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FROM Veritas Prep Blog: 3 Formats for GMAT Inequalities Questions You Need to Know 
As if solving inequalities wasn’t already hard enough, sometimes the way a GMAT 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:
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. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.) 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. (Details on how to solve such inequalities are discussed here.) (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 x > 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. Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter! Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! The post 3 Formats for GMAT Inequalities Questions You Need to Know appeared first on Veritas Prep Blog. 
FROM Veritas Prep Blog: A 700+ GMAT Quant Question on Races 
This week we will look at the question on races that we gave you last week. Question 3: A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds. Next, A gives B a head start of 3 mins and is beaten by 1000 m. Find the time in minutes in which A and B can run the race separately? (A) 8, 10 (B) 4, 5 (C) 5, 9 (D) 6, 9 (E) 7, 10 Solution: Now this question is a little tougher than the previous ones we saw last week. There are two scenarios given: 1 – A gives B a head start of 200 m and beats him by 30 seconds. 2 – A gives B a head start of 3 mins and is beaten by 1000m. Let’s study both of them and see what we can derive from them. Scenario 1: A gives B a start of 200m and beats him by 30 seconds. As we suggested before, we will start by making a diagram. A runs from the Start line till the finish line i.e. a total distance of 2000 m. A gives B a head start of 200 m so B starts, not from the starting point, but from 200 m ahead. A still beats him by 30 sec which means that A completes the race while B takes another 30 sec to complete it. So obviously A is much faster than B. In this race, A covers 2000m. In the same time, B covers the distance shown by the red line. Since B needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance. The green line distance is given by (1/2)*s where s is the speed of B in meters per minute. The distance B has actually covered in the same time as A is the distance shown by the red line. This distance will be (1/2)*s less than 1800 i.e. it will be [1800  (1/2)*s]. Scenario 2: A gives B a head start of 3mins and is beaten by 1000m. A gives B a head start of 3 mins means B starts running first while A sits at the starting point. After 3 mins, B covers the distance shown by the red line which we do not know yet. Now, A starts running too. B beats A by 1000 m which means that B reaches the end point while A is still 1000 m away from the end i.e. at the mid point of the 2000 m track. In this race, A covers a distance of 1000 m only. In that time, B covers the distance shown by the green line. The distance shown by the red line was covered by B in his first 3 mins i.e. this distance is 3*s. This distance shown by the green line is given by (2000 – 3s). Now you see that in the first race, A covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, B would also have covered half the previous distance. Distance covered by B in first race = 2*Distance covered by B in second race 1800 – (1/2)*s = 2*(2000 – 3s) (where s is the speed of B in meters/min) s = 400 meters/min Time taken by B to run a 2000 m race = Distance/Speed = 2000/400 = 5 min Only one option has time taken by B as 5 mins and that must be the answer. If required, you can easily calculate the time required by A too. Distance covered by B in scenario 1 = 1800 – (1/2)*s = 1600 m In the same time, A covers 2000 m which is a ratio of A:B = 5:4. Hence time taken by A:B will be 4:5. Answer (B) Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! 
FROM Veritas Prep Blog: A 700+ GMAT Quant Question on Races 
This week we will look at the question on races that we gave you last week. Question 3: A and B run a race of 2000 m. First, A gives B a head start of 200 m and beats him by 30 seconds. Next, A gives B a head start of 3 mins and is beaten by 1000 m. Find the time in minutes in which A and B can run the race separately? (A) 8, 10 (B) 4, 5 (C) 5, 9 (D) 6, 9 (E) 7, 10 Solution: Now this question is a little tougher than the previous ones we saw last week. There are two scenarios given: 1 – A gives B a head start of 200 m and beats him by 30 seconds. 2 – A gives B a head start of 3 mins and is beaten by 1000m. Let’s study both of them and see what we can derive from them. Scenario 1: A gives B a start of 200m and beats him by 30 seconds. As we suggested before, we will start by making a diagram. A runs from the Start line till the finish line i.e. a total distance of 2000 m. A gives B a head start of 200 m so B starts, not from the starting point, but from 200 m ahead. A still beats him by 30 sec which means that A completes the race while B takes another 30 sec to complete it. So obviously A is much faster than B. In this race, A covers 2000m. In the same time, B covers the distance shown by the red line. Since B needs another 30 sec ( i.e. 1/2 min) to cover the distance, he has not covered the green line distance. The green line distance is given by (1/2)*s where s is the speed of B in meters per minute. The distance B has actually covered in the same time as A is the distance shown by the red line. This distance will be (1/2)*s less than 1800 i.e. it will be [1800  (1/2)*s]. Scenario 2: A gives B a head start of 3mins and is beaten by 1000m. A gives B a head start of 3 mins means B starts running first while A sits at the starting point. After 3 mins, B covers the distance shown by the red line which we do not know yet. Now, A starts running too. B beats A by 1000 m which means that B reaches the end point while A is still 1000 m away from the end i.e. at the mid point of the 2000 m track. In this race, A covers a distance of 1000 m only. In that time, B covers the distance shown by the green line. The distance shown by the red line was covered by B in his first 3 mins i.e. this distance is 3*s. This distance shown by the green line is given by (2000 – 3s). Now you see that in the first race, A covers 2000m while in the second race, he covers only 1000m. So in the second race, he must have run for only half the time. Therefore, in half the time, B would also have covered half the previous distance. Distance covered by B in first race = 2*Distance covered by B in second race 1800 – (1/2)*s = 2*(2000 – 3s) (where s is the speed of B in meters/min) s = 400 meters/min Time taken by B to run a 2000 m race = Distance/Speed = 2000/400 = 5 min Only one option has time taken by B as 5 mins and that must be the answer. If required, you can easily calculate the time required by A too. Distance covered by B in scenario 1 = 1800 – (1/2)*s = 1600 m In the same time, A covers 2000 m which is a ratio of A:B = 5:4. Hence time taken by A:B will be 4:5. Answer (B) Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! 
FROM Veritas Prep Blog: 3 Formats for GMAT Inequalities Questions You Need to Know 
As if solving inequalities wasn’t already hard enough, sometimes the way a GMAT 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:
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. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.) 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. (Details on how to solve such inequalities are discussed here.) (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 x > 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. Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter! Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! The post 3 Formats for GMAT Inequalities Questions You Need to Know appeared first on Veritas Prep Blog. 

