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# Is (2^(y+z))(3^x)(5^y)(7^z) < (90^y)(14^z)

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Intern
Joined: 14 Mar 2013
Posts: 42
Location: United States
GMAT Date: 12-03-2013
WE: General Management (Retail)

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28 Sep 2013, 08:59
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Difficulty:

85% (hard)

Question Stats:

54% (02:19) correct 46% (02:30) wrong based on 154 sessions

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Is $$(2^{(y+z)})(3^x)(5^y)(7^z) < (90^y)(14^z)$$

(1) y and z are positive integers; x = 1

(2) x and z are positive integers; y = 1
Manager
Joined: 29 Apr 2013
Posts: 88
Location: India
Concentration: General Management, Strategy
GMAT Date: 11-06-2013
WE: Programming (Telecommunications)
Re: Is (2^(y+z))(3^x)(5^y)(7^z) < (90^y)(14^z)  [#permalink]

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28 Sep 2013, 10:25
7
Format the question properly using <Math> tag.

Is$$(2^(^y^+^z^))(3^x)(5^y)(7^z) < (90^y)(14^z)$$

(1) y and z are positive integers; x = 1
(2) x and z are positive integers; y = 1

Lets first break the right hand side of the equation into prime factors
$$(90^y)(14^z) = (2*3^2*5)^y*(2*7)^z = (2^y*3^2^y*5^y)*(2^z*7^z) = 2^(^y^+^z^)*3^2^y*5^y*7^z$$

So now the question becomes:
Is $$2^(^y^+^z^)*3^x*5^y*7^z < 2^(^y^+^z^)*3^2^y*5^y*7^z$$

By Simplifying LHS and RHS, we get:
Is $$3^x < 3^2^y$$

Statement 1:
y and z are positive integers; x = 1
So, x will always be less than 2y because $$y>1$$
Thus, $$3^x < 3^2^y$$ ... SUFFICIENT

Statement 2:
x and z are positive integers; y = 1
If x = 1 then $$3^x < 3^2^y$$
But if x = 2 then $$3^x$$ is not less than $$3^2^y$$ ... Hence INSUFFICIENT

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##### General Discussion
Intern
Joined: 29 Sep 2013
Posts: 45
Re: Is (2^(y+z))(3^x)(5^y)(7^z) < (90^y)(14^z)  [#permalink]

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20 Oct 2013, 12:40
TirthankarP wrote:
Format the question properly using <Math> tag.

Is$$(2^(^y^+^z^))(3^x)(5^y)(7^z) < (90^y)(14^z)$$

(1) y and z are positive integers; x = 1
(2) x and z are positive integers; y = 1

Lets first break the right hand side of the equation into prime factors
$$(90^y)(14^z) = (2*3^2*5)^y*(2*7)^z = (2^y*3^2^y*5^y)*(2^z*7^z) = 2^(^y^+^z^)*3^2^y*5^y*7^z$$

So now the question becomes:
Is $$2^(^y^+^z^)*3^x*5^y*7^z < 2^(^y^+^z^)*3^2^y*5^y*7^z$$

By Simplifying LHS and RHS, we get:
Is $$3^x < 3^2^y$$

Statement 1:
y and z are positive integers; x = 1
So, x will always be less than 2y because $$y>1$$
Thus, $$3^x < 3^2^y$$ ... SUFFICIENT

Statement 2:
x and z are positive integers; y = 1
If x = 1 then $$3^x < 3^2^y$$
But if x = 2 then $$3^x$$ is not less than $$3^2^y$$ ... Hence INSUFFICIENT

Kudos plz if my reply helped. Need to unlock G M A T Club Tests

Just a minor question, wont the red portion be contradicting the first statement?
Manager
Joined: 29 Apr 2013
Posts: 88
Location: India
Concentration: General Management, Strategy
GMAT Date: 11-06-2013
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Re: Is (2^(y+z))(3^x)(5^y)(7^z) < (90^y)(14^z)  [#permalink]

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20 Oct 2013, 19:07
suk1234 wrote:
Just a minor question, wont the red portion be contradicting the first statement?

While considering the second statement, we should not even think of the first statement.
While solving the question using statement 2 alone, we don't bother what information we got earlier using statement 1.
However, if both statement 1 and 2 alone are not sufficient, then only we have to consider the information from both statement 1 and 2.

Thats the crux of data sufficiency questions.
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Collection of some good questions on Number System
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Re: Is (2^(y+z))(3^x)(5^y)(7^z) < (90^y)(14^z)  [#permalink]

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20 Dec 2018, 13:35
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Re: Is (2^(y+z))(3^x)(5^y)(7^z) < (90^y)(14^z)   [#permalink] 20 Dec 2018, 13:35
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