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Re: If x, y and z are integers, what is y – z? [#permalink]

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21 Jul 2013, 07:41

1

This post received KUDOS

If x, y and z are integers, what is y – z?

(1) \(100^x = 2^y5^z\) \(2^{2x}5^{2x}=2^y5^z\) so \(y-z=2x-2x=0\). Sufficient

(2) \(10^y = 20^x5^{z+1}\) \(2^y5^y=2^{2x}5^x5^{z+1}\) so \(y=2x\) and \(y=x+z+1\). We cannot determine y-z. Consider y=4,x=2 and z=1 so y-z=3; or y=8,x=4 and y=3 so y-z=5. Not sufficient

A _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If x, y and z are integers, what is y – z? [#permalink]

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01 May 2016, 09:54

Bunuel wrote:

vishalrastogi wrote:

I could not get the explanation here, can anybody explain this, please ?

If x, y and z are integers, what is y – z?

(1) \(100^x = 2^y5^z\) --> \(2^{2x}5^{2x}=2^y5^z\) --> equate the exponents: \(2x=y\) and \(2x=z\) --> thus \(2x-2x=y-z=0\). Sufficient.

(2) \(10^y = 20^x5^{z+1}\) --> \(2^y5^y=2^{2x}*5^{x+z+1}\) --> \(y=2x\) and \(y=x+z+1\). We cannot get the value of y-z from this. Not sufficient,

Answer: A.

Hope it's clear.

How can 1 be sufficient??? In the given statement, its 5 raise to the power 2. And the solution you have provided considers it as 5 raise to the power z.

I could not get the explanation here, can anybody explain this, please ?

If x, y and z are integers, what is y – z?

(1) \(100^x = 2^y5^z\) --> \(2^{2x}5^{2x}=2^y5^z\) --> equate the exponents: \(2x=y\) and \(2x=z\) --> thus \(2x-2x=y-z=0\). Sufficient.

(2) \(10^y = 20^x5^{z+1}\) --> \(2^y5^y=2^{2x}*5^{x+z+1}\) --> \(y=2x\) and \(y=x+z+1\). We cannot get the value of y-z from this. Not sufficient,

Answer: A.

Hope it's clear.

How can 1 be sufficient??? In the given statement, its 5 raise to the power 2. And the solution you have provided considers it as 5 raise to the power z.

It's 5^z both in the question and in the solution.
_________________

Re: If x, y and z are integers, what is y – z? [#permalink]

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02 May 2016, 15:58

I think what's confusing some folks is the second equation is giving y = x+z+1 bringing z to the left side. y-z = x+1. This still doesn't give a value for y-z. Question is asking for a value for y-z and not if you can deduce an expression for y-z. I made this silly mistake once in the heat of the moment so sharing it here.
_________________

Re: If x, y and z are integers, what is y – z? [#permalink]

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17 Nov 2016, 17:58

Bunuel wrote:

vishalrastogi wrote:

I could not get the explanation here, can anybody explain this, please ?

If x, y and z are integers, what is y – z?

(1) \(100^x = 2^y5^z\) --> \(2^{2x}5^{2x}=2^y5^z\) --> equate the exponents: \(2x=y\) and \(2x=z\) --> thus \(2x-2x=y-z=0\). Sufficient.

Answer: A.

Hope it's clear.

If y-z=0, then y=z. If i put the value of y=z, how can we legitimate the statement 1? statement 1: \(100^x = 2^y5^z\) \(2^{2x}5^{2x}=2^z5^z\) To legitimate the statement 1 we still need the value of x and z. But, they are still unknown here. How can you make known it for all? Then, how can we conclude it? Bunuel Thank you...
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

I could not get the explanation here, can anybody explain this, please ?

If x, y and z are integers, what is y – z?

(1) \(100^x = 2^y5^z\) --> \(2^{2x}5^{2x}=2^y5^z\) --> equate the exponents: \(2x=y\) and \(2x=z\) --> thus \(2x-2x=y-z=0\). Sufficient.

Answer: A.

Hope it's clear.

If y-z=0, then y=z. If i put the value of y=z, how can we legitimate the statement 1? statement 1: \(100^x = 2^y5^z\) \(2^{2x}5^{2x}=2^z5^z\) To legitimate the statement 1 we still need the value of x and z. But, they are still unknown here. How can you make known it for all? Then, how can we conclude it? Bunuel Thank you...

The question asks the value of y - z, not the individual value of x, y, and z. From the solution we got that y - z = 0.
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

x, y, and z are given to be integers. We have \(2^{2x}5^{2x}=2^y5^z\) --> equate the exponents of 2 and 5: \(2x=y\) and \(2x=z\). Thus 2x = y = z.
_________________

x, y, and z are given to be integers. We have \(2^{2x}5^{2x}=2^y5^z\) --> equate the exponents of 2 and 5: \(2x=y\) and \(2x=z\). Thus 2x = y = z.

This is the first time i learn that i can equate the exponent after having multiple variables on both side. I, normally, equate the exponent when i have only one part in the right hand side and the other one in the left hand side. like below... 2^{2x}=2^y --> 2x=y it is ok. But when it is something like below then it is the first time i learn. \(2^{2x}5^{2x}=2^y5^z\) \(2x=y\) and \(2x=z\). Anyway, many many thanks with 'kudos'
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

x, y, and z are given to be integers. We have \(2^{2x}5^{2x}=2^y5^z\) --> equate the exponents of 2 and 5: \(2x=y\) and \(2x=z\). Thus 2x = y = z.

This is the first time i learn that i can equate the exponent after having multiple variables on both side. I, normally, equate the exponent when i have only one part in the right hand side and the other one in the left hand side. like below... 2^{2x}=2^y --> 2x=y it is ok. But when it is something like below then it is the first time i learn. \(2^{2x}5^{2x}=2^y5^z\) \(2x=y\) and \(2x=z\). Anyway, many many thanks with 'kudos'

We can only do this here because we know that x, y, and z are integers.
_________________

Re: If x, y and z are integers, what is y – z? [#permalink]

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20 Nov 2016, 02:30

Bunuel wrote:

iMyself wrote:

Bunuel wrote:

x, y, and z are given to be integers. We have \(2^{2x}5^{2x}=2^y5^z\) --> equate the exponents of 2 and 5: \(2x=y\) and \(2x=z\). Thus 2x = y = z.

This is the first time i learn that i can equate the exponent after having multiple variables on both side. I, normally, equate the exponent when i have only one part in the right hand side and the other one in the left hand side. like below... 2^{2x}=2^y --> 2x=y it is ok. But when it is something like below then it is the first time i learn. \(2^{2x}5^{2x}=2^y5^z\) \(2x=y\) and \(2x=z\). Anyway, many many thanks with 'kudos'

We can only do this here because we know that x, y, and z are integers.

That means: we can't equate this type of things if the variable is NOT integer, right Bunuel?
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

Re: If x, y and z are integers, what is y – z? [#permalink]

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20 Nov 2016, 02:40

Bunuel wrote:

iMyself wrote:

That means: we can't equate this type of things if the variable is NOT integer, right Bunuel?

_______________________ Yes...

Thank you Brother with kudos!
_________________

“The heights by great men reached and kept were not attained in sudden flight but, they while their companions slept, they were toiling upwards in the night.” ― Henry Wadsworth Longfellow

Re: If x, y and z are integers, what is y – z? [#permalink]

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29 Nov 2016, 15:33

kingflo wrote:

If x, y and z are integers, what is y – z?

(1) \(100^x = 2^y5^z\)

(2) \(10^y = 20^x5^{z+1}\)

We need to determine the value of y – z.

Statement One Alone:

100^x = 2^y * 5^z

Notice that 100^x = (2^2 * 5^2)^x = 2^(2x) * 5^(2x). Equate this with 2^y * 5^z and we have:

2^(2x) * 5^(2x) = 2^y * 5^z

Therefore, 2^(2x) = 2^y and 5^(2x) = 5^z.

Thus, 2x = y and 2x = z. Therefore, y - z = 2x - 2x = 0.

Statement one alone is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

10^y = 20^x * 5^(z+1)

Since 20^x = (2^2 * 5)^x = 2^(2x) * 5^x, that means 20^x * 5^(z+1) = 2^(2x) * 5^x * 5^(z+1) = 2^(2x) * 5^(x+z+1). Notice that 10^y = (2 * 5)^y = 2^y * 5^y, so we have:

2^y * 5^y = 2^(2x) * 5^(x+z+1)

Therefore, 2^y = 2^(2x) and 5^y = 5^(x+z+1).

Thus, y = 2x and y = x + z + 1. From the second equation, we have y - z = x + 1. However, since we do not know the value of x, we cannot determine the value of y - z. Statement two alone is not sufficient to answer the question.

Answer: A
_________________

Jeffrey Miller Scott Woodbury-Stewart Founder and CEO

gmatclubot

Re: If x, y and z are integers, what is y – z?
[#permalink]
29 Nov 2016, 15:33

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