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Joined: 02 Sep 2009
Posts: 44599

Re S9803 [#permalink]
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16 Sep 2014, 01:52
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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Joined: 26 Sep 2016
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Re: S9803 [#permalink]
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15 Apr 2018, 12:24
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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Posts: 44599

Re: S9803 [#permalink]
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16 Apr 2018, 00:50
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.
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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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