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If y rounded to the nearest thousandth is .00x, is x > 2?

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If y rounded to the nearest thousandth is .00x, is x > 2?  [#permalink]

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New post 04 Feb 2015, 07:46
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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.


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Re: If y rounded to the nearest thousandth is .00x, is x > 2?  [#permalink]

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New post 04 Feb 2015, 09:14
1
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?  [#permalink]

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New post 09 Feb 2015, 05:16
1
2
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?  [#permalink]

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New post 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?  [#permalink]

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New post 12 Aug 2017, 07:41
1
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)

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Re: If y rounded to the nearest thousandth is .00x, is x > 2?  [#permalink]

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New post 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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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

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Re: If y rounded to the nearest thousandth is .00x, is x > 2?  [#permalink]

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New post 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?  [#permalink]

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New post 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.
Last edited by sahilvijay on 17 Aug 2017, 09:03, edited 1 time in total.
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Re: If y rounded to the nearest thousandth is .00x, is x > 2?  [#permalink]

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New post 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 Bunuel

The 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?  [#permalink]

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