NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities

Question 1(i): Solve $24x < 100$ , when $x$ is a natural number.

Answer:

Given : $24x < 100$

$\Rightarrow$ $24x < 100$

Divide by 24 from both sides

$\Rightarrow \, \, \, \frac{24}{24}x< \frac{100}{24}$

$\Rightarrow \, \, \, x< \frac{25}{6}$

$\Rightarrow \, \, \, x< 4.167$

$x$ is a natural number which is less than 4.167.

Hence, values of x can be $\left \{ 1,2,3,4 \right \}$

Question 1(ii): Solve $24x< 100$ , when

$x$ is an integer.

Answer:

Given : $24x < 100$

$\Rightarrow$ $24x < 100$

Divide by 24 from both sides

$\Rightarrow \, \, \, \frac{24}{24}x< \frac{100}{24}$

$\Rightarrow \, \, \, x< \frac{25}{6}$

$\Rightarrow \, \, \, x< 4.167$

$x$ is an integer which is less than 4.167.

Hence, values of x can be $\left \{..........-3,-2,-1,0, 1,2,3,4 \right \}$

Question 2(i): Solve $-12x>30$ , when
x is a natural number.

Answer:

Given : $-12x>30$

$\Rightarrow$ $-12x>30$

Divide by -12 from both sides

$\Rightarrow \, \, \, \frac{-12}{-12}x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< -2.5$

$x$ is a natural number which is less than - 2.5.

Hence, the values of x do not exist for the given inequality.

Question 2(ii): Solve $- 12x > 30$ , when $x$ is an integer.

Answer:

Given : $-12x>30$

$\Rightarrow$ $-12x>30$

Divide by -12 from both sides

$\Rightarrow \, \, \, \frac{-12}{-12}x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< -2.5$

$x$ are integers less than - 2.5.

Hence, values of x can be $\left \{ .............,-6,-5,-4,-3 \right \}$

Question 3(i): Solve $5x - 3 < 7$ , when $x$ is an integer.

Answer:

Given : $5x - 3 < 7$

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

$\Rightarrow \, \, \, 5x< 10$

Divide by 5 from both sides

$\Rightarrow \, \, \, \frac{5}{5}x< \frac{10}{5}$

$\Rightarrow \, \, \, x< 2$

$x$ is an integer less than 2

Hence, values of x can be $\left \{.........-3,-2-1,0,1,\right \}$

Question 3(ii): Solve $5x - 3 < 7$ , when $x$ is a real number.

Answer:

Given : $5x - 3 < 7$

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

$\Rightarrow \, \, \, 5x< 10$

Divide by 5 from both sides

$\Rightarrow \, \, \, \frac{5}{5}x< \frac{10}{5}$

$\Rightarrow \, \, \, x< 2$

$x$ are real numbers less than 2

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

Question 4(i): Solve $3x + 8 >2$ , when
x is an integer.

Answer:

Given : $3x + 8 >2$

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

$\Rightarrow \, \, \, 3x> -6$

Divide by 3 from both sides

$\Rightarrow \, \, \, \frac{3}{3}x> \frac{-6}{3}$

$\Rightarrow \, \, \, x> - 2$

$x$ are integers greater than -2

Hence, the values of x can be $\left \{-1,0,1,2,3,4...............\right \}$ .

Question 4(ii): Solve $3x + 8 >2$ , when $x$ is a real number. )

Answer:

Given : $3x + 8 >2$

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

$\Rightarrow \, \, \, 3x> -6$

Divide by 3 from both sides

$\Rightarrow \, \, \, \frac{3}{3}x> \frac{-6}{3}$

$\Rightarrow \, \, \, x> - 2$

$x$ are real numbers greater than -2

Hence , values of x can be as $x\in (-2,\infty )$

Question 5: Solve the inequality for real $x$.

$4x + 3 < 5x + 7$

Answer:

Given : $4x + 3 < 5x + 7$

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

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

$\Rightarrow \, \, \, x> -4$

$x$ are real numbers greater than -4.

Hence, values of x can be as $x\in (-4 ,\infty )$

Question 6: Solve the inequality for real $x$

$3x - 7 > 5x -1$

Answer:

Given : $3x - 7 > 5x -1$

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

$\Rightarrow \, \, \, -2x> 6$

$\Rightarrow \, \, \, x< \frac{6}{-2}$

$\Rightarrow \, \, \, x< -3$

$x$ are real numbers less than -3.

Hence, values of x can be $x\in (-\infty ,-3)$

Question 7: Solve the inequality for real $x$.

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

Answer:

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

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

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

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

$\Rightarrow \, \, \, x\leq -3$

$x$ are real numbers less than or equal to -3

Hence , values of x can be as , $x\in (-\infty ,-3]$

Question 8: Solve the inequality for real $x$

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

Answer:

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

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

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

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

$\Rightarrow \, \, \, 4\geq x$

$x$ are real numbers less than or equal to 4

Hence, values of x can be as $x\in (-\infty ,4]$

Question 9: Solve the inequality for real $x$

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

Answer:

Given : $x + \frac{x}{2} + \frac{x}{3} < 11$

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

$\Rightarrow \, \, \, x(1+\frac{1}{2}+\frac{1}{3})< 11$

$\Rightarrow \, \, \, x(\frac{11}{6})< 11$

$\Rightarrow \, \, \, 11 x< 11\times 6$

$\Rightarrow \, \, \, x< 6$

$x$ are real numbers less than 6

Hence, values of x can be as $x\in (-\infty ,6)$

Question 10: Solve the inequality for real $x$.

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

Answer:

Given : $\frac{x}{3} > \frac{x}{2} + 1$

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

$\Rightarrow \, \, \, \frac{x}{3}-\frac{x}{2}> 1$

$\Rightarrow \, \, \,x (\frac{1}{3}-\frac{1}{2})> 1$

$\Rightarrow \, \, \,x (-\frac{1}{6})> 1$

$\Rightarrow \, \, \, -x > 6$

$\Rightarrow \, \, \, x< -6$

$x$ are real numbers less than -6

Hence, values of x can be as $x\in (-\infty ,-6)$

Question 11: Solve the inequality for real $x$

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

Answer:

Given : $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$

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

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

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

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

$\Rightarrow \, \, \, 34x\leq 68$

$\Rightarrow \, \, \, x\leq 2$

$x$ are real numbers less than or equal to 2.

Hence, values of x can be as $x\in (-\infty ,2]$

Question 12: Solve the inequality for real $x$

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

Answer:

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

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

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

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

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

$\Rightarrow \, \, 24 \geq \frac{x}{5}$

$\Rightarrow \, \, 120 \geq x$

$x$ are real numbers less than or equal to 120.

Hence, values of x can be as $x\in (-\infty,120 ]$ .

Question 13: Solve the inequality for real $x$

$2(2x + 3) - 10 < 6(x-2)$

Answer:

Given : $2(2x + 3) - 10 < 6(x-2)$

$\Rightarrow$$2(2x + 3) - 10 < 6(x-2)$

$\Rightarrow \, \, \, 4x+6-10 < 6x-12$

$\Rightarrow \, \, \, 6-10+12 < 6x-4x$

$\Rightarrow \, \, \, 8 < 2x$

$\Rightarrow \, \, \, 4 < x$

$x$ are real numbers greater than 4

Hence , values of x can be as $x\in (4,\infty )$

Question 14: Solve the inequality for real $x$

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

Answer:

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

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

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

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

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

$\Rightarrow \, \, \, 8 \geq 4x$

$\Rightarrow \, \, \, 2\geq x$

$x$ are real numbers less than or equal to 2.

Hence , values of x can be as $x\in (-\infty ,2]$

Question 15: Solve the inequality for real x

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

Answer:

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

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

$\Rightarrow \, \, \, \, 15x< 20(5x-2)-12(7x-3)$

$\Rightarrow \, \, \, \, 15x< 100x-40-84x+36$

$\Rightarrow \, \, \, \, 15x< 16x-4$

$\Rightarrow \, \, \, \, 4< x$

$x$ are real numbers greater than 4.

Hence, values of x can be as $x\in (4,\infty)$

Question 16: Solve the inequality for real $x$

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

Answer:

Given : $\frac{(2x - 1)}{3} \geq \frac{3x-2}{4} - \frac{(2-x)}{5}$

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

$\Rightarrow \, \, \, 20(2x - 1) \geq 15(3x-2) - 12(2-x)$

$\Rightarrow \, \, \, 40x - 20 \geq 45x-30 - 24+12x$

$\Rightarrow \, \, \, 30+24 - 20 \geq 45x-40x+12x$

$\Rightarrow \, \, \, 34 \geq 17x$

$\Rightarrow \, \, \, 2 \geq x$

$x$ are real numbers less than or equal to 2.

Hence, values of x can be as $x\in (-\infty,2 ]$ .

Question 17: Solve the inequality and show the graph of the solution on number line $3x - 2 < 2x + 1$

Answer:

Given : $3x - 2 < 2x + 1$

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

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

$\Rightarrow \, \, \, x< 3$

$x$ are real numbers less than 3

Hence, values of x can be as $x\in (-\infty ,3)$

The graphical representation of solutions of the given inequality is as :

Question 18: Solve the inequality and show the graph of the solution on the number line $5x - 3 \geq 3x -5$

Answer:

Given : $5x - 3 \geq 3x -5$

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

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

$\Rightarrow \, \, \, 2x \geq -2$

$\Rightarrow \, \, \, x \geq -1$

$x$ are real numbers greater than or equal to -1.

Hence, values of x can be as $x\in [-1,\infty )$

The graphical representation of solutions of the given inequality is as :

Question 19: Solve the inequality and show the graph of the solution on number line $3(1-x) < 2 (x +4)$

Answer:

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

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

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

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

$\Rightarrow \, \, \, -5< 5 x$

$\Rightarrow \, \, \, -1< x$

$x$ are real numbers greater than -1

Hence, values of x can be as $x\in (-1,\infty )$

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

Question 20: Solve the inequality and show the graph of the solution on number line $\frac{x}{2} \geq \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

Answer:

Given : $\frac{x}{2} \geq \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

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

$\Rightarrow \, \, \, 15x \geq 10(5x-2) - 6(7x-3)$

$\Rightarrow \, \, \, 15x \geq 50x-20 - 42x+18$

$\Rightarrow \, \, \, 15x+42x-50x \geq 18-20$

$\Rightarrow \, \, \, 7x \geq -2$

$\Rightarrow \, \, \, x \geq \frac{-2}{7}$

$x$ are real numbers greater than equal to $= \frac{-2}{7}$

Hence, values of x can be as $x\in (-\frac{2}{7},\infty )$

The graphical representation of solutions of the given inequality is as :

Question 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.

Answer:

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

The student should have an average of at least 60 marks.

$\therefore \, \, \, \frac{70+75+x}{3}\geq 60$

$\, \, \, 145+x\geq 180$

$x\geq 180-145$

$x\geq 35$

The student should have a minimum of 35 marks to have an average of 60

Question 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.

Answer:

Sunita’s marks in the first four examinations are 87, 92, 94 and 95.

Let x be the marks obtained in the fifth examination.

To receive a Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations.

$\therefore \, \, \, \frac{87+92+94+95+x}{5}\geq 90$

$\Rightarrow \, \, \, \frac{368+x}{5}\geq 90$

$\Rightarrow \, \, \, 368+x\geq 450$

$\Rightarrow \, \, \, x\geq 450-368$

$\Rightarrow \, \, \, x\geq 82$

Thus, Sunita must obtain 82 in the fifth examination to get a grade ‘A’ in the course.

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

Answer:

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

Both integers are smaller than 10.

$\therefore \, \, \, x+2< 10$

$\Rightarrow \, \, \, \, x< 10-2$

$\Rightarrow \, \, \, \, x< 8$

Sum of both integers is more than 11.

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

$\Rightarrow \, \, \, (2x+2)> 11$

$\Rightarrow \, \, \, 2x> 11-2$

$\Rightarrow \, \, \, 2x> 9$

$\Rightarrow \, \, \, x> \frac{9}{2}$

$\Rightarrow \, \, \, x> 4.5$

We conclude $\, \, \, \, x< 8$ and $\, \, \, x> 4.5$ and x is odd integer number.

x can be 5,7.

The two pairs of consecutive odd positive integers are $(5,7)\, \, \, and\, \, \, (7,9)$ .

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

Answer:

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

Both integers are larger than 5.

$\therefore \, \, \, x> 5$

Sum of both integers is less than 23.

$\therefore \, \, \, x+(x+2)< 23$

$\Rightarrow \, \, \, (2x+2)< 23$

$\Rightarrow \, \, \, 2x< 23-2$

$\Rightarrow \, \, \, 2x< 21$

$\Rightarrow \, \, \, x< \frac{21}{2}$

$\Rightarrow \, \, \, x< 10.5$

We conclude $\, \, \, \, x< 10.5$ and $\, \, \, x> 5$ and x is even integer number.

x can be 6,8,10.

The pairs of consecutive even positive integers are $(6,8),(8,10),(10,12)$ .

Question 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.

Answer:

Let the length of the smallest side be x cm.

Then the largest side = 3x cm.

Third side = 3x-2 cm.

Given: The perimeter of the triangle is at least 61 cm.

$\therefore\, \, \, x+3x+(3x-2)\geq 61$

$\Rightarrow \, \, \, 7x-2\geq 61$

$\Rightarrow \, \, \, 7x\geq 61+2$

$\Rightarrow \, \, \, 7x\geq 63$

$\Rightarrow \, \, \, x\geq \frac{63}{7}$

$\Rightarrow \, \, \, x\geq 9$

Minimum length of the shortest side is 9 cm.

Question 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 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 5cm longer than the second?

[ Hint: If x is the length of the shortest board, then $x$, $(x + 3)$ and $2x$ are the lengths of the second and third pieces, respectively. Thus, $x + (x + 3) + 2x \leq 91$ and $2x \geq (x + 3) + 5$ ].

Answer:

Let x be the length of the shortest board,

Then $(x + 3)$ and $2x$ are the lengths of the second and third pieces, respectively.

The man wants to cut three lengths from a single piece of board of length 91cm.

Thus, $x + (x + 3) + 2x \leq 91$

$4x+3\leq 91$

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

$\Rightarrow \, \, \, \, 4x\leq 88$

$\Rightarrow \, \, \, \, x\leq \frac{88}{4}$

$\Rightarrow \, \, \, \, x\leq 22$

If the third piece is to be at least 5cm longer than the second, then

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

$\Rightarrow \, \, \, \, 2x\geq x+8$

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

$\Rightarrow \, \, \, \, x\geq 8$

We conclude that $\, \, \, \, x\geq 8$ and $\, \, \, \, x\leq 22$ .

Thus , $8\leq x\leq 22$ .

Hence, the length of the shortest board is greater than or equal to 8 cm and less than equal to 22 cm.

Question 1: Solve the inequality $2\leq 3x-4\leq5$

Answer:

Given : $2\leq 3x-4\leq5$

$2\leq 3x-4\leq5$

$\Rightarrow\, \, 2+4\leq 3x\leq 5+4$

$\Rightarrow\, \, 6\leq 3x\leq 9$

$\Rightarrow\, \, \frac{6}{3}\leq x\leq \frac{9}{3}$

$\Rightarrow\, \, 2\leq x\leq 3$

Thus, all the real numbers greater than equal to 2 and less than equal to 3 are solutions to this inequality.

Solution set is $[2,3]$

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

Answer:

Given $6 \leq -3(2x - 4) < 12$

$6 \leq -3(2x - 4) < 12$

$\Rightarrow\, \ \frac{6}{3}\leq -(2x-4)< \frac{12}{3}$

$\Rightarrow\, \ -2\geq (2x-4)> -4$

$\Rightarrow\, \ -2+4\geq 2x> -4+4$

$\Rightarrow\, \ 2\geq 2x> 0$

$\Rightarrow\, \ 1\geq x> 0$

Solution set is $(01]$

Question 3: Solve the inequality $-3 \leq 4 - \frac{7x}{2}\leq 18$

Answer:

Given $-3 \leq 4 - \frac{7x}{2}\leq 18$

$\Rightarrow \, \, -3 \leq 4 - \frac{7x}{2}\leq 18$

$\Rightarrow \, \, -3-4 \leq - \frac{7x}{2}\leq 18-4$

$\Rightarrow \, \, -7 \leq - \frac{7x}{2}\leq 14$

$\Rightarrow \, \, 7 \geq \frac{7x}{2} \geq -14$

$\Rightarrow \, \, 7\times 2 \geq 7x\geq -14\times 2$

$\Rightarrow \, \, 14 \geq 7x \geq -28$

$\Rightarrow \, \, \frac{14}{7} \geq x \geq \frac{-28}{7}$

$\Rightarrow \, \, 2 \geq x \geq -4$

Solution set is $[-4,2]$

Question 4: Solve the inequality $-15 < \frac{3(x-2)}{5} \leq 0$

Answer:

Given The inequality

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

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

$\Rightarrow\, \ -15\times 5< 3(x-2)\leq 0\times 5$

$\Rightarrow\, \ -75< 3(x-2)\leq 0$

$\Rightarrow\, \ \frac{-75}{3}< (x-2)\leq \frac{0}{3}$

$\Rightarrow\, \ -25< (x-2)\leq 0$

$\Rightarrow\, \ -25+2< x\leq 0+2$

$\Rightarrow\, \ -23< x\leq 2$

The solution set is $(-23,2]$

Question 5: Solve the inequality $-12<4-\frac{3x}{-5} \leq 2$

Answer:

Given the inequality

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

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

$\Rightarrow\, \, -12-4< -\frac{3x}{-5}\leq 2-4$

$\Rightarrow\, \, -16< -\frac{3x}{-5}\leq -2$

$\Rightarrow\, \, -16< \frac{3x}{5}\leq -2$

$\Rightarrow\, \, -16\times 5< 3x\leq -2\times 5$

$\Rightarrow\, \, -80< 3x\leq -10$

$\Rightarrow\, \, \frac{-80}{3}< 3x\leq \frac{-10}{3}$

Solution set is $(\frac{-80}{3}, \frac{-10}{3}]$

Question 6: Solve the inequality $7 \leq \frac{(3x+ 11)}{2}\leq 11$

Answer:

Given the linear inequality

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

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

$\Rightarrow \, \, 7\times 2 \leq (3x+ 11)\leq 11\times 2$

$\Rightarrow \, \, 14 \leq (3x+ 11)\leq 22$

$\Rightarrow \, \, 14-11 \leq (3x)\leq 22-11$

$\Rightarrow \, \, 3 \leq 3x\leq 11$

$\Rightarrow \, \, 1 \leq x\leq \frac{11}{3}$

The solution set of the given inequality is $[1,\frac{11}{3}]$

Question 7: Solve the inequality and represent the solution graphically on number line. $5x + 1 > -24,\ 5x - 1 <24$

Answer:

Given : $5x + 1 > -24,\ 5x - 1 <24$

$5x + 1 > -24\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x - 1 <24$

$\Rightarrow 5x > -24-1\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x <24+1$

$\Rightarrow 5x > -25\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x <25$

$\Rightarrow x > \frac{-25}{5}\, \, \, \, \, \, \, and\, \, \, \, \, \, \ x <\frac{25}{5}$

$\Rightarrow x > -5\, \, \, \, \, \, \, and\, \, \, \, \, \, \ x <5$

$(-5,5)$

The solution graphically on the number line is as shown :

Question 8: Solve the inequality and represent the solution graphically on number line. $2(x-1)<x+5,\ 3(x+2)> 2 -x$

Answer:

Given : $2(x-1)<x+5,\ 3(x+2)> 2 -x$

$2(x-1)<x+5\, \, \, \, and\, \, \, \, \, \ 3(x+2)> 2 -x$

$\Rightarrow \, \, 2x-2<x+5\, \, \, \, and\, \, \, \, \, \ 3x+6> 2 -x$

$\Rightarrow \, \, 2x-x<2+5\, \, \, \, and\, \, \, \, \, \ 3x+x> 2 -6$

$\Rightarrow \, \, x<7\, \, \, \, and\, \, \, \, \, \ 4x> -4$

$\Rightarrow \, \, x<7\, \, \, \, and\, \, \, \, \, \ x> -1$

$(-1,7)$

The solution graphically on the number line is as shown :

Question 9: Solve the inequality and represent the solution graphically on number line. $3x - 7 > 2(x-6),\ 6-x > 11 - 2x$

Answer:

Given : $3x - 7 > 2(x-6),\ 6-x > 11 - 2x$

$3x - 7 > 2(x-6)\, \, \, \, and\, \, \, \, \, \ 6-x > 11 - 2x$

$\Rightarrow \, \, 3x - 7 > 2x-12\, \, \, \, and\, \, \, \, \, \ 6-x > 11 - 2x$

$\Rightarrow \, \, 3x - 2x >7-12\, \, \, \, and\, \, \, \, \, \ 2x-x > 11 - 6$

$\Rightarrow \, \, x >-5\, \, \, \, and\, \, \, \, \, \ x > 5$

$x\in (5,\infty )$

The solution graphically on the number line is as shown :

Question 10: Solve the inequality and represent the solution graphically on number line.

$5(2x-7)-3(2x+3)\leq 0,\quad 2x + 19 \leq 6x +47$

Answer:

Given : $5(2x-7)-3(2x+3)\leq 0,\quad 2x + 19 \leq 6x +47$

$5(2x-7)-3(2x+3)\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad 2x + 19 \leq 6x +47$

$\Rightarrow \, \, 10x-35-6x-9\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad 2x -6x\leq 47-19$

$\Rightarrow \, \, 4x-44\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad -4x\leq 28$

$\Rightarrow \, \, 4x\leq 44\, \, \, \, \, and\, \, \, \, \, \, \, \quad 4x\geq - 28$

$\Rightarrow \, \, x\leq 11\, \, \, \, \, and\, \, \, \, \, \, \, \quad x\geq - 7$

$x\in [-7,11]$

The solution graphically on the number line is as shown :

Question 11: A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by $F = \frac{9}{5}C + 32$

Answer:

Since the solution is to be kept between 68° F and 77° F.

$68< F< 77$

Putting the value of $F = \frac{9}{5}C + 32$ , we have

$\Rightarrow \, \, \, 68< \frac{9}{5}C + 32< 77$

$\Rightarrow \, \, \, 68-32< \frac{9}{5}C < 77-32$

$\Rightarrow \, \, \, 36< \frac{9}{5}C < 45$

$\Rightarrow \, \, \, 36\times 5< 9C < 45\times 5$

$\Rightarrow \, \, \, 180< 9C < 225$

$\Rightarrow \, \, \, \frac{180}{9}< C < \frac{225}{9}$

$\Rightarrow \, \, \, 20< C < 25$

the range in temperature in degree Celsius (C) is between 20 to 25.

Question 12: 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?

Answer:

Let x litres of 2% boric acid solution be required to be added.

Total mixture = (x+640) litres

The resulting mixture is to be more than 4% but less than 6% boric acid.

$\therefore \, 2\%x+8\%\, of\, 640> 4\%\, of\, (640+x)$ and $2\%x+8\%\, of\, 640< 6\%\, of\, (x+640)$

$\Rightarrow \, 2\%x+8\%\, of\, 640> 4\%\, of\, (640+x)$ and $2\%x+8\%\, of\, 640< 6\%\, of\, (x+640)$

$\Rightarrow \, \frac{2}{100}x+(\frac{8}{100}) 640> \frac{4}{100} (640+x)$ and $ \, \frac{2}{100}x+(\frac{8}{100}) 640< \frac{6}{100} (640+x)$

$\Rightarrow \, 2x+5120> 4x+2560$ and $\, 2x+5120< 6x+3840$

$\Rightarrow \, 5120-2560> 4x-2x$ and $ \, 5120-3840< 6x-2x$

$\Rightarrow \, 2560> 2x$ and $ \, 1280< 4x$

$\Rightarrow \, 1280> x$ and $ \, 320< x$

So, $1280> x > 320$

Thus, the number of litres 2% of boric acid solution that is to be added will have to be more than 320 and less than 1280 litres.

Question 13: 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?

Answer:

Let x litres of water be required to be added.

Total mixture = (x+1125) litres

It is evident that the amount of acid contained in the resulting mixture is 45% of 1125 litres.

The resulting mixture contains more than 25 % but less than 30% acid.

$\therefore \, 30\%\, of\, (1125+x) > 45\%\, of\, (1125)$ and $25\%\, of\, (1125+x)< 45\%\, of\, 1125$

$\Rightarrow \, 30\%\, of\, (1125+x) > 45\%\, of\, (1125)$ and $25\%\, of\, (1125+x)< 45\%\, of\, 1125$

$\Rightarrow \, \frac{30}{100}(1125+x)> \frac{45}{100} (1125)$ and $\, (\frac{25}{100}) (1125+x)< \frac{45}{100} (1125)$

$\Rightarrow \, 30\times 1125+30x> 45\times (1125)$ and $ \, 25 (1125+x)< 45(1125)$

$\Rightarrow \, 30x> (45-30)\times (1125)$ and $ \, 25 x< (45-25)1125$

$\Rightarrow \, 30x> (15)\times (1125)$ and $ \, 25 x< (20)1125$

$\Rightarrow \, x> \frac{15\times 1125}{30}$ and $ \, x< \frac{20\times 1125}{25}$

$\Rightarrow \, x> 562.5$ and $\, x< 900$
So, $900>x>562.5$

Thus, the number of litres of water that is to be added will have to be more than 562.5 and less than 900 litres.

Question 14: IQ of a person is given by the formula $IQ= \frac{MA}{CA}\times 100$, where MA is mental age and CA is chronological age. If $80\leq IQ\leq140$ for a group of 12 years old children, find the range of their mental age.

Answer:

Given that the IQ of group of 12 years old children is:

$80\leq IQ\leq140$

For a group of 12 years old children, CA =12 years

$IQ= \frac{MA}{CA}\times 100$

Putting the value of IQ, we obtain

$80\leq IQ\leq140$

$\Rightarrow \, \, 80\leq \frac{MA}{CA}\times 100\leq140$

$\Rightarrow \, \, 80\leq \frac{MA}{12}\times 100\leq140$

$\Rightarrow \, \, 80\times 12\leq MA\times 100\leq140\times 12$

$\Rightarrow \, \, \frac{80\times 12}{100}\leq MA\leq \frac{140\times 12}{100}$

$\Rightarrow \, \, 9.6\leq MA\leq 16.8$

Thus, the range of mental age of the group of 12 years old children is $\, \, 9.6\leq MA\leq 16.8$.