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Albert was driving on a straight highway, from Edenborough
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28 Oct 2019, 06:39
28 Oct 2019, 06:39
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Question Stats:
78% (02:36) correct
21% (02:23) wrong based on 37 sessions
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Albert was driving on a straight highway, from Edenborough to Megiddo, at 60 mph. Beatrice was driving faster than Albert, on the same highway, in the same direction. She started behind Albert. She passed Albert at 1:20 pm. At 3:20 pm, Beatrice arrived at Megiddo, and Albert arrived 50 minutes later, at 4:10 pm. Assume both drivers kept up constant speeds without stops. What was Beatrice's speed, in mph?
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85
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Re: Albert was driving on a straight highway, from Edenborough
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29 Oct 2019, 09:43
29 Oct 2019, 09:43
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huda wrote:
Albert was driving on a straight highway, from Edenborough to Megiddo, at 60 mph. Beatrice was driving faster than Albert, on the same highway, in the same direction. She started behind Albert. She passed Albert at 1:20 pm. At 3:20 pm, Beatrice arrived at Megiddo, and Albert arrived 50 minutes later, at 4:10 pm. Assume both drivers kept up constant speeds without stops. What was Beatrice's speed, in mph?
Show: :: OA
85
Let's start the clock at 1:20pm, when Al and Bea are at the exact SAME LOCATION
Let d = distance from this location to Megiddo
Let x = how many miles per hour FASTER than Al that Bea is travelling.
So, 60 + x = Bea's speed
And 60 = Al's speed
Starting from that SAME LOCATION, Bea's travel time is 2 hours and Al's travel time is 2 5/6 hours (2 5/6 hours = 2 hours and 50 minutes)
time = distance/speed
Let's deal with Al's trip first
We can we can write: 2 5/6 = d/60
Rewrite as: 17/6 = d/60
Cross multiply to get: (17)(60) = 6d
Divide both sides by 6 to get: (17)(10) = d
So d = 170
Now Bea's trip
We can we can write: 2 = d/(60 + x)
Since we now know d = 170, we get: 2 = 170/(60 + x)
Cross multiply get: 2(60 + x) = 170
Expand: 120 + 2x = 170
So... 2x = 50
Solve: x = 25
So, Bea's speed = 60 + 25 = 85 mph
Answer: 85
Cheers,
Brent
Re: Albert was driving on a straight highway, from Edenborough
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29 Oct 2019, 19:50
29 Oct 2019, 19:50
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huda wrote:
Albert was driving on a straight highway, from Edenborough to Megiddo, at 60 mph. Beatrice was driving faster than Albert, on the same highway, in the same direction. She started behind Albert. She passed Albert at 1:20 pm. At 3:20 pm, Beatrice arrived at Megiddo, and Albert arrived 50 minutes later, at 4:10 pm. Assume both drivers kept up constant speeds without stops. What was Beatrice's speed, in mph?
Show: :: OA
85
Hi,
Have you changed the FORMAT of the QUES? I guess this should be MCQ, because TIME/DISTANCE ques comes as MCQ. Moreover, back solving always comes handy in these type of QUES.
Back to the ques,
We need to find the speed of B (Beatrice), let it be \(S\)
M (Megiddo) speed = 60 m/hr
B pass A at \(1:20\) and reached destination at \(3:20\) - Total time taken = 2hrs = \(120\) minutes
A takes \(1:20\) to \(4:10\) to cover the same distance. Total time taken : \(170\) minutes
Since both cover equal distance, we can write
\(120 * S = 170 * 60\) ( \(120 * S\)= Distance covered by B and \(170 * 60\) = distance covered by A)
or \(S = \frac{{170 *60}}{120} = 85\) mph
