How to solve 3 kinds of inequalities

CAT Exam
As if solving inequalities wasn’t already hard enough, sometimes the way a CAT question is framed will make us wonder which answer option to choose, even after we have already solved solved the problem. Let’s look at three different question formats today to understand the difference between them:
  1. Must Be True
  2. Could Be True
  3. Complete Range
Case 1: Must Be True If |-x/3 + 1| < 2, which of the following must be true? (A) x > 0 (B) x < 8 (C) x > -4 (D) 0 < x < 3 (E) None of the above We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related. We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. |x/3 – 1| < 2 (1/3) * |x – 3| < 2 |x – 3| < 6 The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range. The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question: (A) x > 0 Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2. (B) x < 8 Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5 (C) x > -4 Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true. (D) 0 < x < 3 Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3. Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true). Case 2: Could Be True If −1 < x < 5, which is the following could be true? (A) 2x > 10 (B) x > 17/3 (C) x^2 > 27 (D) 3x + x^2 < −2 (E) 2x – x^2 < 0 Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is  −1 < x < 5 and the other will be the correct answer choice. We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options: (A) 2x > 10 x > 5 No values between -1 and 5 will be greater than 5, so this cannot be true. (B) x > 17/3 x > 5.67 No values between -1 and 5 will be greater than 5.67, so this cannot be true. (C) x^2 > 27 x^2 – 27 > 0 x > 3*√(3) or x < -3*√(3) √(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.   (D) 3x + x^2 < −2 x^2 + 3x + 2 < 0 (x + 1)(x + 2) < 0 -2 < x < -1 No values of x will lie between -2 and -1, so this also cannot be true. (E) 2x – x^2 < 0 x * (x – 2) > 0 x > 2 or x < 0 If -1 < x < 5, then x could lie between -1 and 0 (x < 0 is possible) or between 2 and 5 (x > 2 is possible). Therefore, the correct answer is E. Case 3: Complete Range Which of the following represents the complete range of x over which x^3 – 4x^5 < 0? (A) 0 < |x| < ½ (B) |x| > ½ (C) -½ < x < 0 or ½ < x (D) x < -½ or 0 < x < ½ (E) x < -½ or x > 0 We have two linked inequalities, but the relation between them will be a bit different again. One of the inequalities is  x^3 – 4x^5 < 0 and the other will be the correct answer choice. We are given that x^3 – 4x^5 < 0. This inequality can be solved to: x^3 ( 1 – 4x^2) < 0 x^3*(2x + 1)*(2x – 1) > 0 > 1/2 or -1/2 < x < 0 This is our universe of the values of x. It is given that all values of x lie in this range. Here, the question asks us the complete range of x. So we need to look for exactly this range. This is given in answer choice C, and therefore C is our answer. We hope these practice problems will help you become able to distinguish between the three cases now.

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