we get
$\Rightarrow k=167=9$
Put the value of $k$ in the equation $\left(8k72\right)\text{,}$ we get
$\Rightarrow 8\left(9\right)72=7272=0$
If $43\text{m}=0.086\text{,}$ then the value of m is
a. $0.002$
b. $0.02$
c. $0.2$
d. $2$
(a)
Given equation is $43\text{m}=0.086$
On dividing the given equation by $43\text{,}$ we get
$\Rightarrow m=\frac{0.086}{43}$
If we remove the decimal, we get $1000$ in denominator
$\Rightarrow m=\frac{86}{43}\times \frac{1}{1000}=\frac{2}{1000}=0.002$
$x$ exceeds $3$ by $7\text{,}$ can be represented as
a. $x+3=2$
b. $x+7=3$
c. $x3=7$
d. $x7=3$
The given statement means $x$ is $7$ more than $3.$
So, the equation is $x7=3$
We can also write it as $x3=7.$
The equation having $5$ as a solution is:
a. $4x+1=2$
b. $3x=8$
c. $x5=3$
d. $3+x=8$
(d)
Let us solve the equations:
a. Given equation is $4x+1=2$
$\begin{array}{l}\Rightarrow 4x=21\\ \Rightarrow 4x=1\\ \Rightarrow x=\frac{1}{4}\end{array}$
b. Given equation is $3x=8$
$\begin{array}{l}\Rightarrow x=83\\ \Rightarrow x=5\\ \Rightarrow x=5\end{array}$
c. Given equation is $x5=3$
$\begin{array}{l}\Rightarrow x=3+5\\ \Rightarrow x=8\end{array}$
d. Given equation is $3+x=8$
$\begin{array}{l}\Rightarrow x=83\\ \Rightarrow x=5\end{array}$
The equation having $3$ as a solution is:
a. $x+3=1$
b. $8+2x=3$
c. $10+3x=1$
d. $2x+1=3$
(c)
Let us solve the equations:
a. Given equation is $x+3=1$
$\begin{array}{l}\Rightarrow x=13\\ \Rightarrow x=2\end{array}$
b. Given equation is $8+2x=3$
$\begin{array}{l}\Rightarrow 2x=38\\ \Rightarrow 2x=5\\ \Rightarrow x=\frac{5}{2}\end{array}$
c. Given equation is $10+3x=1$
$\begin{array}{l}\Rightarrow 3x=110\\ \Rightarrow 3x=9\\ \Rightarrow x=3\end{array}$
Further, there is no need to solve for option d as we have obtained correct option.
Which of the following equations can be formed starting with $x=0\text{?}$
a. $2x+1=1$
b. $\frac{x}{2}+5=7$
c. $3x1=1$
d. $3x1=1$
(c)
We have $x=0$
On multiplying both the sides by $3\text{,}$ we get
$\begin{array}{l}\Rightarrow 3\times x=3\times 0\\ \Rightarrow 3x=0\end{array}$
On adding $\left(1\right)$ to both sides, we get
$\begin{array}{l}\Rightarrow 3x+\left(1\right)=0+\left(1\right)\\ \Rightarrow 3x1=1\end{array}$
Which of the following equations cannot be formed using the equation $x=7\text{?}$
a. $2x+1=15$
b. $7x1=50$
c. $x3=4$
d. $\frac{x}{7}1=0$
(b)
We have $x=7$
On multiplying both the sides by $7\text{,}$ we get
$\begin{array}{l}\Rightarrow 7\times x=7\times 7\\ \Rightarrow 7x=49\end{array}$
On adding $\left(1\right)$ both sides, we get
$\begin{array}{l}\Rightarrow 7x+\left(1\right)=49+\left(1\right)\\ \Rightarrow 7x1=491\\ \Rightarrow 7x1=48\end{array}$
If $\frac{x}{2}=3\text{,}$ then the value of $3x+2$ is
a. $20$
b. $11$
c. $\frac{13}{2}$
d. $8$
(a)
Given $\frac{x}{2}=3$
On multiplying both the sides by $2\text{,}$ we get $\frac{x}{2}\times 2=3\times 2$
$\Rightarrow x=3\times 2=6$
Put $x=6$ in equation $3x+2\text{,}$ we get
$\Rightarrow 3\left(6\right)+2=18+2=20$
Which of the following numbers satisfy the equation $6+x=12\text{?}$
a. $2$
b. $6$
c. $6$
d. $2$
(c)
Let us put the values given in the options in equation $6+x=12$
a. Put $x=2$
$\begin{array}{l}\Rightarrow 6+2=12\\ \Rightarrow 4=12\end{array}$
LHS $\ne $ RHS
b. Put $x=6$
$\begin{array}{l}\Rightarrow 6+\left(6\right)=12\\ \Rightarrow 0=12\end{array}$
LHS $\ne $ RHS
c. Put $x=6$
$\begin{array}{l}\Rightarrow 6+\left(6\right)=12\\ \Rightarrow 12=12\end{array}$
LHS $=$ RHS (satisfied)
Now, there is no need to check the next option.
Hence, $x=6$ satisfied the given equation.
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
(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.
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 ________.
a. If the smaller number is $x\text{,}$ then the other number is $\left(60x\right)\text{,}$ since the sum of both numbers is $60.$
b. Given, one number $=x\text{}\left[\text{from(a)}\right]$
Then, other number $=\left(60x\right)$
$\therefore $ Difference $=\left(60x\right)x=602x$
c. Given, difference between two no. is $30$
So, the equation is $602x=30$
d. Let us solve the equation for $x$
$602x=30$
$\begin{array}{l}\Rightarrow 2x=3060\\ \Rightarrow 2x=30\\ \Rightarrow 2x=30\end{array}$
$\Rightarrow 2x=30$
On dividing the equation by $2\text{,}$ we get
$\begin{array}{l}\Rightarrow \frac{2x}{2}=\frac{30}{2}\\ \Rightarrow x=15\end{array}$
Hence, the solution of the equation is $15$
e. The numbers are $x\text{}(60x)$
Now, put the value of $x\text{,}$ we get
First number $=15$
Second number $=6015=45$
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 ________.
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 $x\text{,}$
$\begin{array}{l}\Rightarrow x+2x=81\\ \Rightarrow 3x=81\\ \Rightarrow x=\frac{81}{3}\\ \Rightarrow x=27\end{array}$
Hence, the solution of the equation is $27.$
d. The two numbers are $x=27\text{and}2x=2\times 27=54.$
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 ________.
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 $x\text{,}$
$\begin{array}{l}\Rightarrow 4x+3x=280\\ \Rightarrow 7x=280\\ \Rightarrow x=\frac{280}{7}\\ \Rightarrow x=40\end{array}$
Hence, the solution of the equation is $40$
d. Marks obtained by Abha are $2x$ i.e., $2\times 40=80.$
The length of a rectangle is two times its breadth. Its perimeter is $60\text{cm}\text{.}$
a. If the breadth of rectangle is $x\text{cm,}$ 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 ________.
a. It is given that the length of the rectangle is two times its breadth.
$\therefore $ Length $=2x\text{cm}$
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 $60\text{cm}\text{.}$
So, the equation formed is $2(2x+x)=60$
d. On solving,
$\begin{array}{l}\Rightarrow 2\left(3x\right)=60\\ \Rightarrow 6x=60\end{array}$
On dividing the equation by $6\text{,}$ we get
$\begin{array}{l}\Rightarrow \frac{6x}{6}=\frac{60}{6}\\ \Rightarrow x=10\end{array}$
Hence, the solution of the equation is $10.$
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.
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 $=\text{\hspace{0.17em}}Rs5\times x=\text{\hspace{0.17em}}Rs5x$
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
$\Rightarrow \text{\hspace{0.17em}}5x+2x=70$
d. On solving,
$\begin{array}{l}\Rightarrow \text{\hspace{0.17em}}5x+2x=70\\ \Rightarrow 7x=70\\ \Rightarrow \frac{7x}{7}=\frac{70}{7}\\ \Rightarrow x=10\end{array}$
Therefore, there are 10 coins of both 5 rupee and 2 rupee.
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 ________.
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 $\left(30x\right)\text{,}$ because total number of prizes are $30.$
b. Total values of prizes in terms of $x$ are $2000x+1000(30x)\text{.}$
c. The equation formed is
$2000x+1000(30x)=52000$
d. On solving,
$\begin{array}{l}\Rightarrow 2000x+300001000x=52000\\ \Rightarrow 1000x+30000=52000\end{array}$
$\begin{array}{l}\Rightarrow 1000x=22000\\ \Rightarrow x=\frac{22000}{1000}\end{array}$
The solution of the equation is $22.$
e. So, Number of ${2}^{\text{nd}}$ prizes $=3022=8$.
The number of ${1}^{\text{st}}$ prizes are $22$ & the number of ${2}^{\text{nd}}$ prizes are $8.$
If $z+3=5\text{,}$ then $z=$ ________.
On solving the given equation, we get
$\begin{array}{l}\Rightarrow z+3=5\\ \Rightarrow z=53\\ \Rightarrow z=2\end{array}$
_________ is the solution of the equation $3x2=7.$
On solving the equation for $x\text{,}$
$\begin{array}{l}\Rightarrow 3x2=7\\ \Rightarrow 3x=7+2\\ \Rightarrow 3x=9\\ \Rightarrow x=\frac{9}{3}=3\end{array}$
__________ is the solution of $3x+10=7.$
Solving the equation for $x\text{,}$
$\begin{array}{l}3x+10=7\\ \Rightarrow 3x=\text{}710\\ \Rightarrow 3x=3\\ \Rightarrow x=\frac{3}{3}=1\end{array}$
If $2x+3=5\text{,}$ then value of $3x+2$ is __________.
Solving the equation for $x\text{,}$
$\begin{array}{l}2x+3=5\\ \Rightarrow 2x=53\\ \Rightarrow 2x=2\\ \Rightarrow x=\frac{2}{2}=1\end{array}$
Put the value of $x\text{,}$ in $3x+2\text{,}$ we get
$\begin{array}{l}=3\left(1\right)+2\\ =3+2\\ =5\end{array}$
In integers, $4x1=8$ has __________ solution.
Solving the equation for $x\text{,}$
$\begin{array}{l}4x1=8\\ \Rightarrow 4x=8+1\\ \Rightarrow 4x=9\\ \Rightarrow x=\frac{9}{4}\end{array}$
Since the solution of the equation is not an integer, hence the equation has no solution.
In natural numbers, $4x+5=7$ has ______ solution.
Solve the equation for $x\text{,}$
$\begin{array}{l}4x+5=7\\ \Rightarrow 4x=75\\ \Rightarrow 4x=12\\ \Rightarrow x=\frac{12}{4}=3\end{array}$
Since, the value of $x$ is not a natural number, hence equation has no solution in natural numbers.
In natural numbers, $x5=5$ has ______ solution.
Solving the equation for $x\text{,}$
$\begin{array}{l}x5=5\\ x=5+5\\ x=0\end{array}$
Since, natural numbers do not contain zero, hence the equation has no solution.
In whole numbers, $x+8=124$ has ______ solution.
Solving the equation for $x\text{,}$
$\begin{array}{l}x+8=124\\ x+8=8\\ x=88\\ x=0\end{array}$
Since, zero is in the range of whole numbers, hence the equation has one solution.
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 __________.
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. $4x7.$
So, the equation formed is $5+3x=4x7.$
$x+7=10$ has the solution _________.
Solving the equation for $x\text{,}$
$\begin{array}{l}x+7=10\\ \Rightarrow x=107\\ \Rightarrow x=3\end{array}$
$x0=\text{\_\_\_\_\_\_\_\_\_\_;}$ when $3x=12.$
Given, $3x=12$
$\begin{array}{l}\Rightarrow \frac{3x}{3}=\frac{12}{3}\\ \Rightarrow x=4\\ \Rightarrow x0=40=4\end{array}$
$x1=\text{\_\_\_\_\_\_\_\_\_\_;}$ when $2x=2.$
Given, $2x=2$
$\begin{array}{l}\Rightarrow \frac{2x}{2}=\frac{2}{2}\\ \Rightarrow x=1\\ \Rightarrow x1=11=0\end{array}$
$x\text{\_\_\_\_\_\_\_\_\_\_}=15\text{;}$ when $\frac{x}{2}=6.$
Given, $\frac{x}{2}=6$
$\begin{array}{l}\Rightarrow x=12\\ \Rightarrow 12?=15\\ \Rightarrow ?=1512\\ \Rightarrow ?=3\\ \Rightarrow ?=3\end{array}$
Hence, $x\left(3\right)=15$
The solution of the equation $x+15=19$ is ________.
Solving the equation for $x\text{,}$
$\begin{array}{l}x+15=19\\ \Rightarrow x=1915\\ \Rightarrow x=4\end{array}$
Hence, the solution of the given equation is $4.$
Finding the value of a variable in a linear equation that _______ the equation is called a _______ of the equation.
Finding the value of a variable in a linear equation that satisfies the equation is called a root of the equation.
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.
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.
If $\frac{9}{5}x=\frac{18}{5}\text{,}$ then $x=\text{\_\_\_\_\_\_\_}\text{.}$
Given, $\frac{9}{5}x=\frac{18}{5}$
$\begin{array}{l}\Rightarrow \frac{9}{5}x\xf7\frac{9}{5}=\frac{18}{5}\xf7\frac{9}{5}\\ \Rightarrow x=\frac{18}{5}\times \frac{5}{9}=2\end{array}$
If $3x=4\text{,}$ then $x=\text{\_\_\_\_\_\_\_}\text{.}$
Solving the equation for $x\text{,}$
$\begin{array}{l}3x=4\\ \Rightarrow x=43\\ \Rightarrow x=7\\ \Rightarrow x=7\end{array}$
If $x\frac{1}{2}=\frac{1}{2}\text{,}$ then $x=\text{\_\_\_\_\_\_\_}\text{.}$
Given, $x\frac{1}{2}=\frac{1}{2}$
$\begin{array}{l}\Rightarrow x=\frac{1}{2}+\frac{1}{2}\\ \Rightarrow x=0\end{array}$
If $\frac{1}{6}x=\frac{1}{6}\text{,}$ then $x=\text{\_\_\_\_\_\_\_}\text{.}$
Given, $\frac{1}{6}x=\frac{1}{6}$
$\begin{array}{l}\Rightarrow x=\frac{1}{6}\frac{1}{6}\\ \Rightarrow x=0\\ \Rightarrow x=0\end{array}$
If $10$ less than a number is $65\text{,}$ then the number is _______.
Let the number be $x\text{.}$
Then, the equation will be $x10=65$
Now, solving the equation for $x\text{.}$
$\begin{array}{l}\Rightarrow x=65+10\\ \Rightarrow x=75\end{array}$
Hence, the number is $75.$
If a number is increased by $20\text{,}$ it becomes $45.$ Then the number is _______.
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 $x+20=45$
$\begin{array}{l}\Rightarrow x=4520\\ \Rightarrow x=25\end{array}$
Hence, the number is $25.$
If $84$ exceeds another number by $12\text{,}$ then the other number is _______.
Equation formed: $84x=12$
Solving the equation for $x\text{,}$
$\begin{array}{l}84x=12\\ \Rightarrow x=1284\\ \Rightarrow x=72\\ \Rightarrow x=72\end{array}$
Hence, the other number is $72.$
If $x\frac{7}{8}=\frac{7}{8}\text{,}$ then $x=\text{\_\_\_\_\_\_\_}\text{.}$
Given, $x\frac{7}{8}=\frac{7}{8}$
$\begin{array}{l}\Rightarrow x=\frac{7}{8}+\frac{7}{8}\\ \Rightarrow x=\frac{7+7}{8}\\ \Rightarrow x=\frac{14}{8}\\ \Rightarrow x=\frac{7}{4}\end{array}$
In Questions 49 to 55, state whether the statements are True or False.
$5$ is the solution of the equation $3x+2=17.$
True
Solving the equation for $x\text{,}$
$\begin{array}{l}\Rightarrow 3x+2=17\\ \Rightarrow 3x=172\\ \Rightarrow 3x=15\\ \Rightarrow x=\frac{15}{3}=5\end{array}$
$\frac{9}{5}$ is the solution of the equation $4x1=8.$
False
Solving the equation for $x\text{,}$
$\begin{array}{l}\Rightarrow 4x1=8\\ \Rightarrow 4x=8+1\\ \Rightarrow 4x=9\\ \Rightarrow x=\frac{9}{4}\end{array}$
$4x5=7$ does not have an integer as its solution.
False
Given equation $4x5=7$
$\begin{array}{l}4x=7+5\\ 4x=12\\ x=\frac{12}{4}=3\text{}\left(\text{integer}\right)\end{array}$
One third of a number added to itself gives $10\text{,}$ can be represented as $\frac{x}{3}+10=x\text{.}$
False
Let the number be $x\text{.}$
Then, the equation formed is $\frac{x}{3}+x=10$
$\frac{3}{2}$ is the solution of the equation $8x5=7.$
True
Solving the equation for $x\text{,}$
$\begin{array}{l}8x5=7\\ 8x=7+5\\ 8x=12\\ x=\frac{12}{8}\\ x=\frac{3}{2}\end{array}$
If $4x7=11\text{,}$ then $x=4.$
False
Solving the equation for $x\text{,}$
$\begin{array}{l}4x7=11\\ 4x=11+7\\ 4x=18\\ x=\frac{18}{4}=\frac{9}{2}\end{array}$
If $9$ is the solution of variable $x$ in the equation $\frac{5x7}{2}=y\text{,}$ then the value of $y$ is $28.$
False
Given, $x=9$
Put the value of $x$ in the equation, we get
$\begin{array}{l}\frac{5\left(9\right)7}{2}=y\\ \frac{457}{2}=y\\ \frac{38}{2}=y\\ y=19\end{array}$
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. $x7=4$ 
(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 $\frac{1}{3}\left(13p\right)$ 
(f) $60$ 

(g) $3$ 
i. (c)
Given, $x+5=9$
$\begin{array}{l}x=95\\ x=4\end{array}$
ii. (e)
Given, $x7=4$
$\begin{array}{l}x=4+7\\ x=11\end{array}$
iii. (f)
Given, $\frac{x}{12}=5$
$\begin{array}{l}12\times \frac{x}{12}=5\times 12\\ x=60\end{array}$
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 $=\frac{1}{3}\times \left(13p\right)\text{,}$ we get
$\begin{array}{l}=\frac{1}{3}\times \left(13\times 2\right)\\ =\frac{1}{3}\times \left(16\right)\\ =\frac{1}{3}\times \left(5\right)\\ =\frac{5}{3}\end{array}$
In Questions 57 to 67, express each of the given statements as an equation.
$13$ subtracted from twice of a number gives $3.$
Let the number be $x\text{.}$
$13$ is subtracted from twice of a number i.e, $2x13$ and it results into $3.$
So, the equation formed is $2x13=3$
Onefifth of a number is $5$ less than that number.
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 $\frac{x}{5}=x5.$
A number is $7$ more than onethird of itself.
Let the number be $x$
Then, $\frac{1}{3}\text{rd}$ of the number $=\frac{x}{3}$
So, the equation formed is $x=7+\frac{x}{3}$
Six times a number is $10$ more than the number.
Let the number be $x$
Then, $6$ times of the number $=6x$
So, the equation formed is $6x=10+x$
If $10$ is subtracted from half of a number, the result is $4.$
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 $\frac{x}{2}10=4$
Subtracting $5$ from $p\text{,}$ the result is $2.$
Subtract $5$ from $p\text{,}$ i.e., $p5$ & result is $2.$
Hence, the equation formed is $p5=2$
Five times a number increased by $7$ is $27.$
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 $5x+7=27$
Mohan is $3$ years older than Sohan. The sum of their ages is $43$ years.
Let age of Sohan be $x\text{yr}\text{.}$
Then, the age of Mohan is $\left(x+3\right)\text{yr}\text{.}$
$\therefore $ Sum of their ages $=43$
So, the equation formed is $x+\left\{x+3\right\}=43$
If $1$ is subtracted from a number and the difference is multiplied by $\frac{1}{2}\text{,}$ the result is $7.$
Let the number be $x$
Then, $1$ is subtracted from a number & the difference multiplied by $\frac{1}{2}$ i.e., $\frac{1}{2}\left(x1\right)$ it gives result as $7$
So, the equation formed is $\frac{1}{2}\left(x1\right)=7$
A number divided by $2$ and then increased by $5$ is $9.$
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 $\frac{x}{2}+5=9$
The sum of twice a number and $4$ is $18.$
Let the number be $x$.
Then, sum of twice of a number and $4$ gives result $18.$
Hence, $2x+4=18$ is the equation.
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.
Let $x\text{yr}$ be the age of Amit.
Then, age of Sohan Lal $=4x\text{yr}$
According to question,
$\begin{array}{l}4xx=27\\ 3x=27\\ x=\frac{27}{3}=9\end{array}$
Hence, the age of Amit is $9\text{yr}$.
A number exceeds the other number by $12.$ If their sum is $72\text{,}$ find the numbers.
Let $x$ be a number, then another number will be $x+12$.
According to question, $x+x+12=72$
$\begin{array}{l}2x=7212\\ 2x=60\\ x=\frac{60}{2}=30\end{array}$
Hence, the number are $30$ & $\left(30+12\right)$ i.e., $30\text{}42.$
Seven times a number is $12$ less than thirteen times the same number. Find the number.
Let the number be $x$.
The, $7$ times of this number $=7x$
$\text{\&}13$ times of this number $=13x$
According to question,
$\begin{array}{l}13x7x=12\\ 6x=12\\ x=\frac{12}{6}=2\end{array}$
Hence, the required number is $2.$
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.
Let the interest received by Karim is $Rsx\text{,}$ Then interest received by Ramesh will be $Rs\left(x30\right)\text{.}$ So, the interest received by both will be $Rs\left(x+x30\right)\text{.}$
According to question,
$\begin{array}{l}x+x30=70\\ 2x=70+30\\ 2x=100\\ x=\frac{100}{2}=50\end{array}$
So, the interest received by Ramesh $=\text{\hspace{0.17em}}Rs\left(x30\right)$
$=\text{\hspace{0.17em}}Rs\left(5030\right)=\text{\hspace{0.17em}}Rs20$
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.
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,
$\begin{array}{l}x+x+125=975\\ 2x+125=975\\ 2x=975125\\ 2x=850\\ x=\frac{850}{2}=\text{\hspace{0.17em}}Rs425\end{array}$
Hence, the amount of money donated by Subramaniam is $Rs425.$
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.
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,
$\begin{array}{l}x+\left(x+50\right)=1070\\ 2x+50=1070\\ 2x=107050\\ x=\frac{1020}{2}=510\end{array}$
So, the number of boys in the school $=510$.
Then number of girls in the school $=510+50=560$.
Two times a number increased by $5$ equals $9.$ Find the number.
Let the number be $x$.
It is given that $2$ times this number increased by $5$ equals $9.$
$\begin{array}{l}2x+5=9\\ 2x=95\\ 2x=4\\ x=\frac{4}{2}=2\end{array}$
Hence, the required number is $2.$
$9$ added to twice a number gives $13.$ Find the number.
Let the number be $x$.
It is given that $9$ added to twice this number gives $13.$
$\begin{array}{l}2x+9=13\\ 2x=139\\ 2x=4\\ x=\frac{4}{2}=2\end{array}$
Hence, the required number is $2.$
$1$ subtracted from onethird of a number gives $1.$ Find the number.
Let the number be $x$.
Then, onethird of the number $=\frac{1}{3}x$
According to question,
$\begin{array}{l}\frac{1}{3}x1=1\\ \frac{1}{3}x=1+1\\ \frac{1}{3}x=2\\ x=3\times 2\\ x=6\end{array}$
Hence, the required number is $6.$
After $25$ years, Rama will be $5$ times as old as he is now. Find his present age.
Let Rama’s present age be $x\text{yr}\text{.}$
Then, Rama’s age after $25\text{yr}=\left(x+25\right)\text{yr}$
It is given that after $25\text{yr,}$ Rama’s age will be $5$ times his present age.
Therefore, the equation is
$\begin{array}{l}x+25=5x\\ 25=5xx\\ 25=4x\\ \frac{25}{4}=x\\ 6\frac{1}{4}=x\end{array}$
Hence, the present age of Rama is $6\frac{1}{4}\text{yr}\text{.}$
After $20$ years, Manoj will be $5$ times as old as he is now. Find his present age.
Let the present age of Manoj be $x\text{yr}\text{.}$
Then, Manoj’s age after $20\text{yr}=\left(x+20\right)\text{yr}$
It is given that after $20\text{yr,}$ Manoj’s age will be $5$ times his present age.
Therefore, the equation is
$\begin{array}{l}x+20=5x\\ 20=5xx\\ 20=4x\\ x=\frac{20}{4}=5\end{array}$
Hence, the present age of Manoj is $5\text{yr}\text{.}$
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.
Let age of my younger sister be $x\text{yr}\text{.}$
Then, her age after $3\text{yr}=\left(x+3\right)\text{yr}\text{.}$
Also, her age $3\text{yrago}=\left(x3\right)\text{yr}\text{.}$
It is given that her present age is $3$ times her age after $3\text{yr}$ minus $3$ times her age $3\text{yr}$ age.
Therefore, we obtain the following equation
$\begin{array}{l}x=3\left(x+3\right)3\left(x3\right)\\ x=3x+93x+9\\ x=18\text{yr}\end{array}$
Hence, her present age is $18\text{yr}\text{.}$
If $45$ is added to half a number, the result is triple the number. Find the number.
Let $x$ be the number.
Then, half of the number is $\frac{x}{2}$
According to question,
$\begin{array}{l}\frac{x}{2}+45=3x\\ \frac{x+90}{2}=3x\\ x+90=6x\\ x6x=90\\ 5x=90\\ x=\frac{90}{5}=18\end{array}$
Hence, the number is $18.$
In a family, the consumption of wheat is $4$ times that of rice. The total consumption of the two cereals is $80\text{kg}\text{.}$ Find the quantities of rice and wheat consumed in the family.
Given, total consumption of two cereals $=80\text{kg}$
Let $x$ be the consumption of rice.
Then, consumption of wheat $=4x$
According to question,
$\begin{array}{l}x+4x=80\\ 5x=80\\ x=\frac{80}{5}=16\text{kg}\end{array}$
Consumption of wheat $=4x=4\times 16=64\text{kg}$
Hence, the consumption of rice & wheat are $16\text{kg}64\text{kg,}$ respectively.
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.
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 $=3y=3\times 24=72$
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?
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.
$\begin{array}{l}2x+5=17\\ 2x=175\\ 2x=12\\ x=\frac{12}{2}=6\end{array}$
Hence, the number is $6$ thought by Anamika.
One of the two numbers is twice the other. The sum of the numbers is $12.$ Find the numbers.
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 $x=4\text{}2x=2\times 4=8$.
The sum of three consecutive integers is $5$ more than the smallest of the integers. Find the integers.
Let one number be $x\text{.}$ Then, the next two consecutive numbers will be $x+1\text{}x+2.$
Sum of these three numbers $=x+\left(x+1\right)+\left(x+2\right)=3x+3$
According to question,
$\begin{array}{l}3x+3=x+5\\ 3xx=53\\ 2x=2\\ x=\frac{2}{2}=1\end{array}$
Hence, the number are $1\text{,}1+1\text{,}1+2\text{i}\text{.e}\text{.,}1\text{,}2\text{,}3$
A number when divided by $6$ gives the quotient $6.$ What is the number?
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,
$\begin{array}{l}\frac{x}{6}=6\\ \frac{x}{6}\times 6=6\times 6\\ x=36\end{array}$
Hence, the required number is $36.$
The perimeter of a rectangle is $40\text{m}\text{.}$ The length of the rectangle is $4\text{m}$ less than $5$ times its breadth. Find the length of the rectangle.
Given, the perimeter of rectangle is $40\text{m}$
Let $x$ be the breadth of rectangle.
Then, length of rectangle $=5x4$
According to question,
$\begin{array}{l}2\left[x+\left(5x4\right)\right]=40\\ 2x+2\left(5x4\right)=40\\ 2x+10x8=40\\ 12x=48\\ x=\frac{48}{12}=4\end{array}$
Hence, the length of the rectangle $=5x4$
$\begin{array}{l}=\left(5\times 4\right)4\\ =204\\ =16\text{m}\end{array}$
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 $30\text{cm,}$ find the length of each side of the triangle.
Let third side of an isosceles triangle be $x$
Then, two other equal sides are twice.
So, the both equal sides are $2x\text{}2x$
Perimeter of triangle is sum of all sides.
According to question,
$\begin{array}{l}x+2x+2x=30\\ 5x=30\\ x=\frac{30}{5}=6\text{cm}\end{array}$
Third side $=x=6\text{cm}$
So, the other equal sides are $2x=2\times 6=12\text{cm}$ and $2x=2\times 6=12\text{cm}$.
The sum of two consecutive multiples of $2\text{is}18.$ Find the numbers.
Let two consecutive multiplies of $2$ be $2x\text{and}2x+2$.
According to question,
$\begin{array}{l}2x+2x+2=18\\ 4x+2=18\\ 4x=182\\ 4x=16\\ x=\frac{16}{4}=4\end{array}$
Hence, the required numbers are $2x=2\times 4=8$ and $2x+2=2\times 4+2=10.$
Two complementary angles differ by $20\xb0\text{.}$ Find the angles.
Let one of the angle be x, then other will be $x20$
According to question,
$\begin{array}{l}x+(x20)=90\xb0\\ x+x20=90\xb0\\ 2x20=90\xb0\\ 2x=90+20\\ x=\frac{110}{2}=55\xb0\end{array}$
Hence, the required angles are $55\xb0$ and $\left(5520\right)\xb0$ i.e., $55\xb0$ and $35\xb0$.
$150$ has been divided into two parts such that twice the first part is equal to the second part. Find the parts.
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 $2\times 50=100$
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.
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,
$x+\frac{x}{3}=60$
$\frac{3x+x}{3}=60$
$\frac{4x}{3}=60$
$4x=60\times 3$
$4x=180$
$x=\frac{180}{4}=45$
Hence, the number of boys in the class is $45$ & the number of girls in the class is $\frac{45}{3}=15$.
Twothird of a number is greater than onethird of the number by $3.$ Find the number.
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,
$\frac{2}{3}x=\frac{1}{3}x+3$
$\frac{2}{3}x\frac{1}{3}x=3$
$\frac{2xx}{3}=3$
$\frac{x}{3}=3$
$\frac{x}{3}\times 3=3\times 3$
$x=9$
Hence, the required number is $\text{9}\text{.}$
A number is as much greater than $27$ as it is less than $73.$ Find the number.
Let number be ( $x$ greater than $27$ ) and ( $x$ less than $73$ ).
Now according to question,
$\begin{array}{l}27+x=73x\\ x+x=7327\\ 2x=46\\ x=\frac{46}{2}=23\end{array}$
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 $50.$
A man travelled two fifth of his journey by train, onethird by bus, onefourth by car and the remaining $3\text{km}$ on foot. What is the length of his total journey?
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 $=\frac{2}{5}x+\frac{1}{3}x+\frac{1}{4}x$
$\begin{array}{l}=\frac{12\times 2x+20\times x+15\times x}{60}\\ =\frac{24x+20x+15x}{60}\\ =\frac{59x}{60}\end{array}$
Remaining journey $=\frac{x}{1}\frac{59x}{60}=\frac{60x59x}{60}=\frac{x}{60}$
According to question, remaining journey is $3\text{km}$
$\begin{array}{l}\frac{x}{60}=3\\ x=3\times 60=180\text{km}\end{array}$
Hence, the total journey is of $180\text{km}\text{.}$
Twice a number added to half of itself equals $24.$ Find the number.
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 $9.6$
Thrice a number decreased by $5$ exceeds twice the number by $1.$ Find the number.
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 $\left(3x5\right)$
According to question,
$\begin{array}{l}\left(3x5\right)\left(2x\right)=1\\ 3x52x=1\\ x5=1\\ x=1+5=6\end{array}$
Hence, the required number is $6.$
A girl is $28\text{years}$ younger than her father. The sum of their ages is $50\text{years}\text{.}$ Find the ages of the girl and her father.
Let $x$ years be the age of girl.
Then, age of her father $=\left(x+28\right)\text{yr}$
According to question,
$\begin{array}{l}x+\left(x+28\right)=50\\ 2x+28=50\\ 2x=5028\\ 2x=22\\ x=\frac{22}{2}=11\end{array}$
Hence, age of the girl is $11\text{yr}$
& her father’s age is $\left(11+28\right)=39\text{yr}\text{.}$
The length of a rectangle is two times its width. The perimeter of the rectangle is $180\text{cm}\text{.}$ Find the dimensions of the rectangle.
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,
$\begin{array}{l}2\left(x+2x\right)=180\\ 2\left(3x\right)=180\\ 6x=180\\ x=\frac{180}{6}=30\text{cm}\end{array}$
Hence, width of the rectangle is $30\text{cm}$ & length of the rectangle is $2\times 30=60\text{cm}$.
Look at this riddle?
If she answers the riddle correctly however will she pay for the pencils?
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
$\begin{array}{l}7x5x=6\\ 2x=6\\ x=\frac{6}{2}=3\end{array}$
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.
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 $2006\text{by}34.$ How many students got first division in $2006\text{?}$
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,
$\begin{array}{l}x+\left(x+\text{}34\right)=3768\\ 2x+34=3768\\ 2x=376834\\ 2x=3734\\ x=\frac{3734}{2}=1867\end{array}$
Hence, $1867$ students got first division in year $2006.$
Radha got $Rs17\text{,}480$ as her monthly salary and overtime. Her salary exceeds the overtime by $Rs10\text{,}000.$ What is her monthly salary?
Radha’s monthly salary & overtime $=\text{\hspace{0.17em}}Rs17\text{,}480$
Let $x$ be the her monthly salary.
Then, overtime $\left(x10000\right)$
$\begin{array}{l}17480x=x10000\\ xx=1000017480\\ 2x=27480\\ 2x=27480\\ x=13740\end{array}$
Hence, her monthly salary is $Rs13740.$
If one side of a square is represented by $18x20$ and the adjacent side is represented by $4213x\text{,}$ find the length of the side of the square.
Given, one side of a square is $18x20$ and adjacent side is $4213x\text{,}$
We know that, all the sides of a square are always equal
$\begin{array}{l}18x20=4213x\\ 18x+13x=42+20\\ 31x=62\\ x=\frac{62}{31}=2\text{units}\end{array}$
Hence, side of the square is $\left(18\times 2\right)20=3620=16\text{units}\text{.}$
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.
a. Given, $\text{IX}\text{VI}=\text{V}$
according to question, we have to remove two matchsticks to make a valid equation.
Hence, $\text{X}\text{V}=\text{V}$
$105=5$
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. $\text{VI+V}=\text{XI}$
$6+5=11$
What does a duck do when it flies upside down? The answer to this riddle is hidden in the equation given below:
$\text{If}i+69=70\text{,then}i=?\text{If}8u=6u+8\text{,then}u=?$
$\text{If}4a=5a+45\text{,then}a=?\text{if}4q+5=17\text{,then}q=?$
$\text{If}5t60=70,\text{then}t=?\text{If}\frac{\text{1}}{\text{4}}s+98=100\text{,then}s=?$$\text{If}\frac{5}{3}p+9=24\text{,then}p=\_\_\_\_\_\_\_?$
$\text{If}3c=c+12\text{,then}c=\_\_\_\_\_\_\_?$
$\text{If}3(k+1)=24\text{,then}k=\_\_\_\_\_\_\_?$
For riddle answer : substitute the number for the letter it equals
$\frac{}{1}\text{}\frac{}{2}\text{}/\text{}\frac{}{3}\text{}\frac{}{4}\text{}\frac{}{5}\text{}\frac{}{6}\text{}\frac{}{7}\text{}\frac{}{8}\text{}/\text{}\frac{}{4}\text{}\frac{}{9}$
We have, $i+69=70$
$\begin{array}{l}i=7069\\ i=1.\\ \text{And}8u=6u+8\\ 8u6u=8\\ 2u=8\\ u=\frac{8}{2}=4\end{array}$
We have, $4a=5a+45$
$\begin{array}{l}4a+5a=45\\ 9a=45\\ a=\frac{45}{9}=5\\ \text{And}4q+5=17\\ 4q=175\\ 4q=12\\ q=\frac{12}{4}=3\end{array}$
We have, $5t60=70$
$\begin{array}{l}5t=70+60\\ 5t=10\\ t=\frac{10}{5}=2\\ \text{And}\frac{1}{4}s+98=100\\ \frac{1}{4}s=10098\\ s=2\times 4=8\end{array}$
We have, $\frac{5}{3}p+9=24$
$\begin{array}{l}\frac{5}{3}p=249\\ \frac{5}{3}p=15\\ p=15\times \frac{3}{5}=9\end{array}$
We have, $3c=c+12$
$\begin{array}{l}3cc=12\\ 2c=12\\ c=\frac{12}{2}=6\end{array}$
We have, $3\left(k+1\right)=24$
$\begin{array}{l}\frac{3\left(k+1\right)}{3}=\frac{24}{3}\\ k+1=8\\ k=81\\ k=7\end{array}$
By substituting the number for the letter it equals, we get
$\frac{\text{i}}{1}\frac{\text{t}}{2}\text{}/\text{}\frac{\text{q}}{3}\frac{\text{u}}{4}\frac{\text{a}}{5}\frac{\text{c}}{6}\frac{\text{k}}{7}\frac{\text{s}}{8}\text{}/\text{}\frac{\text{u}}{4}\frac{\text{p}}{9}$
The three scales below are perfectly balanced if $\u2022=3.$ What are the values of $\ufffd$ and $\text{*?}$
a.
b.
c.
Let the value of $\Delta $ & $\text{*}$ be $x$ and $y$ & it is given that $\u2022=3.$
From (a), $y+y+y+y+y=x+x+3+3$
$\begin{array}{l}5y=2x+6\\ 5y2x=6\\ 2x5y=6\text{}\mathrm{....}\left(\text{i}\right)\end{array}$
From (b), $x+x=y+y+3+3$
$\begin{array}{l}2x=2y+6\\ 2x2y=6\\ xy=3\text{}\mathrm{....}\text{(ii)}\end{array}$
From (c), $y+y+y+3+3+3=x+x+x$
$\begin{array}{l}3y+9=3x\\ 3x3y=9\\ xy=3\text{}\mathrm{....}\text{(iii)}\end{array}$
From eq. $\text{(ii)}xy=3$
$x=y+3$
On putting $x=y+3$ in eq. (i), we get
$\begin{array}{l}2(y+3)5y=6\\ 2y+65y=6\\ 3y+6=6\\ 3y=66=12\\ y=\frac{12}{3}=4\end{array}$
On putting $y=4$ in eq. (ii), we get
$\begin{array}{l}xy=3\\ x4=3\\ x=3+4=7\\ x=7\end{array}$
Value of $\Delta $ $=x=7$ & value of $\text{*}=y=4.$
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.
Weight on LHS $=20+20=40$ kg
Weight on RHS $=14+4=18$ kg
Weight should be equal,
Therefore, circle weight $=4018=22$ kg