Quote:
K, which is a 15% increase from the price of the same item in January 2006
If price of the item in Jan 2006 is S, then K = 1.15 S
We have to compare S and 0.85K.
Therefore, let's convert S into K.
\(S = \frac{K}{1.15}\)
\(S = K * \frac{100 }{ 115} \)
\(S = K * \frac{20 }{ 23}\)
\(0.85 * K = \frac{85 }{ 100} * K = \frac{17 }{ 20 }* K\)
To compare fractions \(\frac{20}{23}, and \frac{17 }{ 20}\), cross-multiply.
\(20 * 20 > 17 * 20\)
So, \(\frac{20 }{ 23} * K > \frac{17 }{ 20} * K\)
or S > 0.85 K
Hence, A is the correct answer.