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If y rounded to the nearest thousandth is .00x, is x > 2?
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04 Feb 2015, 07:46
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Re: If y rounded to the nearest thousandth is .00x, is x > 2?
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04 Feb 2015, 09:14
ans C.. 1) statement one does not satisfy as z=3 gives a value .008, which is >.002 but z=4 gives a value .001,which is not >.002.. 2) statement two tells us that z is perfect square<9 so z=4 and nothing else ..insufficient combined, one specific value of z will provide an answer whether x>2, sufficient
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Re: If y rounded to the nearest thousandth is .00x, is x > 2?
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09 Feb 2015, 05:16
Bunuel wrote: If y rounded to the nearest thousandth is .00x, is x > 2?
(1) y = 1/(5^z) (2) z has exactly three unique factors and is a positive integer less than 9.
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Solution: C Start with the easier of the two statements. If we only have the stimulus and statement (2), we don’t know what z refers to, so statement (2) is NOT sufficient. If we have only the stimulus and statement (1), we don’t know the value of z. This is usually enough to render a statement insufficient, but if you’re nervous, try a few numbers. If z is 3, y = 1/125 = .008; that says that x = 8. If z is 4, y is 1/625 = .0016; that says that x = 2. The statements conflict, so (1) is NOT sufficient. Together, z must be 4 – any number that has exactly three unique factors is a prime number squared, and the only number less than 9 that is a prime number squared is 4. (C).
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Re: If y rounded to the nearest thousandth is .00x, is x > 2?
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12 Aug 2017, 05:58
Bunuel wrote: Bunuel wrote: If y rounded to the nearest thousandth is .00x, is x > 2?
(1) y = 1/(5^z) (2) z has exactly three unique factors and is a positive integer less than 9.
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Solution: C Start with the easier of the two statements. If we only have the stimulus and statement (2), we don’t know what z refers to, so statement (2) is NOT sufficient. If we have only the stimulus and statement (1), we don’t know the value of z. This is usually enough to render a statement insufficient, but if you’re nervous, try a few numbers. If z is 3, y = 1/125 = .008; that says that x = 8. If z is 4, y is 1/625 = .0016; that says that x = 2. The statements conflict, so (1) is NOT sufficient. Together, z must be 4 – any number that has exactly three unique factors is a prime number squared, and the only number less than 9 that is a prime number squared is 4. (C). How come z=6 r 8? also. 6 has unique 3 factors 6=2*3*1



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Re: If y rounded to the nearest thousandth is .00x, is x > 2?
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12 Aug 2017, 07:41
Bunuel wrote: Bunuel wrote: If y rounded to the nearest thousandth is .00x, is x > 2? (1) y = 1/(5^z) (2) z has exactly three unique factors and is a positive integer less than 9. Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION: Solution: C Start with the easier of the two statements. If we only have the stimulus and statement (2), we don’t know what z refers to, so statement (2) is NOT sufficient. If we have only the stimulus and statement (1), we don’t know the value of z. This is usually enough to render a statement insufficient, but if you’re nervous, try a few numbers. If z is 3, y = 1/125 = .008; that says that x = 8. If z is 4, y is 1/625 = .0016; that says that x = 2. The statements conflict, so (1) is NOT sufficient. Together, z must be 4 – any number that has exactly three unique factors is a prime number squared, and the only number less than 9 that is a prime number squared is 4. (C). How come z=6 r 8? also. 6 has unique 3 factors 6=2*3*1 As per the question (2) z has exactly three unique factors and is a positive integer less than 9. Hence Z=4 , factors are 1, 2, 4 a is factor b means , b completely divides a . if you check for 6 . Its factors are 1, 2, 3, 6 Also other way to calculate factors ( a^b * c^d ) where a and c are prime , then the number of factors = (b+1) (d+1) Regards, Press Kudos if you like the post .
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Re: If y rounded to the nearest thousandth is .00x, is x > 2?
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12 Aug 2017, 08:17
tejasridarsi wrote: Bunuel wrote: Bunuel wrote: If y rounded to the nearest thousandth is .00x, is x > 2?
(1) y = 1/(5^z) (2) z has exactly three unique factors and is a positive integer less than 9.
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:Solution: C Start with the easier of the two statements. If we only have the stimulus and statement (2), we don’t know what z refers to, so statement (2) is NOT sufficient. If we have only the stimulus and statement (1), we don’t know the value of z. This is usually enough to render a statement insufficient, but if you’re nervous, try a few numbers. If z is 3, y = 1/125 = .008; that says that x = 8. If z is 4, y is 1/625 = .0016; that says that x = 2. The statements conflict, so (1) is NOT sufficient. Together, z must be 4 – any number that has exactly three unique factors is a prime number squared, and the only number less than 9 that is a prime number squared is 4. (C). How come z=6 r 8? also. 6 has unique 3 factors 6=2*3*1 6 has four factors not three: 1, 2, 3, and 6. 8 has four factors not three: 1, 2, 4, and 8. Remember a positive integer is a factor of itself.
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Re: If y rounded to the nearest thousandth is .00x, is x > 2?
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12 Aug 2017, 08:41
Bunuel  Pardon me for this silly doubt. Don't we need numbers after thousandth to do rounding. I marked this as A as i thought we cannot round off y = 0.008 as it does not have anything after 8 to round off and rest all values of y tell x < 2 Can you please clarify the rounding concept?
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If y rounded to the nearest thousandth is .00x, is x > 2?
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Updated on: 17 Aug 2017, 09:03
1) take z=1, y=.02, take z= 3 then y= .008 , hence A,D ruled out 2) z can be anything so B ruled out combine 1,2 z has 3 unique factors i.e. z is a perfect square hence its is 4 which is less than 9 as given , the factors are 1,2,4 so the value of y when z=4 is .0016 and when rounded off gives value .002 which is x = 2 and clearly is x>2 we have definite NO it is not hence C is answer.
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Originally posted by sahilvijay on 12 Aug 2017, 10:27.
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Re: If y rounded to the nearest thousandth is .00x, is x > 2?
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13 Aug 2017, 03:38
Bunuel wrote: If y rounded to the nearest thousandth is .00x, is x > 2?
(1) y = 1/(5^z) (2) z has exactly three unique factors and is a positive integer less than 9.
Kudos for a correct solution. Dear BunuelThe question says ''if y is rounded'. In statement 1, if z= 2....then y =0.008. I do not see any rounding process. 'Y' naturally without rounding is 0.008. How do we assume that 0.008 is rounded?



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Re: If y rounded to the nearest thousandth is .00x, is x > 2?
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29 Nov 2018, 06:30
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