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p+q >= 2 (3^m + 4^m) and m > 0 , integer.

a p+q = 3* 3^m + 2* 4^m

for m= 1, 2,3 LHS p+q = 9+8 = 17 RHS = 7*2 = 14

LHS = 27 + 32 = 59 RHS = 9 + 16 = 25

LHS = 81 + 128 = 209 RHS = 27+ 64 = 91 All the cases LHS > 2 * RHS

Sufficient.

b tells nothing about p+q. Not sufficient.

Hence A.
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If m is a positive integer, is the value of p + q at least twice the value of 3^m + 4^m?
(1) p = 3^(m+1) and q = 2^(2m+1)
(2) m = 4

Given: m>0,
Is p+q>= 2(3^m+4^m) ?
=>P+q >= 3^m*2 + 4^m * 2

Stmt:1 p=3^(m+1)=>3^m*3
q=4^m+2
p+q=3^m*3+4^m+2.
p+q=3^m*(2+1) + 4^m+2
p+q=3^m*2 + 3^m + 4^m+2
p+q=[3^m*2 + 4^m * 2] + 3^m
Hence p+q is definitely greater than 3^m*2+4^m*2
Sufficient.

Stmt2: m=4. p+q>=3^4*2+4^4*2
Doesn't say anything about p and q. Insufficient.

OA A.
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gmatpapa
If m is a positive integer, is the value of p + q at least twice the value of 3^m + 4^m?

(1) p = 3^(m+1) and q = 2^(2m+1)
(2) m = 4

This is a copy of an OG question: if-n-is-a-positive-integer-is-the-value-of-b-a-at-least-144344.html
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Since common bases with one variable can be solved, providing the values for p and q under the same common bases allows for the solution, and therefore is sufficient. A
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gmatpapa
If m is a positive integer, is the value of p + q at least twice the value of 3^m + 4^m?

(1) p = 3^(m+1) and q = 2^(2m+1)
(2) m = 4

from 1

is

3^(m+1) + 2^(2m+1) >= 2(3^m +4^m) , i.e. is 3(m+1) + 2^(2m+1) >= 2*3^m + 2^(2m+1)

is 3^(m+1) >= 2* 3^m , is 3 >=2 sure ... suff
from 2 clearly insuff

A
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