Linear Inequalities Important Questions

Very Short Answers [1 Mark Questions]

Ques: Solve \(\frac{x - 4}{2} \ge \frac{x + 1}{4} - 1\)

Ans: \(\frac{x - 4}{2} \ge \frac{x + 1}{4} - 1\)

\(\frac{x - 4}{2} \ge \frac{x + 1 - 4}{4}\)

\(\frac{x - 4}{2} \ge \frac{x - 3}{4}\)……multiplying by 4

2(x - 4) \(\ge\) x - 3

2x - 8 \(\ge\) x - 3

x ≥ 5

Hence the solution (5, ∞)

Ques: Solve 5x + 6 > 1 when x is real number

Ans: 5x + 6 > 1

5x > -5

x > -1 ……Dividing by 5

Hence the solution (-1, ∞)

Ques: If 6x > -12 then x ……. -2

Ans: 6x > -12

x > -2 ……dividing by 6

x > -2

Ques: When x is an integer. Solve 5x – 3 ≤ 3x + 1

Ans: 5x-3 ≤ 3x+1

5x-3x ≤ 1+3

2x ≤ 4

x ≤ 2

Hence the solution {..., -3, -2, -1, 0, 1, 2}

Ques: 30x < 150, where x is the natural number

Ans: 

Given: 30x < 150

Dividing both sides by 30

x < 5

Hence the solution set {1,2,3,4}

Ques: If 6x > -36 then x …… -6

Ans: 6x > -36

x > -6 ………Divide by 6

x > -6

Ques: 50x < 540, where x is the natural number

Ans: 

Given: 50x < 540

Dividing both sides by 50

x < \(\frac{540}{50}\)

x < \(\frac{54}{5}\)

Hence the solution set {1,2,3,4,5,6,7,8,9,10}

Short Answer Questions [2 Marks Questions]

Ques: Solve: \(\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)\)

Ans: \(\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)\)

\(\frac{3 x}{10}+2 \geq \frac{x}{3}-2\)

\(\frac{3 x}{10}-\frac{x}{3} \geq-4\)

\(\frac{9 x-10 x}{30} \geq-4\)

\(\frac{-x}{30} \geq-4\)

\(-x \geq-4 \times 30\)

\(-x \geq-120\)

\(x \leq 120\)

Hence the solution (-∞, 120]

Ques: Solve the equation, 4x-2 ≤ 6 and 9x+3 ≥ -15

Ans: 4x-2 ≤ 6 and 9x+3 ≥ -15

First,

4x-2 ≤ 6

2x -1 ≤ 3……divide by 2

2x ≤ 4

x ≤ 2 ………divide 2

Second,

9x+3 ≥ -15

3x+1 ≥ -5 ……….Divide by 3

3x ≥ -6

x ≥ -2 ………divide by 3

From both the solutions, -2 ≤ x ≤ 2

Hence the solution [-2, 2]

Ques: Solve the inequality \(\frac{x}{2}<\frac{(5 x-3)}{3}-\frac{(7 x-2)}{5}\)

Ans: \(\frac{x}{2}<\frac{(5 x-3)}{3}-\frac{(7 x-2)}{5}\)

\(\frac{x}{2} < \frac{5(5 x-3)-3(7 x-2)}{15}\)

\(\frac{x}{2}<\frac{25 x-15-21 x+6}{15}\)

\(\frac{x}{2}<\frac{4 x-9}{15}\)

15x < 2(4x-9)

15x < 8x-18

7x < -18

x < \(-\frac{18}{7}\)

Hence the solution is (- ∞, \(-\frac{18}{7}\))

Ques: Consider a triangle, the longest side of the triangle is three times the shortest side. Third side of the triangle is 3cm smaller than the longest side. Parameter of the triangle is at least 61cm. What will be the minimum length of the shortest side?

Ans: Let, Shortest side of the triangle be x.

Given: Longest side of the triangle = 3x

Third side of the triangle = 3x-2

Perimeter ≥ 61

Therefore

(3x)+(x)+(3x-2) ≥ 61

7x-2 ≥ 61

7x ≥ 63

x ≥ 9

Therefore, the shortest side of the triangle is at least 9 cm.

Ques: For the pool, the average water acidity is considered normal when in the range of 7.2 and 7.8 pH when measured for three days continuously. What will be the pH for the third day when acidity for the first two days is 7.52 and 7.1. Consider the pH of the pool in normal.

Ans: Let, reading for third day be x

Given, reading for first two days: 7.52 and 7.1

Therefore,

7.2 < 7.52 + 7.1 + x₃ < 7.8

21.6 < 14.62 + x < 23.4 ……….Multiplying by three

(21.6 – 14.62) < x < (23.4 – 14.62)

6.98 < x < 8.78

Long Answers Questions [3 Marks Questions]

Ques: Solve \(\frac{2 x-1}{3} \geq \frac{3 x-2}{4}-\frac{2-x}{5}\)

Ans: \(\frac{2 x-1}{3} \geq \frac{3 x-2}{4}-\frac{2-x}{5}\)

\(\frac{2 x-1}{3} \geq \frac{5(3 x-2)-4(2-x)}{20}\)

\(\frac{2 x-1}{3} \geq \frac{15 x-10-8+4 x}{20}\)

\(\frac{2 x-1}{3} \geq \frac{19 x-18}{20}\)

20(2x – 1) ≥ 3(19x – 18)

40x – 20 ≥ 57x – 54

-17x ≥ -34

-x ≥ -2 ………Dividing by 17

x ≤ 2

Hence the solution (- ∞,2]

Ques: Rahul scored 70 and 65 marks in two of his unit tests. Calculate the minimum marks required in his third text to form an average of at least 70 marks.

Ans: Let, x be the marks obtained by Rahul in the third test.

We know: students should obtain an average of at least 70 marks

Therefor linear equation

\(\frac{70+65+x}{3}\) ≥ 70

135+x ≥ 210

x ≥ 210 – 135

x ≥ 75

Hence, Rahul should obtain 75 marks to obtain an average at least of 70 marks.

Ques: In T20 series Raj scored 30 and 85 runs in the first two matches. Calculate the minimum runs required in his third match to form an average of at least 50 runs.

Ans: Let, x be the runs scored by Raj in the third match.

We know: Raj's average score for three matches is average of at least 50 runs

Therefor linear equation

\(\frac{30+85+x}{3}\) ≥ 50

115+x ≥ 150

x ≥ 150 – 115

x ≥ 35

Hence, Hence Raj should score at least 35 runs to have an average of 50 runs.

Ques: Solve graphically, 3x-6≥0

Ans: 3x-6≥0 ……….(1)

3x-6=0

x=2

Now, Put (0,0) in equation (1)

Therefore, 

0-6≥0

0>6

0>6 is not possible

Therefore graphical representation

Very Long Answers [5 Marks Questions]

Ques: A sol of 10% acid is to be diluted by adding a 2% acid solution to it. The resulting mixture should be more than 4% and less than 6% of acid.If we have 700 litres of 10% solutions. How much litre of the 2% solution should be used.

Ans: Let x be added

Therefore from the given data:

2% of +10% of 700 > 4% of (700+x)

\(\frac{2 x}{100}+\frac{10 \times 700}{100}>\frac{4}{100}(700+x)\)

2x + 7000 > 2800 + 4x

-2x > -4200

x < 2100 ……..(1)

Now,

2% of + 10% of 700 < 6% of (700+x)

\(\frac{2 x}{100}+\frac{10 \times 700}{100}>\frac{6}{100}(700+x)\)

2x + 7000 < 4200 + 6x

-4x < -2800

x > 700 ……(2)

From eq.1 and eq.2

700 < x < 2100

Ques: Find the pairs of consecutive odd natural numbers both of which are larger than 10 such that their sum is less than 50.

Ans: Let two consecutive odd numbers be x and x+2

Given (1)

The numbers are larger than 10

therefore , x >10 ………(1)

Given (2)

The sum of a consecutive odd number is less than 50 and the numbers are greater than 10

Therefore, x+(x+2)<50

2x+2 < 50

2x < 48

x < 24 ………(2)

Therefore from eq.1 and eq.2 pairs obtained are

(11,13), (13,15), (15,17), (17,19), (19,21), (21,23)

Ques: In a Lamp manufacturing company, cost of equation for a week is C = 200+1.5x and the revenue equation is R = 2x. Lamps sold per week is denoted by 'x'. How many lamps must be sold to obtain profit?

Ans:

Given: 'x' = number of lamps sold per week

C=200+1.5x

R=2x

We know,

Profit(P) = Revenue(R) - Cost(C)

Therefore to obtain profit Revenue should be higher than Cost

So, R > C

Therefore,

2x > 200+1.5x

0.5x > 200

x > 400

Therefore, to obtain profit the company has to sell more than 400 lamps per week.

Ques: While drilling hole it was found that temperature (T) changes at (x) km below the surface. It was given by T = 30+25(x-3) and 3<x<15. Determine the depth at which temperature will be between 400°C and 700°C.

Ans: Given: T=30+25(x-3) and 3<x<15

Let, x be the distance where temperature between 400°C and 700°C

Therefore,

400<30+25(x-3)<700

400<30+25x-90<700

400<25x-60<700

460<25x<760

\(\frac{460}{25}\)<x<\(\frac{760}{25}\) ……….Dividing by 25

18.4 < x < 30.3

Therefore between 400°C and 700°C, the depth is more than 18.4 Km and less than 30.3 Km.

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