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 $ \{1,2,3,4\} $.

(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 $ \{\ldots - 3 , - 2 , -1, 0,1,2,3,4\} $.

Answer :

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

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

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

(i) There is no natural number less than

$ (-\dfrac{5}{2}) $

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

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

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

…, -5, -4, -3.

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

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 $\{-4, - 3 , - 2, - 1,0,1\}$.

(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 $x \in(- \infty, 2)$.

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 -2are -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 $\{- 1,0,1,2, \ldots\}$.

(ii) When xis 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 (- $2, \infty$ ).

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 $(-4, \infty)$.

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 (-$\infty$, - 3 ).

Answer :

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

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

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

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

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

$\Rightarrow x \leq -3$

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 $(-\infty,-3]$.

Answer :

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

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

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

$\Rightarrow 6-x \geq 2$

$ \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 $(-\infty,4]$.

Answer :

$ \begin{aligned} & x+\dfrac{x}{2}+\dfrac{x}{3}<11 \\ & \Rightarrow x(1+\dfrac{1}{2}+\dfrac{1}{3})<11 \\ & \Rightarrow x(\dfrac{6+3+2}{6})<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 (-$\infty$, 6).

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 (-$\infty$, -6).

Answer :

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

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

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

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

$\Rightarrow 34 x-18 \leq 50$

$\Rightarrow 34 x \leq 50+18$

$\Rightarrow 34 x \leq 68$

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

$\Rightarrow x \leq 2$

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 [ - $\infty, 2 $].

Answer :

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

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

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

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

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

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

$\Rightarrow 120 \geq x$

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 $(-\infty ,120]$

Answer :

$ \begin{aligned} & 2(2 x+3)-10<6(x-2) \\ & \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 $(4, \infty)$.

Answer :

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

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

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

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

$\Rightarrow 8 \geq 4 x$

$\Rightarrow 2 \geq x$

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 $(- \infty, 2]$.

Answer :

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

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

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

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

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

$\Rightarrow 15 x<16 x-4$

$\Rightarrow 4<16 x-15 x$

$\Rightarrow 4<x$

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 $(4, \infty)$.

Answer :

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

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

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

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

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

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

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

$\Rightarrow 34 \geq 17 x$

$\Rightarrow 2 \geq x$

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 $(-\infty ,2]$

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 :

$3(1-x)<2(x+4)$

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

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

$\Rightarrow-5<5 x$

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

$\Rightarrow-1<x$

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 :

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

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

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

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

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

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

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

$\Rightarrow 7 x \geq-2$

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

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 ‘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

$\Rightarrow x<8 \ldots$ (i)

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

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

$\Rightarrow 2 x+2>11$

$\Rightarrow 2 x>11$ - 2

$\Rightarrow 2 x>9$

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

$\Rightarrow x>4.5$

From (i) and (ii), we obtain

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

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

Answer

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

2. Set up the conditions: We know two things:

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

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

3. Solve the inequality: Combine like terms in the inequality: 2x + 2 < 23 Subtract 2 from both sides:2x < 21 Divide both sides by 2 (since we’re dealing with even integers):x < 10.5

4. Consider the limitations: We found that x must be less than 10.5. However, x also needs to be greater than 5 and an integer (since it represents an even number). Therefore, valid values for x are 6, 8, and 10.

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: (6, 8), (8, 10)and (10,12).

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 $=(3 x - 2) cm$

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

$x cm+3 x cm+(3 x-2) cm \geq 61 cm$

$\Rightarrow 7 x-2 \geq 61$

$\Rightarrow 7 x \geq 61+2$

$\Rightarrow 7 x \geq 63$

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

$\Rightarrow x \geq 9$

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 $(x+3) cm$ and $2 x cm$ respectively.

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

$x cm+(x+3) cm+2 x cm \leq 91 cm$

$\Rightarrow 4 x+3 \leq 91$ $\Rightarrow 4 x \leq 91$ - 3

$\Rightarrow 4 x \leq 88$

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

$\Rightarrow x \leq 22$

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

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

$\Rightarrow 2 x \geq x+8$

$\Rightarrow x \geq 8 \ldots(2)$

From (1) and (2), 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 $22 cm$.