Quantity A

Increasing P by 1% gives us

\(P(1+\frac{1}{100}) \)

\(P(1 + 0.01)\)

\(P(1.01)\)

And then decreasing the result by 1% gives us

\(P(1.01)(1-\frac{1}{100})\)

\(P(1.01)(1-0.01)\)

\(P(1.01)(0.99)\)

\(P(0.9999)\)

Or by applying the formula for successive percents,

When \(P\) is increased by 1% and decreased by 1%, the final value for \(P\) is given by

\(P(1.01)(0.99) = P(0.9999)\)

Quantity B

Increasing \(P\) by 2% gives us

\(P(1+\frac{2}{100}) \)

\(P(1 + 0.02)\)

\(P(1.02)\)

And then decreasing the result by 2% gives us

\(P(1.02)(1-\frac{2}{100})\)

\(P(1.02)(1-0.02)\)

\(P(1.02)(0.98)\)

\(P(0.9996)\)

Or by applying the formula for successive percents,

When \(P\) is increased by 2% and decreased by 2%, the final value for \(P\) is given by

\(P(1.02)(0.98) = P(0.9996)\)

Clearly Quantity A is larger than Quantity B