NCERT Solutions Class 11 Maths Chapter 6

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

Ans

$\begin{array}{l}\text{We are given}\\ \text{\hspace{0.17em}24x}<\text{1}00\\ \mathrm{or}\frac{\text{24x}}{24}<\frac{100}{24}\left[\mathrm{Dividing}\text{both sides by 24}\right]\\ \mathrm{x}<\frac{25}{6}\\ \left(\mathrm{i}\right)\mathrm{When}\text{x is a natural number, in this case following values}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of x make the statement true.}\\ \text{x}=1,2,3,4.\\ \mathrm{The}\text{solution set of the inequality is}\{1,2,3,4\}.\\ \left(\mathrm{ii}\right)\mathrm{When}\text{x is an integer, the solutions of the given inequlity are}\\ ...\text{,}-3,-2,-1\text{, 0, 1, 2, 3,4.}\\ \mathrm{The}\text{solution set of the inequality is}\{...\text{,}-3,-2,-1\text{, 0, 1, 2, 3,4}\}.\end{array}$

Q.2 Solve – 12x > 30, when
(i) x is a natural number.
(ii) x is an integer.

Ans

$\begin{array}{l}\text{We are given}\\ -12\mathrm{x}>30\\ \mathrm{or}\frac{-12\mathrm{x}}{-12}<\frac{30}{-12}[\mathrm{Dividing}\text{both sides by}-12]\\ \mathrm{x}<-\frac{5}{2}\\ \left(\mathrm{i}\right)\mathrm{When}\text{x is a natural number, in this case following no value}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of x make the statement true.}\\ \therefore \mathrm{There}\text{is no solution.}\\ \left(\mathrm{ii}\right)\mathrm{When}\text{x is an integer, the solutions of the given inequality are}\\ ...\text{,}-4,-3.\\ \mathrm{The}\text{solution set of the inequality is}\{...\text{,}-4,-3\}.\end{array}$

Q.3 Solve 5x – 3 < 7, when
(i) x is an integer.
(ii) x is a real number.

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5x}-\text{3}<\text{7}\\ \text{or 5x}-\text{3}+\text{3}<\text{7}+3\left[\mathrm{Adding}\text{3 both sides}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5x}<10\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{5x}}{5}<\frac{10}{5}\left[\mathrm{Divide}\text{both sides by 5}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<2\\ \left(\mathrm{i}\right)\mathrm{When}\mathrm{x}\text{is an integer, the solutions of given inequality are}\\ ...,-3,-2,-1,0,1.\\ \mathrm{The}\text{solution set of the inequality is}\{...,-3,-2,-1,0,1\}.\\ \left(\mathrm{ii}\right)\mathrm{When}\mathrm{x}\text{is a real number, the solutions of given inequality are}\\ \text{given by x}<\text{2, i.e., all real numbers x which are less than 2.\hspace{0.17em}}\\ \text{Therefore, the solution set of the inequality is x}\in (-\mathrm{\infty },2).\end{array}$

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

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3x}+\text{8}>\text{2}\\ \text{or 3x}+\text{8}-\text{8}>\text{2}-8\left[\mathrm{Adding}(-8)\text{both sides}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3x}>-6\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{3x}}{3}>\frac{-6}{3}\left[\mathrm{Divide}\text{both sides by 3}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}>-2\\ \left(\mathrm{i}\right)\mathrm{When}\mathrm{x}\text{is an integer, the solutions of given inequality are}\\ -1,0,\; 1,2,3,...\\ \mathrm{The}\text{solution set of the inequality is}\{-1,0,\; 1,2,3,...\}.\\ \left(\mathrm{ii}\right)\mathrm{When}\mathrm{x}\text{is a real number, the solutions of given inequality are}\\ \text{given by x}>-\text{2, i.e., all real numbers x which are greater than\hspace{0.17em}\hspace{0.17em}}-\text{2.\hspace{0.17em}}\\ \text{Therefore, the solution set of the inequality is x}\in (-2,\mathrm{\infty }).\end{array}$

Q.5 4x + 3 < 5x + 7

Ans

$\begin{array}{l}\text{We are given}\\ \text{4x}+\text{3}<\text{5x}+\text{7}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}4x}+\text{3}-\text{4x}<\text{5x}+\text{7}-\text{4x}\left[\mathrm{Subtracting}\text{4x from both sides}\right]\\ \mathrm{or}3<\mathrm{x}+7\\ \mathrm{or}3-7<\mathrm{x}+7-7\left[\mathrm{Subtracting}\text{7 from both sides}\right]\\ \mathrm{or}-4<\mathrm{x}\\ \text{i.e., all the real numbers which are greater than}-\text{4, are the}\\ \text{solutions of the given inequality. Hence, the solution set is}\\ \text{(}-\text{4,}\mathrm{\infty }\text{).}\end{array}$

Q.6 3x – 7 > 5x – 1

Ans

$\begin{array}{l}\text{We are given}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3x}-\text{7}>\text{5x}-\text{1}\\ \mathrm{or}\text{\hspace{0.33em}3x}-\text{7}-\text{3x}>\text{5x}-\text{1}-\text{3x\hspace{0.33em}\hspace{0.33em}}\left[\mathrm{Subtracting}\text{3x from both sides}\right]\\ \mathrm{or}-\text{7}>\text{2x}-\text{1}\\ \mathrm{or}-\text{7}+\text{1}>\text{2x}-\text{1}+1\left[\mathrm{Adding}\text{1 both sides}\right]\\ \mathrm{or}-6>\text{2x}\\ \mathrm{or}-\frac{6}{2}>\frac{\text{2x}}{2}\left[\mathrm{Dividing}\text{both sides by 2}\right]\\ \mathrm{or}-3>\text{x}\\ \text{i.e., all the real numbers which are less than}-\text{3, are the}\\ \text{solutions of the given inequality. Hence, the solution set is}\\ \text{(}-\infty \text{,}-\text{3).}\end{array}$

Q.7 3(x – 1) ≤ 2 (x – 3)

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3}\left(\text{x}-\text{1}\right)\le \text{2}\left(\text{x}-\text{3}\right)\\ \mathrm{or}3\text{x}-3\le 2\text{x}-6\\ \mathrm{or}3\text{x}-3-2\text{x}\le 2\text{x}-6-2\text{x}\left[\mathrm{Subtracting}\text{2x from both sides}\right]\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}x}-3\le -6\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}x}-3+3\le -6+3\left[\mathrm{Adding}\text{3 both sides}\right]\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}x}\le -3\\ \text{Thus, all real numbers x which are less than or equal to}-3\\ \text{are the solutions of the given inequality, i.e., x}\in (-\infty \text{,}-\text{3].}\end{array}$

Q.8 3(2 – x) ≥ 2 (1 – x)

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{3}(\text{2}-\text{x})\ge \text{2}(\text{1}-\text{x})\\ \mathrm{or}6-3\mathrm{x}\ge 2-2\mathrm{x}\\ \mathrm{or}6-3\mathrm{x}+3\mathrm{x}\ge 2-2\mathrm{x}+3\mathrm{x}\left[\mathrm{Adding}\text{3x both sides}\right]\\ \mathrm{or}6\ge 2+\mathrm{x}\\ \mathrm{or}6-2\ge 2+\mathrm{x}-2\left[\mathrm{Subtracting}\text{2 from both sides}\right]\\ \mathrm{or}4\ge \mathrm{x}\\ \text{Thus, all real numbers x which are less than or equal to}4\\ \text{are the solutions of the given inequality, i.e., x}\in (-\mathrm{\infty }\text{,\hspace{0.17em}}4\text{].}\end{array}$

Q.9

$\mathrm{x}+\frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{3}<11$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{are}\mathrm{given}\\ \mathrm{x}+\frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{3}<11\\ \mathrm{or}\frac{6\mathrm{x}+3\mathrm{x}+2\mathrm{x}}{6}<11\\ \mathrm{or}\frac{11\mathrm{x}}{6}<11\\ \mathrm{or}\frac{11\mathrm{x}}{6}\times \frac{6}{11}<11\times \frac{6}{11}\left[\mathrm{Multliplying}\text{a both sides by}\frac{6}{11}\right]\\ \mathrm{or}\mathrm{x}<6\\ \text{Thus, all real numbers x which are less than}6\text{are the solutions}\\ \text{of the given inequality, i.e., x}\in (-\mathrm{\infty }\text{,\hspace{0.17em}}6)\text{.}\end{array}$

Q.10

$\frac{\mathbf{x}}{\mathbf{3}}\mathbf{>}\frac{x}{2}\mathbf{+}\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{\mathrm{x}}{3}>\frac{\mathrm{x}}{2}+1\\ \mathrm{or}\frac{\mathrm{x}}{3}-\frac{\mathrm{x}}{2}>\frac{\mathrm{x}}{2}+1-\frac{\mathrm{x}}{2}\left[\mathrm{Subtracting}\text{}\frac{\mathrm{x}}{2}\text{from both sides}\right]\\ \mathrm{or}-\frac{\mathrm{x}}{6}>1\\ \mathrm{or}-\frac{\mathrm{x}}{6}\times -6<1\times -6[\mathrm{Multiplying}\text{both sides by}-\text{6}]\\ \mathrm{or}\mathrm{x}<-6\\ \text{Thus, all real numbers x which are greater than}6\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in (-\mathrm{\infty },-6)\text{.}\end{array}$

Q.11

$\frac{3(\mathrm{x}-2)}{5}\le \frac{5(2-\mathrm{x})}{3}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{3(\mathrm{x}-2)}{5}\le \frac{5(2-\mathrm{x})}{3}\\ \mathrm{or}9(\mathrm{x}-2)\le 25(2-\mathrm{x})\\ \mathrm{or}9\mathrm{x}-18\le 50-25\mathrm{x}\\ \mathrm{Adding}\text{25x both sides, we get}\\ 9\mathrm{x}-18+25\mathrm{x}\le 50-25\mathrm{x}+25\mathrm{x}\\ \mathrm{or}34\mathrm{x}-18\le 50\\ \mathrm{Adding}\text{18 both sides, we get}\\ 34\mathrm{x}-18+18\le 50+18\\ \mathrm{or}34\mathrm{x}\le 68\\ \mathrm{Dividing}\text{both sides by 34, we get}\\ \frac{34\mathrm{x}}{34}\le \frac{68}{34}\\ \Rightarrow \mathrm{x}\le 2\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}all real numbers x which are less than or equal to}2\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in (-\mathrm{\infty },2]\text{.}\end{array}$

Q.12

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\frac{\mathbf{3}\mathbf{x}}{\mathbf{5}}\mathbf{+}\mathbf{4}\mathbf{)}\mathbf{\ge }\frac{\mathbf{1}}{\mathbf{3}}\mathbf{(}\mathbf{x}\mathbf{–}\mathbf{6}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{1}{2}(\frac{3\mathrm{x}}{5}+4)\ge \frac{1}{3}(\mathrm{x}-6)\\ \mathrm{or}\frac{1}{2}\left(\frac{3\mathrm{x}+20}{5}\right)\ge \frac{1}{3}(\mathrm{x}-6)\\ \mathrm{or}\left(\frac{3\mathrm{x}+20}{10}\right)\ge \frac{1}{3}(\mathrm{x}-6)\\ \mathrm{or}3(3\mathrm{x}+20)\ge 10(\mathrm{x}-6)\\ \mathrm{or}9\mathrm{x}+60\ge 10\mathrm{x}-60\\ \Rightarrow 60+60\ge 10\mathrm{x}-9\mathrm{x}\\ \Rightarrow 120\ge \mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than or equal to 1}20\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in (-\mathrm{\infty },120]\text{.}\end{array}$

Q.13 2 (2x + 3) – 10 < 6 (x – 2)

Ans

$\begin{array}{l}\text{We are given}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}(\text{2x}+\text{3})-\text{1}0<\text{6}(\text{x}-\text{2})\\ \mathrm{or}4\mathrm{x}+6-10<6\mathrm{x}-12\\ \mathrm{or}4\mathrm{x}-4<6\mathrm{x}-12\\ \mathrm{or}12-4<6\mathrm{x}-4\mathrm{x}\\ \mathrm{or}8<2\mathrm{x}\\ \mathrm{or}4<\mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than}4\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in (4,\mathrm{\infty }).\end{array}$

Q.14

$\mathbf{37}\mathbf{–}\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{5}\mathbf{)}\mathbf{\ge }\mathbf{9}\mathbf{x}\mathbf{–}\mathbf{8}\mathbf{(}\mathbf{x}\mathbf{–}\mathbf{3}\mathbf{)}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ 37–(3\mathrm{x}+5)\ge 9\mathrm{x}–8(\mathrm{x}–3)\\ \mathrm{or}37-3\mathrm{x}-5\ge 9\mathrm{x}-8\mathrm{x}+24\\ \mathrm{or}32-3\mathrm{x}\ge \mathrm{x}+24\\ \mathrm{or}32-24\ge \mathrm{x}+3\mathrm{x}\\ \mathrm{or}8\ge 4\mathrm{x}\\ \mathrm{or}2\ge \mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than or equal to}2\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in (-\mathrm{\infty },2]\text{.}\end{array}$

Q.15

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

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{\mathrm{x}}{4}<\frac{(5\mathrm{x}-2)}{3}-\frac{(7\mathrm{x}-3)}{5}\\ \mathrm{or}\frac{\mathrm{x}}{4}<\frac{5(5\mathrm{x}-2)-3(7\mathrm{x}-3)}{15}\\ \mathrm{or}\frac{\mathrm{x}}{4}<\frac{25\mathrm{x}-10-21\mathrm{x}+9}{15}\\ \mathrm{or}\frac{\mathrm{x}}{4}<\frac{4\mathrm{x}-1}{15}\\ \mathrm{or}15\mathrm{x}<16\mathrm{x}-4\\ \mathrm{or}4<16\mathrm{x}-15\mathrm{x}\\ \mathrm{or}4<\mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than}4\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in (4,\mathrm{\infty })\text{.}\end{array}$

Q.16

$\frac{\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}{\mathbf{3}}\mathbf{\ge }\frac{\mathbf{(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}}{\mathbf{4}}\mathbf{-}\frac{\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{x}\mathbf{)}}{\mathbf{5}}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{(2\mathrm{x}-1)}{3}\ge \frac{(3\mathrm{x}-2)}{4}-\frac{(2-\mathrm{x})}{5}\\ \mathrm{or}\frac{(2\mathrm{x}-1)}{3}\ge \frac{5(3\mathrm{x}-2)-4(2-\mathrm{x})}{20}\\ \mathrm{or}\frac{(2\mathrm{x}-1)}{3}\ge \frac{15\mathrm{x}-10-8+4\mathrm{x}}{20}\\ \mathrm{or}\frac{(2\mathrm{x}-1)}{3}\ge \frac{19\mathrm{x}-18}{20}\\ \mathrm{or}20(2\mathrm{x}-1)\ge 3(19\mathrm{x}-18)\\ \mathrm{or}40\mathrm{x}-20\ge 57\mathrm{x}-54\\ \mathrm{or}54-20\ge 57\mathrm{x}-40\mathrm{x}\\ \mathrm{or}34\ge 17\mathrm{x}\\ \mathrm{or}2\ge \mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than or equal to}2\\ \text{are the solutions of the given inequality, i.e., x}\in (-\mathrm{\infty },2]\text{.}\end{array}$

Q.17 3x – 2 < 2x + 1

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.17em} \hspace{0.17em}3x}-\text{2}<\text{2x}+\text{1}\\ \text{or 3x}-2\mathrm{x}<1+2\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<3\\ \text{The graphical representation of solutions is given in Fig}\mathrm{ure}.\end{array}$

Q.18

$5\mathrm{x}-3\ge 3\mathrm{x}-5$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ 5\mathrm{x}-3\ge 3\mathrm{x}-5\\ \mathrm{or}5\mathrm{x}-3\mathrm{x}\ge -5+3\\ \mathrm{or}2\mathrm{x}\ge -2\\ \mathrm{or}\mathrm{x}\ge -1\\ \mathrm{The}\mathrm{graphical}\mathrm{representation}\mathrm{of}\mathrm{solutions}\mathrm{is}\mathrm{given}\mathrm{in}\mathrm{Figure}.\end{array}$

Q.19 3(1 – x) < 2(x + 4)

Ans

We are given
3(1 – x) < 2(x + 4)
or 3 – 3x < 2x + 8
or 3 – 8 < 2x + 3x
or – 5 < 5x
or – 1 < x
The graphical representation of solution is given below in the figure:

Q.20

$\frac{\mathbf{x}}{\mathbf{2}}\mathbf{\ge }\frac{\mathbf{(}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{(}\mathbf{7}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{)}}{\mathbf{5}}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{\mathrm{x}}{2}\ge \frac{(5\mathrm{x}-2)}{3}-\frac{(7\mathrm{x}-3)}{5}\\ \mathrm{or}\frac{\mathrm{x}}{2}\ge \frac{5(5\mathrm{x}-2)-3(7\mathrm{x}-3)}{15}\\ \mathrm{or}\frac{\mathrm{x}}{2}\ge \frac{25\mathrm{x}-10-21\mathrm{x}+9}{15}\\ \mathrm{or}\frac{\mathrm{x}}{2}\ge \frac{4\mathrm{x}-1}{15}\\ \mathrm{or}15\mathrm{x}\ge 8\mathrm{x}-2\\ \mathrm{or}15\mathrm{x}-8\mathrm{x}\ge -2\\ \mathrm{or}7\mathrm{x}\ge -2\\ \mathrm{or}\mathrm{x}\ge -\frac{2}{7}\end{array}$

The graphical representation of solution is given below in the figure:

Q.21 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.

Ans

$\begin{array}{l}\mathrm{Let}\text{Ravi obtained marks in third test}=\text{x}\\ \text{Ravi obtained marks in first two tests}=70\text{and 75}\\ \text{The average marks obtained by Ravi}=\frac{70+75+\mathrm{x}}{3}\\ \mathrm{According}\text{ to given condition,}\\ \frac{70+75+\mathrm{x}}{3}\ge 60\\ 145+\mathrm{x}\ge 180\\ \mathrm{x}\ge 180-145\\ \mathrm{x}\ge 35\\ \mathrm{Thus},\text{ Ravi should get greater or equal to 35 marks in}\\ \text{third unit test.}\end{array}$

Q.22 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.

Ans

$\begin{array}{l}\mathrm{Let}\text{the minimum marks obtained by Sunita in}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}fifth examination}=\text{x}\\ \text{Marks obtained by Sunita in first four}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}examinations}=87,92,94,95\\ \mathrm{Average}\text{marks obtained by Sunita}=\frac{87+92+94+95+\mathrm{x}}{5}\\ \mathrm{According}\text{to given condition,}\\ \frac{87+92+94+95+\mathrm{x}}{5}\ge 90\\ \mathrm{or}\frac{368+\mathrm{x}}{5}\ge 90\\ \mathrm{or}368+\mathrm{x}\ge 450\\ \mathrm{or}\mathrm{x}\ge 450-368\\ \mathrm{or}\mathrm{x}\ge 82\\ \mathrm{Thus},\text{ Sunita will get grade ‘A’ in examination if she will get}\\ \text{marks greater than or equal to 82.}\end{array}$

Q.23 Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}x be the smaller number of the two consecutive}\\ \text{odd positive integers, so that the other one is x+2. Then}\\ \text{we should have}\\ \text{x}+\text{2}<\text{10}\\ \mathrm{or}\mathrm{x}<8\\ \mathrm{and}\mathrm{x}+(\text{x}+\text{2})>11\\ \mathrm{or}2\mathrm{x}>11-2\\ \mathrm{or}\mathrm{x}>\frac{9}{2}\\ \mathrm{x}>4.5\\ \mathrm{So},4.5<\mathrm{x}<8\\ \mathrm{Since},\text{x is odd number. So, x will be 5 and 7.}\\ \text{Second number is x}+\text{2, so it wil be 7 and 9.}\\ \text{Thus the required pair of possible numbers will}\\ \text{be\hspace{0.17em}\hspace{0.17em}}(5,7)\text{and}(7,9).\end{array}$

Q.24 Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}x be the smaller number of the two consecutive}\\ \text{even positive integers, so that the other one is x+2.}\\ \text{Then we should have}\\ \text{x}+\text{2}>\text{5}\\ \mathrm{or}\mathrm{x}>3\\ \mathrm{and}\mathrm{x}+(\text{x}+\text{2})<23\\ \mathrm{or}2\mathrm{x}<23-2\\ \mathrm{or}\mathrm{x}<\frac{21}{2}\\ \mathrm{x}<10.5\\ \mathrm{So},3<\mathrm{x}<10.5\\ \mathrm{Since},\text{x is even number. So, x will be 6, 8 and 10.}\\ \text{Second number is x}+\text{2, so it wil be 8, 10 and 12.}\\ \text{Thus the required pair of possible numbers will}\\ \text{be\hspace{0.17em}}(6,8)\text{,}(8,10)\text{and}(10,12).\end{array}$

Q.25 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.

Ans

$\begin{array}{l}\mathrm{Let}\text{the shortest side of triangle}=\text{x\hspace{0.17em}cm}\\ \text{The longest side of triangle}=3\mathrm{x}\text{\hspace{0.17em}cm}\\ \mathrm{Third}\text{side of triangle}=(3\mathrm{x}-2)\mathrm{cm}\\ \mathrm{The}\text{perimeter of the triangle}=\mathrm{x}+3\mathrm{x}+(3\mathrm{x}-2)\\ =(7\mathrm{x}-2)\mathrm{cm}\\ \mathrm{According}\text{to given condition,}\\ (7\mathrm{x}-2)\ge 61\\ \mathrm{or}7\mathrm{x}\ge 63\\ \mathrm{or}\mathrm{x}\ge \frac{63}{7}\\ \mathrm{or}\mathrm{x}\ge 9\\ \mathrm{Thus},\text{the minimum length of the shortest side is 9 cm.}\end{array}$

Q.26 A man wants to cut three lengths from a single piece of board of length 91cm.
The second length is to be 3cm longer than the shortest and the third length is to 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?

Ans

$\begin{array}{l}\text{The length of the single piece of board}=91\mathrm{cm}\\ \mathrm{Let}\text{t}\mathrm{he}\text{length of the smaller piece of board}=\mathrm{x}\mathrm{cm}\\ \mathrm{The}\text{ length of the second piece of board}=(\mathrm{x}+3)\mathrm{cm}\\ \mathrm{The}\text{length of the third piece of board}=2\mathrm{x}\mathrm{cm}\\ \mathrm{According}\text{to given conditions:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2x}\ge (\mathrm{x}+3)+\text{5}\\ \mathrm{or}\mathrm{x}\ge 8...\left(1\right)\\ \mathrm{Since},\text{these the total length of these three pieces will at most}\\ \text{equal to 91 cm. Then,}\\ \mathrm{x}+(\mathrm{x}+3)+2\mathrm{x}\le 91\\ 4\mathrm{x}\le 91-3\\ 4\mathrm{x}\le 88\\ \mathrm{x}\le 22...\left(2\right)\\ \mathrm{From}\text{}\left(1\right)\text{ and}\left(2\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}8}\le \text{x}\le \text{22}\\ \mathrm{Thus},\text{the length of the shortest piece of wood is greater or equal}\\ \text{to 8 cm but less than or equal to 22 cm.}\end{array}$

Q.27 x + y < 5

Ans

Graph of x + y = 5 is given as dotted line in the figure.

This line divides the xy-plane in two half planes I and II.
We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that
(0) + (0) < 5
or 0 < 5 , which is true.
Hence, half plane I is the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane I excluding the points on the line is the solution region of the inequality.

Q.28

$2\mathrm{x}+\mathrm{y}\ge 6$

Ans

Graph of 2x + y = 6 is given as line in the figure.

This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that 2(0) + (0) = 6 or 0 = 6 , which is false.
Hence, half plane II is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that
2(0) + (0) ≥ 6
or 0 ≥ 6 , which is false.
Hence, half plane II is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.

Q.29

$3\mathrm{x}+4\mathrm{y}\le 12$

Ans

Graph of 3x + 4y = 12 is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that

3(0) + 4(0) ≤ 12
or 0 ≤ 12, which is true.
Hence, half plane I is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane I including the points on the line is the solution region of the inequality.

Q.30

$\mathrm{y}+8\ge 2\mathrm{x}$

Ans

Graph of y + 8 = 2x is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that

(0) + 8 ≥ 2(0)
or 8 ≥ 0, which is true.
Hence, half plane I is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane I including the points on the line is the solution region of the inequality.

Q.31

$\mathrm{x}-\mathrm{y}\le 2$

Ans

Graph of x – y = 2 is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that

(0) – (0) ≤ 2
or 0 ≤ 2, which is true.
Hence, half plane I is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane I including the points on the line is the solution region of the inequality.

Q.32

$2\mathrm{x}-3\mathrm{y}>6$

Ans

Graph of 2x – 3y = 6 is given as dotted line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that

2(0) – 3(0) > 6
or 0 > 6, which is false.
Hence, half plane II is the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.

Q.33

$-3\mathrm{x}+2\mathrm{y}\ge -6$

Ans

Graph of –3x+2y = – 6 is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0),
which lies in one of the half planes and determine if this point satisfies the given inequality,
we note that

– 3 (0) + 2(0) ≥ – 6
or 0 ≥ – 6, which is true.
Hence, half plane I is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane I including the points on the line is the solution region of the inequality.

Q.34

$3\mathrm{y}-5\mathrm{x}<30$

Ans

Graph of 3y – 5x = 30 is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0),
which lies in one of the half planes and determine if this point satisfies the given inequality,
we note that

3(0) – 5(0) < 30
or 0 < 30, which is true.
Hence, half plane I is the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane I excluding the points on the line is the solution region of the inequality.

Q.35 y < – 2

Ans

Graph of y = – 2 is given as line in the figure.
This line divides the xy-plane in two half planes I and II.
We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality,
we note that

(0) < – 2, which is false.
Hence, half plane II is the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.

Q.36 x > – 3

Ans

Graph of x = – 3 is given as line in the figure.
This line divides the xy-plane in two half planes I and II.
We select a point (not on the line), say (0, 0),
which lies in one of the half planes and determine if this point satisfies the given inequality,
we note that

0 > – 2, which is true.
Hence, half plane I is the solution region of the given inequality.
Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane I excluding the points on the line is the solution region of the inequality.

Q.37 Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{\ge }\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{2}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \mathrm{x}\ge 3...\left(\mathrm{i}\right)\\ \mathrm{y}\ge 2...\left(\mathrm{ii}\right)\\ \mathrm{We}\text{draw the graph of linear equations x}=\text{3 and y}=\text{2 in xy-plane}\\ \text{as shown in figure. Solution of inequality}\left(\text{1}\right)\text{is represented by}\\ \text{the}\text{shaded region right the line\hspace{0.17em}\hspace{0.17em}x}=\text{3},\text{including the points on}\\ \text{the line and solution of inequality}\left(\text{2}\right)\text{is represented by the}\\ \text{shaded region above the line y}=\text{2},\text{including the points on the}\\ \text{line}.\text{\hspace{0.17em}The common region of these two lines is shaded and it is the}\\ \text{required solution region of the given system of inequalities}.\end{array}$

Q.38 Solve the following system of inequalities graphically:

$\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\le }\mathbf{12}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{The}\text{given inequalities are:}\\ \text{3x}+\text{2y}\le \text{12 …}\left(1\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}x}\ge \text{1 …}\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}\ge \text{1 …}\left(3\right)\\ \mathrm{The}\text{graphs of following lines are drawn in the figure:}\\ \text{3x}+2\mathrm{y}=1\text{2,\hspace{0.17em}\hspace{0.17em}x}=\text{1 and y}=\text{1.}\end{array}$ $\begin{array}{l}\text{The inequality}\left(1\right)\text{shows shaded region below the line}\\ \text{3x + 2y = 12, inequality}\left(2\right)\text{shows the shaded region right of}\\ \text{the line x = 1 and inequality}\left(3\right)\text{shows the shaded region}\\ \text{above the line y}=\text{1.Hence, the common region of all these}\\ \text{lines is the solution of the given system of the linear inequalities.}\end{array}$

Q.39

$\mathbf{\text{Solve the following system of inequalities graphically:}}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{6}\mathbf{,}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\le }\mathbf{12}$

Ans

$\begin{array}{l}\text{The given inequalities are:}\\ \mathrm{x}+\mathrm{y}\ge 4...\left(1\right)\\ 2\mathrm{x}-\mathrm{y}\ge 0...\left(2\right)\\ \text{The graphs of linear equations}\mathrm{x}+\mathrm{y}=4\text{and}2\mathrm{x}-\mathrm{y}=0\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x + y = 6, including the points on the line.
On the same set of axes, we draw the graph of the equation 3x + 4y = 12 as shown in Figure.
Then we see that inequality (2) represents the shaded region below the line 3x + 4y = 12, including the points on the line.

In this way, the double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.40

$\mathbf{\text{Solve the following system of inequalities graphically:}}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{\ge }\mathbf{0}$

Ans

$\begin{array}{l}\text{The given inequalities are:}\\ \mathrm{x}+\mathrm{y}\ge 4...\text{(1)}\\ 2\mathrm{x}-\mathrm{y}\ge 0...\left(2\right)\\ \text{The graphs of linear equations}\mathrm{x}+\mathrm{y}=4\text{and}2\mathrm{x}-\mathrm{y}=0\\ \text{are drawn in}\mathrm{xy}\text{-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line x + y = 4, including the points on the line.
On the same set of axes, we draw the graph of the equation 2x – y = 0 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line 2x – y =0, including the points on the line.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.41 Solve the following system of inequalities graphically:

$\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{>}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{<}\mathbf{-}\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given\hspace{0.33em}inequalities\hspace{0.33em}are:}\\ 2\text{x}-\text{y}>1...\left(1\right)\\ \text{x}-2\text{y}<-1...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations 2x}-\text{y}=1\text{and x}-2\text{y}=-1\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x – y = 1, excluding the points on the line.
On the same set of axes, we draw the graph of the equation 2x – y = 1 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line x – 2y = –1.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.42 Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le }\mathbf{6}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{4}$

Ans

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given\hspace{0.33em}inequalities\hspace{0.33em}are:}\\ \text{x}+\text{y}\le 6...\left(1\right)\\ \text{x}+\text{y}\ge 4...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations x}+\text{y}=6\text{and x}+\text{y}=4\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line x + y = 6, including the points on the line.
On the same set of axes, we draw the graph of the equation x + y = 4 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line x + y = 4, including the points on the line.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.43 Solve the following system of inequalities graphically:

$\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{8}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\ge }\mathbf{10}$

Ans

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given\hspace{0.33em}inequalities\hspace{0.33em}are:}\\ 2\text{x}+\text{y}\ge 8...\left(1\right)\\ \text{x}+2\text{y}\ge 10...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations 2x}+\text{y}=8\text{and x}+2\text{y}=10\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x + y = 8,
including the points on the line.
On the same set of axes, we draw the graph of the equation x + 2y = 10 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line x + 2y = 10,
including the points on the line.

In this way, the double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.44 Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le }\mathbf{9}\mathbf{,}\mathbf{y}\mathbf{>}\mathbf{x}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{0}$

Ans

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given\hspace{0.33em}inequalities\hspace{0.33em}are:}\\ \text{x}+\text{y}\le 9...\left(1\right)\\ \text{y}>\text{x\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(2\right)\\ \text{x}\ge 0...\left(3\right)\\ \mathrm{The}\text{graphs of linear equations 2x}+\text{y}=8\text{and x}+2\text{y}=10\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line x + y = 9, including the points on the line.
On the same set of axes, we draw the graph of the equation y = x and x = 0 as shown in Figure.
Then we see that inequality (2) and (3) represent the shaded region above the line y = x and right of the line and
x = 0 excluding the points on the line respectively.

In this way, the triple shaded region, common to the above three shaded regions, is the required solution region of the given system of inequalities.

Q.45

$\mathbf{\text{Solve the following system of inequalities graphically:}}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\le }\mathbf{60}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{2}$

Ans

$\begin{array}{l}\text{The given inequalities are:.}\\ 5\mathrm{x}+4\mathrm{y}\le 60...\left(1\right)\\ \mathrm{x}\ge 1...\left(2\right)\\ \mathrm{y}\ge 2...\left(3\right)\\ \text{The graphs of linear equations}5\mathrm{x}+4\mathrm{y}=60,\mathrm{x}=1\text{and}\mathrm{y}=2\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line 5x + 4y = 60, including the points on the line.
On the same set of axes, we draw the graph of the equation x = 1 and y = 2 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line x = 1 on the line including the points on the line and (3) represents the shaded region above the line y = 2 on the line including the points on the line.

In this way, green region is the common region. Which is the required solution region of the given system of inequalities.

Q.46

$\begin{array}{l}\mathbf{\text{Solve the following system of inequalities graphically:}}\\ \mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\le }\mathbf{60}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\le }\mathbf{30}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{0}\end{array}$

Ans

$\begin{array}{l}\text{The given inequalities are:}\\ 3\mathrm{x}+4\mathrm{y}\le 60...\left(1\right)\\ \mathrm{x}+3\mathrm{y}\le 30...\left(2\right)\\ \mathrm{x}\ge 0...\left(3\right)\\ \mathrm{y}\ge 0...\left(4\right)\\ \text{The graphs of linear equations}3\mathrm{x}+4\mathrm{y}=60,\mathrm{x}+3\mathrm{y}=30\text{,}\\ \mathrm{x}=0\text{and}\mathrm{y}=0\text{are drawn in}\mathrm{xy}\text{-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line 3x + 4y = 60, including the points on the line.
On the same set of axes, we draw the graph of the equation x + 3y = 30, x = 0 and y = 0 as shown in Figure. Then we see that inequality (2), (3) and (4) represent the shaded region below the line x + 3y = 30,
x = 0 and y = 0 including the points on the line.

In this way, the green shaded region is the required solution region of the given system of inequalities.

Q.47

$\mathbf{\text{Solve the following system of inequalities graphically:}}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{4}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le }\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{x}\mathbf{–}\mathbf{3}\mathbf{y}\mathbf{\le }\mathbf{6}$

Ans

$\begin{array}{l}\text{The given inequalities are:.}\\ 2\mathrm{x}+\mathrm{y}\ge 4...\left(1\right)\\ \mathrm{x}+\mathrm{y}\le 3...\left(2\right)\\ 2\mathrm{x}-3\mathrm{y}\le 6...\left(3\right)\\ \text{The graphs of linear equations}2\mathrm{x}+\mathrm{y}=4,\mathrm{x}+\mathrm{y}=3,\\ \mathrm{and}2\mathrm{x}-3\mathrm{y}=6\text{\hspace{0.33em}are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.33em}}\left(1\right)\text{is false for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is above the line of equation}\\ \text{2x}+\text{y}=4\text{including the points on\hspace{0.33em}the line.}\\ \text{Inequality\hspace{0.33em}}\left(2\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below the line of equation}\\ \text{x}+\text{y}=3\text{including the points on\hspace{0.33em}the line.}\\ \text{Inequality\hspace{0.33em}}\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below the line of equation}\\ \text{2x}-\text{3y\hspace{0.33em}=\hspace{0.33em}6 including the points on\hspace{0.33em}the line.}\\ \text{The common shaded area of the given three inequalities is\hspace{0.33em}the solution area.}\end{array}$

Q.48

$\mathbf{Solve}\mathbf{}\mathbf{the}\mathbf{}\mathbf{following}\mathbf{}\mathbf{system}\mathbf{}\mathbf{of}\mathbf{}\mathbf{inequalities}\mathbf{}\mathbf{graphically}\mathbf{:}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\le }\mathbf{3}\mathbf{,}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\ge }\mathbf{12}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{The}\text{\hspace{0.33em}given inequalities are:}\\ \text{\hspace{0.33em}\hspace{0.33em}x}-2\text{y}\le 3...\left(1\right)\\ 3\text{x}+4\text{y}\ge 12...\left(2\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x}\ge 0...\left(3\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}y}\ge 1...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations x}-2\text{y}=3\text{,\hspace{0.33em}}3\text{x}+4\text{y}=12,\text{x}=0\\ \text{and y}=\text{1 are drawn in xy-plane.}\\ \text{Inequality\hspace{0.33em}}\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is left of the line of equation}\\ \text{x}-2\text{y}=3\text{including the points on\hspace{0.33em}}\mathrm{the}\mathrm{line}.\\ \text{Inequality\hspace{0.33em}}\left(2\right)\text{is false for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is above the line of equation}\\ 3\text{x}+4\text{y}=12\text{including the points on\hspace{0.33em}the\hspace{0.33em}line.}\\ \text{Inequality\hspace{0.33em}}\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is right of the line of equation}\\ \text{x}=0\text{including the points on the\hspace{0.33em}line.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.33em}}\left(4\right)\text{is false for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is above the line of equation}\\ \text{2x}-3\text{y}=6\text{including the points on \hspace{0.33em}}\mathrm{the}\mathrm{line}.\\ \text{Thus, the common green area of the given three inequalities is \hspace{0.33em}the solution area.}\end{array}$

Q.49

$\mathbf{Solve}\mathbf{}\mathbf{the}\mathbf{}\mathbf{following}\mathbf{}\mathbf{system}\mathbf{}\mathbf{of}\mathbf{}\mathbf{inequalities}\mathbf{}\mathbf{graphically}\mathbf{:}\mathbf{}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\le }\mathbf{60}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{2}\mathbf{x}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{3}\mathbf{,}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{0}$

Ans

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ 4\mathrm{x}+3\mathrm{y}\le 60...\left(1\right)\\ \mathrm{y}\ge 2\mathrm{x}...\left(2\right)\\ \mathrm{x}\ge 3...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations 4}\mathrm{x}+3\mathrm{y}=60\text{,\hspace{0.17em}}\mathrm{y}=2\mathrm{x}\text{,x}=\text{3,}\mathrm{x}=0\\ \text{and y}=\text{0 are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.33em}}\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below of the line of equation}\\ \text{4x}+3\text{y}=60\text{including the points\hspace{0.33em}on the line.}\\ \text{Inequality\hspace{0.33em}}\left(2\right)\text{is true for the point}\left(1,3\right),\text{so its solution area\hspace{0.33em}is above the line of equation}\\ \text{y}=2\text{x including the points on\hspace{0.33em}the\hspace{0.33em}line.}\\ \text{Inequality\hspace{0.33em}}\left(3\right)\text{is false for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is right of the line of equation}\\ \text{x}=3\text{including the points on the\hspace{0.33em}line.}\\ \text{Inequality\hspace{0.33em}}\left(4\right)\text{is true for the values of x and y equal to zero\hspace{0.33em}or greater than zero.}\\ \text{So, its solution area is right of the x}-\mathrm{axis}\text{\hspace{0.33em}and above to the y-axis.}\\ \text{Thus, the green shaded area of the given three inequalities is\hspace{0.33em}the solution area.}\end{array}$

Q.50 

$\begin{array}{l}\text{Solve the following system of inequalities graphically:}\\ 3\text{x}+2\text{y}\le 150,\text{\hspace{0.33em}x}+4\text{y}\le 80,\text{x}\le 15,\text{x}\ge 0,\text{\hspace{0.33em}y}\ge 0\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ 3\mathrm{x}+2\mathrm{y}\le 150...\left(1\right)\\ \mathrm{x}+4\mathrm{y}\le 80...\left(2\right)\\ \mathrm{x}\le 15...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations}3\mathrm{x}+2\mathrm{y}=150\text{,\hspace{0.17em}}\mathrm{x}+4\mathrm{y}=80\text{,\hspace{0.17em}x}=\text{15,}\\ \mathrm{x}=0\text{\hspace{0.17em}\hspace{0.17em}and y}=\text{0 are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.33em}}\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below of the line of equation}\\ \text{3x}+2\text{y}=150\text{including the points\hspace{0.33em}on the line.}\\ \text{Inequality\hspace{0.33em}}\left(2\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is below the line of equation}\\ \text{x}+4\text{y}=80\text{including the points on\hspace{0.33em}the line.}\\ \text{Inequality\hspace{0.33em}}\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area\hspace{0.33em}is left of the line of equation}\\ \text{x}=15\text{including the points on the\hspace{0.33em}}\mathrm{line}.\\ \text{Inequality\hspace{0.33em}}\left(4\right)\text{is true for the values of x and y equal to zero\hspace{0.33em}or greater than zero.}\\ \text{So, its solution area is right of the x}-\mathrm{axis}\text{\hspace{0.33em}and above to the y-axis.}\\ \text{Thus, the green shaded area of the given four inequalities is\hspace{0.33em}}\mathrm{the}\mathrm{solution}\mathrm{area}.\end{array}$

Q.51 

$\begin{array}{l}\text{Solve the following system of inequalities graphically:}\\ \mathrm{x}+2\mathrm{y}\le 10,\mathrm{x}+\mathrm{y}\ge 1,\mathrm{x}-\mathrm{y}\le 0,\mathrm{x}\ge 0,\mathrm{y}\ge 0\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ \mathrm{x}+2\mathrm{y}\le 10...\left(1\right)\\ \mathrm{x}+\mathrm{y}\ge 1...\left(2\right)\\ \mathrm{x}-\mathrm{y}\le 0...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations}\mathrm{x}+2\mathrm{y}=10\text{, \hspace{0.17em}}\mathrm{x}+\mathrm{y}=1\text{,\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-\mathrm{y}=\text{0,}\\ \mathrm{x}=0\text{\hspace{0.17em}\hspace{0.17em}and y}=\text{0 are drawn in xy-plane.}\\ \text{Inequality\hspace{0.17em}}\left(1\right)\text{is true for the point}(0,0),\text{so its solution area}\\ \text{is below of the line of equation}\mathrm{x}+2\mathrm{y}=10\text{including the points}\\ \text{on the line.}\\ \text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(2\right)\text{is false for the point}(0,0),\text{so its solution area}\\ \text{is above the line of equation}\mathrm{x}+\mathrm{y}=1\text{including the points on}\\ \text{the line.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(3\right)\text{is true for the point}(2,1),\text{so its solution area}\\ \text{is above the line of equation}\mathrm{x}-\mathrm{y}=0\text{including the points on the}\\ \text{line.}\\ \text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(4\right)\text{is true for the values of x and y equal to zero}\\ \text{or greater than zero. So, its solution area is right of the}\mathrm{x}-\mathrm{axis}\\ \text{and above to the y-axis.}\\ \text{Thus, the blue shaded area of the given four inequalities is}\\ \text{the solution area.}\end{array}$

Q.52

$2\le 3\mathrm{x}–4\le 5$

Ans

$\begin{array}{l}\mathrm{We}\text{have}\\ 2\le 3\mathrm{x}-4\le 5\end{array}$ $\begin{array}{l}\mathrm{or}6\le 3\mathrm{x}\le 9\\ \mathrm{or}\frac{6}{3}\le \mathrm{x}\le \frac{9}{3}\\ \mathrm{or}2\le \mathrm{x}\le 3\\ \Rightarrow \mathrm{x}\in [2,3]\end{array}$

Q.53

$6\le -\mathbf{3}(2\mathrm{x}–4)<12$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ 6\le -3(2\mathrm{x}-4)<12\\ \mathrm{or}\frac{6}{3}\le -(2\mathrm{x}-4)<\frac{12}{3}\\ \mathrm{or}2\le -2\mathrm{x}+4<4\\ \mathrm{or}-2\le -2\mathrm{x}<0\\ \mathrm{or}1\le \mathrm{x}<0\\ \Rightarrow \mathrm{x}\in (0,1]\end{array}$

Q.54

$-3\le 4-\frac{7\mathrm{x}}{2}\le 18$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ -3\le 4-\frac{7\mathrm{x}}{2}\le 18\\ \mathrm{or}-3-4\le -\frac{7\mathrm{x}}{2}\le 18-4\\ \mathrm{or}-7\le -\frac{7\mathrm{x}}{2}\le 14\\ \mathrm{or}-14\le -7\mathrm{x}\le 28\\ \mathrm{or}-\frac{14}{7}\le -\mathrm{x}\le \frac{28}{7}\end{array}$ $\begin{array}{l}\mathrm{or}-2\le -\mathrm{x}\le 4\\ \mathrm{or}2\ge \mathrm{x}\ge -4\\ \Rightarrow \mathrm{x}\in [-4,2]\end{array}$

Q.55

$-15<\frac{3(\mathrm{x}-2)}{5}\le 0$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ -15<\frac{3(\mathrm{x}-2)}{5}\le 0\\ \mathrm{or}-75<3(\mathrm{x}-2)\le 0\\ \mathrm{or}-\frac{75}{3}>(\mathrm{x}-2)\le \frac{0}{3}\\ \mathrm{or}-25>\mathrm{x}-2\le 0\\ \mathrm{or}-25+2>\mathrm{x}\le 0+2\\ \mathrm{or}-23>\mathrm{x}\le 2\\ \Rightarrow \mathrm{x}\in (-23,2]\end{array}$

Q.56

$-12<\mathbf{4}-\frac{3\mathbf{x}}{-5}\le 2$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ -12<4-\frac{3\mathrm{x}}{-5}\le 2\\ \mathrm{or}-12-4<\frac{3\mathrm{x}}{5}\le 2-4\\ \mathrm{or}-16<\frac{3\mathrm{x}}{5}\le -2\\ \mathrm{or}-16\times 5<3\mathrm{x}\le -2\times 5\end{array}$ $\begin{array}{l}\mathrm{or}-80<3\mathrm{x}\le -10\\ \mathrm{or}-\frac{80}{3}<\mathrm{x}\le -\frac{10}{3}\\ \Rightarrow \mathrm{x}\in (-\frac{80}{3},-\frac{10}{3}]\end{array}$

Q.57

$7\le \frac{(3\mathrm{x}+11)}{2}\le 11$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ 7\le \frac{(3\mathrm{x}+11)}{2}\le 11\\ \mathrm{or}14\le 3\mathrm{x}+11\le 22\\ \mathrm{or}14-11\le 3\mathrm{x}\le 22-11\\ \mathrm{or}3\le 3\mathrm{x}\le 11\\ \mathrm{or}1\le \mathrm{x}\le \frac{11}{3}\\ \Rightarrow \mathrm{x}\in [1,\frac{11}{3}]\end{array}$

Q.58 Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.

7. 5x + 1 > – 24, 5x – 1 < 24
8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
9. 3x – 7 > 2(x – 6), 6 – x >11–2x
10.

$\mathbf{}\mathbf{5}\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{–}\mathbf{7}\mathbf{)}\mathbf{–}\mathbf{3}\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{\le }\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{19}\mathbf{\le }\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{47}\mathbf{.}$

Ans

7.

The given inequalities are:
5x + 1 > – 24 …(1)
5x – 1 < 24 …(2)
From inequality (1), we have
5x + 1 > – 24
or 5x > – 25
or x > – 5 …(3)
From inequality (2), we have
5x – 1 < 24
or 5x < 25
or x < 5 …(4)
If we draw the graph of inequalities (3) and (4) on the number line, we see that the values of x,
which are common to both, are shown by bold line in Figure.

Thus, solution of the system is real numbers x lying between – 5 and 5,
i.e., – 5 < x < 5,
or x ∈ (– 5, 5).

8.

The given inequalities are:
2(x – 1) < x + 5 …(1)
3(x + 2) > 2 – x …(2)
From inequality (1), we have
2(x – 1) < x + 5
or 2x – 2 < x + 5
or x < 7 …(3)
From inequality (2), we have
3(x + 2) > 2 – x
or 3x + 6 > 2 – x
or 4x > – 4
or x > – 1 …(4)
If we draw the graph of inequalities (3) and (4) on the number line, we see that the values of x,
which are common to both, are shown by bold line in Figure.

Thus, solution of the system is real numbers x lying between – 1 and 7,
i.e., –1< x < 7,
or x ∈ (– 1, 7).

9.

3x – 7 > 2(x – 6) …(1)
6 – x > 11–2x …(2)
From inequality (1), we have
3x – 7 > 2(x – 6)
or 3x – 7 > 2x – 12
or x > – 5 …(3)
From inequality (2), we have
6 – x > 11–2x
or x > 5 …(4)
From equation (3) and equation (4), we can say that
x > 5

Thus, solution of the system is real numbers x lying between 5 and infinity,
i.e., 5 < x < ∞,
or x ∈ (5, ∞).

10.

$\begin{array}{l}5\left(2\text{x}-7\right)-3\left(2\text{x}+3\right)\le 0,2\text{x}+19\le 6\text{x}+47.\\ 5\left(2\text{x}-7\right)-3\left(2\text{x}+3\right)\le 0...\left(1\right)\\ 2\text{x}+19\le 6\text{x}+47...\left(2\right)\\ \text{From inequality}\left(\text{1}\right),\text{we have}\\ 10\text{x}-35-6\text{x}-9\le 0\\ 4\text{x}-44\le 0\\ 4\text{x}\le 44\\ \text{From inequality}\left(\text{1}\right),\text{we have}\\ 10\text{x}-35-6\text{x}-9\le 0\\ 4\text{x}-44\le 0\\ 4\text{x}\le 44\\ \text{x}\le 11...\left(3\right)\\ \text{From inequality}\left(\text{2}\right),\text{we have}\\ 19-47\le 6\text{x}-2\text{x}\\ -28\le 4\text{x}\\ -7\le \text{x\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}...\left(4\right)\end{array}$

If we draw the graph of inequalities (3) and (4) on the number line, we see that the values of x,
which are common to both, are shown by bold line in Figure.

$\begin{array}{l}\text{Thus},\text{solution of the system is real numbers x lying \hspace{0.33em}between}–\text{7 and 11},\\ \text{i}.\text{e}.,–\text{7}\le \text{x}\le \text{11},\\ \text{or}\text{x}\in \left[–\text{7},\text{11}\right].\end{array}$

Q.59

$\begin{array}{l}\text{A solution is to be kept between 68° F and 77° F. What is}\\ \text{the range in temperature in degree Celsius (C) if the}\\ \text{Celsius / Fahrenheit (F) conversion formula is given by}\\ \text{F =}\frac{\text{9}}{\text{5}}\text{C + 32 ?}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{solution}\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{kept}\mathrm{between}68\mathrm{°}\mathrm{F}\mathrm{}\mathrm{and}77\mathrm{°}\mathrm{F}.\\ \mathrm{So},68<\mathrm{F}<77\\ \Rightarrow 68<\frac{9}{5}\mathrm{C}+32<77\\ \Rightarrow 68-32<\frac{9}{5}\mathrm{C}<77-32\\ \Rightarrow 36<\frac{9}{5}\mathrm{C}<45\\ \Rightarrow 36\times \frac{5}{9}<\mathrm{C}<45\times \frac{5}{9}\\ \Rightarrow 20<\mathrm{C}<25\\ \Rightarrow \mathrm{C}\in (20,25)\\ \mathrm{Thus},\text{range of temperature is 20°C to 25°C.}\end{array}$

Q.60 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?

Ans

$\begin{array}{l}\mathrm{Let}\text{added 2\% boric acid solution}=\text{x litres}\\ \text{Volume of 8\% boric acid solution}=\text{640 litres}\\ \text{Total solution of boric acid}=(640+\mathrm{x})\text{\hspace{0.17em}litres}\\ \mathrm{Therefore},\\ 2\mathrm{\%}\mathrm{of}\mathrm{x}+8\mathrm{\%}\mathrm{of}640>4\mathrm{\%}(640+\mathrm{x})\\ \mathrm{and}2\mathrm{\%}\mathrm{of}\mathrm{x}+8\mathrm{\%}\mathrm{of}640<6\mathrm{\%}(640+\mathrm{x})\\ \mathrm{or}\frac{2}{100}\times \mathrm{x}+\frac{8}{100}\times 640>\frac{4}{100}\times (640+\mathrm{x})\end{array}$ $\begin{array}{l}\mathrm{and}\frac{2}{100}\times \mathrm{x}+\frac{8}{100}\times \mathrm{of}640<\frac{6}{100}\times (640+\mathrm{x})\\ \mathrm{or}2\mathrm{x}+8\times 640>4(640+\mathrm{x})\\ \mathrm{and}2\mathrm{x}+8\times 640<6\times (640+\mathrm{x})\\ \mathrm{or}2\mathrm{x}+5120>2560+4\mathrm{x}\\ \mathrm{and}2\mathrm{x}+5120<3840+6\mathrm{x}\\ \mathrm{or}2560>2\mathrm{x}\\ \mathrm{and}1280<4\mathrm{x}\\ \mathrm{or}1280>\mathrm{x}\\ \mathrm{and}320<\mathrm{x}\\ \Rightarrow \mathrm{x}\in (320,1280)\\ \mathrm{Thus},\text{2\% solution of boric acid should be more than}\\ \text{320 litres but less than 1280 litres.}\end{array}$

Q.61 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?

Ans

$\begin{array}{l}\mathrm{Let}\text{Volume of added water to acid}=\text{x litres}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Volume of 45\% solution of acid}=\text{1125 litres}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Total solution of acid}=\left(1125+\text{x}\right)\text{\hspace{0.33em}litres}\\ \mathrm{Therefore},\\ 45\text{\%\hspace{0.33em}}\mathrm{of}1125>25\text{\%}\left(1125+\text{x}\right)\\ \mathrm{and}45\text{\%\hspace{0.33em}}\mathrm{of}1125<30\text{\%}\left(1125+\text{x}\right)\\ \mathrm{or}\frac{45}{100}\times 1125>\frac{25}{100}\times \left(1125+\text{x}\right)\\ \mathrm{and}\frac{45}{100}\times 1125<\frac{30}{100}\times \left(1125+\text{x}\right)\\ \mathrm{or}45\times 1125>25\times \left(1125+\text{x}\right)\\ \mathrm{and}445\times 1125<30\times \left(1125+\text{x}\right)\\ \mathrm{or}50625>28125+25\text{x}\\ \mathrm{and}50625<33750+30\text{x}\\ \mathrm{or}50625-28125>25\text{x}\\ \mathrm{and}50625-33750<30\text{x}\\ \mathrm{or}22500>25\text{x}\\ \mathrm{and}16875<30\text{x}\\ \mathrm{or}900>\text{x}\\ \mathrm{and}562.5<\text{x}\\ \Rightarrow \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x}\in \left(562.5,900\right)\\ \mathrm{Thus},\text{added water should be more than}562.5\text{litres \hspace{0.33em}but less than 900 litres.}\end{array}$

Q.62

$\begin{array}{l}\text{IQ of a person is given by the formula}\\ \text{IQ=}\frac{\text{MA}}{\text{CA}}\text{×100,}\\ \begin{array}{l}\text{where MA is mental age and CA is chronological age.}\\ \text{If 80}\le \text{IQ}\le \text{140 for a group of 12 years old children,}\\ \text{find the range of their mental age.}\end{array}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{IQ}\mathrm{of}\text{a}\mathrm{person}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{formula}.\\ \mathrm{IQ}=\frac{\mathrm{MA}}{\mathrm{CA}}\times 100,\\ \mathrm{For}\text{a}\mathrm{group}\mathrm{of}12\mathrm{years}\mathrm{old}\mathrm{children}\\ 80\le \mathrm{IQ}\le 140\\ 80\le \frac{\mathrm{MA}}{\mathrm{CA}}\times 100\le 140\\ 80\le \frac{\mathrm{MA}}{12}\times 100\le 140\\ 80\times \frac{12}{100}\le \mathrm{MA}\le 140\times \frac{12}{100}\\ \frac{48}{5}\le \mathrm{MA}\le \frac{84}{5}\\ 9.6\le \mathrm{MA}\le 16.8\\ \text{Thus, the range of mental age of a group of 12 years old}\\ \text{children is 9.6 to 16.8 years.}\end{array}$