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Jake rides his bike for the first 2/3 of the distance from home to sch

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Jake rides his bike for the first 2/3 of the distance from home to sch  [#permalink]

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New post 13 Jun 2018, 12:33
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Jake rides his bike for the first 2/3 of the distance from home to school, traveling at 10 miles per hour. He then walks the remaining 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

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Jake rides his bike for the first 2/3 of the distance from home to sch  [#permalink]

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New post 13 Jun 2018, 12:59
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Bunuel wrote:
Jake rides his bike for the first 2/3 of the distance from home to school, traveling at 10 miles per hour. He then walks the remaining 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


Formula used: Average speed = \(\frac{(p+q)*(ab)}{(qa + pb)}\)

where the ratio of the distances traveled is p:q
speeds at which the respective distances are traveled are a and b


Jake rides his bike for the first \(\frac{2}{3}\)rd of the distance and the remaining
\(\frac{1}{3}\)rd of the distance on foot, making the ratio of the distances 2:1(p:q)

Jake travels by bike at speed(a)=10mph and travels by foot at speed(b)=3mph

Substituting values of a,b,p, and q, we get the value for the average speed, as follows

Average speed = \(\frac{(2+1)*(10*3)}{(10 + 6)} = \frac{3*30}{16} = \frac{45}{8}\)

Therefore, the distance traveled by Jake is \(\frac{40}{60}*\frac{45}{8}\) = \(\frac{15}{4}\)(Option B)

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Re: Jake rides his bike for the first 2/3 of the distance from home to sch  [#permalink]

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New post 13 Jun 2018, 14:48
1
B
D is miles from home to school
(2/3 D) / 10mph + (1/3 D) / 3mph = 40 / 60
2/30 D + 1/9 D = 40/60
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Re: Jake rides his bike for the first 2/3 of the distance from home to sch  [#permalink]

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New post 13 Jun 2018, 21:46

Solution



Given:
    • Jake rides his bike for the first \(\frac{2}{3}\) of the distance from home to school at 10 mph
    • He walks the remaining \(\frac{1}{3}\) distance at 3 mph
    • Total travel time is 40 minutes or \(\frac{2}{3}\) hours

To find:
    • The distance between Jake’s home to his school

Approach and Working:

If we assume the total distance to be x miles, we can write
    • Distance travelled in bike = \(\frac{2x}{3}\) miles
    • Therefore, time for bike travel = (\(\frac{2x}{3}\)/10) hours = \(\frac{x}{15}\) hours
    • Distance travelled in walk = \(\frac{x}{3}\) miles
    • Therefore, time for walk = (\(\frac{x}{3}\)/3) hours = \(\frac{x}{9}\) hours

As the total travel time is \(\frac{2}{3}\) hours,
    • \(\frac{x}{15} + \frac{x}{9}\) = \(\frac{2}{3}\)
    Or, \(\frac{8x}{45}\) = \(\frac{2}{3}\)
    Or, x = \(\frac{90}{24}\) = \(\frac{15}{4}\) miles

Hence, the correct answer is option B.

Answer: B
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Re: Jake rides his bike for the first 2/3 of the distance from home to sch  [#permalink]

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New post 13 Jun 2018, 21:53
Answer is B for the explanations given above.
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Re: Jake rides his bike for the first 2/3 of the distance from home to sch  [#permalink]

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New post 17 Jun 2018, 19:47
Bunuel wrote:
Jake rides his bike for the first 2/3 of the distance from home to school, traveling at 10 miles per hour. He then walks the remaining 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


We can let d = the distance from Jake’s home to his school. Thus,

The time spent riding the bike is (2d/3)/10 = 2d/30 = d/15.

The time spent walking is (1d/3)/3 = d/9.

Since 40 minutes = 2/3 of an hour, we can create the equation for time:

d/15 + d/9 = 2/3

3d/45 + 5d/45 = 2/3

8d/45 = 2/3

24d = 90

d = 90/24 = 30/8 = 15/4 miles.

Answer: B
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Re: Jake rides his bike for the first 2/3 of the distance from home to sch  [#permalink]

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Re: Jake rides his bike for the first 2/3 of the distance from home to sch   [#permalink] 03 Dec 2019, 11:49
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