Difference between revisions of "2009 Grade 8 CEMC Gauss Problems/Problem 16"

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-==Problem==
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-When it is 3:00 p.m. in Victoria, it is 6:00 p.m. in Timmins. Stefan’s flight departed at 6:00 a.m. local Victoria time and arrived at 4:00 p.m. local Timmins time. How long, in hours, was his flight?
-<math> \text{ (A) }\ 5 \qquad\text{ (B) }\ 9 \qquad\text{ (C) }\ 13 \qquad\text{ (D) }\ 7 \qquad\text{ (E) }\ 8 </math>
-==Solution 1==
-From the first sentence of the problem, we can realize that Timmins is <math>3</math> hours ahead. This means that <math>6:00</math> am in local Victoria time is the same thing as <math>9:00</math> am in local Timmins time.
-<math>3</math> hours after <math>9:00</math> am is <math>12:00</math> pm, and <math>4</math> hours after that is <math>4:00</math> pm, so his flight was <math>3 + 4 = \boxed {\textbf {(D) } 7}</math> hours long.
-~anabel.disher
-==Solution 1.1==
-We can also use the time conversion, but convert Timmins time to Victoria time to solve the problem.
-<math>4:00</math> pm in local Timmins time would be <math>1:00</math> pm in local Victoria time≥
-<math>6</math> hours after <math>6:00</math> am is <math>12:00</math> pm, and <math>1</math> hour after that <math>1:00</math> pm, so his flight was <math>6 + 1 = \boxed {\textbf {(D) } 7}</math> hours long.
-~anabel.disher