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23 Sep 2024, 06:24
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sarthak1701
"MUST BE TRUE" ("ALWAYS TRUE"/"IS TRUE") questions:
- These questions ask which statement is always true for every valid set of numbers. If you can find just one valid set of numbers where a statement is not true, it means the statement is not always true and therefore not the correct answer. So, for 'MUST BE TRUE' questions, the plug-in method is good for eliminating an option, but it does not provide a 100% guarantee that an option is always true. For "MUST BE TRUE" questions, when using the plug-in method, if you find that more than one option appears to be correct for a particular number or set of numbers, try using different numbers to double-check. Reevaluate only those options that were previously considered correct.
"COULD BE TRUE" questions:
- The questions that ask which of the following statements could be true are different. If you can demonstrate that a statement is true for a specific set of numbers, it implies that the statement could be true and therefore is a correct answer.
You can practice more questions of this type HERE.
Hope it helps.
11 Nov 2024, 06:31
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Solution:
We start by inferring from the question stem, which states that X > Y^2 > Z^4. This implies:
• X must be greater than Y^2, and Y^2 must be greater than Z^4, X > 0 , Y^2 > 0
• Since squares and fourth powers of numbers between 0 and 1 are smaller than the original numbers, we should consider values for Y and Z that are between 0 and 1 for some cases to verify each statement.
Checking Each Statement:
1. Statement I: X > Y > Z
values: X = 2, Y = 1, Z = 0
Check X > Y^2 > Z^4:
so 2 > 1 > 0, which satisfies X > Y^2 > Z^4.
Conclusion: Statement I could be true.
2. Statement II: Z > Y > X
Values: X = 1/4, Y = 1/3, Z = 1/2
Check X > Y^2 > Z^4:
Y^2 = (1/3)^2 = 1/9 and Z^4 = (1/2)^4 = 1/16, so 1/4 > 1/9 > 1/16, which satisfies X > Y^2 > Z^4.
Check Z > Y > X:
1/2 > 1/3 > 1/4, which holds.
Conclusion: Statement II could be true.
3. Statement III: X > Z > Y
Values: X = 1, Z = 1/3, Y = 1/4
Check X > Y^2 > Z^4:
Y^2 = (1/4)^2 = 1/16 and Z^4 = (1/3)^4 = 1/81, so 1 > 1/16 > 1/81, which satisfies X > Y^2 > Z^4.
Check X > Z > Y:
1 > 1/3 > 1/4, which holds.
Conclusion: Statement III could be true.
Final Conclusion: Answer: E
We start by inferring from the question stem, which states that X > Y^2 > Z^4. This implies:
• X must be greater than Y^2, and Y^2 must be greater than Z^4, X > 0 , Y^2 > 0
• Since squares and fourth powers of numbers between 0 and 1 are smaller than the original numbers, we should consider values for Y and Z that are between 0 and 1 for some cases to verify each statement.
Checking Each Statement:
1. Statement I: X > Y > Z
values: X = 2, Y = 1, Z = 0
Check X > Y^2 > Z^4:
so 2 > 1 > 0, which satisfies X > Y^2 > Z^4.
Conclusion: Statement I could be true.
2. Statement II: Z > Y > X
Values: X = 1/4, Y = 1/3, Z = 1/2
Check X > Y^2 > Z^4:
Y^2 = (1/3)^2 = 1/9 and Z^4 = (1/2)^4 = 1/16, so 1/4 > 1/9 > 1/16, which satisfies X > Y^2 > Z^4.
Check Z > Y > X:
1/2 > 1/3 > 1/4, which holds.
Conclusion: Statement II could be true.
3. Statement III: X > Z > Y
Values: X = 1, Z = 1/3, Y = 1/4
Check X > Y^2 > Z^4:
Y^2 = (1/4)^2 = 1/16 and Z^4 = (1/3)^4 = 1/81, so 1 > 1/16 > 1/81, which satisfies X > Y^2 > Z^4.
Check X > Z > Y:
1 > 1/3 > 1/4, which holds.
Conclusion: Statement III could be true.
Final Conclusion: Answer: E
04 Apr 2025, 00:06
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Thank you for very nice explanations:
my doubt is in case 3:
III. x>z>y
in order to make this condition true we are assuming y>y^2,
can we also make the same condition true by assuming y<y^2 ?
my doubt is in case 3:
III. x>z>y
in order to make this condition true we are assuming y>y^2,
can we also make the same condition true by assuming y<y^2 ?
