NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities

Question:1 Solve $24x<100$, when

(i)$x$ is a natural number.

(ii) $x$ is an integer.

Answer(i):

Given: $24x<100$

$\Rightarrow$ $24x<100$

Divide by 24 on 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 $1,2,3,4$.

Answer(ii):

Given : $24x<100$

$\Rightarrow$ $24x<100$

Divide by 24 on both sides

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

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

$\Rightarrow x<4.167$

$x$ are integers which are less than 4.167.

Hence, values of x can be $..........-3,-2,-1,0,1,2,3,4$.

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

(ii) $x$ is an integer.

Answer(i) :

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 natural number which is less than - 2.5.

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

Answer(ii):

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 - 25.

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

Question:3 Solve $5x-3<7$ , when

(i) $x$ is an integer.

(ii) $x$ is a real number.

Answer(i):

Given : $5x-3<7$

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

$\Rightarrow 5x<10$

Divide by 5 on 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 $\{.........-3,-2-1,0,1,\}$

Answer(ii) :

Given : $5x-3<7$

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

$\Rightarrow 5x<10$

Divide by 5 on both sides

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

$\Rightarrow x<2$

$x$ are real numbers less than 2.

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

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

(ii)$x$ is a real number.

Answer(i) :

Given : $3x+8>2$

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

$\Rightarrow 3x>-6$

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

Answer(ii):

Given : $3x+8>2$

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

$\Rightarrow 3x>-6$

Divide by 3 on 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 $x\in (-2,\mathrm{\infty })$

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 equal numbers greater than -4.

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

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 equal numbers less than -3.

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

Question:7 Solve the inequality for real $x$. $3(x-1)\le 2(x-3)$

Answer:

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

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

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

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

$\Rightarrow x\le -3$

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

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

Question:8 Solve the inequality for real $x$ $3(2-x)\ge 2(1-x)$

Answer:

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

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

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

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

$\Rightarrow 4\ge x$

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

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

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$

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

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

$\Rightarrow 11x<11\times 6$

$\Rightarrow x<6$

$x$ are real numbers less than 6.

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

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 $x\in (-\mathrm{\infty },-6)$

Question:11 Solve the inequality for real $x$ $\frac{3(x-2)}{5}\le \frac{5(2-x)}{3}$

Answer:

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

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

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

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

$\Rightarrow 9x+25x\le 50+18$

$\Rightarrow 34x\le 68$

$\Rightarrow x\le 2$

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

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

Question:12 Solve the inequality for real $x$ $\frac{1}{2}(\frac{3x}{5}+4)\ge \frac{1}{3}(x-6)$

Answer:

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

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

$\Rightarrow 3(\frac{3x}{5}+4)\ge 2(x-6)$

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

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

$\Rightarrow 24\ge \frac{x}{5}$

$\Rightarrow 120\ge x$

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

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

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 $x\in (4,\mathrm{\infty })$

Question:14 Solve the inequality for real $x$ $37-(3x+5)\ge 9x-8(x-3)$

Answer:

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

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

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

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

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

$\Rightarrow 8\ge 4x$

$\Rightarrow 2\ge x$

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

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

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 l numbers greater than 4.

Hence, values of x can be as $x\in (4,\mathrm{\infty })$.

Question:16 Solve the inequality for real $x$ $\frac{(2x-1)}{3}\ge \frac{3x-2}{4}-\frac{(2-x)}{5}$

Answer:

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

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

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

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

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

$\Rightarrow 34\ge 17x$

$\Rightarrow 2\ge x$

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

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

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 $x\in (-\mathrm{\infty },3)$
The graphical representation of solutions to the given inequality is as :

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

Answer:
Given : $5x-3\ge 3x-5$

$\Rightarrow$$5x-3\ge 3x-5$

$\Rightarrow 5x-3x\ge 3-5$

$\Rightarrow 2x\ge -2$

$\Rightarrow x\ge -1$

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

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

The graphical representation of solutions to 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)$

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

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

$\Rightarrow -5<5x$

$\Rightarrow -1<x$

$x$ are real numbers greater than -1.

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

The graphical representation of solutions to given inequality is as :

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

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

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

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

$\Rightarrow 15x\ge 50x-20-42x+18$

$\Rightarrow 15x+42x-50x\ge 18-20$

$\Rightarrow 7x\ge -2$

$\Rightarrow x\ge \frac{-2}{7}$

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

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

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

Question:21 Ravi obtained 70 and 75 marks in the first two unit tests 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.

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

⇒ $145+x\ge 180$

⇒ $x\ge 180-145$

⇒ $x\ge 35$

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

Question 22 To receive a Ga grade of ‘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 the first four examinations are 87, 92, 94 and 95, find the minimum marks that Sunita must obtain in the first examination to get a de ‘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 a Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations.

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

$\Rightarrow \frac{368+x}{5}\ge 90$

$\Rightarrow 368+x\ge 450$

$\Rightarrow x\ge 450-368$

$\Rightarrow x\ge 82$

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

$\therefore x+2<10$

$\Rightarrow x<10-2$

$\Rightarrow x<8$

The 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.

$\therefore x>5$

The 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.

The highest side = 3x cm.

Third side = 3x-2 cm.

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

$\therefore x+3x+(3x-2)\ge 61$

$\Rightarrow 7x-2\ge 61$

$\Rightarrow 7x\ge 61+2$

$\Rightarrow 7x\ge 63$

$\Rightarrow x\ge \frac{63}{7}$

$\Rightarrow x\ge 9$

The 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, $x+(x+3)+2x\le 91$ and $2x\ge (x+3)+5$ ].

Answer:
Let x 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, $x+(x+3)+2x\le 91$

$4x+3\le 91$

$\Rightarrow 4x\le 91-3$

$\Rightarrow 4x\le 88$

$\Rightarrow x\le \frac{88}{4}$

$\Rightarrow x\le 22$

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

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

$\Rightarrow 2x\ge x+8$

$\Rightarrow 2x-x\ge 8$

$\Rightarrow x\ge 8$

We conclude that $x\ge 8$ and $x\le 22$ .

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

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

Question:1 Solve the inequality $2\le 3x-4\le 5$

Answer:
Given : $2\le 3x-4\le 5$

$2\le 3x-4\le 5$

$\Rightarrow 2+4\le 3x\le 5+4$

$\Rightarrow 6\le 3x\le 9$

$\Rightarrow \frac{6}{3}\le x\le \frac{9}{3}$

$\Rightarrow 2\le x\le 3$

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

The solution set is $[2,3]$.

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

Answer:
Given $6\le -3(2x-4)<12$

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

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

$\Rightarrow -2\ge (2x-4)>-4$

$\Rightarrow -2+4\ge 2x>-4+4$

$\Rightarrow 2\ge 2x>0$

$\Rightarrow 1\ge x>0$

The solution set is $(01]$.

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

Answer:
Given $-3\le 4-\frac{7x}{2}\le 18$

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

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

$\Rightarrow -7\le -\frac{7x}{2}\le 14$

$\Rightarrow 7\ge \frac{7x}{2}\ge -14$

$\Rightarrow 7\times 2\ge 7x\ge -14\times 2$

$\Rightarrow 14\ge 7x\ge -28$

$\Rightarrow \frac{14}{7}\ge x\ge \frac{-28}{7}$

$\Rightarrow 2\ge x\ge -4$

Solution set is $[-4,2]$.

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

Answer:
Given The inequality

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

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

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

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

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

$\Rightarrow -25<(x-2)\le 0$

$\Rightarrow -25+2<x\le 0+2$

$\Rightarrow -23<x\le 2$
The solution set is $(-23,2]$.

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

Answer:
Given the inequality
$-12<4-\frac{3x}{-5}\le 2$

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

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

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

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

$\Rightarrow -16\times 5<3x\le -2\times 5$

$\Rightarrow -80<3x\le -10$

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

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

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

Answer:
Given the linear inequality

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

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

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

$\Rightarrow 14\le (3x+11)\le 22$

$\Rightarrow 14-11\le (3x)\le 22-11$

$\Rightarrow 3\le 3x\le 11$

$\Rightarrow 1\le x\le \frac{11}{3}$

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

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

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

$5x+1>-24and5x-1<24$

$\Rightarrow 5x>-24-1and5x<24+1$

$\Rightarrow 5x>-25and5x<25$

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

$\Rightarrow x>-5andx<5$

The solution graphically on the number line is as shown :

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

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

$2(x-1)<x+5and3(x+2)>2-x$

$\Rightarrow 2x-2<x+5and3x+6>2-x$

$\Rightarrow 2x-x<2+5and3x+x>2-6$

$\Rightarrow x<7and4x>-4$

$\Rightarrow x<7andx>-1$

$(-1,7)$

The solution graphically on the number line is as shown :

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

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

$3x-7>2(x-6)and6-x>11-2x$

$\Rightarrow 3x-7>2x-12and6-x>11-2x$

$\Rightarrow 3x-2x>7-12and2x-x>11-6$

$\Rightarrow x>-5andx>5$

$x\in (5,\mathrm{\infty })$

The solution graphically on the number line is as shown :

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

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

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

$5(2x-7)-3(2x+3)\le 0and2x+19\le 6x+47$

$\Rightarrow 10x-35-6x-9\le 0and2x-6x\le 47-19$

$\Rightarrow 4x-44\le 0and-4x\le 28$

$\Rightarrow 4x\le 44and4x\ge -28$

$\Rightarrow x\le 11andx\ge -7$

$\Rightarrow x\in [-7,11]$

The solution graphically on the number line is as shown :