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02 Mar 2021, 20:19
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C
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86% (01:04) correct
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If \(3^{m + n} = 27\) and \(3^{m – n} = \frac{1}{3}\) then \(\frac{m}{n} =\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{2}\)
(D) 1
(E) 2
PS05884
Source: GMAT Official Practice
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{2}\)
(D) 1
(E) 2
PS05884
Source: GMAT Official Practice
02 Mar 2021, 20:42
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3^(m+n)=27 should be written as 3^(m+n) = 3^3
3^(m-n)=1/3 should be written as 3^(m-n)=3^-1
Therefore we know:
From the first statement
m+n=3
and from the second statement
m-n=-1
solving for m
m=3-n
plug that into the second equation
3-n-n=-1
3-2n=-1
-2n=-4
n=2
which means m=1
Answer: m/n = 1/2
3^(m-n)=1/3 should be written as 3^(m-n)=3^-1
Therefore we know:
From the first statement
m+n=3
and from the second statement
m-n=-1
solving for m
m=3-n
plug that into the second equation
3-n-n=-1
3-2n=-1
-2n=-4
n=2
which means m=1
Answer: m/n = 1/2
02 Mar 2021, 20:45
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quynhtran2209
\(If 3^{m + n} = 27 and 3^{m – n} = \frac{1}{3} then \frac{m}{n} =\)
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{2}\)
(D) 1
(E) 2
Source: GMAT Official Practice
(A) \(\frac{1}{5}\)
(B) \(\frac{1}{4}\)
(C) \(\frac{1}{2}\)
(D) 1
(E) 2
Source: GMAT Official Practice
3^(m+n)=27-3^3 and 3^(m-2) = 1/3=3^(-1). Add the equations together: 3^(m+m+n-n)=3^3-1 =3^2. Now, if X^m=X^n, m=n so 3^2m=3^2 so 2m=2 and m=1. so n must equal 2 so m/n=1/2 OA is C
