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1. Solve 24x < 100, when(i) x is a natural number.(ii) x is an integer.2. Solve – 12x > 30, when(i) x is a natural number.(ii) x is an integer.3. 3 (1 – x) < 2 (x + 4)4. 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.5. 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. |
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Answer» Q1. (i) \(24x < 100\) \(⇒\) \(x < \frac{100}{24}\) (Dividing both sides by 24) \(⇒\) \(x < \frac{25}{6} \) The natural numbers less than 4 are 1, 2, 3 ∴ The solution set is {1, 2, 3} (ii) The integers less than 4 are … -3, -2, -1, 0, 1, 2, 3, 4 ∴ The solution set is {… -3, -2, -1, 0, 1, 2, 3, 4}. Q2. (i) \(-12x > 30 \) \(⇒\) \(x > \frac{-30}{12} \) \(⇒\) \(x > \frac{-5}{2} \) There is no natural number less than, \(\frac{-5}{2} \) thus there is no solution for this inequality. (ii) The integers less than \(\frac{-5}{2} \) are … -5, -4, -3 The solution set is {…-5, -4, -3}. Q3. Sol:- \( 3(1 – x)< 2 (x + 4)\) \(⇒\) \(3 - 3x < 2x + 8\) \(⇒\) \(3 < 5x + 8\) \(⇒\) \(-5 < 5x\) \(⇒\) \(x > - 1\) \(∴ x ∈ (-1, ∞)\) Q4. Let the minimum marks be x. Then, \(⇒\) \(\frac{70 + 75 + x}{3} ≥ 60 \) \(⇒\) \(\frac{145 + x}{3} ≥ 60 \) \(⇒\) \(145 + x ≥ 180\) \(⇒\) \(x ≥ 35\) Ravi should get minimum 35 marks in the third set to have an average of at least 60 marks. Q5. Let the minimum marks scored by Sunita be x. Then, \(⇒\) \(\frac{87 + 92 + 94 + 95 + x}{5} ≥ 90 \) \(⇒\) \(\frac{368 + x}{5} ≥ 90 \) \(⇒\) \(368 + x ≥ 450 \) \(⇒\) \(x ≥ 82\) The minimum marks scored by Sunita must be 82. |
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