NCERT Solutions Class 11 Mathematics Linear Inequalities Exercise 6-Misc

Question 10:
Solve the inequalities and represent the solution graphically on number line:
5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

Answer:
5(2x – 7) – 3(2x + 3) ≤ 0
⇒ 10x – 35 – 6x – 9 ≤ 0
⇒ 4x – 44 ≤ 0
⇒ 4x ≤ 44
⇒ x ≤ 11     ………………1
Again, 2x + 19 ≤ 6x + 47
⇒ 19 – 47 ≤ 6x – 2x
⇒ –28 ≤ 4x
⇒ –7 ≤ x      ……………..2
From (1) and (2), it can be concluded that the solution set for the given system of inequalities
is [–7, 11]. The solution of the given system of inequalities can be represented on number line
as:
               

Question 11:
A solution is to be kept between 68°F and 77°F. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by F = 9C/5 + 32

Answer:
Given F = (C * 9)/5 + 32
⇒ F – 32 = 9C/5
⇒ C = {5 * (F – 32)}/9 ………….1
Now put F = 68 in equation 1, we get
      C = {5 * (68 – 32)}/9 
⇒ C = (5 * 36)/9
⇒ C = 5 * 4
⇒ C = 20
Again put F = 77 in equation 1, we get
      C = {5 * (77 – 32)}/9 
⇒ C = (5 * 45)/9
⇒ C = 5 * 5
⇒ C = 25
So, the range of solution is kept it between 20° C and 25° C

Question 12:
A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Answer:
Given mixture = 640 litres
Let 2% boric acid solution is x litre.
So, total amount of mixture = 640 + x.
Now given in the question
       2% of x + 8% of 640 > 4% of (640 + x)
⇒ 2x/100 + (8*640)/100 > {4(640 + x)}/100
⇒ 2x + 5120 > 2560 + 4x
⇒ 2x – 4x > 2560 – 5120
⇒ -2x > -2560
⇒ 2x < 2560
⇒ x < 2560/2
⇒ x < 1280
Again from the question
      2% of x + 8% of 640 < 6% of (640 + x)
⇒ 2x/100 + (8 * 640)/100 < {6 * (640 + x)}/100
⇒ 2x + 5120 < 3840 + 6x
⇒ 2x – 6x < 3840 – 5120
⇒ -4x < -1280
⇒ 4x > 1280
⇒ x > 1280/4
⇒ x > 320
So, 320 < x < 1289
Thus, the number of litres of 2% of boric acid solution that is to be added will have to be more
than 320 litres but less than 1280 litres.

Question 13:
How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Answer:
Let amount of water is added = x litre
Given solution is 1125 litre
Percentage of acid in solution = 45%
So percentage of water in solution = 55%
Case 1: Resulting mixture will contain more than 25% acid
      x + 1125 * 55% = (x + 1125) * 75%
⇒ x + (1125 * 55)/100 = (x + 1125) * (75/100)
⇒ 100x + 1125 * 55 =(x + 1125) * 75
⇒ 100x + 1125 * 55 = 75x + 1125 * 75
⇒ 100x – 75x = 1125 * 75 – 1125 * 55
⇒ 25x = 1125 * (75 – 55)
⇒ 25x = 1125 * 20
⇒ x = (1125 * 20)/25
⇒ x = (1125 * 4)/5
⇒ x = 225 * 4
⇒ x = 900
Case 2: Resulting mixture will contain more than 30% acid
      x + 1125 * 55% = (x + 1125) * 70%
⇒ x + (1125 * 55)/100 =(x + 1125) * (70/100)
⇒ 100x + 1125 * 55 =(x + 1125) * 70
⇒ 100x + 1125 * 55 = 70x + 1125 * 70
⇒ 100x – 70x = 1125 * 70 – 1125 * 55
⇒ 30x = 1125 * (70-55)
⇒ 30x = 1125 * 15
⇒ x = (1125 * 15)/30
⇒ x = (1125 * 5)/10
⇒ x = (1125)/2
⇒ x = 562.5
Now, 562.5 < x < 900
Thus, the required number of litres of water that is to be added will have to be more than
562.5 but less than 900.

Question 14:
IQ of a person is given by the formula IQ = (MA/CA) * 100
Where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.

Answer:
It is given that for a group of 12 years old children,
80 ≤ IQ ≤ 140    ……………..1
For a group of 12 years old children, CA = 12 years
IQ = (MA/CA) * 100
Putting this value of IQ in 1, we get
      80 ≤ (MA/12) * 100 ≤ 140
⇒ 80 * (12/100) ≤ MA ≤ 140 * (12/100)
⇒ 960/100 ≤ MA ≤ 1680/100
⇒ 9.6 ≤ MA ≤ 16.8
Thus, the range of mental age of the group of 12 years old children is 9.6 ≤ MA ≤ 16.8