*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 −3≤4−7x2≤18*

Sol:

=

=

=

=

=

= 2 ≥ x ≥ -4

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

*Question 4*

*Solve: −15<4−3(x−2)5≤18*

Sol:

=

= -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<4−3x−5≤2*

Sol:

=

=

=

= -80 ˂ 3x ≤ -10

=

Thus, solution set for the above mentioned inequality is:

*Question 6*

*Solve: 7<3x+112≤11*

Sol:

=

= 14 ≤ 3x +11 ≤ 22

= 14 – 11 ≤ 3x ≤ 22 – 11

=

Thus, solution set for the above mentioned inequality is:

*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 (

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 68 ^{0}F and 77^{0}F. Predict the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) formula is given by:*

Sol:

According to the question the solution is to be maintained in temperature 68^{0}F and 77^{0}F, 68 ˂ F ˂ 77

We know,

=

=

=

= 20 ˂ C ˂ 25

The range of temperature in degree Celsius is between 20^{0}C and 25^{0}C

*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)

=

= 2x + 5120 ˃ 4x + 2560

= 5120 – 2560 ˃ 2x

= 2560 ˃ 2 x

= 1280 ˃ x

2% of x+ 8% of 640 ˂ (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

=

= 30(1125 + x) ˃ 45(1125)

= 30(1125) +30x ˃ 45(1125)

= 30x ˃ 45(1125) – 30(1125)

= x ˃

= x ˃ 562.5

25% of (1125 + x) ˂ 45% of 1125

=

= 25(1125 + x) ˂ 45(1125)

= 25(1125) +30x ˂ 45(1125)

= 25x ˃ 45(1125) – 25(1125)

= x ˃

= 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*

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 =

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

= 80 ≤

= 80 (

= 9.6 ≤ MA ≤ 16.8

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