NCERT Solutions for Exercise 6.1 Class 11 Maths Chapter 6 - Linear Inequalities

Access Linear Inequalities Class 11 Chapter 6- Exercise: 6.1

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 $\{1,2,3,4\}$

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

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 $\{.............,-6,-5,-4,-3\}$

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 $\{.........-3,-2-1,0,1,\}$

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

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

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

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 real 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 real 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 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 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 real numbers less than 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 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 real numbers less than 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 real 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 equal 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 of 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 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)$

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

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

$\Rightarrow -5<5x$

$\Rightarrow -1<x$

$x$ are real numbers greater than -1

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

The graphical representation of solutions of 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 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.

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

$145+x\ge 180$

$x\ge 180-145$

$x\ge 35$

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.

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

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

$\Rightarrow 368+x\ge 450$

$\Rightarrow x\ge 450-368$

$\Rightarrow x\ge 82$

Thus, Sunita must obtain 82 in the fifth examination to get 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.

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

$\Rightarrow 7x-2\ge 61$

$\Rightarrow 7x\ge 61+2$

$\Rightarrow 7x\ge 63$

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

$\Rightarrow x\ge 9$

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 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, $x+(x+3)+2x\le 91$

$4x+3\le 91$

$\Rightarrow 4x\le 91-3$

$\Rightarrow 4x\le 88$

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

$\Rightarrow x\le 22$

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

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

$\Rightarrow 2x\ge x+8$

$\Rightarrow 2x-x\ge 8$

$\Rightarrow x\ge 8$

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.