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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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Answer is B for the explanations given above.
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Bunuel
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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in the question they have said that "He then walks the remaining 1/3 of the distance" shouldnt the remaining 1/3rd distance be (1/3)*(2/3)*(x) as he has already covered (2/3)*(x) on his bike so the remaining should be 1/3rd of (2/3)*(x) ?

if that is not the case then when should we take (1/3)*(2/3)*(x)
like how should we know when something like this "(1/3)*(2/3)*(x)" should be considered
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