Class 11 Maths NCERT Solutions Chapter 5 Miscellaneous Exercise – Linear Inequalities

2\le3x-4\le5

Adding 4 to all three parts:

6\le3x\le9

Dividing all three parts by 3 (positive, sign unchanged):

2\le x\le3

Expanding -3(2x-4)=-6x+12:

6\le-6x+12\lt12

Subtracting 12 from all three parts:

-6\le-6x\lt0

Dividing all three parts by −6 (negative, so both signs flip and the order reverses):

1\ge x\gt0, i.e. 0\lt x\le1

-3\le4-\dfrac{7x}{2}\le18

Subtracting 4 from all three parts:

-7\le-\dfrac{7x}{2}\le14

Multiplying all three parts by -\dfrac{2}{7} (negative, so both signs flip and the order reverses):

2\ge x\ge-4, i.e. -4\le x\le2

-15\lt\dfrac{3(x-2)}{5}\le0

Multiplying all three parts by 5:

-75\lt3(x-2)\le0

Dividing all three parts by 3:

-25\lt x-2\le0

Adding 2 to all three parts:

-23\lt x\le2

Simplify the middle term first: 4-\dfrac{3x}{-5}=4+\dfrac{3x}{5} (dividing by −5 and then subtracting is the same as adding).

So the inequality becomes:

-12\lt4+\dfrac{3x}{5}\le2

Subtracting 4 from all three parts:

-16\lt\dfrac{3x}{5}\le-2

Multiplying all three parts by \dfrac{5}{3} (positive, sign unchanged):

-\dfrac{80}{3}\lt x\le-\dfrac{10}{3}

7\le\dfrac{3x+11}{2}\le11

Multiplying all three parts by 2:

14\le3x+11\le22

Subtracting 11 from all three parts:

3\le3x\le11

Dividing all three parts by 3:

1\le x\le\dfrac{11}{3}

Solving the first inequality:

5x+1\gt-24\ \Rightarrow\ 5x\gt-25\ \Rightarrow\ x\gt-5

Solving the second inequality:

5x-1\lt24\ \Rightarrow\ 5x\lt25\ \Rightarrow\ x\lt5

The values of x satisfying both conditions lie between −5 and 5:

Solving the first inequality:

2(x-1)\lt x+5\ \Rightarrow\ 2x-2\lt x+5\ \Rightarrow\ x\lt7

Solving the second inequality:

3(x+2)\gt2-x\ \Rightarrow\ 3x+6\gt2-x\ \Rightarrow\ 4x\gt-4\ \Rightarrow\ x\gt-1

The values of x satisfying both conditions lie between −1 and 7:

Solving the first inequality:

3x-7\gt2(x-6)\ \Rightarrow\ 3x-7\gt2x-12\ \Rightarrow\ x\gt-5

Solving the second inequality:

6-x\gt11-2x\ \Rightarrow\ 2x-x\gt11-6\ \Rightarrow\ x\gt5

Both conditions must hold. Since x\gt5 is the stricter requirement (every value satisfying it also satisfies x\gt-5), the combined solution is x\gt5:

Solving the first inequality:

5(2x-7)-3(2x+3)\le0\ \Rightarrow\ 10x-35-6x-9\le0\ \Rightarrow\ 4x-44\le0\ \Rightarrow\ x\le11

Solving the second inequality:

2x+19\le6x+47\ \Rightarrow\ 19-47\le6x-2x\ \Rightarrow\ -28\le4x\ \Rightarrow\ x\ge-7

The values of x satisfying both conditions lie between −7 and 11:

It is given that 68\lt F\lt77.

Substituting F=\dfrac{9}{5}C+32:

68\lt\dfrac{9}{5}C+32\lt77

Subtracting 32 from all three parts:

36\lt\dfrac{9}{5}C\lt45

Multiplying all three parts by \dfrac{5}{9}:

20\lt C\lt25

Let x litres of the 2% solution be added. The total mixture is then (640+x) litres.

The acid content in the mixture must satisfy both:

8\%(640)+2\%(x)\gt4\%(640+x)   and   8\%(640)+2\%(x)\lt6\%(640+x)

Multiplying every term by 100 to clear the percentages:

8(640)+2x\gt4(640+x)\ \Rightarrow\ 5120+2x\gt2560+4x\ \Rightarrow\ 2560\gt2x\ \Rightarrow\ x\lt1280

8(640)+2x\lt6(640+x)\ \Rightarrow\ 5120+2x\lt3840+6x\ \Rightarrow\ 1280\lt4x\ \Rightarrow\ x\gt320

Combining both conditions: 320\lt x\lt1280

Let x litres of water be added. Since water contains no acid, the total acid amount stays fixed at 45\% of 1125 litres, while the total mixture becomes (1125+x) litres.

The acid content must satisfy both:

25\%(1125+x)\lt45\%(1125)   and   45\%(1125)\lt30\%(1125+x)

Multiplying every term by 100:

25(1125+x)\lt45(1125)\ \Rightarrow\ 28125+25x\lt50625\ \Rightarrow\ 25x\lt22500\ \Rightarrow\ x\lt900

45(1125)\lt30(1125+x)\ \Rightarrow\ 50625\lt33750+30x\ \Rightarrow\ 16875\lt30x\ \Rightarrow\ x\gt562.5

Combining both conditions: 562.5\lt x\lt900

Here CA=12 years. Substituting into the given range 80\le IQ\le140:

80\le\dfrac{MA}{12}\times100\le140

Dividing all three parts by 100:

0.8\le\dfrac{MA}{12}\le1.4

Multiplying all three parts by 12:

9.6\le MA\le16.8