MathRevolution wrote:
[GMAT math practice question]
Which of the following is equal to \(2^{k-1}3^{k+1}\)?
\(A. 2^2(6^{k-1})\)
\(B. 3^2(6^{k-1})\)
\(C. 6(2^{k-1})\)
\(D. 6(3^{k-1})\)
\(E. 6^{k-1}\)
Approach 01 : From Q-stem to Answer: \(2^{k-1}3^{k+1}\) = \(3^2*(2^{k-1}3^{k-1})\) = \(3^2(6^{k-1})\) ..........
Option B is the answerApproach 02 : From Answer to Q-stem:
Option 01:\(A. 2^2(6^{k-1})\) = \(2^2*(2^{k-1}*3^{k-1})\) = \(2^{k+1}*3^{k-1}\)
Option 02:\(3^2(6^{k-1})\) = \(3^2*(2^{k-1}*3^{k-1})\) = \(2^{k-1}*3^{k+1}\) ..........
Option B is the answerOption 03:\(6(2^{k-1})\) = \(3*2*(2^{k-1})\) = \(3*2^k\)
Option 04:\(6(3^{k-1})\) = \(3*2*(3^{k-1})\) = \(2*3^k\)
Option 05:\(6^{k-1}\) = \(2^{k-1}*3^{k-1}\)
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Please let me know if I am going in wrong direction.
Thanks in appreciation.