**1** **Solve $24x < 100,$ when$\\$ (i) x is a natural number $\\$ (ii) x is an integer.**

The given inequality is $24x < 100$$\\$ $24x < 100 \\ \implies \dfrac{24x}{24} < \dfrac{100}{24} \\ [\text{Dividing both sides by same positive number}] \\ \implies x < \dfrac{25}{6}$$\\$ (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 $\{- 3, - 2, - 1,0,1,2,3,4 \}$

**2** **Solve $- 12 x > 30,$ when $\\$ (i) x is a natural number $\\$ (ii) x is an integer**

The given inequality is $-12 x > 30.$$\\$ $-12x >30.\\ \implies \dfrac{-12x}{-12} < \dfrac{30}{-12}\\ [\text{Dividing both sides by same negative number}]\\ \implies x< -\dfrac{5}{2}$$\\$ (i) There is no natural number less than $(-\dfrac{5}{2}).$ Thus, when x is a natural number, there is no solution of the given inequality.$\\$ (ii)The integers less than $(-\dfrac{5}{2})$ are......$-5,-4,-3.$$\\$ Thus, when x is an integer, the solutions of the given inequality are $...., - 5, - 4, - 3$$\\$ Hence, in this case, the solution set is $\{ ...., - 5, -4, -3 \}.$

**3** **Solve $5 x - 3 < 7 ,$ when $\\$(i) x is an integer$\\$ (ii) x is a real number**

The given inequality is $5 x - 3 < 7 .$$\\$ $5 x - 3 < 7 \\ \implies 5x-3+3 < 7+3\\ \implies 5x < 10\\ \implies \dfrac{5x}{5} < \dfrac{10}{5}\\ \implies x < 2$$\\$ (i) The integers less than $2$ are $..., - 4, - 3, - 2, - 1,0,1 .$$\\$ Thus, when $x$ is an integer, the solutions of the given inequality are $..., -4, - 3, - 2, - 1,0,1 .$$\\$ 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 $x \in (-\infty , 2 ).$

**4** **Solve $3 x + 8 > 2 ,$ when$\\$ (i) x is an integer $\\$ (ii) x is a real number**

The given inequality is $ 3 x + 8 > 2$$\\$ $3 x + 8 > 2\\ \implies 3 x + 8 - 8 > 2 -8\\ \implies 3 x >- 6\\ \implies \dfrac{3x}{3}> \dfrac{-6}{3}\\ \implies x > -2$$\\$ (i) The integers greater than $- 2$ are $- 1,0,1, 2,...$$\\$ Thus, when x is an integer, the solutions of the given inequality are $- 1,0,1, 2,...$$\\$ Hence, in this case, the solution set is $\{ -1,0,1,2,... \}$ .$\\$ (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 $(-2, \infty)$ .

**5** **Solve the given inequality for real $x : 4 x + 3 < 5 x + 7$**

$4 x + 3 < 5 x + 7\\ \implies 4 x + 3 - 7 < 5 x + 7 - 7 \\ \implies 4 x - 4 < 5 x\\ \implies 4 x - 4 - 4 x < 5 x - 4 x\\ \implies -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 $(-4, \infty).$

**6** **Solve $24 x < 100$ , when (i) x is a natural number (ii) x is an integer.**

The given inequality is $24 x < 100$ $\\$ $24 x < 100 \\ \Rightarrow \dfrac{24 x}{24} < \dfrac{100}{24 } \quad [\text{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 $\{-3, - 2, -1,0,1,2,3,4 \}$

**7** **Solve $-12 x > 30$, when$\\$ (i) x is a natural number $\\$ (ii) x is an integer**

The given inequality is $-12 x > 30.$$\\$ $-12x > 30\\ \Rightarrow \dfrac{-12x}{-12} < \dfrac{30}{-12} [\text{Dividing both sides by same negative number}]\\ \Rightarrow x < -\dfrac{5}{2}$$\\$ (i) There is no natural number less than $(-\dfrac{5}{2})$$\\$ Thus, when x is a natural number, there is no solution of the given inequality.$\\$ (ii) The integers less than $(-\dfrac{5}{2})$ are .... $-5,-4,-3.$$\\$ Thus, when $x$ is an integer, the solutions of the given inequality are ....,$- 5, - 4, - 3$$\\$ Hence, in this case, the solution set is $\{...., - 5, - 4, - 3 \}$ .

**8** **Solve $5 x -3 < 7$ , when$\\$ (i) x is an integer$\\$ (ii) x is a real number**

The given inequality is $5 x - 3 < 7$$\\$ $ 5x-3 < 7\\ \Rightarrow 5x -3+3 < 7+3\\ \Rightarrow 5x < 10\\ \Rightarrow \dfrac{5x}{5} < \dfrac{10}{5}\\ \Rightarrow x< 2$$\\$ (i) The integers less than $2$ are ..., $-4, - 3, -2, - 1,0,1 .$$\\$ Thus, when x is an integer, the solutions of the given inequality are ..., $- 4, - 3, - 2, - 1,0,1 .$$\\$ 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 $x \in (- \infty, 2 )$ .

**9** **Solve $3 x + 8 > 2$ , when$\\$ (i) x is an integer$\\$ (ii) x is a real number**

The given inequality is $3 x + 8 > 2$$\\$ $3x+8 > 2\\ \Rightarrow 3x+8-8 > 2-8\\ \Rightarrow 3x > -6\\ \Rightarrow \dfrac{3x}{3} > \dfrac{-6}{3}\\ \Rightarrow x > -2$$\\$ (i) The integers greater than $-2$ are $-1,0,1, 2,...$$\\$ Thus, when x is an integer, the solutions of the given inequality are $- 1,0,1, 2,...$$\\$ Hence, in this case, the solution set is $\{-1,0,1,2,... \}$ .$\\$ (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 $(-2, \infty)$ .

**10** **Solve the given inequality for real $x : 4 x + 3 < 5 x + 7$**

$4x+3 < 5x +7\\ \Rightarrow 4x+3-7 < 5x +7 -7\\ \Rightarrow 4x -4 < 5x\\ \Rightarrow 4x - 4 - 4x < 5x -4x \\ \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 $(-4, \infty )$ .

**11** **Solve the given inequality for real x : 3 x - 7 > 5 x - 1**

3 x - 7 > 5 x - 1$\\$ 3 x - 7 + 7 > 5 x - 1 + 7$\\$ 3 x >5 x + 6$\\$ 3 x - 5 x > 5 x +6 - 5 x$\\$ - 2 x > 6$\\$ $\dfrac{-2x}{-2}<\dfrac{6}{-2}$$\\$ 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 )

**12** **Solve the given inequality for real x : 3 (x -1 )$\leq$ 2 ( x - 3 )**

3 ( x - 1 )$\leq$ 2 ( x - 3 )$\\$ 3 x - 3 $\leq$ 2 x - 6$\\$ 3 x - 3 + 3 $\leq$ 2 x - 6 + 3$\\$ 3 x $\leq$ 2 x - 3$\\$ 3 x - 2 x $\leq$ 2 x - 3 - 2 x$\\$ 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 ($-\infty,- 3 ]$ .

**13** **Solve the given inequality for real $x : 3 ( 2 - x ) \geq 2( 1 - x )$**

$3 ( 2 - x ) \geq 2( 1 - x )\\ \implies 6 - 3 x \geq 2 - 2 x\\ \implies 6 - 3 x + 2 x \geq 2 - 2 x + 2 x\\ \implies 6 - x \geq 2\\ \implies 6 - x - 6 \geq 2 - 6\\ \implies - x \geq - 4\\ \implies x \leq 4$ $\\$ 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 $ ( -\infty, 4]$ .

**14** **Solve the given inequality for real $x:x+\dfrac{x}{2}+\dfrac{x}{3 } < 11$**

$x+\dfrac{x}{2}+\dfrac{x}{3 } < 11\\ \implies x(1+\dfrac{1}{2}+\dfrac{1}{3})<11\\ \implies x(\dfrac{6+3+2}{6}) < 11\\ \implies \dfrac{11x}{6} < 11\\ \implies \dfrac{11x}{6*11}< \dfrac{11}{11}\\ \implies \dfrac{x}{6} < 1\\ \implies x < 6$ $\\$ 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 $(-\infty , 6).$

**15** **Solve the given inequality for real $ x:\dfrac{x}{3} > \dfrac{x}{2}+1$**

$ \dfrac{x}{3} > \dfrac{x}{2}+1\\ \implies \dfrac{x}{3} -\dfrac{x}{2} >1\\ \implies \dfrac{2x-3x}{6 } > 1\\ \implies -\dfrac{x}{6} > 1\\ \implies -x > 6\\ \implies 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 $(-\infty,-6).$

**16** **Solve the given inequality for real $ x:\dfrac{3(x-2)}{5} \leq \dfrac{5(2-x)}{3}$**

$\dfrac{3(x-2)}{5} \leq \dfrac{5(2-x)}{3}\\ \implies 9(x-2)\leq 25(2-x)\\ \implies 9x-18 \leq 50-25x\\ \implies 9x-18+25x \leq 50\\ \implies 34x-18 \leq 50\\ \implies 34x \leq 50+18\\ \implies 34x \leq 68\\ \implies \dfrac{34x}{34}\leq \dfrac{68}{34}\\ \implies x \leq 2 $ $\\$ Thus, all real numbers x, which are less than or equal to 2, are the solutions of the given hence the solution set of the given inequality is $( - \infty , 2] $.

**17** **Solve the given inequality for real $ x:\dfrac{1}{2}(\dfrac{3x}{5}+4)\geq \dfrac{1}{3}(x-6)$**

$\dfrac{1}{2}(\dfrac{3x}{5}+4)\geq \dfrac{1}{3}(x-6)\\ \implies \dfrac{9x}{5}+12 \geq 2x -12\\ \implies 12+12 \geq 2x-\dfrac{9x}{5}\\ \implies 24 \geq \dfrac{10x-9x}{5}\\ \implies 24 \geq \dfrac{x}{5}\\ \implies 120 \geq x$ $\\$ 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 $(-\infty, 120).$

**18** **Solve the given inequality for real $x : 2 ( 2 x + 3 )- 10 < 6 (x - 2 )$**

$ 2 ( 2 x + 3 )- 10 < 6 (x - 2 )\\ \implies 4 x + 6 - 10 < 6 x - 12\\ \implies 4 x- 4 < 6 x - 12\\ \implies - 4 + 12 < 6 x - 4 x\\ \implies 8 < 2x\\ \implies 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 ) .$

**19** **Solve the given inequality for real $x : 37 -( 3 x + 5 )\geq 9 x - 8 ( x - 3 )$**

$37 -( 3 x + 5 )\geq 9 x - 8 ( x - 3 )\\ \implies 37 - 3 x - 5 \geq 9 x - 8 x + 24\\ \implies 32 - 3 x \geq x + 24\\ \implies 32 - 24 \geq x + 3 x\\ \implies 8 \geq 4x\\ \implies 2 \geq x$ $\\$ 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 $(-\infty,2)$

**20** **Solve the given inequality for real $x: \dfrac{x}{4}< \dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5}$**

$ \dfrac{x}{4}< \dfrac{(5x-2)}{3}-\dfrac{(7x-3)}{5} \\ \implies \dfrac{x}{4} < \dfrac{5(5x-2)-3(7x-3)}{15}\\ \implies \dfrac{x}{4} < \dfrac{25x-10-21x+9}{15}\\ \implies \dfrac{x}{4} < \dfrac{4x-1}{15}\\ \implies 15x < 4(4x-1)\\ \implies 15x < 16x-4\\ \implies 4 < 16x -15x\\ \implies 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 $(4,\infty)$

**21** **Solve the given inequality for real $x:\dfrac{(2x-1)}{3} \geq \dfrac{(3x-2)}{4}-\dfrac{2-x}{5}$**

$\dfrac{(2x-1)}{3} \geq \dfrac{(3x-2)}{4}-\dfrac{2-x}{5}\\ \implies \dfrac{2x-1}{3} \geq \dfrac{5(3x-2)-4(2-x)}{20}\\ \implies \dfrac{2x-1}{3} \geq \dfrac{15x-10-8+4x}{20}\\ \implies \dfrac{2x-1}{3} \geq \dfrac{19x-18}{20}\\ \implies 20(2x-1) \geq 3(19x-18)\\ \implies 40x-20 \geq 57x-54\\ \implies -20+54 \geq 57x -40x\\ \implies 34 \geq 17x \\ \implies 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 $(-\infty,2)$.

**22** **Solve the given inequality and show the graph of the solution on number line: $3 x - 2 < 2 x + 1$**

$3 x - 2 < 2 x + 1\\ \implies 3x -2x < 1+2 \\ \implies x < 3 $ $\\$ The graphical representation of the solutions of the given inequality is as follows:

**23** **Solve the given inequality and show the graph of the solution on number line:$\\$ $5 x - 3 \geq 3 x - 5$**

$5 x - 3 \geq 3 x - 5\\ \implies 5x -3x \geq -5+3\\ \implies 2x \geq -2\\ \implies \dfrac{2x}{2}\geq \dfrac{-2}{2}\\ \implies x \geq -1 $ $\\$ The graphical representation of the solutions of the given inequality is as follows.

**24** **Solve the given inequality and show the graph of the solution on number line:$\\$ $3 ( 1 - x ) < 2 ( x + 4 )$**

$3 ( 1 - x ) < 2 ( x + 4 )\\ \implies 3-3x < 2x+8\\ \implies 3-8 < 2x +3x \\ \implies -5 < 5x\\ \implies \dfrac{-5}{5} < \dfrac{5x}{5}\\ \implies -1 < x $ $\\$ The graphical representation of the solutions of the given inequality is as follows:

**25** **Solve the given inequality and show the graph of the solution on number line: $\\$ $\dfrac{x}{2}\geq \dfrac{5x-2}{3}-\dfrac{7x-3}{5}$**

$\dfrac{x}{2}\geq \dfrac{5x-2}{3}-\dfrac{7x-3}{5}\\ \implies \dfrac{x}{2} \geq \dfrac{5(5x-2)-3(7x-3)}{15}\\ \implies \dfrac{x}{2} \geq \dfrac{25x-10-21x+9}{15}\\ \implies \dfrac{x}{2} \geq \dfrac{4x-1}{15}\\ \implies 15x \geq 2(4x-1)\\ \implies 15x \geq 8x-2\\ \implies 15x - 8x \geq 8x-2 -8x\\ \implies 7x \geq -2\\ \implies x \geq -\dfrac{2}{7}$ The graphical representation of the solutions of the given inequality is as follows.

**26** **Ravi obtained 70 and 75 marks in first two-unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.**

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\\ \implies 145+x \geq 180\\ \implies x \geq 180-145\\ \implies x\geq 35 $ Thus, the student must obtain a minimum of 35 marks to have an average of at least 60 marks.

**27** **To receive Grade ‘A’ in a course, one must obtain an average of $90$ marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are $87, 92, 94$ and $95,$ find minimum marks that Sunita must obtain in fifth examination to get grade $‘A’ $ in the course.**

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,$\\$ $ \dfrac{87+92+94+95+x}{5}\geq 90\\ \implies \dfrac{368+x}{5} \geq 90\\ \implies 368+x \geq 450\\ \implies x \geq 450 - 368\\ \implies x \geq 82 $ $\\$ Thus, Sunita must obtain greater than or equal to $82 $marks in the fifth examination.

**28** **Find all pairs of consecutive odd positive integers both of which are smaller than $10$ such that their sum is more than $11.$**

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\\ \implies x < 10-2\\ \implies x < 8......(i)$ $\\$ Also, the sum of the two integers is more than $11.$ $\\$ $\therefore x+(x+2)>11\\ \implies 2x + 2 > 11 \\ \implies 2x > 11 -2 \\ \implies 2x > 9 \\ \implies x > \dfrac{9}{2}\\ \implies x > 4.5....(ii)$ $\\$ 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).$\\$

**29** **Find all pairs of consecutive even positive integers, both of which are larger than $5$ such that their sum is less than $23.$**

Let x be the smaller of the two consecutive even positive integers. Then, the other integer is $x + 2 .$ $\\$ Since both the integers are larger than 5, $x > 5........ (1 )$ $\\$ Also, the sum of the two integers is less than $23$ $\\$ $x+(x+2)<23\\ \implies 2x+2 < 23\\ \implies 2x < 23 -2\\ \implies 2x < 21 \\ \implies x < \dfrac{21}{2}\\ \implies x < 10.5....(2)$ $\\$ From (1) and (2), we obtain $5 < x < 10.5 .$ $\\$ Since x is an even number, x can take the values, 6, 8 and 10.$\\$ Thus, the required possible pairs are ( 6,8 ) , ( 8,10 ) and (10,12)

**30** **The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.**

Let the length of the shortest side of the triangle be $x cm.$$\\$ Then, length of the longest side = $3x cm$$\\$ Length of the third side $ ( 3 x - 2 ) cm$$\\$ Since the perimeter of the triangle is at least $61 cm, xcm + 3 xcm +( 3 x - 2 ) cm \geq 61 cm\\ \implies 7 x - 2 \geq 61\\ \implies 7 x \geq 61 + 2\\ \implies 7 x \geq 63\\ \implies \dfrac{7x}{7}\geq \dfrac{63}{7}\\ \implies x\geq 9$ $\\$ Thus, the minimum length of the shortest side is $9 cm.$

**31** **A man wants to cut three lengths from a single piece of board of length $91 cm$. The second length is to be$ 3 cm$ longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least $5 cm$ longer than the second?$\\$ [Hint: It x is the length of the shortest board, then $x , ( x + 3 )$ and 2x are the lengths of the second and third piece, respectively. Thus, $x =( x + 3 )+ 2 x \leq 91$ and $ 2 x \geq ( x + 3 )+ 5 ]$**

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 $2x \ cm$ respectively.$\\$ Since the three lengths are to be cut from a single piece of board of length $91 cm, x \ cm +( x + 3 ) cm + 2 x \ cm \leq 91 cm$

$\implies 4x+3 \leq 91\\ \implies 4x \leq 91-3\\ \implies 4x \leq 88\\ \implies \dfrac{4x}{4} \leq \dfrac{88}{4}\\ \implies x \leq 22.....(1)$ $\\$ Also, the third piece is at least $5 \ cm$ longer than the second piece. $\\$ $\therefore 2x \geq (x+3)+5\\ \implies 2x \geq x+8\\ \implies x \geq 8.....(2)$ $\\$ 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 $22\ cm.$

**32** **Solve the given inequality graphically in two-dimensional plane:$ x + y < 5$**

The graphical representation of $x + y = 5$ is given as dotted line in the figure below.$\\$ This line divides the xy - plane in two half planes, I and II.$\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.$\\$ We select the point as (0, 0).$\\$ It is observed that, 0 + 0 < 5 or 0 < 5 , which is true Therefore, half plane II is not the solution region of the given inequality. Also, it is evident that any point on the line does not satisfy the given strict inequality.$\\$ Thus, the solution region of the given inequality is the shaded half plane I excluding the points on the line.$\\$ This can be represented as follows.$\\$

**33** **Solve the given inequality graphically in two-dimensional plane: $2 x+ y \geq 6$**

The graphical representation of $ 2 x + y = 6$ is given in the figure below.$\\$ This line divides the xy-plane in two half planes, I and II.$\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.$\\$ We select the point as (0, 0).$\\$ It is observed that,$\\$ $2 ( 0 )+ 0 \geq 6 $or $0 \geq 6,$ which is false$\\$ Therefore, half plane I is not the solution region of the given inequality. Also, it is evident that any point on the line satisfies the given inequality.$\\$ Thus, the solution region of the given inequality is the shaded half plane II including the points on the line.$\\$ This can be represented as follows.$\\$

**34** **Solve the given inequality graphically in two-dimensional plane: $3 x + 4 y \leq 12$**

$3 x + 4 y \leq 12$ $\\$ The graphical representation of $3 x + 4 y = 12$ is given in the figure below. $\\$ This line divides the xy-plane in two half planes, I and II. $\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not. $\\$ We select the point as (0, 0). $\\$ Thus, the solution region of the given inequality is the shaded half plane I including the points on the line. $\\$ This can be represented as follows. $\\$

**35** **Solve the given inequality graphically in two-dimensional plane: $y + 8 \geq 2 x$**

The graphical representation of $y+8=2x$ is given in the figure below. $\\$ This line divides the xy-plane in two half planes. $\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not. $\\$ We select the point as (0, 0). $\\$ It is observed that, $\\$ $0 + 8 \geq 2 ( 0 )$ or $8 \geq 0$ , which is true $\\$ Therefore, lower half plane is not the solution region of the given inequality. Also, it is evident that any point on the line satisfies the given inequality. Thus, the solution region of the given inequality is the half plane containing the point $(0, 0)$ including the line. $\\$ The solution region is represented by the shaded region as follows.

**36** **Solve the given inequality graphically in two-dimensional plane: $x - y \leq 2$**

The graphical representation of $x - y = 2$ is given in the figure below. This line divides the xy-plane in two half planes.$\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.$\\$ We select the point as (0, 0).$\\$ It is observed that,$\\$ $0 - 0 \leq 2$ or $0 \leq 2$, which is true$\\$ Therefore, the lower half plane is not the solution region of the given inequality. Also, it is clear that any point on the line satisfies the given inequality.$\\$ Thus, the solution region of the given inequality is the half plane containing the point$ (0, 0)$ including the line. The solution region is represented by the shaded region as follows.

**37** **Solve the given inequality graphically in two-dimensional plane: $2 x - 3 y > 6$**

The graphical representation of $2 x - 3 y = 6$ is given as dotted line in the figure below. This line divides the xy-plane in two half planes.$\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not. We select the point as $ (0, 0).$ It is observed that, $2 ( 0 )-3 (0) > 6$ or $0 > 6$, which is false$\\$ Therefore, the upper half plane is not the solution region of the given inequality. Also, it is clear that any point on the line does not satisfy the given inequality.$\\$ Thus, the solution region of the given inequality is the half plane that does not contain the point $(0, 0)$ including the line.$\\$ The solution region is represented by the shaded region as follows.

**38** **Solve the given inequality graphically in two-dimensional plane: $- 3 x + 2 y \geq -6$**

The graphical representation of $- 3 x + 2 y =- 6$ is given in the figure below.$\\$ This line divides the xy-plane in two half planes.$\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.$\\$ We select the point as $(0, 0).$$\\$ It is observed that, $- 3 ( 0 )+ 2 ( 0 )\geq - 6$ or $0 \geq - 6,$ which is true$\\$ Therefore, the lower half plane is not the solution region of the given inequality. Also, it is evident that any point on the line satisfies the given inequality.$\\$ Thus, the solution region of the given inequality is the half plane containing the point $(0, 0)$ including the line.$\\$ The solution region is represented by the shaded region as follows.$\\$

**39** **Solve the given inequality graphically in two-dimensional plane: $3 y - 5 x < 30$**

The graphical representation of $3 y - 5 x = 30$ is given as dotted line in the figure below.$\\$ This line divides the xy-plane in two half planes.$\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.$\\$ We select the point as $(0, 0)$.$\\$ It is observed that,$\\$ $3 ( 0 )- 5 ( 0)< 30$ or 0 - 30, which is true Therefore, the upper half plane is not the solution region of the given inequality. Also, it is evident that any point on the line does not satisfy the given inequality.$\\$ Thus, the solution region of the given inequality is the half plane containing the point (0, 0) excluding the line.$\\$ The solution region is represented by the shaded region as follows.

**40** **Solve the given inequality graphically in two-dimensional plane: $y < - 2$**

The graphical representation of $y =- 2$ is given as dotted line in the figure below. This line divides the xy-plane in two half planes.$\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.$\\$ We select the point as (0, 0).$\\$ It is observed that, $0 < - 2,$ which is false$\\$ Also, it is evident that any point on the line does not satisfy the given inequality.$\\$ Hence, every point below the line, $y =- 2$ (excluding all the points on the line), determines the solution of the given inequality.$\\$ The solution region is represented by the shaded region as follows.$\\$

**41** **Solve the given inequality graphically in two-dimensional plane:$x >- 3$**

The graphical representation of $x =- 3 $ is given as dotted line in the figure below. This line divides the xy-plane in two half planes.$\\$ Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.$\\$ We select the point as $(0, 0).$ $\\$ It is observed that,$\\$ $0 < - 3$, which is true $\\$ Also, it is evident that any point on the line does not satisfy the given inequality. $\\$ Hence, every point on the right side of the line, $x =-3$ (excluding all the points on the line), determines the solution of the given inequality.$\\$ The solution region is represented by the shaded region as follows.

**42** **Solve the following system of inequalities graphically: $x \geq 3, y \geq 2 .$**

$x \geq 3....... ( 1 )\\ y \geq 2....... ( 2 )$ $\\$ The graph of the lines, $x = 3$ and $ y = 2,$ are drawn in the figure below.$\\$ Inequality (1) represents the region on the right hand side of the line, x = 3 (including the line x = 3 ), and inequality (2) represents the region above the line, y = 2 (including the line y = 2).$\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

**43** **Solve the following system of inequalities graphically: $3 x + 2 y \leq 12, x \geq 1, y \leq 2$**

$3 x + 2 y \leq 12........ ( 1 )\\ x \geq 1....... ( 2 )\\ y \geq 2....... ( 3 )$ $\\$ The graphs of the lines, $3 x + 2 y = 12, x = 1$ , and $y = 2 $, are drawn in the figure below.$\\$ Inequality (1) represents the region below the line, $3 x + 2 y = 12$ (including the line $3 x + 2 y = 12)$. Inequality (2) represents the region on the right side of the line, x= 1 (including the line x = 1 ). Inequality (3) represents the region above the line, y = 2 (including the line y = 2 ).$\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.$\\$

**44** **Solve the following system of inequalities graphically: $2 x + y \geq 6, 3 x + 4 y \leq 12$**

$2 x + y \geq 6....... (1 )\\ 3 x + 4 y \leq 12....... (2 )$ The graph of the lines, $2 x + y \geq 6$ and $3 x + 4 y = 12$ , are drawn in the figure below. Inequality$\\$ (1) represents the region above the line, $2 x + y = 6 $(including the line $2 x + y = 6 )$, and inequality$\\$ (2) represents the region below the line, $3 x + 4 y = 12$ (including the line $3 x + 4 y = 12 ).$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

**45** **Solve the following system of inequalities graphically:$ x + y \geq 4, 2 x - y > 0$**

$x + y \geq 4....... ( 1 )\\ 2 x - y > 0........ ( 2 )$ $\\$ The graph of the lines, $x+ y= 4 $and $2 x - y = 0$ , are drawn in the figure below.$\\$ Inequality (1) represents the region above the line,$x + y = 4$ (including the line $x + y = 4 $). It is observed that (1, 0) satisfies the inequality,$ 2 x - y > 0 . [ 2 ( 1 )- 0 = 2 > 0 ]$ $\\$ Therefore, inequality (2) represents the half plane corresponding to the line, $2 x - y = 0 ,$ containing the point (1, 0) [excluding the line $ 2 x - y = 0 ]$.$\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on line $x + y = 4$ and excluding the points on line $2 x - y = 0$ as follows.

**46** **Solve the following system of inequalities graphically: $2 x - y > 1, x - 2 y < - 1$**

$2 x - y > 1........ ( 1 )\\ x - 2 y <- 1....... ( 2 )$ $\\$ The graph of the lines, $2 x - y = 1$ and $x - 2 y =- 1 , $ are drawn in the figure below. $\\$ Inequality (1) represents the region below the line, $2 x -y = 1 $(excluding the line $2 x - y = 1$ ), and inequality (2) represents the region above the line, x - 2 y =- 1 (excluding the line x - 2 y =- 1). $\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region excluding the points on the respective lines as follows. $\\$

**47** **Solve the following system of inequalities graphically: $x + y \leq 6, x + y \geq 4$**

$x + y \leq 6 ....... ( 1 )\\ x + y \geq 4...... ( 2 )$ $\\$ The graph of the lines,$ x + y = 6$ and $x + y = 4$ , are drawn in the figure below. $\\$ Inequality (1) represents the region below the line, x +y = 6 (including the line x + y = 6 ), and inequality (2) represents the region above the line, x + y = 4 (including the line x + y = 4 ). $\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

**48** **Solve the following system of inequalities graphically: $2 x + y \geq 8, x + 2 y \geq 10$**

$2 x + y = 8....... ( 1 )\\ x + 2 y = 10........ ( 2 )$$\\$ The graph of the lines, $2 x + y = 8$ and $x + 2 y = 10 $, are drawn in the figure below.$\\$ Inequality (1) represents the region above the line, x + 2 y =8 , and inequality (2) represents the region above the line, x + 2 y = 10 .$\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

**49** **Solve the following system of the inequalities graphically: $x + y \leq 9, y > x , x \geq 0$**

$x + y \leq 9....... ( 1 )\\ y > x ........ ( 2 )\\ x \geq 0 ........ ( 3 )$$\\$ The graph of the lines, $x + y = 9 $ and $y = x $, are drawn in the figure below.$\\$ Inequality (1) represents the region below the line, x + y = 9 (including the line x + y = 9 ). It is observed that (0, 1) satisfies the inequality, y = x . [ 1 > 0 ] . Therefore, inequality (2) represents the half plane corresponding to the line, y = x , containing the point (0, 1) [excluding the line y = x ]. Inequality (3) represents the region on the right hand side of the line, x = 0 or y - axis (including y - axis )$\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the lines, x + y = 9 and x = 0 , and excluding the points on line y = x as follows.

**50** **Solve the following system of inequalities graphically: $5 x + 4 y \leq 20, x \geq 1, y \geq 2$**

$5 x + 4 y \leq 20 ....... ( 1 )\\ x \geq 1........ (2 )\\ y \geq 2....... (3 )$ $\\$ The graph of the lines,$ 5 x + 4 y = 20, x = 1$ and y = 2 , are drawn in the figure below.$\\$ Inequality (1) represents the region below the line, 5 x + 4 y = 20 (including the line 5 x + 4 y = 20). Inequality (2) represents the region on the right hand side of the line, x = 1 $\\$ (including the line x = 1 ). Inequality (3) represents the region above the line, y = 2 (including the line y = 2 ). $\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows. $\\$

**51** **Solve the following system of inequalities graphically: $3 x + 4 y \leq 60, x + 3 y \leq 30, x \geq 0, y \geq 0$**

$3 x + 4 y \leq 60 ....... ( 1 )\\ x + 3 y \leq 30....... ( 2 )$$\\$ The graph of the lines,$3 x + 4 y = 60$ and $ x + 3 y = 30$ , are drawn in the figure below.$\\$ Inequality (1) represents the region below the line, $3 x + 4 y = 60$(including the line $3 x + 4 y = 60$), and inequality (2) represents the region below the line, $x + 3 y = 30$ (including the line$ x + 3 y = 30 ).$$\\$ Since $ x \geq 0$ and$ y \geq 0 $, every point in the common shaded region in the first quadrant including the points on the respective line and the axes represents the solution of the given system of linear inequalities.

**52** **Solve the following system of inequalities graphically: $ \\ 2 x + y \geq 4, x + y \leq 3, 2 x - 3 y \leq 6$**

$2 x + y \geq 4....... ( 1 )\\ x + y \leq 3....... ( 2 )\\ 2 x - 3 y \leq 6....... ( 3 )$ $\\$ The graph of the lines, $2 x + y = 4, x +y = 3$ and 2 x - 3 y = 6 , are drawn in the figure below. $\\$ Inequality (1) represents the region above the line, 2 x + y = 4 (including the line 2 x + y = 4 ). $\\$ Inequality (2) represents the region below the line, x + y = 3 (including the line x + y = 3 ). $\\$ Inequality (3) represents the region above the line, 2 x - 3 y = 6 (including the line 2 x - 3 y = 6 ). $\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows. $\\$

**53** **Solve the following system of inequalities graphically: $\\$ $ x - 2 y \leq 3, 3 x + 4 y \geq 12, x \geq 0, y \geq 1$**

$x - 2 y \leq 3....... ( 1 )\\ 3 x + 4 y \geq 12........ ( 2 )\\ y \geq 1..... ( 3 )$ $\\$ The graph of the lines, $x - 2 = 3, 3 x + 4 y = 12$ and $y =1$ , are drawn in the figure below.$\\$ Inequality (1) represents the region above the line, x - 2 y = 3 (including the line x - 2 y = 3 ). $\\$ Inequality (2) represents region above the line, 3 x + 4 y = 12 (including the line 3 x + 4 y = 12 ). $\\$ Inequality (3) represents the region above the line, y = 1 (including the line y = 1 ). $\\$The inequality,$ x \geq 0 $, represents the region on the right and side of y - axis (including y - axis). $\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines and y - axis as follows.

**54** **Solve the following system of inequalities graphically:$\\$ $4 x + 3 y \leq 60, y \geq 2 x , x \geq 3, x , y \geq 0$**

$4 x + 3 y \leq 60,....(1)\\ y \geq 2 x ,.....(2)\\ x \geq 3,....(3) $$\\$ The graph of the lines, 4 x + 3 y = 60, y = 2 x x = 3 , are drawn in the figure below.$\\$ Inequality (1) represents the region below the line, 4 x + 3 y =60 (including the line 4 x + 3 y = 60). $\\$Inequality (2) represents the region above the line, y = 2 x (including the line y = 2 x ). $\\$Inequality (3) represents the region on the right hand side of the line, x = 3 (including the line x = 3 ).$\\$ Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.

**55** **Solve the following system of inequalities graphically:$\\$ $3 x + 2 y \leq 150, x + 4 y = 80, x \leq 15, y \geq 0, x \geq 0$**

$3 x + 2 y \leq 150 ..... ( 1 )\\ x + 4 y = 80...... ( 2 )\\ x \leq 15..... ( 3 )$$\\$ The graph of the lines, 3 x + 2 y = 150, x + 4 y = 80 and x = 15 , are drawn in the figure below.$\\$ Inequality (1) represents the region below the line, 3 x + 2 y=150 (including the line 3 x + 2 y = 150 ). $\\$Inequality (2) represents the region below the line, x + 4 y = 80 (including the line x + 4 y = 80 ). $\\$Inequality (3) represents the region on the left hand side the line, x = 15 (including the line x = 15 ).$\\$ Since $x \geq 0$ and $y \geq 0$ , every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.

**56** **Solve the following system of inequalities graphically:$\\$ $x + 2 y \leq 10, x+ y \geq 1, x - y \leq 0, x \geq 0, y \geq 0$**

$x + 2 y \leq 10 ....... ( 1 )\\ x + y \geq 1...... ( 2 )\\ x - y \leq 0..... ( 3 )$ $\\$ The graph of the lines, $x + 2 y = 10, x + y = 1$ and x - y = 0 , are drawn in the figure below. $\\$ Inequality (1) represents the region below the line, x + 2 y =10 (including the line x + 2 y = 10 ). $\\$ Inequality (2) represents the region above the line, x + y = 1 (including the line x + y = 1 ). $\\$ Inequality (3) represents the region above the line, x - y = 0 (including the line x - y = 0 ). $\\$ Since $x \geq 0$ and $y \geq 0$ , every point in the common shaded region in the first quadrant including the points on the respective lines and the axes represents the solution of the given system of linear inequalities.

**57** **Solve the inequality $2 \leq 3 x - 4 \leq 5$**

$2 \leq 3 x - 4 \leq 5\\ \implies 2 + 4 \leq 3 x - 4 + 4 \leq 5 + 4\\ \implies 6 \leq 3 x \leq 9\\ \implies 2 \leq x \leq 3$ $\\$ Thus, all the real numbers, x, which are greater than or equal to 2 but less than or equal to 3, are the solutions of the given inequality. The solution set for the given inequality is [2, 3].

**58** **Solve the inequality $6 \leq -3 ( 2 x - 4 )< 12$**

$6 \leq -3 ( 2 x - 4 )< 12\\ \implies 2 \leq -(2x-4)<4\\ \implies -2 \geq 2x-4 >-4\\ \implies 4-2 \geq 2x > 4 - 4\\ \implies 2 \geq 2x >0\\ \implies 1 \geq x> 0$ Thus, the solution set for the given inequality is ( 0,1 ) .

**59** **Solve the inequality $-3 \leq 4-\dfrac{7x}{2}\leq 18$**

$-3 \leq 4-\dfrac{7x}{2}\leq 18\\ \implies -3-4 \leq -\dfrac{7x}{2}\leq 18-4\\ \implies -7 \leq -\dfrac{7x}{2}\leq 14\\ \implies 7 \geq \dfrac{7x}{2}\geq - 14\\ \implies 1 \geq \dfrac{x}{2}\geq -2\\ \implies 2 \geq x \geq -4 $ $\\$ Thus, the solution set for the given inequality is [-4, 2 ] .

**60** **Solve the inequality $-15 < \dfrac{3(x-2)}{5}\leq 0$**

$-15 < \dfrac{3(x-2)}{5}\leq 0\\ \implies -75 < 3(x-2)\leq 0\\ \implies -25 < x-2 \leq 0\\ \implies -25 + 2 < x\leq 2\\ \implies -23 < x \leq 2 $$\\$ Thus, the solution set for the given inequality is (- 23, 2 ) .

**61** **Solve the inequality $-12 < 4-\dfrac{3x}{-5}\leq 2 $**

$-12 < 4-\dfrac{3x}{-5}\leq 2 \\ \implies-12 - 4<\dfrac{-3x}{-5}\leq 2-4\\ \implies -16 < \dfrac{3x}{5} \leq -2\\ \implies -80 < 3x \leq -10\\ \implies \dfrac{-80}{3}< x \leq \dfrac{-10}{3} $ Thus, the solution set for the given inequality is $(\dfrac{-80}{3},\dfrac{-10}{3})$

**62** **Solve the inequality $7\leq \dfrac{(3x+11)}{2}\leq 11$**

$7\leq \dfrac{(3x+11)}{2}\leq 11\\ \implies 14 \leq 3x +11 \leq 22\\ \implies 14 - 11 \leq 3x \leq 22 -11\\ \implies 3 \leq 3x \leq 11\\ \implies 1 \leq x \leq \dfrac{11}{3}$ Thus, the solution set for the given inequality is $[1,\dfrac{11}{3}]$

**63** **Solve the inequalities and represent the solution graphically on number line: $\\$ $5 x + 1 >- 24, 5 x - 1 < 24$**

$5 x + 1 >- 24 \implies 5 x >- 25\\ \implies x >- 5 ..... ( 1 )\\ 5 x - 1 < 24 \implies 5 x < 25\\ \implies x < 5 ...... ( 2 )$ $\\$ From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (- 5,5 ) . The solution of the given system of inequalities can be represented on number line as

**64** **Solve the inequalities and represent the solution graphically on number line:$\\$ $2 ( x - 1 )< x + 5 , 3 ( x + 2 )> 2 - x$**

$2 ( x - 1 )< x + 5\implies 2 x - 2 < x + 5 \implies 2 x - x < 5 + 2\\ \implies x < 7 ....... ( 1 )\\ 3 ( x +2 )> 2 - x \implies 3 x + 6 > 2 - x \implies 3 x + x > 2 - 6\\ \implies 4 x >- 4\\ \implies x >- 1 ....... ( 2)$ From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (-1,7 ) . The solution of the given system of inequalities can be represented on number line as

**65** **Solve the following inequalities and represent the solution graphically on number line: $3 x - 7 > 2 ( x - 6 ) , 6 - x > 11 - 2 x$**

$3 x - 7 > 2 ( x - 6 )\implies 3 x - 7 > 2 x - 12 \implies 3 x - 2 x >- 12 + 7\\ \implies x >- 5 ......... (1)\\ - 6 - x > 11 - 2 x\implies -x + 2 x > 11 - 6\\ \implies x > 5 ...... ( 2 )$ From (1) and (2), it can be concluded that the solution set for the given system of inequalities is $( 5,\infty)$ . The solution of the given system of inequalities can be represented on number line as

**66** **Solve the inequalities and represent the solution graphically on number line:$\\$ $5(2x-7)-3(2x+3)\leq 0, 2x+19 \leq 6x+47$**

$5 ( 2 x - 7 )- 3 ( 2 x + 3 )\leq 0 \implies 10 x - 35 - 6 x -9 \leq 0 \implies 4 x - 44 \leq 0 \implies 4 x \leq 44\\ \implies x \leq 11 ....... ( 1 )\\ 2 x + 19 \leq 6 x + 47 \implies 19 - 47 \leq 6 x - 2 x \implies - 28 \leq 4 x\\ \implies - 7 \leq x ...... (2 )$ From (1) and (2), it can be concluded that the solution set for the given system of inequalities is [-7,11 ]. The solution of the given system of inequalities can be represented on number line as

**67** **A solution is to be kept between $68 ^o F$ and $77 ^o F $. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by $\\$ $F=\dfrac{9}{5}C+32?$**

Since the solution is to be kept between $68 ^o F$ and $77 ^o F ,68 < F < 77 $.$\\$ Putting $ F=\dfrac{9}{5}C+32,$ we obtain $\\$ $68 < \dfrac{9}{5} C+32 < 77 \\ \implies 68-32 < \dfrac{9}{5} C < 77 -32 \\ \implies 36 < \dfrac{9}{5} C < 45 \\ \implies 36 * \dfrac{5}{9} < C < 45* \dfrac{5}{9}\\ \implies 20 < C < 25 $ $\\$ Thus, the required range of temperature in degree Celsius is between $20 ^o C $ and $25 ^o C $.

**68** **A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?**

Let x litres of 2% boric acid solution is required to be added.$\\$ Then, total mixture =( x + 640 ) litres$\\$ This resulting mixture is to be more than 4% but less than 6% boric acid.$\\$ $\therefore 2 \% x + 8 \% of 640 > 4 \% of ( x + 640 )$ and $x + 8 \% of 640 < 6 \% of ( x + 640 )\\ 2 \% x + 8 \% of 640 > 4 \% of ( x + 640 )\\ \implies \dfrac{2}{100}x+\dfrac{8}{100}(640) > \dfrac{4}{100}(x+640)\\ \implies 2x+5120 > 4x + 2560\\ \implies 5120 - 2560 > 4 x - 2 x\\ \implies 5120 - 2560 > 2x\\ \implies 2560 > 2x\\ \implies 1280 > x\\ 2 \% x + 8 \% of 640 < 6\% of ( x +640 )\\ \dfrac{2}{100}x+\dfrac{8}{100}(640) < 6\% \ of \ (x+640)\\ \implies 2 x + 5120 < 6 x + 3840 \\ \implies 5120 - 3840 < 6 x - 2 x \\ \implies 5120 - 3840 < 6 x - 2 x \\ \implies 1280 < 4x \\ \implies 320 < x \\ \therefore 320 < x < 1280$ Thus, the number of litres of 2% of boric acid solution that is to be added will have to be more than 320 litres but less than 1280 litres.

**69** **How many litres of water will have to be added to 1125 litres of the $45\%$ solution of acid so that the resulting mixture will contain more than $25\%$ but less than $30\%$ acid content?**

Let x litres of water is required to be added.$\\$ Then, total mixture =( x + 1125 ) litres$\\$ It is evident that the amount of acid contained in the resulting mixture is $45\%$ of 1125 litres.$\\$ This resulting mixture will contain more than $25\%$ but less than $30\%$ acid content.$\\$ $\therefore 30\% of ( 1125 + x )> 45\% of 1125$ $\\$ And, $25\% of ( 1125 + x )> 45\% of 1125\\ 30\% of ( 1125 + x )> 45\% of 1125\\ \implies \dfrac{30}{100}(1125+x)> \dfrac{45}{100} * 1125\\ \implies 30 ( 1125 + x )>45 * 1125\\ \implies 30 * 1125 + 30 x > 45 * 1125\\ \implies 30 > 45 * 1125 - 30 * 1125\\ \implies 30 x >( 45 - 30 )* 1125\\ \implies x > \dfrac{15*1125}{30} =562.5\\ 25 \% of (1125+x) < 45 \% \ of \ 1125\\ \implies \dfrac{25}{100} (1125 + x) < \dfrac{45}{100}* 1125\\ \implies 25(1125 +x) > 45 * 1125\\ \implies 25 * 1125 +25x > 45 * 1125\\ \implies 25x > 45 * 1125 -25 * 1125\\ \implies 25x > (45 -25)* 1125\\ \implies x > \dfrac{20*1125}{25}=900\\ \therefore 562.5 < x < 900$ Thus, the required number of litres of water that is to be added will have to be more than $562.5$ but less than$ 900.$

**70** **$IQ$ of a person is given by the formula $ IQ =\dfrac{MA}{CA}*100$ $\\$ Where $MA$ is mental age and $CA$ is chronological age. If $80 \leq IQ \leq 140$ for a group of $12$ years old children, find the range of their mental age.**

It is given that for a group of $12$ years old children,$\\$ $80 \leq IQ \leq 140 ....... ( i ) $ $\\$ For a group of $12$ years old children, $CA = 12$ years $\\$ $IQ=\dfrac{MA}{12}*100$ $\\$ Putting this value of $ IQ $ in (i), we obtain $\\$ $ 80 \leq \dfrac{MA}{12} *100 \leq 140\\ \implies 80* \dfrac{12}{100} \leq MA \leq 140 * \dfrac{12}{100}\\ \implies 9.6 \leq MA \leq 16.8$ $\\$ Thus, the range of mental age of the group of $12$ years old children is $\implies 9.6 \leq MA \leq 16.8 .$