Class 11 Maths Ncert Solutions Ex 6.1

Class 11 Maths Ncert Solutions Chapter 6 Ex 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.