Answer :

The given inequality is $24 x < 100$.

$24 x < 100$

$\Rightarrow \dfrac{24 x}{24} < \dfrac{100}{24} \quad$ [Dividing both sides by same positive number]

$\Rightarrow x < \dfrac{25}{6}$

(i) It is evident that $1,2,3$, and $4$ are the only natural numbers less than $\dfrac{25}{6}$.

Thus, when $x$ is a natural number, the solutions of the given inequality are $1,2,3$, and 4 .

Hence, in this case, the solution set is $ \lbrace 1,2,3,4 \rbrace $.

(ii) The integers less than $\dfrac{25}{6}$ are $…-3, -2, -1,$ $0,1,2,3,4$.

Thus, when $x$ is an integer, the solutions of the given inequality are

$…-3, -2, -1, 0, 1, 2, 3, 4.$

Hence, in this case, the solution set is $ \lbrace \ldots - 3 , - 2 , -1, 0,1,2,3,4 \rbrace $.

Answer :

The given inequality is - $12 x > 30$.

$\Rightarrow \dfrac{-12 x}{-12} < \dfrac{30}{-12} \quad$ [Dividing both sides by same negative number]

$\Rightarrow x < -\dfrac{5}{2}$

(i) There is no natural number less than $ \left (-\dfrac{5}{2} \right ) $

Thus, when xis a natural number, there is no solution of the given inequality.

(ii) The integers less than $\left (-\dfrac{5}{2} \right )$ are $…, -5, -4, -3.$

Thus, when $x$ is an integer, the solutions of the given inequality are

$…, -5, -4, -3.$

Hence, in this case, the solution set is $ \lbrace {-5, -4, -3} \rbrace $.

Answer :

The given inequality is $5 x - 3 < 7$.

$5 x-3 < 7$

$\Rightarrow 5 x-3+3 < 7+3$

$\Rightarrow 5 x < 10$

$\Rightarrow \dfrac{5 x}{5} < \dfrac{10}{5}$

$\Rightarrow x < 2$

(i) The integers less than 2are $…, -4, -3, -2, -1, 0,1 .$

Thus, when xis an integer, the solutions of the given inequality are

$…, -4, -3, -2, -1, 0, 1.$

Hence, in this case, the solution set is $…, \lbrace -4, - 3 , - 2, - 1,0,1 \rbrace $.

(ii) When $x$ is a real number, the solutions of the given inequality are given by $x < 2$, that is, all real numbers $x$ which are less than 2.

Thus, the solution set of the given inequality is $x \in\left (- \infty, 2 \right )$.

Answer :

The given inequality is $3 x+8 > 2$.

$3 x+8 > 2$

$\Rightarrow 3 x+8-8 > 2-8$

$\Rightarrow 3 x > -6$

$\Rightarrow \dfrac{3 x}{3} > \dfrac{-6}{3}$

$\Rightarrow x > -2$

(i) The integers greater than $-2$ are $-1, 0, 1, 2, ..$

Thus, when xis an integer, the solutions of the given inequality are

$-1, 0,1,2 \ldots$

Hence, in this case, the solution set is $ \lbrace - 1,0,1,2, \ldots \rbrace $.

(ii) When $x$ is a real number, the solutions of the given inequality are all the real numbers, which are greater than $- 2 .$

Thus, in this case, the solution set is $\left (- 2, \infty \right ).$

Answer :

$4 x+3 < 5 x+7$

$\Rightarrow 4 x+3-7 < 5 x+7-7$

$\Rightarrow 4 x-4 < 5 x$

$\Rightarrow 4 x-4-4 x < 5 x-4 x$

$\Rightarrow-4 < x$

Thus, all real numbers $x$, which are greater than $-4 ,$ are the solutions of the given inequality.

Hence, the solution set of the given inequality is $\left (-4, \infty \right )$.

Answer :

$ \Rightarrow 3 x-7+7 > 5 x - 1+7$

$\Rightarrow 3 x > 5 x+6$

$\Rightarrow 3 x - 5 x > 5 x+6$- $5 x$

$\Rightarrow - 2 x > 6$

$\Rightarrow \dfrac{-2 x}{-2} < \dfrac{6}{-2}$

$\Rightarrow x < -3$

Thus, all real numbers $x$, which are less than $- 3$ , are the solutions of the given inequality.

Hence, the solution set of the given inequality is $\left (-\infty, - 3 \right ).$

Answer :

$3\left (x-1 \right ) \leq 2\left (x-3 \right )$

$\Rightarrow 3 x-3 \leq 2 x-6$

$\Rightarrow 3 x-3+3 \leq 2 x-6+3$

$\Rightarrow 3 x \leq 2 x-3$

$\Rightarrow 3 x-2 x \leq 2 x-3$ - $2 x$

$\Rightarrow x \leq -3$

Thus, all real numbers $x$, which are less than or equal to $-3$ , are the solutions of the given inequality.

Hence, the solution set of the given inequality is $\left (-\infty,-3\right]$.

Answer :

$3\left (2-x \right ) \geq 2\left (1-x \right )$

$\Rightarrow 6-3 x \geq 2-2 x$

$\Rightarrow 6-3 x+2 x \ \geq 2-2 x+2 x$

$\Rightarrow 6-x \geq 2$

$ \begin{aligned} & \Rightarrow 6-x-6 \geq 2-6 \\ \\ & \Rightarrow-x \geq-4 \\ \\ & \Rightarrow x \leq 4 \end{aligned} $

Thus, all real numbers $x$, which are less than or equal to 4 , are the solutions of the given inequality.

Hence, the solution set of the given inequality is $\left (-\infty,4\right]$.

Answer :

$ \begin{aligned} & x+\dfrac{x}{2}+\dfrac{x}{3} < 11 \\ \\ & \Rightarrow x\left (1+\dfrac{1}{2}+\dfrac{1}{3} \right ) < 11 \\ \\ & \Rightarrow x\left (\dfrac{6+3+2}{6} \right ) < 11 \\ \\ & \Rightarrow \dfrac{11 x}{6} < 11 \\ \\ & \Rightarrow \dfrac{11 x}{6 \times 11} < \dfrac{11}{11} \\ \\ & \Rightarrow \dfrac{x}{6} < 1 \\ \\ & \Rightarrow x < 6 \end{aligned} $

Thus, all real numbers $x$, which are less than $6$, are the solutions of the given inequality.

Hence, the solution set of the given inequality is $\left (-\infty, 6 \right ).$

Answer :

$\dfrac{x}{3} > \dfrac{x}{2}+1$

$\Rightarrow \dfrac{x}{3}-\dfrac{x}{2} > 1$

$\Rightarrow \dfrac{2 x-3 x}{6} > 1$

$\Rightarrow-\dfrac{x}{6} > 1$

$\Rightarrow-x > 6$

$\Rightarrow x < -6$

Thus, all real numbers $x$, which are less than $- 6 ,$ are the solutions of the given inequality.

Hence, the solution set of the given inequality is $\left (-\infty, -6 \right ).$

Answer :

$\dfrac{3\left (x-2 \right )}{5} \leq \dfrac{5\left (2-x \right )}{3}$

$\Rightarrow 9\left (x-2 \right ) \leq 25\left (2-x \right )$

$\Rightarrow 9 x-18 \leq 50-25 x$

$\Rightarrow 9 x-18+25 x \leq 50$

$\Rightarrow 34 x-18 \leq 50$

$\Rightarrow 34 x \leq 50+18$

$\Rightarrow 34 x \leq 68$

$\Rightarrow \dfrac{34 x}{34} \leq \dfrac{68}{34}$

$\Rightarrow x \leq 2$

Thus, all real numbers $x$, which are less than or equal to $2 ,$ are the solutions of the given inequality.

Hence, the solution set of the given inequality is $ ~ \left ( - \infty, 2\right ].$

Answer :

$\dfrac{1}{2}\left (\dfrac{3 x}{5}+4 \right ) \geq \dfrac{1}{3}\left (x-6 \right )$

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

$\Rightarrow \dfrac{9 x}{5}+12 \geq 2 x-12$

$\Rightarrow 12+12 \geq 2 x-\dfrac{9 x}{5}$

$\Rightarrow 24 \geq \dfrac{10 x-9 x}{5}$

$\Rightarrow 24 \geq \dfrac{x}{5}$

$\Rightarrow 120 \geq x$

Thus, all real numbers $x$, which are less than or equal to $120 ,$ are the solutions of the given inequality. Hence, the solution set of the given inequality is $120$

Answer :

$ \begin{aligned} & 2\left (2 x+3 \right )-10 < 6\left (x-2 \right ) \\ \\ & \Rightarrow 4 x+6-10 < 6 x-12 \\ \\ & \Rightarrow 4 x-4 < 6 x-12 \\ \\ & \Rightarrow-4+12 < 6 x-4 x \\ \\ & \Rightarrow 8 < 2 x \\ \\ & \Rightarrow 4 < x \end{aligned} $

Thus, all real numbers $x$, which are greater than or equal to $4 ,$ are the solutions of the given inequality. Hence, the solution set of the given inequality is $\left (4, \infty \right )$.

Answer :

$37-\left (3 x+5 \right ) \geq 9 x-8\left (x-3 \right )$

$\Rightarrow 37-3 x-5 \geq 9 x-8 x+24$

$\Rightarrow 32-3 x \geq x+24$

$\Rightarrow 32-24 \geq x+3 x$

$\Rightarrow 8 \geq 4 x$

$\Rightarrow 2 \geq x$

Thus, all real numbers $x$, which are less than or equal to $2 ,$ are the solutions of the given inequality.

Hence, the solution set of the given inequality is $\left (- \infty, 2\right]$.

Answer :

$\dfrac{x}{4} < \dfrac{\left (5 x-2 \right )}{3}-\dfrac{\left (7 x-3 \right )}{5}$

$\Rightarrow \dfrac{x}{4} < \dfrac{5\left (5 x-2 \right )-3\left (7 x-3 \right )}{15}$

$\Rightarrow \dfrac{x}{4} < \dfrac{25 x-10-21 x+9}{15}$

$\Rightarrow \dfrac{x}{4} < \dfrac{4 x-1}{15}$

$\Rightarrow 15 x < 4\left (4 x-1 \right )$

$\Rightarrow 15 x < 16 x-4$

$\Rightarrow 4 < 16 x-15 x$

$\Rightarrow 4 < x$

Thus, all real numbers $x$, which are greater than $4 ,$ are the solutions of the given inequality.

Hence, the solution set of the given inequality is $\left (4, \infty \right )$.

Answer :

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

$\Rightarrow \dfrac{\left (2 x-1 \right )}{3} \geq \dfrac{5\left (3 x-2 \right )-4\left (2-x \right )}{20}$

$\Rightarrow \dfrac{\left (2 x-1 \right )}{3} \geq \dfrac{15 x-10-8+4 x}{20}$

$\Rightarrow \dfrac{\left (2 x-1 \right )}{3} \geq \dfrac{19 x-18}{20}$

$\Rightarrow 20\left (2 x-1 \right ) \geq 3\left (19 x-18 \right )$

$\Rightarrow 40 x-20 \geq 57 x-54$

$\Rightarrow-20+54 \geq 57 x-40 x$

$\Rightarrow 34 \geq 17 x$

$\Rightarrow 2 \geq x$

Thus, all real numbers $x$, which are less than or equal to $2 ,$ are the solutions of the given inequality.

Hence, the solution set of the given inequality is $2$.

Answer :

$3 x-2 < 2 x+1$

$\Rightarrow 3 x-2 x < 1+2$

$\Rightarrow x < 3$

The graphical representation of the solutions of the given inequality is as follows.

Answer :

$3\left (1-x \right ) < 2\left (x+4 \right )$

$\Rightarrow 3-3 x < 2 x+8$

$\Rightarrow 3-8 < 2 x+3 x$

$\Rightarrow-5 < 5 x$

$\Rightarrow \dfrac{-5}{5} < \dfrac{5 x}{5}$

$\Rightarrow-1 < x$

The graphical representation of the solutions of the given inequality is as follows.

Answer :

$5 x-3 \geq 3 x-5$

$\Rightarrow 5 x-3 x \geq-5+3$

$\Rightarrow 2 x \geq-2$

$\Rightarrow \dfrac{2 x}{2} \geq \dfrac{-2}{2}$

$\Rightarrow x \geq-1$

The graphical representation of the solutions of the given inequality is as follows.

Answer :

$\dfrac{x}{2} \geq \dfrac{\left (5 x-2 \right )}{3}-\dfrac{\left (7 x-3 \right )}{5}$

$\Rightarrow \dfrac{x}{2} \geq \dfrac{5\left (5 x-2 \right )-3\left (7 x-3 \right )}{15}$

$\Rightarrow \dfrac{x}{2} \geq \dfrac{25 x-10-21 x+9}{15}$

$\Rightarrow \dfrac{x}{2} \geq \dfrac{4 x-1}{15}$

$\Rightarrow 15 x \geq 2\left (4 x-1 \right )$

$\Rightarrow 15 x \geq 8 x-2$

$\Rightarrow 15 x-8 x \geq 8 x-2-8 x$

$\Rightarrow 7 x \geq-2$

$\Rightarrow x \geq-\dfrac{2}{7}$

The graphical representation of the solutions of the given inequality is as follows.

Answer :

Let $x$ be the marks obtained by Ravi in the third unit test.

Since the student should have an average of at least 60 marks,

$\dfrac{70+75+x}{3} \geq 60$

$\Rightarrow 145+x \geq 180$

$\Rightarrow x \geq 180-145$

$\Rightarrow x \geq 35$

Thus, the student must obtain a minimum of $35$ marks to have an average of at least $60$ marks.

Answer :

Let $x$ be the marks obtained by Sunita in the fifth examination.

In order to receive grade $\mathbf{A}$ in the course, she must obtain an average of $90$ marks or more in five examinations.

Therefore,

$ \begin{aligned} & \dfrac{87+92+94+95+x}{5} \geq 90 \\ \\ & \Rightarrow \dfrac{368+x}{5} \geq 90 \\ \\ & \Rightarrow 368+x \geq 450 \\ \\ & \Rightarrow x \geq 450-368 \\ \\ & \Rightarrow x \geq 82 \end{aligned} $

Thus, Sunita must obtain greater than or equal to $82$ marks in the fifth examination.

Answer :

Let $x$ be the smaller of the two consecutive odd positive integers. Then, the other integer is $x+2$.

Since both the integers are smaller than $10 ,$

$x+2 < 10$

$\Rightarrow x < 10 - 2$

$\Rightarrow x < 8 \qquad\ldots \left (i \right )$

Also, the sum of the two integers is more than $11 .$

$\therefore \ \ x+\left (x+2 \right ) > 11$

$\Rightarrow 2 x+2 > 11$

$\Rightarrow 2 x > 11 - 2$

$\Rightarrow 2 x > 9$

$\Rightarrow x > \dfrac{9}{2}$

$\Rightarrow x > 4.5\qquad … \left (ii \right )$

From \left (i \right ) and \left (ii \right ), we obtain

Since $x$ is an odd number, $x$ can take the values, $5$ and $7 .$

Thus, the required possible pairs are $\left (5,7 \right )$ and $\left (7,9 \right )$.

Answer

Let $x$ be the smaller even integer. Since the consecutive even number is next in order, it will be $x + 2.$

We know that

Both integers must be larger than $5: x > 5$ and $x > 3$

Their sum must be less than $23: x + \left (x + 2 \right ) < 23$ \left (This combines $x$ and $x + 2$ because they represent the consecutive even numbers \right ).

So,combine like terms in the inequality

$2x + 2 < 23$ Subtract $2$ from both sides: $2x < 21$ Divide both sides by $2$ \left (since we’re dealing with even integers \right ) $x < 10.5$

We found that x must be less than $10.5$ . However, $x$ also needs to be greater than $5$ and an integer \left (since it represents an even number \right ). Therefore, valid values for $x$ are $6, 8,$ and $10.$

Now we find the corresponding even integers

If $x = 6,$ then the consecutive even number is $x + 2 = 8$

If $x = 8,$ then the consecutive even number is $x + 2 = 10$

If $x = 10,$ then the consecutive even number is $x + 2 = 12$

Therefore, the valid pairs of consecutive even positive integers are: $\left (6, 8 \right ), \left (8, 10 \right )$ and $\left (10,12 \right ).$

Answer :

Let the length of the shortest side of the triangle be $x cm$.

Then, length of the longest side $=3 x cm$

Length of the third side $=\left (3 x - 2 \right ) cm$

Since the perimeter of the triangle is at least $61 cm$,

$x +3 x +\left (3 x-2 \right ) \geq 61 $

$\Rightarrow 7 x-2 \geq 61$

$\Rightarrow 7 x \geq 61+2$

$\Rightarrow 7 x \geq 63$

$\Rightarrow \dfrac{7 x}{7} \geq \dfrac{63}{7}$

$\Rightarrow x \geq 9$

Thus, the minimum length of the shortest side is $9 cm$.

Answer :

Let the length of the shortest piece be $x cm$. Then, length of the second piece and the third piece are $\left (x+3 \right ) cm$ and $2 x cm$ respectively.

Since the three lengths are to be cut from a single piece of board of length $91 cm$,

$x +\left (x+3 \right ) +2 x \leq 91 $

$\Rightarrow 4 x+3 \leq 91$

$\Rightarrow 4 x \leq 91 - 3$

$\Rightarrow 4 x \leq 88$

$\Rightarrow \dfrac{4 x}{4} \leq \dfrac{88}{4}$

$\Rightarrow x \leq 22\qquad … \left (i \right ) $

Also, the third piece is at least $5 cm$ longer than the second piece.

$\therefore \ \ 2 x \geq\left (x+3 \right )+5$

$\Rightarrow 2 x \geq x+8$

$\Rightarrow x \geq 8 \qquad\ldots\left (ii \right )$

From $\left (i \right )$ and $\left (ii \right ),$ we obtain

$8 \leq x \leq 22$

Thus, the possible length of the shortest board is greater than or equal to $8 cm$ but less than or equal to $22 cm$.