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by sumgb » Mon Dec 12, 2011 8:16 am
IMO A

we need to find b - a >= 2* 3^n - 2*2^n

from I - we get, b - a = 3*3^n - 2*2^n which is more than 2* 3^n - 2*2^n given that n > 0 and an int.

from II - no info abt a or b. Hence insuff.

whats the OA and OE?
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by neelgandham » Mon Dec 12, 2011 9:02 am
If n is a positive integer, is the value of b-a at least twice the value of 3^n-2^n?
Rephrasing the question to Is b-a > 2*(3^n-2^n), i.e Is b-a > (2*(3^n))-(2*(2^n)) ?
1) a=2^n+1 and b=3^n+1
b-a = 3^(n+1) - 2^(n+1) = (3*(3^n))-(2*(2^n)) = (3^n) + (2*(3^n))-(2*(2^n)) > (2*(3^n))-(2*(2^n))
because (3^n) + Number A > Number A
Sufficient!
2) n=3
Is b-a > 3^3-2^3 = 19
Insufficient!, because you don't know the values of b and a

Answer A IMO
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by pemdas » Mon Dec 12, 2011 9:37 am
n>0, Is (b-a)>=2(3^n-2^n)
st(1) a=2^n+1 and b=3^n+1 imply b-3^n=a-2^n, b-a=(3^n-2^n). Since n>0 (3^n-2^n) is also +ve. Sufficient, as we can answer No.
st(2) n=3 doesn't tell anything about a and b, Not Sufficient.

a
romitvsingh wrote:if n is a positive integer, is the value of b-a at least twice the value of 3^n-2^n?

1) a=2^n+1 and b=3^n+1

2) n=3
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by GMATGuruNY » Mon Dec 12, 2011 2:48 pm
romitvsingh wrote:if n is a positive integer, is the value of b-a at least twice the value of 3^n-2^n?
Statement 1: a=2^n+1 and b=3^n+1
Let n=1.
Then a=2² and b=3², so b-a = 3²-2².
Twice the value of 3^n-2^n = 2(3¹-2¹) = 2(3¹) - 2².
Is the value of b-a at least twice the value of 3^n-2^n?
3²-2² ≥ 2(3¹) - 2²
3² ≥ 2(3)
3 ≥ 2.
YES.

Increasing the value of n will simply increase the value of each exponent by the same amount.
If n=2, each exponent will increase by 1:
3³- 2³ ≥ 2(3²) - 2³
3³ ≥ 2(3²)
3 ≥ 2.

Thus, regardless of the value of n, the left-hand side will always be greater than the right-hand side.
SUFFICIENT.

Statement 2: n=3
No information about b-a.
INSUFFICIENT.

The correct answer is A.
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