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Jake rides his bike for the first

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Jake rides his bike for the first [#permalink]  19 Aug 2017, 08:13
Expert's post
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Question Stats:

88% (03:11) correct 11% (00:00) wrong based on 9 sessions

Jake rides his bike for the first $$\frac{2}{3}$$ of the distance from home to school, traveling at 10 miles per hour. He then walks the remaining $$\frac{1}{3}$$ of the distance at 3 miles per hour. If his total trip takes 40 minutes, how many miles is it from Jake's home to his school?

A. $$\frac{5}{4}$$

B. $$\frac{15}{4}$$

C. 5

D. 6

E. 10
[Reveal] Spoiler: OA

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Re: Jake rides his bike for the first [#permalink]  28 Aug 2017, 15:00
Expert's post
Carcass wrote:
Jake rides his bike for the first $$\frac{2}{3}$$ of the distance from home to school, traveling at 10 miles per hour. He then walks the remaining $$\frac{1}{3}$$ of the distance at 3 miles per hour. If his total trip takes 40 minutes, how many miles is it from Jake's home to his school?

A. $$\frac{5}{4}$$

B. $$\frac{15}{4}$$

C. 5

D. 6

E. 10

(time spent on bike) + (time spent walking) = 2/3 hours (= 40 minutes)

Let D = total distance (in miles) from home to school
So, Jake rode his bike for a distance of (2/3)D miles, which equals 2D/3 miles
Then Jake walked for a distance of (1/3)D miles, which equals D/3 miles

Time = distance/rate

So, we can write: (2D/3)/10 + (D/3)/3 = 2/3
Multiply both sides by 30 to get: 6D/3 + 10D/3 = 60/3
Multiply both sides by 3 to get: 6D + 10D = 60
Simplify: 16D = 60
Solve: D = 60/16 = 30/8 = 15/4

[Reveal] Spoiler:
B

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Re: Jake rides his bike for the first   [#permalink] 28 Aug 2017, 15:00
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