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S98-03

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S98-03  [#permalink]

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New post 16 Sep 2014, 00:52
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If \(x\), \(y\), and \(z\) are integers, with \(x \lt y \lt z\), what is the average (arithmetic mean) of \(x\), \(y\), and \(z\)?


(1) \((x + y)z = 5\)

(2) \(x + z \lt 3\)

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Re S98-03  [#permalink]

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


Since the average of \(x\), \(y\), and \(z\) is the sum of the three variables, divided by 3, we can rephrase the question as "what is the value of \(x + y + z\)?" We also note the restrictions on the possible values of \(x\), \(y\), and \(z\) - the variables must be integers in ascending order from \(x\) to \(z\) (not necessarily consecutive). Moreover, they must be different integers, since the inequality \(x \lt y \lt z\) indicates no equality among any of the variables. We note these conditions, but at this stage there is no simple way to apply them in a further rephrasing of the question.

Statement (2): INSUFFICIENT. We start with statement (2), which is the easier statement (it only contains 2 of the 3 variables). We can quickly come up with sets of values that satisfy this statement and the given conditions but that have different sums (or averages). For instance, the set \(x = 0\), \(y = 1\), and \(z = 2\) meets all conditions \((x + z = 2 \lt 3\), all variables are integers and in ascending order), with \(x + y + z = 3\). Another set (\(x = -1\), \(y = 0\), and \(z = 1\)) also meets all conditions but sums to 0. Thus, there is no single value determined by this statement.

Statement (1): INSUFFICIENT. The equation states that \(x + y\) (which must be an integer) multiplied by \(z\) (another integer) equals 5. Since 5 is a prime number, there are only 2 pairs of integers that multiply together to 5: 1 and 5, and -1 and -5. (Don't forget about the negative possibilities). Keeping the conditions that \(x \lt y \lt z\), we can construct the only sets that work:

\(x + y = 1\) and \(z = 5\) (There's no way to assign \(z = 1\) and \(x + y = 5\) while preserving \(x \lt y \lt z\).)

\(x = 0\), \(y = 1\), \(z = 5\), \(sum = 6\)

\(x = -1\), \(y = 2\), \(z = 5\), \(sum = 6\)

\(x = -2\), \(y = 3\), \(z = 5\), \(sum = 6\)

\(x = -3\), \(y = 4\), \(z = 5\), \(sum = 6\)

\(x + y = -5\) and \(z = -1\)

\(x = -3\), \(y = -2\), \(z = -1\), \(sum = -6\)

Since there are 2 possible sums, this statement is insufficient.

Statements (1) and (2) together: INSUFFICIENT. Using the sets determined with Statement (1), we check the value of \(x + z\) for each case, keeping only the cases in which \(x + z\) is less than 3. Two cases remain.

Case 1: \(x = -3\), \(y = 4\), \(z = 5\), \(x + z = 2 \lt 3\), \(x + y + z = 6\)

Case 2: \(x = -3\), \(y = -2\), \(z = -1\), \(x + z = -4 \lt 3\), \(x + y + z = -6\)

Since the two cases yield different sums, we cannot determine a single value for that sum.


Answer: E
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Re: S98-03  [#permalink]

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New post 15 Apr 2018, 11:24
Bunuel, hello!
If I am not mistaken we are looking for average. So if we have that sum everywhere is 6, then average is two, isnt it?



Bunuel wrote:
Official Solution:


Since the average of \(x\), \(y\), and \(z\) is the sum of the three variables, divided by 3, we can rephrase the question as "what is the value of \(x + y + z\)?" We also note the restrictions on the possible values of \(x\), \(y\), and \(z\) - the variables must be integers in ascending order from \(x\) to \(z\) (not necessarily consecutive). Moreover, they must be different integers, since the inequality \(x \lt y \lt z\) indicates no equality among any of the variables. We note these conditions, but at this stage there is no simple way to apply them in a further rephrasing of the question.

Statement (2): INSUFFICIENT. We start with statement (2), which is the easier statement (it only contains 2 of the 3 variables). We can quickly come up with sets of values that satisfy this statement and the given conditions but that have different sums (or averages). For instance, the set \(x = 0\), \(y = 1\), and \(z = 2\) meets all conditions \((x + z = 2 \lt 3\), all variables are integers and in ascending order), with \(x + y + z = 3\). Another set (\(x = -1\), \(y = 0\), and \(z = 1\)) also meets all conditions but sums to 0. Thus, there is no single value determined by this statement.

Statement (1): INSUFFICIENT. The equation states that \(x + y\) (which must be an integer) multiplied by \(z\) (another integer) equals 5. Since 5 is a prime number, there are only 2 pairs of integers that multiply together to 5: 1 and 5, and -1 and -5. (Don't forget about the negative possibilities). Keeping the conditions that \(x \lt y \lt z\), we can construct the only sets that work:

\(x + y = 1\) and \(z = 5\) (There's no way to assign \(z = 1\) and \(x + y = 5\) while preserving \(x \lt y \lt z\).)

\(x = 0\), \(y = 1\), \(z = 5\), \(sum = 6\)

\(x = -1\), \(y = 2\), \(z = 5\), \(sum = 6\)

\(x = -2\), \(y = 3\), \(z = 5\), \(sum = 6\)

\(x = -3\), \(y = 4\), \(z = 5\), \(sum = 6\)

\(x + y = -5\) and \(z = -1\)

\(x = -3\), \(y = -2\), \(z = -1\), \(sum = -6\)

Since there are 2 possible sums, this statement is insufficient.

Statements (1) and (2) together: INSUFFICIENT. Using the sets determined with Statement (1), we check the value of \(x + z\) for each case, keeping only the cases in which \(x + z\) is less than 3. Two cases remain.

Case 1: \(x = -3\), \(y = 4\), \(z = 5\), \(x + z = 2 \lt 3\), \(x + y + z = 6\)

Case 2: \(x = -3\), \(y = -2\), \(z = -1\), \(x + z = -4 \lt 3\), \(x + y + z = -6\)

Since the two cases yield different sums, we cannot determine a single value for that sum.


Answer: E
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Re: S98-03  [#permalink]

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New post 15 Apr 2018, 23:50
Jahfors wrote:
Bunuel, hello!
If I am not mistaken we are looking for average. So if we have that sum everywhere is 6, then average is two, isnt it?



Bunuel wrote:
Official Solution:


Since the average of \(x\), \(y\), and \(z\) is the sum of the three variables, divided by 3, we can rephrase the question as "what is the value of \(x + y + z\)?" We also note the restrictions on the possible values of \(x\), \(y\), and \(z\) - the variables must be integers in ascending order from \(x\) to \(z\) (not necessarily consecutive). Moreover, they must be different integers, since the inequality \(x \lt y \lt z\) indicates no equality among any of the variables. We note these conditions, but at this stage there is no simple way to apply them in a further rephrasing of the question.

Statement (2): INSUFFICIENT. We start with statement (2), which is the easier statement (it only contains 2 of the 3 variables). We can quickly come up with sets of values that satisfy this statement and the given conditions but that have different sums (or averages). For instance, the set \(x = 0\), \(y = 1\), and \(z = 2\) meets all conditions \((x + z = 2 \lt 3\), all variables are integers and in ascending order), with \(x + y + z = 3\). Another set (\(x = -1\), \(y = 0\), and \(z = 1\)) also meets all conditions but sums to 0. Thus, there is no single value determined by this statement.

Statement (1): INSUFFICIENT. The equation states that \(x + y\) (which must be an integer) multiplied by \(z\) (another integer) equals 5. Since 5 is a prime number, there are only 2 pairs of integers that multiply together to 5: 1 and 5, and -1 and -5. (Don't forget about the negative possibilities). Keeping the conditions that \(x \lt y \lt z\), we can construct the only sets that work:

\(x + y = 1\) and \(z = 5\) (There's no way to assign \(z = 1\) and \(x + y = 5\) while preserving \(x \lt y \lt z\).)

\(x = 0\), \(y = 1\), \(z = 5\), \(sum = 6\)

\(x = -1\), \(y = 2\), \(z = 5\), \(sum = 6\)

\(x = -2\), \(y = 3\), \(z = 5\), \(sum = 6\)

\(x = -3\), \(y = 4\), \(z = 5\), \(sum = 6\)

\(x + y = -5\) and \(z = -1\)

\(x = -3\), \(y = -2\), \(z = -1\), \(sum = -6\)

Since there are 2 possible sums, this statement is insufficient.

Statements (1) and (2) together: INSUFFICIENT. Using the sets determined with Statement (1), we check the value of \(x + z\) for each case, keeping only the cases in which \(x + z\) is less than 3. Two cases remain.

Case 1: \(x = -3\), \(y = 4\), \(z = 5\), \(x + z = 2 \lt 3\), \(x + y + z = 6\)

Case 2: \(x = -3\), \(y = -2\), \(z = -1\), \(x + z = -4 \lt 3\), \(x + y + z = -6\)

Since the two cases yield different sums, we cannot determine a single value for that sum.


Answer: E


In once case the sum is 6 and in another case the sum is -6:

Case 1: \(x = -3\), \(y = 4\), \(z = 5\) --> \(x + y + z = 6\) --> the average = 6/3 = 2.

Case 2: \(x = -3\), \(y = -2\), \(z = -1\) --> \(x + y + z = -6\) --> the average = -6/3 = -2.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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: S98-03 &nbs [#permalink] 15 Apr 2018, 23:50
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