The result of increasing, then decreasing, q by 1 %
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22 Jan 2024, 21:29
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