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Algebra Level 3

Three people are driving on the same route to the shop and back.
Jack drives at 10 mph to the shop & 50 mph back.
Bob drives at 20 mph to the shop & 40 mph back.
Mr. Bean drives at 30 mph to the shop & drives back at the same speed.
Suppose they start at the same time and spend the same amount of time at the shop.
Who arrives back earlier?

Jack Bob Mr. Bean They arrive back at the same time.

Jeff Giff by Jeff Giff , 13, China

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2 solutions

Average speed of Jack is 2 × 10 × 50 10 + 50 = 50 3 \dfrac{2\times 10\times 50}{10+50}=\dfrac{50}{3} mph.

Average speed of Bob is 2 × 20 × 40 20 + 40 = 80 3 \dfrac {2\times 20\times 40}{20+40}=\dfrac{80}{3} mph.

Average speed of Mr. Bean is 30 = 90 3 30=\dfrac{90}{3} mph.

So, the average speed of Mr. Bean being the highest, he will be back the earliest.

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Jeff Giff
Jun 3, 2020

Suppose the route is n n miles long.

Jack spends: n 10 \frac{n}{10} + n 50 \frac{n}{50} = 6 n 50 \frac{6n}{50} hrs.
Bob spends: n 20 \frac{n}{20} + n 40 \frac{n}{40} = 3 n 40 \frac{3n}{40} hrs.
Mr. Bean spends: n 30 \frac{n}{30} × \times 2 2 = 2 n 30 \frac{2n}{30} hrs.

Mr. Bean spends the least time so he arrives back earlier.

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You can't compare fractions directly, either you have to make their denominators equal or you have to compare their decimal expansion

Zakir Husain - 1 year ago

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Oh, I forgot to elaborate on that part.

Jeff Giff - 1 year ago

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