[GMAT math practice question]
Is x < y < z ?
1) |x+2| < y < z+2
2) |x-2| < y < z-2
Is x < y < z ?
This topic has expert replies
- Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]
- Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Now, by condition 1), x < x + 2 ≤ | x + 2 | < y < z + 2, and x < y.
By condition 2), |x-2| < y < z-2 < z, and so y < z.
Thus, both conditions together are sufficient since they yield x < y < z, and the unique answer is 'yes'.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
If x = 1, y = 4, and z = 7, then the answer is 'yes'.
If x = 1, y = 4, and z = 3, then the answer is 'no'.
Condition 1) is not sufficient since it doesn't yield a unique answer.
Condition 2)
If x = 1, y = 4, and z = 7, then the answer is 'yes'.
If x = 5, y = 4, and z = 7, then the answer is 'no'.
Condition 2) is not sufficient since it doesn't yield a unique answer.
Therefore, C is the answer.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 3 variables (x, y and z) and 0 equations, E is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Now, by condition 1), x < x + 2 ≤ | x + 2 | < y < z + 2, and x < y.
By condition 2), |x-2| < y < z-2 < z, and so y < z.
Thus, both conditions together are sufficient since they yield x < y < z, and the unique answer is 'yes'.
Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
If x = 1, y = 4, and z = 7, then the answer is 'yes'.
If x = 1, y = 4, and z = 3, then the answer is 'no'.
Condition 1) is not sufficient since it doesn't yield a unique answer.
Condition 2)
If x = 1, y = 4, and z = 7, then the answer is 'yes'.
If x = 5, y = 4, and z = 7, then the answer is 'no'.
Condition 2) is not sufficient since it doesn't yield a unique answer.
Therefore, C is the answer.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]