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The result of increasing, then decreasing, q by 1 %
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20 Dec 2022, 07:55
20 Dec 2022, 07:55
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Question Stats:
81% (01:25) correct
18% (01:33) wrong based on 32 sessions
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\(q >0\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The result of increasing, then decreasing, q by 1 % |
The result of increasing, then decreasing, q by 2 % |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: The result of increasing, then decreasing, q by 1 %
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23 Dec 2022, 08:24
23 Dec 2022, 08:24
1
Qty A = 2 * 1.1 * 0.99 = 217
Qty B = 2 * 1.02 * 0.98 = 1.99
Hence, A
Qty B = 2 * 1.02 * 0.98 = 1.99
Hence, A
The result of increasing, then decreasing, q by 1 %
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22 Jan 2024, 21:29
22 Jan 2024, 21:29
1
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
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
