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As r and s are positive,

\(2r+2s < 2r+3s < 3r+3s\)

or, \(\frac{2r+2s}{r+s} < \frac{2r+3s}{r+s} < \frac{3r+3s}{r+s}\)

or, \(2<\frac{2r+3s}{r+s}<3\)



Hence answer is C
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gmatbusters
If r and s are positive, the value of \(\frac{2r+3s}{r+s}\) is ?


A. 0.15
B. Between 0.5 and 1.5
C. Between 2 and 3
D. Between 3 and 3.5
E. Between 3.5 and 5

Take r = 2 and s = 3

\(\frac{2*2 + 3*3}{(2 + 3)}\) = \(\frac{13}{5}\) = \(2.6\)

Take r = 4 and s = 4

\(\frac{2*4 + 3*4}{(4 + 4)}\) = \(\frac{20}{8}\)= \(2.5\)

Take r = 1.5 and s = 1.5

\(\frac{2*1.5 + 3*1.5}{(1.5 + 1.5)} = \frac{7.5}{3} = 2.5\)

Hence C
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gmatbusters
If r and s are positive, the value of \(\frac{2r+3s}{r+s}\) is ?


A. 0.15
B. Between 0.5 and 1.5
C. Between 2 and 3
D. Between 3 and 3.5
E. Between 3.5 and 5


This question is designed to test weighted average method.

Between weight 2 and 3 , 3 is greater integer.

Approximate result will be closer to 3 and greater than 2.

C is the correct answer.
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For this question, why can't we just assume r = 1, and s = 1? Then it's 5/2 =2.5 which is C...

GMATBusters
If r and s are positive, the value of \(\frac{2r+3s}{r+s}\) is ?


A. 0.15
B. Between 0.5 and 1.5
C. Between 2 and 3
D. Between 3 and 3.5
E. Between 3.5 and 5
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eaat
For this question, why can't we just assume r = 1, and s = 1? Then it's 5/2 =2.5 which is C...
You can, your approach is valid too.
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