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S98-15

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S98-15  [#permalink]

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New post 16 Sep 2014, 00:52
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A
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D
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Difficulty:

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Question Stats:

54% (02:24) correct 46% (02:09) wrong based on 24 sessions

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If \(a\), \(b\), and \(c\) are positive integers, with \(a \lt b \lt c\), are \(a\), \(b\), and \(c\) consecutive integers?


(1) \(\frac{1}{a} - \frac{1}{b} = \frac{1}{c}\)

(2) \(a + c = b^2 - 1\)

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Re S98-15  [#permalink]

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New post 16 Sep 2014, 00:52
Official Solution:


The question can be rephrased "Is \(b = a + 1\) and is \(c = a + 2\)?"

One way to approach the statements is to substitute these expressions involving \(a\) and solve for \(a\). Since this could involve a lot of algebra at the start, we can just substitute \(a + 1\) for \(b\) and test whether \(c = a + 2\), given that both are integers.

Statement 1: SUFFICIENT.

Following the latter method, we have
\(\frac{1}{a} - \frac{1}{a + 1} = \frac{1}{c}\)
\(\frac{a + 1}{a(a + 1)} - \frac{a}{a(a + 1)} = \frac{1}{c}\)
\(\frac{1}{a(a + 1)} = \frac{1}{c}\)
\(a^2 + a = c\)

Now we substitute a + 2 for c and examine the results:
\(a^2 + a = a + 2\)
\(a^2 = 2\)

\(a\) is the square root of 2. However, since \(a\) is supposed to be an integer, we know that our assumptions were false, and \(a\), \(b\), and \(c\) cannot be consecutive integers.

We can now answer the question with a definitive "No," making this statement sufficient.

We could also test numbers. Making \(a\) and \(b\) consecutive positive integers, we can solve the original equation \((\frac{1}{a} - \frac{1}{b} = \frac{1}{c})\). The first 4 possibilities are as follows:
\(\frac{1}{1} - \frac{1}{2} = \frac{1}{2}\)
\(\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\)
\(\frac{1}{3} - \frac{1}{4} = \frac{1}{12}\)
\(\frac{1}{4} - \frac{1}{5} = \frac{1}{20}\)

Examining the denominators, we can see that \(c = ab\). None of these triples so far are consecutive, and as \(a\) and \(b\) get larger, \(c\) will become more and more distant, leading us to conclude that \(a\), \(b\), and \(c\) are not consecutive.

Statement 2: SUFFICIENT.

Let's try substituting \((a + 1)\) for \(b\) and \((a + 2)\) for \(c\).
\(a + a + 2 = (a + 1)^2 - 1\)
\(2a + 2 = a^2 + 2a\)
\(2 = a^2\)

Again, we get that \(a\) must be the square root of 2. However, we know that \(a\) is an integer, so the assumptions must be false. We can answer the question with a definitive "No," and so the statement is sufficient.


Answer: D
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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: S98-15  [#permalink]

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New post 06 Feb 2019, 01:54
Why can't I just put the consecutive values like a=1, b=2 and c=3
and then try to find out if LHS=RHS
But that doesn't work, please explain where m i missing?
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Re: S98-15   [#permalink] 06 Feb 2019, 01:54
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