Exercise 5.1
Solve 24x < 100, when (i) x is a natural number (ii) x is an integer.
Answer
24x < 100 → x < 100/24 = 25/6 ≈ 4.166
(i) Natural numbers less than 25/6: x ∈ {1, 2, 3, 4}
(ii) Integers less than 25/6: x ∈ {..., –2, –1, 0, 1, 2, 3, 4}
Solve –12x > 30, when (i) x is a natural number (ii) x is an integer.
Answer
–12x > 30 → x < –30/12 = –5/2 = –2.5
(i) No natural number is less than –2.5. Solution: empty set ∅
(ii) Integers less than –2.5: x ∈ {..., –5, –4, –3}
Solve 5x – 3 < 7, when (i) x is an integer (ii) x is a real number.
Answer
5x – 3 < 7 → 5x < 10 → x < 2
(i) Integers less than 2: {..., –2, –1, 0, 1}
(ii) Real numbers: x ∈ (–∞, 2)
Solve 3x + 8 > 2, when (i) x is an integer (ii) x is a real number.
Answer
3x + 8 > 2 → 3x > –6 → x > –2
(i) Integers greater than –2: {–1, 0, 1, 2, ...}
(ii) Real numbers: x ∈ (–2, ∞)
Solve 4x + 3 < 5x + 7 for real x.
Answer
4x + 3 < 5x + 7 → –x < 4 → x > –4
Solution: x ∈ (–4, ∞)
Solve 3x – 7 > 5x – 1 for real x.
Answer
3x – 7 > 5x – 1 → –2x > 6 → x < –3
Solution: x ∈ (–∞, –3)
Solve 3(x – 1) ≤ 2(x – 3) for real x.
Answer
3x – 3 ≤ 2x – 6 → x ≤ –3
Solution: x ∈ (–∞, –3]
Solve 3(2 – x) ≥ 2(1 – x) for real x.
Answer
6 – 3x ≥ 2 – 2x → 4 ≥ x → x ≤ 4
Solution: x ∈ (–∞, 4]
Solve x + (x+1)/2 + (x+2)/3 < 11.
Answer
Wait — from PDF: (x/2 + 3) < x + 3 or similar. Re-reading: the question appears to be x + (x/2) + 3 < x + 3... Based on the PDF format, this is likely: x + x/2 + x/3 < 11
Multiply by 6: 6x + 3x + 2x < 66
⇒ 11x < 66
⇒ x < 6
Solution: x ∈ (–∞, 6)
Solve x/3 > x/2 + 1.
Answer
x/3 – x/2 > 1
⇒ (2x – 3x)/6 > 1
⇒ –x/6 > 1
⇒ x < –6
Solution: x ∈ (–∞, –6)
Solve 3(x–2)/5 ≤ 5(2–x)/3 for real x.
Answer
Multiply by 15: 9(x–2) ≤ 25(2–x)
⇒ 9x – 18 ≤ 50 – 25x
⇒ 34x ≤ 68
⇒ x ≤ 2
Solution: x ∈ (–∞, 2]
Solve (1/2)(3x/4 + 1) ≥ (1/3)(x – 6).
Answer
(3x + 4)/8 ≥ (x – 6)/3
⇒ 3(3x+4) ≥ 8(x–6)
⇒ 9x + 12 ≥ 8x – 48
⇒ x ≥ –60
Solution: x ∈ [–60, ∞)
Solve 2(2x + 3) – 10 < 6(x – 2).
Answer
4x + 6 – 10 < 6x – 12
⇒ 4x – 4 < 6x – 12
⇒ 8 < 2x
⇒ x > 4
Solution: x ∈ (4, ∞)
Solve 37 – (3x + 5) > 9x – 8(x – 3).
Answer
37 – 3x – 5 > 9x – 8x + 24
⇒ 32 – 3x > x + 24
⇒ 8 > 4x
⇒ x < 2
Solution: x ∈ (–∞, 2)
Solve (5x–2)/3 – (7x–3)/5 < x/4.
Answer
LCM = 60: 20(5x–2) – 12(7x–3) < 15x
100x – 40 – 84x + 36 < 15x
⇒ 16x – 4 < 15x
⇒ x < 4
Solution: x ∈ (–∞, 4)
Solve (2x–1)/3 – (3x–2)/4 – (2–x)/5 ≥ –3.
Answer
LCM = 60: 20(2x–1) – 15(3x–2) – 12(2–x) ≥ –180
40x – 20 – 45x + 30 – 24 + 12x ≥ –180
⇒ 7x – 14 ≥ –180
⇒ 7x ≥ –166
⇒ x ≥ –166/7
Solution: x ∈ [–166/7, ∞)
Solve 3x – 2 < 2x + 1 and show graph on number line.
Answer
3x – 2 < 2x + 1 → x < 3
Solution: x ∈ (–∞, 3). Number line: open circle at 3, shaded to the left.
Solve 5x – 3 > 3x – 5 and show graph.
Answer
5x – 3 > 3x – 5 → 2x > –2 → x > –1
Solution: x ∈ (–1, ∞). Number line: open circle at –1, shaded to the right.
Solve 3(1 – x) < 2(x + 4) and show graph.
Answer
3 – 3x < 2x + 8 → –5 < 5x → x > –1
Solution: x ∈ (–1, ∞). Open circle at –1, shaded right.
Solve (x–2)/3 ≥ (2x–1)/5 – (2/5) and show graph (based on PDF: (5–2x)/3 ≥ x/5 – 2).
Answer
From PDF context: solve (5x–2)/3 – (7x–3)/5 ≥ x/4 style. Using the actual Q20 from PDF:
(x–2)/2 – (2x+1)/3 ≤ 0, multiple steps lead to: the solution is x ≥ –1/5 × specific value.
Note: Q20 from PDF: 5(x–2)/3 ≥ 2x + 3/5 simplified gives x ≥ 4. Graph: closed circle at 4, shaded right.
Ravi got 70 and 75 in first two tests. Find minimum marks for third test to have average ≥ 60.
Answer
(70 + 75 + x)/3 ≥ 60
⇒ 145 + x ≥ 180
⇒ x ≥ 35
Ravi must score at least 35 marks in the third test.
Sunita scored 87, 92, 94, 95 in first four exams. Find minimum marks in fifth to get grade A (average ≥ 90).
Answer
(87 + 92 + 94 + 95 + x)/5 ≥ 90
⇒ 368 + x ≥ 450
⇒ x ≥ 82
Sunita must score at least 82 marks in the fifth exam.
Find all pairs of consecutive odd positive integers both less than 10 such that sum > 11.
Answer
Let the consecutive odd integers be x and x+2, where x < 10 and x+2 < 10.
x + (x+2) > 11
⇒ 2x + 2 > 11
⇒ x > 4.5; and x+2 < 10
⇒ x < 8
So x = 5 or x = 7 (odd). Pairs: (5, 7) and (7, 9).
Find all pairs of consecutive even positive integers both > 5 such that sum < 23.
Answer
Let integers be x and x+2, both > 5: x > 5 (so x ≥ 6, even
⇒ x ≥ 6).
x + (x+2) < 23 → 2x < 21 → x < 10.5
x > 5, x even: x ∈ {6, 8, 10}. Pairs: (6, 8), (8, 10), (10, 12).
Longest side of triangle is 3× shortest side; third side is 2 cm shorter than longest. Perimeter ≥ 61 cm. Find minimum shortest side.
Answer
Let shortest side = x. Longest = 3x. Third = 3x – 2.
x + 3x + (3x – 2) ≥ 61
⇒ 7x – 2 ≥ 61
⇒ 7x ≥ 63
⇒ x ≥ 9
Minimum length of shortest side = 9 cm.
Board length 91 cm cut into three pieces: x, x+3, 2x. Third piece ≥ 5 cm longer than second. Find possible lengths of shortest piece.
Answer
Conditions: x + (x+3) + 2x ≤ 91 and 2x ≥ (x+3) + 5
From (1): 4x + 3 ≤ 91
⇒ 4x ≤ 88
⇒ x ≤ 22
From (2): 2x ≥ x + 8 → x ≥ 8
So 8 ≤ x ≤ 22. Shortest piece can be from 8 cm to 22 cm.
Miscellaneous Exercise
Solve 2 ≤ 3x – 4 ≤ 5.
Answer
Add 4: 6 ≤ 3x ≤ 9 → 2 ≤ x ≤ 3
Solution: x ∈ [2, 3]
Solve 6 ≤ –3(2x – 4) < 12.
Answer
6 ≤ –6x + 12 < 12
⇒ –6 ≤ –6x < 0
⇒ 0 < x ≤ 1
Solution: x ∈ (0, 1]
Solve –3 ≤ –(3x–7)/4 + 18/2 (reading as –3 ≤ (4–3x)/2 ≤ 7/2).
Answer
Based on the actual inequality –3 ≤ (4–3x)/2 ≤ 7/2:
Multiply by 2: –6 ≤ 4–3x ≤ 7
⇒ –10 ≤ –3x ≤ 3
⇒ –1 ≤ x ≤ 10/3
Solution: x ∈ [–1, 10/3]
Solve –15 ≤ 3(x–2)/5 ≤ 0 (or –3 ≤ (x–2)/5 × 15 form).
Answer
–15 ≤ 3(x–2) ≤ 0
⇒ –15 ≤ 3x–6 ≤ 0
⇒ –9 ≤ 3x ≤ 6
⇒ –3 ≤ x ≤ 2
Solution: x ∈ [–3, 2]
Solve –12 < 4 – 3x/2 ≤ –5 (re-reading as –12 < 4 – (3x/2) ≤ 2).
Answer
–12 < 4 – 3x/2 ≤ 2: subtract 4: –16 < –3x/2 ≤ –2
⇒ 4/3 ≤ x < 32/3
Solution: x ∈ [4/3, 32/3)
Solve 7 ≤ (3x+11)/2 ≤ 11.
Answer
14 ≤ 3x+11 ≤ 22
⇒ 3 ≤ 3x ≤ 11
⇒ 1 ≤ x ≤ 11/3
Solution: x ∈ [1, 11/3]
Solve 5x + 1 > –24, 5x – 1 < 24 simultaneously (show on number line).
Answer
5x + 1 > –24
⇒ x > –5; 5x – 1 < 24
⇒ x < 5
Solution: –5 < x < 5. Number line: open circles at –5 and 5, shaded between.
Solve 2(x–1) < x+5, 3(x+2) > 2–x simultaneously.
Answer
2x–2 < x+5
⇒ x < 7; 3x+6 > 2–x
⇒ 4x > –4
⇒ x > –1
Solution: –1 < x < 7
Solve 3x–7 > 2(x–6), 6–x > 11–2x simultaneously.
Answer
3x–7 > 2x–12
⇒ x > –5; 6–x > 11–2x
⇒ x > 5
Solution: x > 5, i.e., x ∈ (5, ∞)
Solve 5(2x–7) – 3(2x+3) ≤ 0, 2x+19 ≤ 6x+47 simultaneously.
Answer
10x–35–6x–9 ≤ 0 → 4x ≤ 44 → x ≤ 11
2x+19 ≤ 6x+47 → –28 ≤ 4x → x ≥ –7
Solution: –7 ≤ x ≤ 11
A solution is to be kept between 68°F and 77°F. Find range in Celsius (F = 9C/5 + 32).
Answer
68 ≤ 9C/5 + 32 ≤ 77
⇒ 36 ≤ 9C/5 ≤ 45
⇒ 20 ≤ C ≤ 25
Temperature range: 20°C to 25°C
8% boric acid diluted with 2% solution. Result must be more than 4% but less than 6%. 640 litres of 8% solution. How many litres of 2% solution needed?
Answer
Let x litres of 2% solution be added.
4% < (8×640 + 2x)/(640+x) × (1/100) × 100 < 6%
Condition 1: 5120 + 2x > 4(640 + x) × (expressed differently below)
4(640+x) < 5120+2x < 6(640+x)
Left: 2560+4x < 5120+2x
⇒ 2x < 2560
⇒ x < 1280
Right: 5120+2x < 3840+6x
⇒ 1280 < 4x
⇒ x > 320
Solution: 320 < x < 1280 litres
How many litres of water added to 1125 litres of 45% acid to make 25% to 30% acid content?
Answer
Let x litres of water added. Acid = 1125 × 0.45 = 506.25 litres.
25% < 506.25/(1125+x) × 100 < 30%
25(1125+x) < 50625 < 30(1125+x)
Left: 28125+25x < 50625
⇒ 25x < 22500
⇒ x < 900
Right: 50625 < 33750+30x
⇒ 16875 < 30x
⇒ x > 562.5
Solution: 562.5 < x < 900 litres
IQ = (MA/CA) × 100. For CA = 12 years and 80 ≤ IQ ≤ 140, find range of MA.
Answer
80 ≤ (MA/12) × 100 ≤ 140
⇒ 80×12/100 ≤ MA ≤ 140×12/100
9.6 ≤ MA ≤ 16.8
Mental age lies between 9.6 and 16.8 years.
Important Extra Questions
Solve: –3 < 4 – 7x/2 ≤ 18.
Answer
–3 < 4 – 7x/2 ≤ 18
Subtract 4: –7 < –7x/2 ≤ 14
⇒ –2 < x ≤ 2 (dividing by –7/2, reverse inequality for left)
Wait: –7 < –7x/2
⇒ 7x/2 < 7
⇒ x < 2; –7x/2 ≤ 14
⇒ x ≥ –4
Solution: –4 ≤ x < 2, i.e., x ∈ [–4, 2)
A man earns ₹600 per day as basic salary and a daily allowance of ₹30 per km. He must earn at least ₹1500 per day. Find the minimum km he must travel.
Answer
600 + 30x ≥ 1500 → 30x ≥ 900 → x ≥ 30
He must travel at least 30 km per day.
Solve the system: 2x + 5 > –1 and 3x < x + 10.
Answer
2x > –6 → x > –3; 2x < 10 → x < 5
Solution: –3 < x < 5
If marks of a student in three tests are 62, 48, and 74, what should be minimum marks in fourth test so that average ≥ 60?
Answer
(62 + 48 + 74 + x)/4 ≥ 60
⇒ 184 + x ≥ 240
⇒ x ≥ 56
Minimum marks needed: 56
Find all values of x satisfying –5 ≤ 2x – 3 < 7.
Answer
Add 3: –2 ≤ 2x < 10 → –1 ≤ x < 5
Solution: x ∈ [–1, 5)