Question

In a 1-mile race, the winner crosses the finish line 10 feet ahead of the second-place runner and 23 feet ahead of the third-place runner. Assuming that each runner maintains a constant speed throughout the race, by how many feet does the second-place runner beat the third-place runner? (5280 feet in 1 mile.)

A. -13.06 ft

B. 13.02 ft

C. 3.01 ft

D. -10.04 ft

Answer 1

@Directrix

Answer 2

Let's think.

Answer 3

how do we start @Directrix

Answer 4

I am writing two equations and made an error I am correcting.

These others are writing 23 - 10 as the answer but I am not yet convinced of that.

Answer 5

ok take your time

Answer 6

F = 23 + T where F is the first place winner and T is the third place winner
F = 10 + S where S is the second place winner.

Answer 7

So, S = F - 10
T = F - 23
-----------

Subtracting to get S - T (distance from second to third), the two Fs add out, leaving this: -10 - (-23) = ? @Yacoub1993

I guess it should have been obvious that those others were correct. :)

Answer 8

A. -13.06 ft

B. 13.02 ft

C. 3.01 ft

D. -10.04 ft

Answer 9

@Yacoub1993 What did you get for the answer option?

Answer 10

there is some minor mistakes you did i think

Answer 11

I didn't see a minor mistake. I think the answer options are off.
But, there is only one answer option which could be anywhere near correct.

Do you see it?

Answer 12

yes 13.02

Answer 13

It is the answer to this: -10 - (-23) =

Answer 14

yes 13.02 is the nearest answer

Answer 15

I agree with that, the B option.