A shirt and a jacket are each for sale at a price that is discounted from the item's original price. Which has the lower

What we see: a Data Sufficiency percent question. Which item ends up cheaper after its discount, the shirt or the jacket?

Step 1. Set up the prices.
What we see: each sale price is the original price times what is left after the discount.
What it means: call the shirt's original price \(s\) and its discount \(p\%\). Then the shirt sells for \(s(1 - \frac{p}{100})\).

Step 2. Each statement alone.
What we see: (1) gives the jacket's original price, \(2s\), but nothing about the discounts. A huge jacket discount could make it cheaper; a tiny one leaves it pricier. Not sufficient.
What we see: (2) gives the jacket's discount as \(3p\%\), but nothing about the original prices. Not sufficient.

Step 3. Both together.
What we see: jacket sells for \(2s(1 - \frac{3p}{100})\), shirt for \(s(1 - \frac{p}{100})\).
What it means: divide both by \(s\) and compare \(2 - \frac{6p}{100}\) with \(1 - \frac{p}{100}\). The jacket is cheaper only when \(1 < \frac{5p}{100}\), that is, when \(p > 20\).
What we do and why: we still do not know \(p\), so test two numbers with \(s = 100\).
\(p = 10\): shirt \(90\), jacket \(200 \times 0.70 = 140\). The shirt is cheaper.
\(p = 30\): shirt \(70\), jacket \(200 \times 0.10 = 20\). The jacket is cheaper.
Two different answers, so even both statements together are not sufficient.

The trap: (C). With both statements, everything is "in terms of" \(s\) and \(p\), so it feels solved. But the winner flips at a 20 percent shirt discount, and we never learn which side of 20 we are on.
Takeaway: when the statements give only relationships (twice, three times), plug in two sets of numbers. If the answer flips, it is not sufficient.

Answer: E