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Difficulty:
95%
(hard)
Question Stats:
47% (03:00) correct
53%
(03:08)
wrong
based on 1478
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Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?
A. \(4\pi – 1.6\)
B. \(4\pi + 8.4\)
C. \(4\pi + 10.4\)
D. \(2\pi – 1.6\)
E. \(2\pi – 0.8\)
A. \(4\pi – 1.6\)
B. \(4\pi + 8.4\)
C. \(4\pi + 10.4\)
D. \(2\pi – 1.6\)
E. \(2\pi – 0.8\)
The OA is pretty long and even solving it that way takes me +2 mins. Hopefully someone can offer a fast solution.
11 Nov 2009, 03:12
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Car B begins moving at 2 mph around a circular track with a radius of 10 miles. Ten hours later, Car A leaves from the same point in the opposite direction, traveling at 3 mph. For how many hours will Car B have been traveling when car A has passed and moved 12 miles beyond Car B?
A. \(4\pi – 1.6\)
B. \(4\pi + 8.4\)
C. \(4\pi + 10.4\)
D. \(2\pi – 1.6\)
E. \(2\pi – 0.8\)
It's possible to write the whole formula right away but I think it would be better to go step by step:
B speed: 2 mph;
A speed: 3 mph (travelling in the opposite direction);
Track distance: \(2*\pi*r=20*\pi\);
What distance will cover B in 10h: \(10*2=20\) miles
Distance between B and A by the time, A starts to travel: \(20*\pi-20\)
Time needed for A and B to meet distance between them divided by the relative speed: \(\frac{20*\pi-20}{2+3}= \frac{20*\pi-20}{5}=4*\pi-4\), as they are travelling in opposite directions relative speed would be the sum of their rates;
Time needed for A to be 12 miles ahead of B: \(\frac{12}{2+3}=2.4\);
So we have three period of times:
Time before A started travelling: 10 hours;
Time for A and B to meet: \(4*\pi-4\) hours;
Time needed for A to be 12 miles ahead of B: 2.4 hours;
Total time: \(10+4*\pi-4+2.4=4*\pi+8.4\) hours.
Answer: B.
A. \(4\pi – 1.6\)
B. \(4\pi + 8.4\)
C. \(4\pi + 10.4\)
D. \(2\pi – 1.6\)
E. \(2\pi – 0.8\)
It's possible to write the whole formula right away but I think it would be better to go step by step:
B speed: 2 mph;
A speed: 3 mph (travelling in the opposite direction);
Track distance: \(2*\pi*r=20*\pi\);
What distance will cover B in 10h: \(10*2=20\) miles
Distance between B and A by the time, A starts to travel: \(20*\pi-20\)
Time needed for A and B to meet distance between them divided by the relative speed: \(\frac{20*\pi-20}{2+3}= \frac{20*\pi-20}{5}=4*\pi-4\), as they are travelling in opposite directions relative speed would be the sum of their rates;
Time needed for A to be 12 miles ahead of B: \(\frac{12}{2+3}=2.4\);
So we have three period of times:
Time before A started travelling: 10 hours;
Time for A and B to meet: \(4*\pi-4\) hours;
Time needed for A to be 12 miles ahead of B: 2.4 hours;
Total time: \(10+4*\pi-4+2.4=4*\pi+8.4\) hours.
Answer: B.
10 Jan 2011, 20:01
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ajit257
Make the diagram. You will be able to see how to solve it.
Attachment:
Ques1.jpg [ 16.63 KiB | Viewed 66901 times ]
B starts from X and travels for 10 hrs clockwise at 2 mph i.e. it travels 20 miles. Now car A starts from X counter clockwise. Distance between A and B is 20*pi - 20.
Now, to meet, they have to together cover this distance plus 12 miles more which they have to put between them.
Time taken to cover this distance by them = (20*pi - 20 + 12)/(3 + 2) = 4*pi - 1.6 hrs
Car B has been traveling for 10 + 4*pi - 1.6 = (4*pi + 8.4) hrs
