Q1
Solve \( 24x < 100 \), when
(i) \( x \) is a natural number.
(ii) \( x \) is an integer.
(i) \( x \) is a natural number.
(ii) \( x \) is an integer.
Answer:
Given inequality: \( 24x < 100 \)
Dividing both sides by 24:
\( x < \frac{100}{24} \Rightarrow x < \frac{25}{6} \Rightarrow x < 4.16 \)
(i) When \( x \) is a natural number:
Natural numbers less than 4.16 are 1, 2, 3, 4.
Solution set: \( \{1, 2, 3, 4\} \)
(ii) When \( x \) is an integer:
Integers less than 4.16 are ..., -2, -1, 0, 1, 2, 3, 4.
Solution set: \( \{\dots, -3, -2, -1, 0, 1, 2, 3, 4\} \)
Q2
Solve \( -12x > 30 \), when
(i) \( x \) is a natural number.
(ii) \( x \) is an integer.
(i) \( x \) is a natural number.
(ii) \( x \) is an integer.
Answer:
Given inequality: \( -12x > 30 \)
Dividing both sides by -12 (inequality sign reverses):
\( x < \frac{30}{-12} \Rightarrow x < -\frac{5}{2} \Rightarrow x < -2.5 \)
(i) When \( x \) is a natural number:
There are no natural numbers less than -2.5.
Solution set: \( \phi \) (Empty set)
(ii) When \( x \) is an integer:
Integers less than -2.5 are -3, -4, -5, ...
Solution set: \( \{\dots, -5, -4, -3\} \)
Q3
Solve \( 5x - 3 < 7 \), when
(i) \( x \) is an integer.
(ii) \( x \) is a real number.
(i) \( x \) is an integer.
(ii) \( x \) is a real number.
Answer:
\( 5x - 3 < 7 \Rightarrow 5x < 10 \Rightarrow x < 2 \)
(i) When \( x \) is an integer:
Integers less than 2 are 1, 0, -1, -2, ...
Solution set: \( \{\dots, -2, -1, 0, 1\} \)
(ii) When \( x \) is a real number:
Solution interval: \( (-\infty, 2) \)
Q4
Solve \( 3x + 8 > 2 \), when
(i) \( x \) is an integer.
(ii) \( x \) is a real number.
(i) \( x \) is an integer.
(ii) \( x \) is a real number.
Answer:
\( 3x + 8 > 2 \Rightarrow 3x > 2 - 8 \Rightarrow 3x > -6 \Rightarrow x > -2 \)
(i) When \( x \) is an integer:
Integers greater than -2 are -1, 0, 1, 2, ...
Solution set: \( \{-1, 0, 1, 2, \dots\} \)
(ii) When \( x \) is a real number:
Solution interval: \( (-2, \infty) \)
Solve the inequalities in Exercises 5 to 16 for real \( x \).
Q5
\( 4x + 3 < 5x + 7 \)
Answer:
\( 4x - 5x < 7 - 3 \)
\( -x < 4 \)
Multiplying by -1 (sign reverses): \( x > -4 \).
Solution: \( (-4, \infty) \)
Q6
\( 3x - 7 > 5x - 1 \)
Answer:
\( 3x - 5x > -1 + 7 \)
\( -2x > 6 \)
Dividing by -2 (sign reverses): \( x < -3 \).
Solution: \( (-\infty, -3) \)
Q7
\( 3(x - 1) \le 2(x - 3) \)
Answer:
\( 3x - 3 \le 2x - 6 \)
\( 3x - 2x \le -6 + 3 \)
\( x \le -3 \)
Solution: \( (-\infty, -3] \)
Q8
\( 3(2 - x) \ge 2(1 - x) \)
Answer:
\( 6 - 3x \ge 2 - 2x \)
\( 6 - 2 \ge -2x + 3x \)
\( 4 \ge x \) or \( x \le 4 \)
Solution: \( (-\infty, 4] \)
Q9
\( x + \frac{x}{2} + \frac{x}{3} < 11 \)
Answer:
Taking LCM of 2 and 3, which is 6:
\( \frac{6x + 3x + 2x}{6} < 11 \)
\( \frac{11x}{6} < 11 \)
\( 11x < 66 \Rightarrow x < 6 \)
Solution: \( (-\infty, 6) \)
Q10
\( \frac{x}{3} > \frac{x}{2} + 1 \)
Answer:
\( \frac{x}{3} - \frac{x}{2} > 1 \)
\( \frac{2x - 3x}{6} > 1 \)
\( \frac{-x}{6} > 1 \Rightarrow -x > 6 \Rightarrow x < -6 \)
Solution: \( (-\infty, -6) \)
Q11
\( \frac{3(x-2)}{5} \le \frac{5(2-x)}{3} \)
Answer:
Cross multiplying by positive numbers 3 and 5:
\( 3 \times 3(x-2) \le 5 \times 5(2-x) \)
\( 9(x-2) \le 25(2-x) \)
\( 9x - 18 \le 50 - 25x \)
\( 9x + 25x \le 50 + 18 \)
\( 34x \le 68 \Rightarrow x \le 2 \)
Solution: \( (-\infty, 2] \)
Q12
\( \frac{1}{2}(\frac{3x}{5} + 4) \ge \frac{1}{3}(x - 6) \)
Answer:
Multiply both sides by 6 (LCM of 2 and 3):
\( 3(\frac{3x}{5} + 4) \ge 2(x - 6) \)
\( \frac{9x}{5} + 12 \ge 2x - 12 \)
\( \frac{9x}{5} - 2x \ge -12 - 12 \)
\( \frac{9x - 10x}{5} \ge -24 \)
\( \frac{-x}{5} \ge -24 \Rightarrow -x \ge -120 \Rightarrow x \le 120 \)
Solution: \( (-\infty, 120] \)
Q13
\( 2(2x + 3) - 10 < 6(x - 2) \)
Answer:
\( 4x + 6 - 10 < 6x - 12 \)
\( 4x - 4 < 6x - 12 \)
\( 12 - 4 < 6x - 4x \)
\( 8 < 2x \Rightarrow 4 < x \Rightarrow x > 4 \)
Solution: \( (4, \infty) \)
Q14
\( 37 - (3x + 5) \ge 9x - 8(x - 3) \)
Answer:
\( 37 - 3x - 5 \ge 9x - 8x + 24 \)
\( 32 - 3x \ge x + 24 \)
\( 32 - 24 \ge x + 3x \)
\( 8 \ge 4x \Rightarrow 2 \ge x \Rightarrow x \le 2 \)
Solution: \( (-\infty, 2] \)
Q15
\( \frac{x}{4} < \frac{5x-2}{3} - \frac{7x-3}{5} \)
Answer:
\( \frac{x}{4} < \frac{5(5x-2) - 3(7x-3)}{15} \)
\( \frac{x}{4} < \frac{25x - 10 - 21x + 9}{15} \)
\( \frac{x}{4} < \frac{4x - 1}{15} \)
Cross multiplying (since 4 and 15 are positive):
\( 15x < 4(4x - 1) \)
\( 15x < 16x - 4 \)
\( 4 < 16x - 15x \Rightarrow 4 < x \Rightarrow x > 4 \)
Solution: \( (4, \infty) \)
Q16
\( \frac{2x-1}{3} \ge \frac{3x-2}{4} - \frac{2-x}{5} \)
Answer:
\( \frac{2x-1}{3} \ge \frac{5(3x-2) - 4(2-x)}{20} \)
\( \frac{2x-1}{3} \ge \frac{15x - 10 - 8 + 4x}{20} \)
\( \frac{2x-1}{3} \ge \frac{19x - 18}{20} \)
\( 20(2x-1) \ge 3(19x-18) \)
\( 40x - 20 \ge 57x - 54 \)
\( 54 - 20 \ge 57x - 40x \)
\( 34 \ge 17x \Rightarrow 2 \ge x \Rightarrow x \le 2 \)
Solution: \( (-\infty, 2] \)
Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.
Q17
\( 3x - 2 < 2x + 1 \)
Answer:
\( 3x - 2x < 1 + 2 \Rightarrow x < 3 \).
Solution: \( (-\infty, 3) \).
(Paste image URL for the graph of x < 3 here)
Q18
\( 5x - 3 \ge 3x - 5 \)
Answer:
\( 5x - 3x \ge -5 + 3 \Rightarrow 2x \ge -2 \Rightarrow x \ge -1 \).
Solution: \( [-1, \infty) \).
(Paste image URL for the graph of x >= -1 here)
Q19
\( 3(1 - x) < 2(x + 4) \)
Answer:
\( 3 - 3x < 2x + 8 \Rightarrow 3 - 8 < 2x + 3x \Rightarrow -5 < 5x \Rightarrow -1 < x \Rightarrow x > -1 \).
Solution: \( (-1, \infty) \).
(Paste image URL for the graph of x > -1 here)
Q20
\( \frac{x}{2} \ge \frac{5x-2}{3} - \frac{7x-3}{5} \)
Answer:
\( \frac{x}{2} \ge \frac{5(5x-2) - 3(7x-3)}{15} \)
\( \frac{x}{2} \ge \frac{25x - 10 - 21x + 9}{15} \)
\( \frac{x}{2} \ge \frac{4x - 1}{15} \)
\( 15x \ge 2(4x - 1) \Rightarrow 15x \ge 8x - 2 \Rightarrow 7x \ge -2 \Rightarrow x \ge -\frac{2}{7} \).
Solution: \( [-\frac{2}{7}, \infty) \).
(Paste image URL for the graph of x >= -2/7 here)
Q21
Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Answer:
Let \( x \) be the marks in the third test.
Average \( \ge 60 \Rightarrow \frac{70 + 75 + x}{3} \ge 60 \)
\( 145 + x \ge 180 \)
\( x \ge 180 - 145 \Rightarrow x \ge 35 \).
Minimum marks = 35
Q22
To receive Grade 'A' in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita's marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade 'A' in the course.
Answer:
Let \( x \) be the marks in the 5th exam.
Average \( \ge 90 \Rightarrow \frac{87 + 92 + 94 + 95 + x}{5} \ge 90 \)
\( \frac{368 + x}{5} \ge 90 \)
\( 368 + x \ge 450 \)
\( x \ge 450 - 368 \Rightarrow x \ge 82 \).
Minimum marks = 82
Q23
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Answer:
Let the two consecutive odd positive integers be \( x \) and \( x+2 \).
Condition 1: Both are smaller than 10 \(\Rightarrow x < 10\) and \( x+2 < 10 \Rightarrow x < 8 \).
Condition 2: Sum is more than 11 \(\Rightarrow x + (x+2) > 11 \Rightarrow 2x + 2 > 11 \Rightarrow 2x > 9 \Rightarrow x > 4.5 \).
So, \( 4.5 < x < 8 \).
Since \( x \) is an odd integer, possible values for \( x \) are 5 and 7.
Pairs are: (5, 7) and (7, 9).
Q24
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Answer:
Let consecutive even integers be \( x \) and \( x+2 \).
Condition 1: Both larger than 5 \(\Rightarrow x > 5 \).
Condition 2: Sum less than 23 \(\Rightarrow x + (x+2) < 23 \Rightarrow 2x + 2 < 23 \Rightarrow 2x < 21 \Rightarrow x < 10.5 \).
So, \( 5 < x < 10.5 \).
Possible even values for \( x \) are 6, 8, 10.
Pairs are: (6, 8), (8, 10), and (10, 12).
Q25
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
Answer:
Let the length of the shortest side be \( x \) cm.
Longest side = \( 3x \).
Third side = \( 3x - 2 \).
Perimeter \( \ge 61 \Rightarrow x + 3x + (3x - 2) \ge 61 \)
\( 7x - 2 \ge 61 \Rightarrow 7x \ge 63 \Rightarrow x \ge 9 \).
Minimum length of shortest side = 9 cm.
Q26
A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?
Answer:
Let length of shortest piece be \( x \).
Second piece = \( x + 3 \).
Third piece = \( 2x \).
Total length constraint: Sum of lengths \( \le 91 \)
\( x + (x + 3) + 2x \le 91 \Rightarrow 4x + 3 \le 91 \Rightarrow 4x \le 88 \Rightarrow x \le 22 \).
Constraint 2: Third piece \(\ge\) Second piece + 5
\( 2x \ge (x + 3) + 5 \Rightarrow 2x \ge x + 8 \Rightarrow x \ge 8 \).
Combining both: \( 8 \le x \le 22 \).
Possible lengths of shortest board are between 8 cm and 22 cm (inclusive).
Solve the inequalities in Exercises 1 to 6.
Q1
\( 2 \le 3x - 4 \le 5 \)
Answer:
Add 4 to all parts:
\( 2 + 4 \le 3x \le 5 + 4 \)
\( 6 \le 3x \le 9 \)
Divide by 3:
\( 2 \le x \le 3 \)
Solution: \( [2, 3] \)
Q2
\( 6 \le -3 (2x - 4) < 12 \)
Answer:
Divide by -3 (inequality signs reverse):
\( -2 \ge 2x - 4 > -4 \)
\( -4 < 2x - 4 \le -2 \)
Add 4:
\( 0 < 2x \le 2 \)
Divide by 2:
\( 0 < x \le 1 \)
Solution: \( (0, 1] \)
Q3
\( -3 \le 4 - \frac{7x}{2} \le 18 \)
Answer:
Subtract 4:
\( -7 \le -\frac{7x}{2} \le 14 \)
Multiply by -2 (inequality signs reverse):
\( 14 \ge 7x \ge -28 \)
\( -28 \le 7x \le 14 \)
Divide by 7:
\( -4 \le x \le 2 \)
Solution: \( [-4, 2] \)
Q4
\( -15 < \frac{3(x-2)}{5} \le 0 \)
Answer:
Multiply by 5:
\( -75 < 3(x-2) \le 0 \)
Divide by 3:
\( -25 < x-2 \le 0 \)
Add 2:
\( -23 < x \le 2 \)
Solution: \( (-23, 2] \)
Q5
\( -12 < 4 - \frac{3x}{-5} \le 2 \)
Answer:
\( -12 < 4 + \frac{3x}{5} \le 2 \)
Subtract 4:
\( -16 < \frac{3x}{5} \le -2 \)
Multiply by 5:
\( -80 < 3x \le -10 \)
Divide by 3:
\( -\frac{80}{3} < x \le -\frac{10}{3} \)
Solution: \( (-\frac{80}{3}, -\frac{10}{3}] \)
Q6
\( 7 \le \frac{3x + 11}{2} \le 11 \)
Answer:
Multiply by 2:
\( 14 \le 3x + 11 \le 22 \)
Subtract 11:
\( 3 \le 3x \le 11 \)
Divide by 3:
\( 1 \le x \le \frac{11}{3} \)
Solution: \( [1, \frac{11}{3}] \)
Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.
Q7
\( 5x + 1 > -24, 5x - 1 < 24 \)
Answer:
Inequality 1: \( 5x > -25 \Rightarrow x > -5 \).
Inequality 2: \( 5x < 25 \Rightarrow x < 5 \).
Combined Solution: \( -5 < x < 5 \).
(Paste image URL for the graph of -5 < x < 5 here)
Q8
\( 2(x - 1) < x + 5, \quad 3(x + 2) > 2 - x \)
Answer:
Inequality 1: \( 2x - 2 < x + 5 \Rightarrow x < 7 \).
Inequality 2: \( 3x + 6 > 2 - x \Rightarrow 4x > -4 \Rightarrow x > -1 \).
Combined Solution: \( -1 < x < 7 \).
(Paste image URL for the graph of -1 < x < 7 here)
Q9
\( 3x - 7 > 2(x - 6), \quad 6 - x > 11 - 2x \)
Answer:
Inequality 1: \( 3x - 7 > 2x - 12 \Rightarrow x > -5 \).
Inequality 2: \( 6 - x > 11 - 2x \Rightarrow x > 5 \).
Combined Solution (Intersection): \( x > 5 \).
Solution: \( (5, \infty) \).
(Paste image URL for the graph of x > 5 here)
Q10
\( 5(2x - 7) - 3(2x + 3) \le 0, \quad 2x + 19 \le 6x + 47 \)
Answer:
Inequality 1: \( 10x - 35 - 6x - 9 \le 0 \Rightarrow 4x - 44 \le 0 \Rightarrow 4x \le 44 \Rightarrow x \le 11 \).
Inequality 2: \( 2x + 19 \le 6x + 47 \Rightarrow 19 - 47 \le 6x - 2x \Rightarrow -28 \le 4x \Rightarrow -7 \le x \).
Combined Solution: \( -7 \le x \le 11 \).
(Paste image URL for the graph of -7 <= x <= 11 here)
Q11
A solution is to be kept between \( 68^\circ \) F and \( 77^\circ \) F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by \( F = \frac{9}{5} C + 32 \)?
Answer:
Given \( 68 \le F \le 77 \).
Substitute \( F = \frac{9}{5} C + 32 \):
\( 68 \le \frac{9}{5} C + 32 \le 77 \)
Subtract 32:
\( 36 \le \frac{9}{5} C \le 45 \)
Multiply by 5/9:
\( 36 \times \frac{5}{9} \le C \le 45 \times \frac{5}{9} \)
\( 20 \le C \le 25 \).
Range: Between 20°C and 25°C.
Q12
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:
Let \( x \) litres of 2% solution be added.
Total mixture volume = \( (640 + x) \) litres.
Total acid = \( 8\% \) of 640 + \( 2\% \) of \( x = \frac{8}{100}(640) + \frac{2}{100}(x) \).
Condition 1: Mixture > 4% acid
\( \frac{8(640) + 2x}{100} > \frac{4}{100}(640 + x) \)
\( 5120 + 2x > 2560 + 4x \)
\( 2560 > 2x \Rightarrow x < 1280 \).
Condition 2: Mixture < 6% acid
\( \frac{8(640) + 2x}{100} < \frac{6}{100}(640 + x) \)
\( 5120 + 2x < 3840 + 6x \)
\( 1280 < 4x \Rightarrow x > 320 \).
Range: More than 320 litres but less than 1280 litres.
Q13
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 \( x \) litres of water (0% acid) be added.
Total mixture = \( (1125 + x) \).
Total acid = \( 45\% \) of 1125 = \( 0.45 \times 1125 = 506.25 \).
Condition 1: Mixture > 25%
\( 506.25 > 0.25(1125 + x) \)
\( 506.25 > 281.25 + 0.25x \Rightarrow 225 > 0.25x \Rightarrow x < 900 \).
Condition 2: Mixture < 30%
\( 506.25 < 0.30(1125 + x) \)
\( 506.25 < 337.5 + 0.3x \Rightarrow 168.75 < 0.3x \Rightarrow x > 562.5 \).
Range: Between 562.5 and 900 litres.
Q14
IQ of a person is given by the formula \( IQ = \frac{MA}{CA} \times 100 \), where MA is mental age and CA is chronological age. If \( 80 \le IQ \le 140 \) for a group of 12 years old children, find the range of their mental age.
Answer:
Given \( CA = 12 \) and \( 80 \le IQ \le 140 \).
\( 80 \le \frac{MA}{12} \times 100 \le 140 \)
Divide by 100:
\( 0.8 \le \frac{MA}{12} \le 1.4 \)
Multiply by 12:
\( 9.6 \le MA \le 16.8 \).
Range: \( 9.6 \le MA \le 16.8 \).