NCERT Solutions for Class 11 Maths Chapter 5 Exercise 5.1 - Linear Inequalities

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

Answer:

Given : $24x < 100$

$\Rightarrow$ $24x < 100$

Divide by 24 from both sides

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

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

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

$x$ i s 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

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

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

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

$x$ i s are integers which are 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 side

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

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

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

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

Hence, the values of x do not exist for 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 side

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

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

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

$x$ 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

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

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

$x$ are integers 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 side

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

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

$x$ 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 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 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 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 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 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 equal 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 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 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 solutions of given inequality is as :

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 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 minimum marks of 35 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 marks obtained in the fifth examination.

To receive 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 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 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 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 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 piece, respectively. Thus, $x + (x + 3) + 2x \leq 91$ and $2x \geq (x + 3) + 5$ ].

Answer:

Let x is the length of the shortest board,

then $(x + 3)$ and $2x$ are the lengths of the second and third piece, 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, than

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

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

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

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

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 equal to 8 cm and less than equal to 22 cm.