Class 11 Maths Ncert Solutions Ex 6.4

Class 11 Maths Ncert Solutions Chapter 6 Ex 6.4

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

Sol:

= 347x218

= 347x2184

= 77x214

= 77x214

= 1x22

= 2 ≥ x ≥ -4

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

 

Question 4

Solve: 15<43(x2)518

Sol:

= 15<43(x2)518

= -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

Sol:

= 12<43x52

= 124<3x524

= 16<3x52

= -80 ˂ 3x ≤ -10

= 803<x103

Thus, solution set for the above mentioned inequality is: 803,103

 

Question 6

Solve: 7<3x+11211

Sol:

= 1243x52

= 14 ≤ 3x +11 ≤ 22

= 14 – 11 ≤ 3x ≤ 22 – 11

= 1x113

Thus, solution set for the above mentioned inequality is: 1,113

 

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:

1

 

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:

2

 

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,).

The number is as follows:

3

 

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:

4

 

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

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

= 6832<95C<7732

= 36<95C<45

= 3659<C<4559

= 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)

= 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)

= 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)

= 30(1125 + x) ˃ 45(1125)

= 30(1125) +30x ˃ 45(1125)

= 30x ˃ 45(1125) – 30(1125)

= x ˃ 15(1125)30

= x ˃ 562.5

25% of (1125 + x) ˂ 45% of 1125

= 25100(1125+x)˂45100(1125)

= 25(1125 + x) ˂ 45(1125)

= 25(1125) +30x ˂ 45(1125)

= 25x ˃ 45(1125) – 25(1125)

= x ˃ 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

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

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

= 80 ≤ MA12×100 ≤ 140

= 80 ( 12100) ≤ MA ≤ 140 ( 12100)

= 9.6 ≤ MA ≤ 16.8

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