Is x > 10^10 1) x > 2^34 2) x = 2^35

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Is x > 10^10 1) x > 2^34 2) x = 2^35 [#permalink]

06 Sep 2007, 15:19

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Is x > 10^10

1) x > 2^34
2) x = 2^35

Please explain your answer and approach

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Re: Exponent DS [#permalink]

06 Sep 2007, 16:06

tinman1412 wrote:

Is x > 10^10

1) x > 2^34
2) x = 2^35

Please explain your answer and approach

S2:

x=2^35. 2^35>10^10? 2^35>2^10*5^10. Divide both sides by 2^10.

2^25>5^10? Obviously 2^25 is greater than 5^10. while 5 might seem greater than 2. the ^25 significantly makes up the difference.

S1:
Just put this: greater than 2^34> 10^10? Solve similar to above.

Greater than 2^24>5^10? again. 2^24 is def. greater than 5^10.

Ans D.

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06 Sep 2007, 19:07

St1:
2^34 > 2^10(5^10) ?
2^24 > 5^10 ?? --> yes. so x > 10^10

St2:
2^35 > 10^10? --> yes. so x > 10^10.

Ans D

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06 Sep 2007, 20:29

Anyone know what's the fastest method to compare 10^10 with 2^34???

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06 Sep 2007, 22:55

x>10^10 is equal to x> (2^30)*1.25^10

1) x>2^34 here we can simplify it by dividing both inequalities by 2^30
so x>2^4 and x>1.25^10 = (1.25^(5/2))^4
if 2 > (1.25)^(5/2) than it sufficient, which is true

2) 4 is also higer than (1.25)^(5/2) so again sufficient

Ans; D

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Re: Exponent DS [#permalink]

07 Sep 2007, 08:00

tinman1412 wrote:

Is x > 10^10

1) x > 2^34
2) x = 2^35

Please explain your answer and approach

got D,

2st- gives specific value of X, so it is suff to say Yes or No.

as for 1st statement I used the same method like those posted above.

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08 Sep 2007, 05:25

My approach:

Stmt1: x > 2^34
10^10 = 10* (1000)^3
2^34 = 2^4 * (2^10)^3 =16 * (1024)^3
16 * (1024)^3 > 10* (1000)^3
So SUFF

Stmt2: Suff as above

So 'D' is the answer.

[#permalink] 08 Sep 2007, 05:25

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Is x > 10^10 1) x > 2^34 2) x = 2^35

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Is x > 10^10 1) x > 2^34 2) x = 2^35

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