Exercise 6.1 (Linear Inequalities)

Ex 6.1 Linear Inequalities ncert math solution class 11

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

(i) $x$ is a natural number.

(ii) $x$ is an integer.

Solution: Given that: $24 x<100$

$\Rightarrow x<\frac{100}{24} \quad \Rightarrow x<\frac{25}{6}=4.25$

(i) $x$ is a natural number.

Hence, in this case, the solution set $x=\{1,2,3,4\}$ .

(ii) $x$ is an integer.

The integers less than $\frac{25}{6}$ are …-3, $-2,-1,0,1,2,3,4$ .

Hence, in this case, the solution set $x=\{\ldots-3,-2,-1,0,1,2,3,4\}$ .

Question 2: Solve $-12 x>30$ , when

(i) $x$ is a natural number.

(ii) $x$ is an integer.

Solution : Given that: $-12 x>30$

$\Rightarrow-x>\frac{30}{12}$

$\Rightarrow-x>\frac{5}{2}$

$\Rightarrow x<-\frac{5}{2}=-2.5$

(i) $x$ is a natural number.

We know that there is no natural number less than $-\frac{5}{2}$ .

Thus, when $x$ is a natural number, there is no solution of the given inequality.

(ii) $x$ is an integer.

The integers less than $-\frac{5}{2}$ are $\ldots,-5,-4,-3$ .

Hence, in this case, the solution set $x=\{\ldots,-5,-4,-3\}$ .

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

(i) $x$ is an integer.

(ii) $x$ is a real number.

Solution : Given that: $5 x-3<7$

$\Rightarrow 5 x<10 \quad \Rightarrow x<\frac{10}{5} \quad \Rightarrow x<2$

(i) $x$ is an integer.

The integers less than 2 are $\ldots,-4,-3,-2,-1,0,1$ .

Hence, in this case, the solution set $x= \{\ldots,-4,-3,-2,-1,0,1\}$ .

(ii) $x$ is a real number.

Thus, the solution set of the given inequality is $x \in(-\infty, 2)$ .

Question 4: Solve $3 x+8>2$ , when

(i) $x$ is an integer.

(ii) $x$ is a real number.

Solution : Given that: $3 x+8>2$

$\Rightarrow 3 x>-6$

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

$\Rightarrow x>-2$

(i) $x$ is an integer.

The integers greater than -2 are $-1,0,1,2, \ldots$

Hence, in this case, the solution set $x=\{-1,0,1,2, \ldots\}$ .

(ii) $x$ is a real number.

Thus, in this case, the solution set is $x\in(-2, \infty)$ .

Solve the inequalities in Exercises 5 to 16 for real x.

Question 5: Solve: $4 x+3<6 x+7$

Solution : Given that $4 \mathrm{x}+3<5 \mathrm{x}+7$

$\Rightarrow 4 \mathrm{x}+3-7<5 \mathrm{x}+7-7$

$\Rightarrow 4 \mathrm{x}-4<5 \mathrm{x}$

$\Rightarrow 4 \mathrm{x}-4-4 \mathrm{x}<5 \mathrm{x}-4 \mathrm{x}$

$\Rightarrow-4<\mathrm{x}$

Hence, the solution set of the given inequality is $x\in(-4, \infty)$ .

Question 6: $3 x-7>5 x-1$

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

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

$\Rightarrow 3 x>5 x+6$

$\Rightarrow 3 x-5 x>5 x+6-5 x$

$\Rightarrow-2 x>6$

$\Rightarrow-x>\frac{6}{2}$

$\Rightarrow-x>3$

$\Rightarrow x<-3$

Hence, the solution set of the given inequality is $x\in(-\infty,-3)$ .

Question 7: $3(x-1) \leq 2(x-3)$

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

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

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

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

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

$\Rightarrow x \leq-3$

Hence, the solution set of the given inequality is $x\in(-\infty,-3]$ .

Question 8: $3(2-x) \geq 2(1-x)$

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

$\Rightarrow 6-3 \mathrm{x} \geq 2-2 \mathrm{x}$

$\Rightarrow 6-3 \mathrm{x}+2 \mathrm{x} \geq 2-2 \mathrm{x}+2 \mathrm{x}$

$\Rightarrow 6-x \geq 2$

$\Rightarrow 6-\mathrm{x}-6 \geq 2-6$

$\Rightarrow-x \geq-4$

$\Rightarrow \mathrm{x} \leq 4$

Hence, the solution set of the given inequality is $(-\infty, 4]$ .

Question 9: $x+\frac{x}{2}+\frac{x}{3}<11$

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

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

$\Rightarrow x\left(\frac{6+3+2}{6}\right)<11$

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

$\Rightarrow x<11 \times \frac{6}{11}$

$\Rightarrow x<6$

Hence, the solution set of the given inequality is $x\in(-\infty, 6)$ .

Question 10: $\frac{x}{3}>\frac{x}{2}+1$

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

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

$\Rightarrow x\left(\frac{2-3}{6}\right)>1$

$\Rightarrow-\frac{1}{6} x>1$

$\Rightarrow-x>6$

$\Rightarrow x<-6$

Hence, the solution set of the given inequality is $x\in(-\infty,-6)$ .

Question 11: $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$

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

$\Rightarrow \frac{3 x}{5}-\frac{6}{5} \leq \frac{10}{3}-\frac{5 x}{3}$

$\Rightarrow \frac{3 x}{5}+\frac{5 x}{3} \leq \frac{10}{3}+\frac{6}{5}$

$\Rightarrow \frac{9 x+25 x}{15} x \leq \frac{50+18}{15}$

$\Rightarrow \frac{34 x}{15} \leq \frac{68}{15}$

$\Rightarrow 34 x \leq 68$

$\Rightarrow x \leq \frac{68}{34}$

$\Rightarrow x \leq 2$

Hence, the solution set of the given inequality is $x\in(-\infty, 2]$ .

Question 12: $\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$

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

$\Rightarrow \frac{3 x}{10}+2 \geq \frac{x}{3}-2$

$\Rightarrow \frac{3 x}{10}-\frac{x}{3} \geq-2-2$

$\Rightarrow \frac{9 x-10 x}{30} \geq-4$

$\Rightarrow-\frac{x}{30} \geq-4$

$\Rightarrow-x \geq-120$

$\Rightarrow x \leq 120$

Hence, the solution set of the given inequality is $x\in(-\infty, 120]$ .

Question 13: $2(2 x+3)-10<6(x-2)$

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

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

$\Rightarrow 4 x-4<6 x-12$

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

$\Rightarrow-2 x<-8$

$\Rightarrow-x<-4$

$\Rightarrow x>4$

Hence, the solution set of the given inequality is $x\in[4, \infty)$ .

Question 14: $37-(3 x+5) \geq 9 x-8(x-3)$

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

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

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

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

$\Rightarrow-4 x \geq-8$

$\Rightarrow-x \geq-2$

$\Rightarrow x \leq 4$

Hence, the solution set of the given inequality is $x\in(-\infty, 2]$ .

Question 15: $\frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$

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

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

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

$\Rightarrow \frac{15 x-100 x+84 x}{60}<\frac{-10+9}{15}$

$\Rightarrow-\frac{x}{60}<-\frac{1}{15}$

$\Rightarrow-x<-\frac{60}{15}$

$\Rightarrow x>4$

Hence, the solution set of the given inequality is $x\in(4, \infty)$ .

Question 16: $\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}$

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

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

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

$\Rightarrow \frac{40 x-45 x-12 x}{60} \geq \frac{-30-24+20}{60}$

$\Rightarrow-17 x \geq-34$

$\Rightarrow-x \geq-2$

$\Rightarrow x \leq 2$

Hence, the solution set of the given inequality is $x\in(-\infty, 2]$ .

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line

Question 17: $3 x-2<2 x+1$

$3 x-2<2 x+1$

$\Rightarrow 3 \mathrm{x}-2 \mathrm{x}<1+2$

$\Rightarrow \mathrm{x}<3$

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

Ex 6.1 Linear Inequalities ncert math solution class 11

Question 18: $5 x-3 \geq 3 x-5$

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

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

$\Rightarrow 2 x \geq-2$

$\Rightarrow x \geq-1$

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

Ex 6.1 Linear Inequalities ncert math solution class 11

Question 19: $3(1-x)<2(x+4)$

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

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

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

$\Rightarrow-5 x>5 \quad \Rightarrow-x>1$

$\Rightarrow x<-1$

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

Ex 6.1 Linear Inequalities ncert math solution class 11

Question 20: $\frac{x}{2} \geq \frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$

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

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

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

$\Rightarrow \frac{15 x-50 x+42 x}{30} \geq \frac{-10+9}{15}$

$\Rightarrow \frac{7 x}{30} \geq-\frac{1}{15}$

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

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

Ex 6.1 Linear Inequalities ncert math solution class 11

Question 21: Ravi obtained 70 and 75 marks in first two unit test. Find the number if minimum marks he should get in the third test to have an average of at least 60 marks.

Solution: Let $x$ be the marks obtained by Ravi in the third unit test.

Since the student should have an average of at least 60 marks, so

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

$\Rightarrow 145+x \geq 180$

$\Rightarrow x \geq 35$

Hence, the student must obtain a minimum of 35 marks to have an average of at least 60 marks.

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.

Solution : Let $x$ be the marks obtained by Sunita in the fifth examination.

In order to receive grade ‘ $A$ ‘ in the course, she must obtain an average of 90 marks or more in five examinations. Therefore,

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

$\Rightarrow 368+x \geq 450$

$\Rightarrow x \geq 82$

Hence, Sunita must obtain greater than or equal to 82 marks in the fifth examination.

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.

Solution : Let $\mathrm{</strong></em></span>x}$ be the smaller of the two consecutive odd positive integers. Then, the other integer is $\mathrm{x}+2$ .

Since both the integers are smaller than 10 , therefore

$x+2<10$

$\Rightarrow \mathrm{x}<10-2$

$\Rightarrow \mathrm{x}<8--(i)$

Also, the sum of the two integers is more than 11.

$\therefore \mathrm{x}+(\mathrm{x}+2)>11$

$\Rightarrow 2 \mathrm{x}+2>11$

$\Rightarrow 2 x>11-2$

$\Rightarrow 2 x>9$

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

$\Rightarrow x>4.5--(ii)$

From (i) and (ii), we obtain that the value of $x$ can be 4, 5, 6 or 7 . .

Since $x$ is an odd number, $x$ can take the values, 5 and 7 .

Hence, the required possible pairs of numbers 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.

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

Since both the integers are larger than 5 ,

$x>5--(i)$

Also, the sum of the two integers is less than 23.

$x+(x+2)<23$

$\Rightarrow 2 x+2<23$

$\Rightarrow 2 x<23-2$

$\Rightarrow 2 x<21$

$\Rightarrow x<10.5$

$\Rightarrow x<\frac{21}{2}--(ii)$

From (1) and (2), we obtain $5<x<10.5$ .

Since $x$ is an even number, so $x$ can take the values, 6,8 and 10 .

Hence, the required possible pairs are $(6,8),(8,10)$ and $(10,12)$ .

Question 25: The longest side of a triangle is 3 times the shortest side and the third side is $2 \mathrm{~cm}$ shorter than the longest side. If the perimeter of the triangle is at least $61 \mathrm{~cm}$ , find the minimum length of the shortest side.

Solution: Let the length of the shortest side of the triangle= $\mathrm{x} \mathrm{cm}$ .

Then, length of the longest side $=3 x \mathrm{~cm}$ and the length of the third side $=(3 \mathrm{x}-2) \mathrm{cm}$

Since the perimeter of the triangle is at least $61 \mathrm{~cm}$ ,

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

$\Rightarrow 7x-2\geq61$

$\Rightarrow 7x\geq63$

$\Rightarrow x\geq 9$

Hence, the minimum length of the shortest side is $9 \mathrm{~cm}$ .

Question 26: A man wants to cut three lengths from a single piece of board of length $91 \mathrm{~cm}$ . The second length is to be $3 \mathrm{~cm}$ 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 $5 \mathrm{~cm}$ longer than the second?

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

Solution: Let the length of the shortest piece = $\mathrm{x} \mathrm{~cm}$ . Then, length of the second piece and the third piece are $(\mathrm{x}+3) \mathrm{~cm}$ and $2 \mathrm{ x} \mathrm{~cm}$ respectively.

Since the three lengths are to be cut from a single piece of board of length $91 \mathrm{~cm}$ ,

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

$\Rightarrow 4 x+3 \leq 91$

$\Rightarrow 4 x \leq 91-3$

$\Rightarrow 4 x \leq 88$

$\Rightarrow x \leq 22$

Also, the third piece is at least $5 \mathrm{~cm}$ longer than the second piece.

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

$\Rightarrow 2 x \geq x+8$

$\Rightarrow x \geq 8$

From (1) and (2), we obtain, $8 \leq x \leq 22$

Thus, the possible length of the shortest board is greater than or equal to $8 \mathrm{~cm}$ but less than or equal to $22 \mathrm{~cm}$ .

Exercise 5.1 complex no. ncert math solution class 11

Exercise 5.2 complex no. ncert math solution class 11

Exercise 5.3 complex no. ncert math solution class 11

Chapter 5 Miscellaneous ncert math solution class 11

gmath.in

Exercise 6.1 (Linear Inequalities)

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line

Leave a Comment Cancel reply