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Re: Davis drove from Amityville to Beteltown at 50 miles per [#permalink]
1
sandy wrote:
Davis drove from Amityville to Beteltown at 50 miles per hour, and returned by the same route at 60 miles per hour.

Quantity A
Quantity B
Davis’s average speed for the round trip, in miles per hour
\(55\)


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.


GIVEN: Davis drove from Amityville to Beteltown at 50 miles per hour, and returned by the same route at 60 miles per hour.

Let's determine Davis’s average speed for the round trip, in miles per hour (Quantity A)
Average speed = (total distance)/(total time)

Let d = distance from Amityville to Beteltown

So, 2d = total distance

Total time = time spent driving 50 mph + time spent driving 60 mph
time = distance/speed
So, Total time = d/50 + d/60
= 6d/300 + 5d/300
= 11d/300

So, Average speed = (2d)/(11d/300)
= 600d/11d
= 600/11
= 54 6/11

We get:
Quantity A: 54 6/11
Quantity B: 55

Answer: B

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Re: Davis drove from Amityville to Beteltown at 50 miles per [#permalink]
2
sandy wrote:
Davis drove from Amityville to Beteltown at 50 miles per hour, and returned by the same route at 60 miles per hour.

Quantity A
Quantity B
Davis’s average speed for the round trip, in miles per hour
\(55\)


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.


When the distances are half and the speeds are (let us say) \(x\) and \(y\), Average speed is given by \(\frac{2xy}{(x+y)}\)

Here,
Average speed = \(\frac{2(50)(60)}{(50+60)} = 54.54mph\)

Col. A: \(54.54\)
Col. B: \(55\)

Hence, option B
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Re: Davis drove from Amityville to Beteltown at 50 miles per [#permalink]
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Re: Davis drove from Amityville to Beteltown at 50 miles per [#permalink]
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