# NCERT Solutions For Class 11 Maths Chapter 6

## NCERT Solutions Class 11 Maths Linear Inequalities

NCERT solutions for class 11 maths chapter 6 linear inequalities includes all the solved questions provided in class 11 maths NCERT textbooks. Students can practice these questions for their board and other competitive exams. NCERT solutions for class 11 maths chapter 6 are prepared by the team of subject experts to help students in their preparations.

Those students who are finding difficulties in solving linear inequalities problems can download NCERT solutions for class 11 maths chapter 6 pdf files for free and practice more questions on this topic. Apart from the textbook questions, this study material also covers all the important questions from exam point of view.

### NCERT Solutions Class 11 Maths Chapter 6 Exercises

Exercise 6.1

Question 1

Solve 26x<200$26x < 200$ when:

(i) The variable(x) is a natural number

(ii) The variable(x) is an integer.

Sol:

Given, 26x<200$26x< 200$

Dividing both sides of the equation by the same non – negative number, we get:

= 26x26<20026$\frac{26x}{26}< \frac{200}{26}$

= x<10013$x < \frac{100}{13}$

(i). Numbers 1,2,3,4,5,6,7 are the natural numbers smaller than the given fraction 26x<200$26x< 200$.

The solution of 26x<200$26x< 200$ in the form of solution set  {1,2,3,4,5,6,}

(ii). Integers smaller than 10013$\frac{100}{13}$ in the form of solution set is {-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7}

Question 2

Solve 40x<6$-40x < 6$ when:

(i) The variable(x) is a natural number

(ii) The variable(x) is an integer.

Sol:

Given, 40x<6$-40x < 6$

Dividing both sides of the equation by the negative number results interchanging the inequality symbol, we get:

= 40x6<66$\frac{-40x}{-6} < \frac{6}{-6}$

= x<640$x< \frac{-6}{40}$

= x<320$x< \frac{-3}{20}$

(i) There is no natural number less than 320$\frac{-3}{20}$

The integers smaller than 320$\frac{-3}{20}$ are -1, -2, -3………

Solution set :{ -1, -2, -3……… }

Question 3

Solve the following inequality:

5x3<7$5x-3 < 7$, when:

(i) The variable(x) is an integer.

(ii) The variable(x) is a real number.

Sol:

Given, 5x3<7$5x-3 < 7$

= 5x3+3<7+3$5x-3+3 < 7+3$

= 5x<10$5x< 10$

Dividing both sides of the equation by the non-negative number. We get:

= 5x5<105$\frac{5x}{5} < \frac{10}{5}$

= x<2$x < 2$

(i) lntegers smaller than 2 are 0,1,-1,-2,-3,-4…………

Solution set {0, 1,-1,-2,-3,-4,…..}

(ii) Considering the variable(x) is a real number, the solution set of the given inequality is  x,2$x\in {-\infty ,2}$

Question 4

Solve the following inequality:

3x+8 > 2, when:

(i) The variable(x) is an integer.

(ii) The variable(x) is a real number.

Sol:

Given, 3x+8˃2$3x+8 ˃ 2$

= 3x+88˃28$3x+8-8 ˃ 2-8$

= 3x˃6$3x ˃ -6$

Dividing both sides of the equation by the non-negative number. We get:

= 3x3˃63$\frac{3x}{3} ˃ \frac{-6}{3}$

= x˃2$x ˃ -2$

(i) lntegers smaller than -2 are 0,1,2,……..-1

Solution set {0, 1,2,……….-1,-3,-4}

(ii) Considering the variable(x) is a real number, the solution set of the given inequality is x2,$x\in {-2,\infty }$

Question 5

Solve: 4x+3 <  5x+7

= 4x+37<5x+77$4x+3-7< 5x+7-7$

= 4x4<5x$4x-4 < 5x$

= 4x44x<5x4x$4x-4-4x < 5x-4x$

= x>4$x > -4$

Therefore, x is greater than -4, is she solution of the given problem.

Solution set: { 4,$-4,\infty$}

Question 6

Solve: 3x-7 > 5x-1

Sol:

= 3x7+7˃5x1+7$3x-7+7 ˃ 5x-1+7$

= 3x˃5x+6$3x ˃ 5x+6$

= 3x5x˃5x+65x$3x-5x ˃ 5x+6-5x$

= 2x>6$-2x > 6$

= 2x2˂62$\frac{-2x}{-2} ˂ \frac{6}{-2}$

= x˂3$x ˂ -3$

Therefore, real number(x) is less than -3, is the solution of the given problem.

Solution set: { ,3$-\infty, -3$}

Question 7

Solve for x: 3(x1)2(x3)$3(x-1)\leq 2(x-3)$

Sol:

= 3(x1)2(x3)$3(x-1)\leq 2(x-3)$

= 3x32x6$3x-3\leq 2x-6$

= 3x3+32x6+3$3x-3+3\leq 2x-6+3$

= 3x2x3$3x\leq 2x-3$

= 3x2x2x32x$3x-2x\leq 2x-3-2x$

= x3$x\leq -3$

Therefore, all real number(x) less than or equal to -3, are the solution to the given problem.

Solution set: { ,3$-\infty,-3$}

Question 8

Solve for x: 3(2x)2(1x)$3(2-x)\geq 2(1-x)$

Sol:

= 63x22x$6-3x\geq 2-2x$

= 63x+2x22x+2x$6-3x+2x\geq 2-2x+2x$

= 6x2$6-x\geq 2$

= 6x626$6-x-6\geq 2-6$

= x4$-x\geq -4$

= x4$x\leq 4$

Therefore, all real number(x) less than or equal to 4, are the solution to the given problem.

Solution set: { ,4$-\infty,4$}

Question 9

Solve for x: x+x2+x3<11$x+\frac{x}{2}+\frac{x}{3}< 11$

Sol:

= x+x2+x3<11$x+\frac{x}{2}+\frac{x}{3}< 11$

= x(1+12+13)<11$x(1+\frac{1}{2}+\frac{1}{3})< 11$

= x(6+3+26)<11$x(\frac{6+3+2}{6})< 11$

= x(116)<11$x(\frac{11}{6})< 11$

Dividing by a non-negative number, that is, 11 on the both sides of the inequality, we get:

= x(116(11))<1111$x(\frac{11}{6(11)})< \frac{11}{11}$

= x(16)<1$x(\frac{1}{6})< 1$

= x<6$x< 6$

Therefore, all real number(x) less than 6, are the solution to the given problem.

Solution set: { ,6$-\infty,6$}

Question 10

Solve for x: x3>x2+1$\frac{x}{3}> \frac{x}{2}+1$

Sol:

= x3>x2+1$\frac{x}{3}> \frac{x}{2}+1$

= x3x2>+1$\frac{x}{3}-\frac{x}{2}> +1$

= 2x3x6>1$\frac{2x-3x}{6}> 1$

= x6>1$\frac{-x}{6}> 1$

= x>6$-x> 6$

= x<6$x< -6$

Therefore, all real number(x) less than -6, are the solution to the given problem.

Solution set: { ,6$-\infty,-6$}

Question 11

Solve for x: 3(x2)55(2x)3$\frac{3(x-2)}{5}\leq \frac{5(2-x)}{3}$

Sol:

= 3(x2)55(2x)3$\frac{3(x-2)}{5}\leq \frac{5(2-x)}{3}$

= 9(x2)25(2x)$9(x-2)\leq 25(2-x)$

= 9x185025x$9x-18\leq 50-25x$

= 9x18+25x5025x+25x$9x-18+25x\leq 50-25x+25x$

= 34x1850$34x-18\leq 50$

= 34x50+18$34x\leq 50+18$

= 34x68$34x\leq 68$

Dividing by a non-negative number that is 34 on the both sides of the inequality, we get:

= 34x346834$\frac{34x}{34}\leq \frac{68}{34}$

= x2$x\leq 2$

Therefore, all real number(x) less than 2, are the solution to the given problem.

Solution set: { ,2$-\infty,2$}

Question 12

Solve for x : 12(3x5+4)13(x6)$\frac{1}{2}(\frac{3x}{5}+4)\geq \frac{1}{3}(x-6)$

Sol:

= 12(3x5+4)13(x6)$\frac{1}{2}(\frac{3x}{5}+4)\geq \frac{1}{3}(x-6)$

= 3(3x5+4)2(x6)$3(\frac{3x}{5}+4)\geq 2(x-6)$

= 9x5+122x12$\frac{9x}{5}+12\geq 2x-12$

= 12+122x95$12+12\geq 2x-\frac{9}{5}$

= 242x9x5$24\geq 2x-\frac{9x}{5}$

= 2410x9x5$24\geq \frac{10x-9x}{5}$

= 24x5$24\geq \frac{x}{5}$

= 120x$120\geq x$

Therefore, all real number(x) less than or equals to 120, are the solution to the given problem.

Solution set: {,120$-\infty,120$}

Question 13

Solve the inequality for x: 2(2x+3)10<6(x2))$2(2x+3)-10< 6(x-2))$

Sol:

= 2(2x+3)10<6(x2))$2(2x+3)-10< 6(x-2))$

= 4x+610<6x12$4x+6-10< 6x-12$

= 4x4<6x12$4x-4< 6x-12$

= 4+12<6x4x$-4+12< 6x-4x$

= 8<2x$8< 2x$

= 4<x$4< x$

Therefore, all real number(x) greater than or equal to 4, are the solution to the given problem.

Solution set: {4,$4,\infty$}

Question 14

Solve for x: 37(3x+5)9x8(x3)$37-(3x+5)\geq 9x-8(x-3)$

Sol:

= 37(3x+5)9x8(x3)$37-(3x+5)\geq 9x-8(x-3)$

= 373x59x8x+3$37-3x-5\geq 9x-8x+3$

= 323xx+24$32-3x\geq x+24$

= 3224x+3x$32-24\geq x+3x$

= 84x$8\geq 4x$

= 2x$2\geq x$

Therefore, all real number(x)  less than or equal to 2, are the solution to the given problem.

Solution set: {,2$-\infty,2$}

Question 15

Solve for x: x4<5x237x35$\frac{x}{4}< \frac{5x-2}{3}-\frac{7x-3}{5}$

Sol:

= x4<5x237x35$\frac{x}{4} < \frac{5x-2}{3}-\frac{7x-3}{5}$

= x4<5(5x2)3(7x2)15$\frac{x}{4} < \frac{5(5x-2)-3(7x-2)}{15}$

= x4<25x1021x+915$\frac{x}{4} < \frac{25x-10-21x+9}{15}$

= x4<4x115$\frac{x}{4} < \frac{4x-1}{15}$

= x4<4(4x1)$\frac{x}{4} < 4(4x-1)$

= 15x<4(4x1)$15x < 4(4x-1)$

= 15x<16x4$15x < 16x-4$

= 4<16x15x$4 < 16x-15x$

= 4<x$4 < x$

Therefore, all real number(x) greater than or equal to 4, are the solution to the given problem.

Solution set: {4,$4,\infty$}

Question 16

Solve for x: 2x133x242x5$\frac{2x-1}{3}\geq \frac{3x-2}{4}-\frac{2-x}{5}$

Sol:

= 2x133x242x5$\frac{2x-1}{3}\geq \frac{3x-2}{4}-\frac{2-x}{5}$

= 2x135(3x2)4(2x)20$\frac{2x-1}{3}\geq \frac{5(3x-2)-4(2-x)}{20}$

= 2x1315x108+4x20$\frac{2x-1}{3}\geq \frac{15x-10-8+4x}{20}$

= 2x1319x1820$\frac{2x-1}{3}\geq \frac{19x-18}{20}$

= 20(2x1)3(19x18)$20(2x-1) \geq 3(19x-18)$

= 40x2057x54$40x-20 \geq 57x-54$

= 20+5457x40x$-20+54 \geq 57x-40x$

= 3417x$34 \geq 17x$

= 2x$2\geq x$

Therefore, all real number(x)  less than or equal to 2, are the solution to the given problem.

Solution set: {,2$-\infty,2$}

Question 17

Solve the inequality and draw the solution on the number line: 3x2<2x+1$3x-2< 2x+1$

Sol:

= 3x2<2x+1$3x-2 < 2x+1$

= 3x2x<1+2$3x-2x < 1+2$

= x<3$x < 3$

Graphical representation on the number line is:

Question 18

Solve the inequality and draw the solution on the number line: 5x33x5$5x-3\geq 3x-5$

Sol:

= 5x33x5$5x-3\geq 3x-5$

= 5x3x5+3$5x-3x\geq -5+3$

= 2x2$2x\geq -2$

Dividing by a non-negative number, that is, 2 on the both sides of the inequality, we get:

= 2x222$\frac{2x}{2}\leq \frac{-2}{2}$

= x1$x\leq -1$

Graphical representation on the number line is:

Question 19

Solve the inequality and draw the solution on the number line: 3(1x)<2(x+4)$3(1-x)< 2(x+4)$

Sol:

= 3(1x)<2(x+4)$3(1-x)< 2(x+4)$

= 33x<2x+8$3-3x< 2x+8$

Taking the variables in one side of the equation. We get:

= 38<2x+3x$3-8< 2x+3x$

= 5<5x$-5< 5x$

Dividing by a non negative number, that is, 5 on the both sides of the inequality, we get:

= 55<5x5$\frac{-5}{5} < \frac{5x}{5}$

= 1<x$-1 < x$

Graphical representation on the number line is:

Question 20

Solve the inequality and draw the solution on the number line: x25x237x35$\frac{x}{2}\geq \frac{5x-2}{3}-\frac{7x-3}{5}$

Sol:

= x25x237x35$\frac{x}{2}\geq \frac{5x-2}{3}-\frac{7x-3}{5}$

= x25(5x2)3(7x3)15$\frac{x}{2}\geq \frac{5(5x-2)-3(7x-3)}{15}$

= x225x1021x+915$\frac{x}{2}\geq \frac{25x-10-21x+9}{15}$

= x24x115$\frac{x}{2}\geq \frac{4x-1}{15}$

= 15x2(4x1)$15x\geq 2(4x-1)$

= 15x8x2$15x\geq 8x-2$

= 15x8x8x28x$15x-8x\geq 8x-2-8x$

= 7x2$7x\geq -2$

= x27$x\geq \frac{-2}{7}$

Question 21

Marks scored by Ravi in his first two unit test are 70 and 75 respectively. In order to score an average of 60 marks in his three unit tests what is the minimum mark he should obtain in this third unit test?

Sol:

Let y be the mark scored by Ravi in the third unit test.

According to the question, Ravi has to maintain an average of 60 marks in his three unit tests.

= 70+75+y360$\frac{70 + 75 + y}{3}\geq 60$

= 145+y180$145 + y\geq 180$

= y180145$y \geq 180 -145$

= y35$y \geq 35$

In order to have an average of 60 marks in his three unit tests, Ravi must obtain a minimum of 35 marks.

Question 22

In order to obtain Grade A in the course, one should maintain an average of 90 marks or more in a series of five examinations each of 100 marks. Marks obtained by Sunita in her first four examinations are 87, 92, 94 and 95 respectively. Find the minimum marks Sunita should obtain in her fifth examination in order to get a Grade A in the course.

Sol:

Let y be the mark obtained by Sunita in the fifth examination.

She must get an average of 90 marks in her five exams in order to obtain grade ‘A’ in the course.

According to the question,

= 87+92+94+95+y590$\frac{87 + 92 + 94 + 95 + y}{5}\geq 90$

= 368+y590$\frac{368 + y}{5}\geq 90$

= 368+y450$368 + y\geq 450$

= y450368$y \geq 450 – 368$

= y82$y \geq 82$

In order to get an average of 90 marks in five examinations, Sunita must score which is equal to or more than 82 marks in her fifth examination.

Question 23

Find all sets of consecutive odd positive integers both of which are less than 10 but the sum of the numbers in the sets is more than 11.

Sol:

Let y be the smaller of the two consecutive odd positive integers. Then the integer be y+2

= y+2 ˂ 10

= y ˂ 10-2

= y ˂ 8 ………………………..(i)

According to the question, the sum of the two integers in the set is more than 11.

Therefore,

= y + (y+2) ˃ 11

= 2x + 2 ˃ 11

= 2y ˃ 9

= y = 92$\frac{9}{2}$

= y = 4.5   …………………………..(ii)

From equation (i) and (ii) , we get:

Since y is an odd number, y can take values such as 5 and 7

Thus, the possible sets are (5,7) and (7,9)

Question 24

Obtain  all the sets of consecutive positive integers, both of which are larger than 5 such that their sum is lesser than 23.

Sol:

Let y be the smaller of the two consecutive even positive integers.

Therefore, the next integer in the set is y+2

According to the question, each of the integer in the set should be larger than 5

Therefore, y ˃ 5 …………………….(i)

Also, the sum of the integer in the set should be smaller than 23.

= y + (y+2) ˂ 23

= 2y + 2 ˂ 23

= 2y ˂ 21

= x = 212$\frac{21}{2}$

= y = 10.5 ………………………… (ii)

From equation (i) and (ii), we get:

= 5 ˂ y ˂ 10.5

According to the question, y should be an even number. So, it can take values such as 6,8,10

Hence the obtained sets are (6, 8), (8, 10) and (10, 12) respectively

Question 25

It is noticed that the longest side of the triangle is three times the shortest side also the third side of the triangle is 2 centimeter shorter than the longest side. If it is found that the perimeter of the triangle is at least 61 centimeter, find out the minimum length of the shortest side.

Sol:

Let ‘y’ cm be the length of the shortest side of the triangle.

Therefore, length of the longest side = 3y cm

Length of the third side = (3y-2) cm

According to the question, perimeter of the triangle is at least 61 cm

Therefore,

= y + 3y + (3y-2) ≥ 61

= 7x – 2 ≥ 61 + 2

= 7x ≥ 63

Dividing by a non-negative integer to the both the sides of the inequality,

= 7x7$\frac{7x}{7}$637$\frac{63}{7}$

= x ≥ 9

Therefore, the minimum length of the shortest side of the triangle is 9 cm.

Question 26

A woman wants to cut 3 lengths from a single piece of cloth of length 91 cm. The 2nd piece of cloth is found that it is 3 cm longer than the shortest and the third piece of cloth is twice as long as the third piece of cloth. Find the possible lengths of 3 pieces if the 3rd piece of cloth is at least 5 cm longer than the 2nd piece.

Sol:

Let ‘y’ be the piece of the shortest piece of cloth.

Therefore, the lengths of the 2nd and 3rd piece of cloth are (y+3) cm and 2y cm respectively.

All the 3 pieces of cloth are cut from a single piece of cloth of length 91 cm

Therefore,

= y + (y+3) + 2y ≤ 91

= 4y ≤ 91 -3

= 4y ≤ 88

Dividing by a non-negative integer to the both the sides of the inequality, we get:

= 4y4$\frac{4y}{4}$884$\frac{88}{4}$

= x ≤ 22 ………………………..(i)

3rd piece of cloth is 5 cm longer than the 2nd piece.

Therefore,

= 2y ≥ (y+3) +5

= 2y ≥ y + 8

= y ≥ 8

From equation (i) and (ii), we get:

= 8 ≤ y ≤ 22

Hence, the possible length of the shortest piece of cloth is found to be greater than 8 cm but less than or equals to 22 cm.

Exercise 6.2

Question 1

Solve the inequality x + y 5 graphically in two dimensional plane.

Sol:

The given inequality x + y ≤ 5 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts, that is I and II, respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

= 0 + 0 ˂ 5

= 0 ˂ 5

This is true.

Therefore, plane II is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane I.

Graphical representation:

Question 2

Solve the inequality 2x + y 5 graphically in two dimensional plane.

Sol:

The given inequality 2x + y ≥ 5 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts that, is I and II, respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

= 2(0) + 0 ≥ 6

= 0 ≥ 6

This is false.

Therefore, plane I is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane II.

Graphical representation:

Question 3

Solve the inequality 3x+4y 12 graphically in two dimensional plane.

Sol:

The given inequality 3x + 4y ≤ 12 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts, that is, I and II respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

= 3(0) + 4(0) ≤ 12

= 0 ≤ 12

This is true.

Therefore, plane II is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane I.

Graphical representation:

Question 4

Solve the inequality y+8 2x graphically in two dimensional plane.

Sol:

The given inequality y+8 ≥ 2x is graphically represented with the help of the blue line which divides the Cartesian plane into two parts that is, I and II, respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

= (0) + (8) ≥ 2(0)

= 8 ≥ 0

This is true.

Therefore, plane II is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane I.

Graphical representation:

Question 5

Solve the inequality x-y 2 graphically in two dimensional plane.

Sol:

The given inequality x-y ≤ 2 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts that is, I and II, respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

= 0 – 0 ≤ 2

= 0 ≤ 2

This is true.

Therefore, lower half of the plane is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane I.

Graphical representation:

Question 6

Solve the inequality 2x – 3y ˃ 6 graphically in two dimensional plane.

Sol:

The given inequality 2x – 3y ˃ 6 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts, that is, I and II respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

=2(0) – 3(0) ≥ 6

= 0 ≥ 6

This is false.

Therefore, the upper part of the plane is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane II.

Graphical representation:

Question 7

Solve the inequality -3x + 2y -6 graphically in two dimensional plane.

Sol:

The given inequality -3x + 2y ≥ -6 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts that is, I and II, respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

=-3(0) + 2(0) ≥ -6

= 0 ≥ -6

This is true.

Therefore, the lower part of the plane is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane II.

Graphical representation:

Question 8

Solve the inequality 3y – 5x ˂ 30 graphically in two dimensional plane.

Sol:

The given inequality 3y – 5x ˂ 30 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts that is, I and II, respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

=3(0) – 5(0) ˂ 30

= 0 ˂ 30

This is true.

Therefore, the upper part of the plane is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane II.

Graphical representation:

Question 9

Solve the inequality y ˂ -2 graphically in two dimensional plane.

Sol:

The given inequality y ˂ -2 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts that is, I and II, respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

= (0) ˂ -2

This is false.

Therefore, the upper part of the plane is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane II.

Graphical representation:

Question 10

Solve the inequality x ˃ -3 graphically in two dimensional plane.

Sol:

The given inequality x ˃ -3 is graphically represented with the help of the blue line which divides the Cartesian plane into two parts that is, I and II, respectively.

Select a point a in any one of the parts but not on the line to determine whether the point satisfies the given inequality or not.

Let us consider the point be O (0, 0)

Putting the values of x and y in the given inequality. We have:

= (0) ˂ -3

= 0 ˂ -3

This is true

Therefore, the upper part of the plane is not the graphical solution of the given inequality. Also, if we take any point on the line it will not satisfy the given inequality.

Thus the feasible solution of the given inequality is the shaded portion of the in the Cartesian plane excluding the points on the line and that is plane II.

Graphical representation:

Exercise 6.3

Question 1

Solve the system of inequalities graphically:

= x 3 and y 2

Sol:

= x ≥ 3 ……………………(i)

= y ≥ 2……………………..(ii)

In order to draw the graph of the following inequalities, let us consider x = 3 and y = 2

In case of equation (i) the feasible region is the right hand side of the line including the points on the line x = 3 , whereas in case of equation (ii) the feasible region is above the line including the points on the line y = 2.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 2

Solve graphically the system of inequalities:

3x + 2y 12, x 1 and y 2

Sol:

3x + 2y ≤ 12 …………… (i)

x ≥ 1 ……………………… (ii)

y ≥ 2 ……………………….(iii)

In order to draw the graph of the following inequalities, let us consider 3x + 2y = 12, x = 1 and y = 2

In case of equation (i) the feasible region is below the line including the points on the line 3x + 2y = 12. In case equation (ii) the feasible region is on the right hand side including the points on the line x = 1 and in case of equation (iii) the feasible region is above the line including the points on the line y = 2.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 3

Solve graphically the system of inequalities:

2x + y 6 and 3x + 4y 12.

Sol:

2x + y ≥ 6 …………………………. (i)

3x + 4y ≤ 12……………………… (ii)

In order to draw the graph of the following inequalities, let us consider 2x + y = 6 and 3x + 4y = 12.

In case of equation (i) the feasible region is above the line including the points on the line 2x + y  = 6 , whereas in case of equation (ii) the feasible region is below the line including the points on the line 3x + 4y = 12

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 4

Solve graphically the system of inequalities:

x + y 4 and 2x – y ˃ 0

Sol:

x + y ≥ 4 …………………………. (i)

2x – y ˃ 0……………………… (ii)

In order to draw the graph of the following inequalities, let us consider x + y = 4 and 2x – y = 0.

In case of equation (i) the feasible region is above the line including the points on the line x + y  = 4, whereas in case of equation (ii) the feasible region is the right hand side of  the line including the points on the line 2x – y = 0.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 5

Solve graphically the system of inequalities:

2x – y ˃ 1 and x – 2y ˂ -1

Sol:

2x – y ˃ 1 …………………………. (i)

x – 2y ˂ -1……………………… (ii)

In order to draw the graph of the following inequalities, let us consider 2x – y = 1 and x – 2y = -1.

In case of equation (i) the feasible region is below the line including the points on the line 2x – y = 1, whereas in case of equation (ii) the feasible region is the above   the line including the points on the line x – 2y = -1.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 6

Solve graphically the system of inequalities:

x + y 6 and x + y 4

Sol:

x + y ≤ 6 …………………………. (i)

x + y ≥ 4……………………… (ii)

In order to draw the graph of the following inequalities, let us consider x + y = 6 and x + y = 4.

In case of equation (i) the feasible region is below the line including the points on the line x + y = 6, whereas in case of equation (ii) the feasible region is the above   the line including the points on the line x + y = 4.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 7

Solve graphically the system of inequalities:

2x + y 8 and x + 2y 10

Sol:

2x + y ≥ 8 …………………………. (i)

x + 2y ≥ 10……………………… (ii)

In order to draw the graph of the following inequalities, let us consider 2x + y = 8 and x + 2y = 10.

In case of equation (i) the feasible region is above the line including the points on the line 2x + y = 8 , whereas in case of equation (ii) the feasible region is the above   the line including the points on the line x + 2y = 10.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 8

Solve graphically the system of inequalities:

x + y 9, y ˃ x and x 0

Sol:

x + y ≤ 9 …………………………. (i)

y ˃ x ……….……………………… (ii)

x ≥ 0 ………………………………… (iii)

In order to draw the graph of the following inequalities, let us consider x + y = 9, y = x and x = 0.

In case of equation (i) the feasible region is below the line including the points on the line x + y = 9. In case of equation (ii) the feasible region is above the line including the points on the line y =x  , whereas in case of equation (iii) the feasible region is the right hand side of the line  including the points on the line x = 0

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 9

Solve graphically the system of inequalities:

5x + 4y 20, x 1 and y 2

Sol:

5x + 4y ≤ 20 …………………………. (i)

x ≥ 1 ……….……………………… (ii)

y ≥ 2 ………………………………(iv)

In order to draw the graph of the following inequalities, let us consider 5x + 4y = 20, x = 1, x = 0 and y = 2

In case of equation (i) the feasible region is below the line including the points on the line 5x + 4y = 20. In case of equation (ii) the feasible region is on the right hand side of the line including the points on the line x = 1, where as in equation (iv) the feasible region is above line including the points on the line y = 2

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 10

Solve graphically the system of inequalities:

3x + 4y 60, x + 3y 30 and x 0

Sol:

3x + 4y ≤ 60 …………………………. (i)

x + 3y ≤ 30          ……………………… (ii)

x ≥ 0 …………………………………….(iii)

In order to draw the graph of the following inequalities, let us consider 3x + 4y = 60, x + 3y = 30 and x = 0

In case of equation (i) the feasible region is below the line including the points on the line 3x + 4y = 60. In case of equation (ii) the feasible region is the below the line including the points on the line x + 3y = 30, whereas in case of equation (iii) the feasible region is right and side of the line.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 11

Solve graphically the system of inequalities:

2x + y 4, x + y 3 and 2x -3y 6

Sol:

2x + y ≥ 4 …………………………. (i)

x + y ≤ 3 ……….……………………… (ii)

2x – 3y ≤ 6 ………………………………… (iii)

In order to draw the graph of the following inequalities, let us consider 2x +y = 4, x + y = 3 and 2x – 3y = 6.

In case of equation (i) the feasible region is above the line including the points on the line 2x + y = 4. In case of equation (ii) the feasible region is below the line including the points on the line x + y = 3, whereas in case of equation (iii) the feasible region is above line including the points on the line 2x- 3y = 6

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 12

Solve graphically the system of inequalities:

5x + 4y 3, 3x + 4y 12, x 0 and y 1

Sol:

x – 2y ≤ 3 …………………………. (i)

3x + 4y ≥ 12 ……….……………………… (ii)

x ≥ 0 ………………………………… (iii)

y ≥ 1 ………………………………(iv)

In order to draw the graph of the following inequalities, let us consider x – 2y = 3, 3x + 4y = 12, x = 0 and y = 1

In case of equation (i) the feasible region is above the line including the points on the line x – 2y = 3. In case of equation (ii) the feasible region is above the line including the points on the line 3x + 4y = 12, whereas in case of equation (iii) the feasible region is in the right hand side of the axis including the points on the line x = 0 and (iv) the feasible region is above line including the points on the line y = 1

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 13

Solve graphically the system of inequalities:

4x + 3y 60, y 2x, x 3 and x, y 0

Sol:

4x + 3y ≤ 60 …………………………. (i)

y ≥ 2x ……….……………………… (ii)

x ≥ 3 ………………………………… (iii)

x ≥ 0 and y ≥ 0 …………………(iv)

In order to draw the graph of the following inequalities, let us consider 4x + 3y = 60, y = 2x, x = 3 and x, y = 0

In case of equation (i) the feasible region is below the line including the points on the line 4x + 3y = 60. In case of equation (ii) the feasible region is above the line including the points on the line y = 2x, whereas in case of equation (iii) the feasible region is in the right hand side of the axis including the points on the line x = 3 and (iv) the feasible region is above the line and right hand side of the line including the points on the line y = 0 and x =0 respectively.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 14

Solve graphically the system of inequalities:

3x + 2y 150, x + 4y 80, x 15

Sol:

3x + 2y ≤ 150 …………………………. (i)

x + 4y ≤ 80 ……….……………………… (ii)

x ≤ 15 ………………………………… (iii)

In order to draw the graph of the following inequalities, let us consider 3x + 2y = 150, x + 4y = 80, x = 15.

In case of equation (i) the feasible region is below the line including the points on the line 3x + 2y = 150. In case of equation (ii) the feasible region is below the line including the points on the line x + 4y = 80 , whereas in case of equation (iii) the feasible region is in the left hand side of the axis including the points on the line x = 15.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Question 15

Solve graphically the system of inequalities:

x + 2y 10, x + y 1, x – y  0

Sol:

x + 2y ≤ 10 …………………………. (i)

x + y ≥ 1 ……….……………………… (ii)

x – y ≤ 0 ………………………………… (iii)

In order to draw the graph of the following inequalities, let us consider x + 2y = 10, x + 4y = 1, x – y = 0.

In case of equation (i) the feasible region is below the line including the points on the line x + 2y = 10. In case of equation (ii) the feasible region is above the line including the points on the line x + y = 1, whereas in case of equation (iii) the feasible region is above the line including the points on the line x – y =0.

Therefore, the solution of the given set of linear inequalities is the common shaded area including the points on the line as shown in the graph below.

Miscellaneous Exercise

Question 1

Solve 2 3x-4 5

Sol:

= 2 ≤ 3x-4 ≤ 5

= 2 + 4≤ 3x-4+4 ≤ 5+4

= 2 ≤ x ≤ 3

Real numbers which are greater than or equal to 2 but less than or equal to 3 is the solution of the above mentioned inequality. Hence the solution set is [2, 3].

Question 2

Solve 6 -3(2x-4) 12

Sol:

= 6 ≤ -3(2x-4) ˂ 12

= 2 ≤ -(2x-4) ˂ 4

= -2 ≥ 2x-4 ˃ -4

= 4-2 ≥ 2x ˃ 4-4

=2 ≥ 2x ˃ 0

= 1 ≥ x ˃ 0

The Solution set of the above mentioned inequality is [0, 1].

Question 3

Solve 347x218$-3\leq 4-\frac{7x}{2}\leq 18$

Sol:

= 347x218$-3\leq 4-\frac{7x}{2}\leq 18$

= 347x2184$-3-4\leq -\frac{7x}{2}\leq 18-4$

= 77x214$-7\leq -\frac{7x}{2}\leq 14$

= 77x214$7\leq \frac{7x}{2}\leq -14$

= 1x22$1\leq \frac{x}{2}\leq -2$

= 2 ≥ x ≥ -4

Thus, the solution set for the above mentioned inequality is: [-4, 2]

Question 4

Solve: 15<43(x2)518$-15< 4-\frac{3(x-2)}{5}\leq 18$

Sol:

= 15<43(x2)518$-15< 4-\frac{3(x-2)}{5}\leq 18$

= -75 ˂ 3(x-2) ≤ 0

= -25 ˂ (x-2) ≤ 0

= -25+2 ˂ 2 ≤ 2

= -23 ˂ x ≤ 2

Thus, the solution set for the above mentioned inequality is [-23, 2].

Question 5

Solve: 12<43x52$-12< 4-\frac{3x}{-5}\leq 2$

Sol:

= 12<43x52$-12< 4-\frac{3x}{-5}\leq 2$

= 124<3x524$-12-4< \frac{-3x}{-5}\leq 2-4$

= 16<3x52$-16< \frac{3x}{5}\leq -2$

= -80 ˂ 3x ≤ -10

= 803<x103$\frac{-80}{3}< x \leq \frac{-10}{3}$

Thus, solution set for the above mentioned inequality is: 803,103$\frac{-80}{3} , \frac{-10}{3}$

Question 6

Solve: 7<3x+11211$7< \frac{3x+11}{2}\leq 11$

Sol:

= 1243x52$-12\leq 4-\frac{3x}{-5}\leq 2$

= 14 ≤ 3x +11 ≤ 22

= 14 – 11 ≤ 3x ≤ 22 – 11

= 1x113$1\leq x\leq \frac{11}{3}$

Thus, solution set for the above mentioned inequality is: 1,113$1 , \frac{11}{3}$

Question 7

Solve the inequalities on the number line:

5 x+1 ˃ -24    = 5x – 1 ˂ 24

Sol:

= 5x + 1 ˃ -24

= 5x – 1 ˂ 24

= x ˃ -5 …………….(i)

= 5x – 1 ˂ 24

= 5x ˂ 25

= x ˂ 5 …………….(ii)

From (i) and (ii) it is observed that the solution set of the above mentioned inequality is (-5, 5).

The number is as follows:

Question 8

Solve the inequalities on the number line:

2(x-1) ˂ x + 5    = 3(x+2) ˃ 2-x

Sol:

= 2(x-1) ˂ x + 5

= 2x – 2 ˂ x +5

= x ˂7 …………….(i)

= 3(x+2) ˃ 2-x

= 3x + 6 ˃ 2 – x

= 4x ˃ -4

= x ˃ -1 …………….(ii)

From (i) and (ii) it is observed that the solution set of the above mentioned inequality is (-1, 7).

The number is as follows:

Question 9

Solve the inequalities on the number line:

3x – 7 ˃ 2(x-6), 6 – x ˃ 11 – 2x

Sol:

= 3x – 7 ˃ 2(x – 6)

= 3x – 7 ˃ 2x – 12

= 3x – 2x ˃ -12 + 7

= x ˃ -5 ……………………(i)

= 6 – x ˃ 11 – 2x

= -x + 2x ˃ 11 – 6

= x ˃ 5 …………………… (ii)

From (i) and (ii) it is observed that the solution set of the above mentioned inequality is (5,$5,\infty$).

The number is as follows:

Question 10

Solve the inequalities on the number line:

5(2x – 7) – 3(2x + 3) 0, 2x + 19 6x +47

Sol:

5(2x – 7) – 3(2x + 3) ≤ 0

= 10 – 35 – 6x – 9 ≤ 0

= 4x – 44 ≤ 0

= 4x ≤ 44

= x ≤ 11 ……………………(i)

2x + 19 ≤ 6x + 47

= 19 – 47 ≤ 6x – 2x

= -28 ≤ 4x

= x ≥ -7 …………………….. (ii)

From (i) and (ii) it is observed that the solution set of the above mentioned inequality is (-7, 11).

The number is as follows:

Question 11

A solution is to be maintained between 680F and 770F. Predict the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) formula is given by:

F=((9/8)C)+32$F = ((9/8)C) +32$

Sol:

According to the question the solution is to be maintained in temperature 680F and 770F, 68 ˂ F ˂ 77

We know, 68<95C+32<77$68< \frac{9}{5}C+32< 77$

= 6832<95C<7732$68 – 32< \frac{9}{5}C< 77 – 32$

= 36<95C<45$36< \frac{9}{5}C< 45$

= 3659<C<4559$36\frac{5}{9}< C< 45\frac{5}{9}$

= 20 ˂ C ˂ 25

The range of temperature in degree Celsius is between 200C and 250C

Question 12

2 % boric acid solution is added to 8% of boric acid in order to dilute. The resultant mixture so obtained contains more than 4 % but less than 6% of boric acid. How many litres of 2% solution is to be added if you have 640 litres of the 8% solution.

Sol:

Let us consider y be the 2% of boric acid solution which is to be added.

Total mixture becomes: (y + 640) litres.

Resultant mixture so obtained:

= 2 % of y + 8% of 640 ˃ 4% of (x+640) and 2% of x+ 8% of 640 ˂ (x+640)

= 2% of x + 8% of 640 ˃ 4 % of (x+ 640)

= 2100x+8100(640)>4100(x+640)$\frac{2}{100}x+\frac{8}{100}(640)> \frac{4}{100}(x+640)$

= 2x + 5120 ˃ 4x + 2560

= 5120 – 2560 ˃ 2x

= 2560 ˃ 2 x

= 1280 ˃ x

2% of x+ 8% of 640 ˂ (x+640)

2100x+8100(640)˂6100(x+640)$\frac{2}{100}x+\frac{8}{100}(640)˂ \frac{6}{100}(x+640)$

= 2x + 5120 ˂ 6x + 3840

= 5120 – 3840 ˂ 6x – 2x

= 1280 ˂ 4x

= 320 ˂ 1x

Therefore the inequality becomes: 320 ˂ x ˂ 1280

If 25 of boric acid is to be added then it would have more than 320 litres but less than 1280 litres.

Question 13

Estimate how many litres of water should be added to 1125 litres of a 45 % solution of an acid so that the resultant mixture so formed should contain more than 25% but less than 30% of the acid content.

Sol:

Let y be the litres of water which is required to be added.

Total mixture obtained = (x+1125) litres.

Acid content in the resultant mixture = 45% of 1125 litres.

The resultant mixture so formed will contain more than 25% but less than 30% of the acid.

Therefore 30% of (1125 + x) ˃ 45% of 1125

25% of (1125 + x) ˂ 45% of 1125

= 30% of (1125 + x) ˃   45% of 1125

= 30100(1125+x)>45100(1125)$\frac{30}{100}(1125+x)> \frac{45}{100}(1125)$

= 30(1125 + x) ˃ 45(1125)

= 30(1125) +30x ˃ 45(1125)

= 30x ˃ 45(1125) – 30(1125)

= x ˃ 15(1125)30$\frac{15(1125)}{30}$

= x ˃ 562.5

25% of (1125 + x) ˂ 45% of 1125

= 25100(1125+x)˂45100(1125)$\frac{25}{100}(1125+x)˂ \frac{45}{100}(1125)$

= 25(1125 + x) ˂ 45(1125)

= 25(1125) +30x ˂ 45(1125)

= 25x ˃ 45(1125) – 25(1125)

= x ˃ 20(1125)25$\frac{20(1125)}{25}$

= x ˃ 900

The required litres of water to be added is 562.5 ˂ x ˂ 900

Question 14

It is considered that that IQ of a person is given by a formula

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

Sol:

Where mental age is termed as MA and chronological age.

If 80 ˂ IQ ˂ 140 ………………..(i)

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

IQ = MA12×100$\frac{MA}{12}\times 100$

Putting this value of IQ in equation (i), we obtain

= 80 ≤ MA12×100$\frac{MA}{12}\times 100$ ≤ 140

= 80 ( 12100)$\frac{12}{100} )$ ≤ MA ≤ 140 ( 12100)$\frac{12}{100} )$

= 9.6 ≤ MA ≤ 16.8

The range of mental age would be 9.6 ≤ MA ≤ 16.8