Is there a secret to these question types?
If 0 < a < b < c, which of the following statements must be true?
I. 2a > b + c
II. c - a > b - a
III. c/a < b/a
A) I only
B) II only
C) III only
D) I and II
E) II and III
B
Number Properties question?
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I does not need to be true : just let a = 1, b = 2, c = 3. In fact, I can never be true, because if a < b and a < c, then by adding those two inequalities together, you find that 2a < b + c.
For II, we know c > b from the question stem. Subtract a on both sides of that inequality, and you get c - a > b - a, so that must be true.
For III, we know c > b from the question stem. We can divide by a on both sides, because we know a is positive. We then get c/a > b/a, so III must be false.
In general for this kind of question, if you see something algebraic you can do to make the information in the stem more closely resemble the things you're trying to prove, try that and see where it leads. If you can't see anything like that to try, and can't see a conceptual reason why something ought to be true or false, then try plugging in two extremely different sets of numbers, and see what happens. Here I'd try simple values very close together and very far apart. If you find one of the roman numeral items is true for both of your very different sets of numbers, then there's a good chance it's always true. If you find it's false for one or both of your sets of numbers, then you've proved that it does not need to be true. That's not a foolproof strategy, because sometimes you'll miss some exceptional case, but if you don't see anything else to try, it gives you a good chance to get a correct answer.
For II, we know c > b from the question stem. Subtract a on both sides of that inequality, and you get c - a > b - a, so that must be true.
For III, we know c > b from the question stem. We can divide by a on both sides, because we know a is positive. We then get c/a > b/a, so III must be false.
In general for this kind of question, if you see something algebraic you can do to make the information in the stem more closely resemble the things you're trying to prove, try that and see where it leads. If you can't see anything like that to try, and can't see a conceptual reason why something ought to be true or false, then try plugging in two extremely different sets of numbers, and see what happens. Here I'd try simple values very close together and very far apart. If you find one of the roman numeral items is true for both of your very different sets of numbers, then there's a good chance it's always true. If you find it's false for one or both of your sets of numbers, then you've proved that it does not need to be true. That's not a foolproof strategy, because sometimes you'll miss some exceptional case, but if you don't see anything else to try, it gives you a good chance to get a correct answer.
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There is no secret. BTG is a forum in which experts will introduce you to all the possible approaches to any question.AbeNeedsAnswers wrote:Is there a secret to these question types?
If 0 < a < b < c, which of the following statements must be true?
I. 2a > b + c
II. c - a > b - a
III. c/a < b/a
A) I only
B) II only
C) III only
D) I and II
E) II and III
B
This one is a Must be True kind of question. In Must be True kind of questions, under all the circumstances, a statement under consideration must be true.
Let's take this one.
We have 0 < a < b < c. What do you make out of this information?
1. a, b and c all are positive numbers.
2. a, b and c are all not necessarily positive integers. One or none of them can be integers.
3. All can be less than 1. For example, a = 1/4; b = 1/3; c = 1/2.
Keeping these in mind, let's take the three statements.
I. 2a > b + c:
=> 2*SMALLEST number > A SMALLER number + A SMALL number
This is not possible as the sum of a SMALLER number & a SMALL number would always be greater than twice the SMALLEST number.
For example, say a = 1, b = 2, and c = 3.
2*1 < 2 + 3 => 2 < 5. This statement is false.
II. c - a > b - a:
=> c > b; cancelling out '-a' from both the sides. This is a must be a true statement. it is given that 0 < a < b < c
III. c/a < b/a:
=> c < b; cancelling 'a' from both the sides. Since 'a' is positive, we can cancel it without reversing the sign of inequality. Had it 'a' been negative, we would have to reverse the sign on the inequality. Had it been not known whether 'a' is positive or negative, we cannot cancel 'a.'
As seen in Statement II, this is false. We know that b < c.
The correct answer: B
Hope this helps!
-Jay
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If by that you mean Roman Numeral questions, then... no. RN's make it impossible to backsolve, estimate, or use a lot of the other tactics we can use on other PS questions. Don't try to game it based on whether I, II, or III shows up most often in answer choices.AbeNeedsAnswers wrote:Is there a secret to these question types?
They come in 2 flavors:
Which of the following MUST be true?
For these, try to DISPROVE each RN option. Try to think of a value / circumstance in which it would be untrue. Make sure to test extremes (very big & small), zero, negatives, fractions, etc, where applicable.
Which of the following COULD be true?
These are less common. For these, you just need to come up with an example that could be true in 1 circumstance.
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For more on Roman Numerals, see:
https://www.manhattanprep.com/gmat/blog ... -problems/
https://www.manhattanprep.com/gmat/blog ... at-part-2/
https://www.manhattanprep.com/gmat/blog ... -problems/
https://www.manhattanprep.com/gmat/blog ... -to-guess/
https://www.manhattanprep.com/gmat/blog ... -problems/
https://www.manhattanprep.com/gmat/blog ... at-part-2/
https://www.manhattanprep.com/gmat/blog ... -problems/
https://www.manhattanprep.com/gmat/blog ... -to-guess/
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Hi AbeNeedsAnswers,
We're told that 0 < A < B < C. We're asked which of the following statements MUST be true (which really means "which of the following is ALWAYS true no matter how many different examples we can come up with?"). We can TEST VALUES to try to prove that a given Roman Numeral is NOT true (and then eliminate the corresponding answers).
Let's start by TESTing the easiest option that we can:
A = 1, B = 2, C = 3
I. 2A > B + C
Here, is 2 > (2+3)? No, it clearly isn't, so Roman Numeral 1 is NOT always true.
Eliminate Answers A and D.
II. C - A> B - A
Here, is (3-1) > (2-1)? Yes. Let's keep this for now (although it's worth noting that we haven't proven that this is always true just yet).
III. C/A < B/A
Here is 3/1 < 2/1. No, it clearly isn't, so Roman Numeral 3 is NOT always true.
Eliminate Answers C and E.
There's only one answer remaining....
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
We're told that 0 < A < B < C. We're asked which of the following statements MUST be true (which really means "which of the following is ALWAYS true no matter how many different examples we can come up with?"). We can TEST VALUES to try to prove that a given Roman Numeral is NOT true (and then eliminate the corresponding answers).
Let's start by TESTing the easiest option that we can:
A = 1, B = 2, C = 3
I. 2A > B + C
Here, is 2 > (2+3)? No, it clearly isn't, so Roman Numeral 1 is NOT always true.
Eliminate Answers A and D.
II. C - A> B - A
Here, is (3-1) > (2-1)? Yes. Let's keep this for now (although it's worth noting that we haven't proven that this is always true just yet).
III. C/A < B/A
Here is 3/1 < 2/1. No, it clearly isn't, so Roman Numeral 3 is NOT always true.
Eliminate Answers C and E.
There's only one answer remaining....
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
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Let's analyze each Roman numeral.AbeNeedsAnswers wrote:Is there a secret to these question types?
If 0 < a < b < c, which of the following statements must be true?
I. 2a > b + c
II. c - a > b - a
III. c/a < b/a
A) I only
B) II only
C) III only
D) I and II
E) II and III
I. 2a > b + c
Since a < b and a < c, a + a = 2a < b + c. So Roman numeral I is false.
II. c - a > b - a
Adding a to both sides, we have c > b and we are given that c > b. So Roman numeral II is true.
III. c/a < b/a
Multiply both sides by a, we have c < b. So Roman numeral III is also false.
Answer: B
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