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23 May 2026, 03:07
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E
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The price of oil fell by \(p% \) the day before yesterday and rose by \(p%\) yesterday. As a result, the price of oil was \(2\frac14%\) percent lower yesterday than it was two days before. If it rises by \(3p%\) today, the price of oil today will be approximately what percent higher than it was two days ago?
(A) \(2%\)%
(B) \(13%\)%
(C) \(20%\)%
(D) \(25%\)%
(E) \(42%\)%
(A) \(2%\)%
(B) \(13%\)%
(C) \(20%\)%
(D) \(25%\)%
(E) \(42%\)%
23 May 2026, 15:36
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The price of oil fell by \(p%\) the day before yesterday and rose by \(p%\) yesterday. As a result, the price of oil was \(2\frac14%\) percent lower yesterday than it was two days before. If it rises by \(3p%\) today, the price of oil today will be approximately what percent higher than it was two days ago?
Let the initial price of oil be \(I\).
Calculate p:
\(I × (1 - \frac{p}{100})(1 + \frac{p}{100}) = I × (1 - \frac{2.25}{100})\)
\((1 - \frac{p}{100})(1 + \frac{p}{100}) = (1 - \frac{2.25}{100})\)
\(1 - \frac{p^2}{10,000} = 1 - \frac{2.25}{100}\)
\(\frac{p^2}{10,000} = 0.0225\)
\(p^2 = 225\)
\(p = 15\)
Answer the question asked:
\(0.975I(1 + \frac{3p}{100}) = 0.975I(1 + \frac{45}{100}) ≈ 1.42I\)
Alternative Approach
Try \(p = 10\).
If \(p = 10\), then a 10% decrease followed by 10% increase would be 100% --> 90% --> 99%, which is a 1% decrease.
We have been told that the net decrease after two days was 2.25%.
So, \(p > 10\).
If \(p > 10\), then on the third day, the price increased by more than 30 percent, becoming higher than \(0.975I + (0.30 × 0.975I) ≈ 1.27I\).
So, the net increase would be greater than around 27%.
The only choice greater than 27% is 42%.
(A) \(2%\)%
(B) \(13%\)%
(C) \(20%\)%
(D) \(25%\)%
(E) \(42%\)%
Correct answer: E
Let the initial price of oil be \(I\).
Calculate p:
\(I × (1 - \frac{p}{100})(1 + \frac{p}{100}) = I × (1 - \frac{2.25}{100})\)
\((1 - \frac{p}{100})(1 + \frac{p}{100}) = (1 - \frac{2.25}{100})\)
\(1 - \frac{p^2}{10,000} = 1 - \frac{2.25}{100}\)
\(\frac{p^2}{10,000} = 0.0225\)
\(p^2 = 225\)
\(p = 15\)
Answer the question asked:
\(0.975I(1 + \frac{3p}{100}) = 0.975I(1 + \frac{45}{100}) ≈ 1.42I\)
Alternative Approach
Try \(p = 10\).
If \(p = 10\), then a 10% decrease followed by 10% increase would be 100% --> 90% --> 99%, which is a 1% decrease.
We have been told that the net decrease after two days was 2.25%.
So, \(p > 10\).
If \(p > 10\), then on the third day, the price increased by more than 30 percent, becoming higher than \(0.975I + (0.30 × 0.975I) ≈ 1.27I\).
So, the net increase would be greater than around 27%.
The only choice greater than 27% is 42%.
(A) \(2%\)%
(B) \(13%\)%
(C) \(20%\)%
(D) \(25%\)%
(E) \(42%\)%
Correct answer: E
