A difficult BMAT question

This was from section 1 of the 2022 Ocotber BMAT paper-question 28. I cannot figure out how to answer this.

Karl is racing against Frank. The race is run over a distance of 800m.

In the first race, Karl allows Frank to start 5 seconds before him, but he still completes the race 3 seconds before Frank.

In the second race, they start to run at the same time, but Karl allows Frank to start 100m ahead of him on the track, so Frank only has to run 700m. Frank completes this race 4.5 seconds before Karl.

In the third race, they again start to run at the same time and Karl allows Frank to start 40m ahead on the track.

Karl runs each race at the same constant speed. Frank runs slower but at the same constant speed for each race.

What is the outcome of the third race?
A) Karl finishes the race 3 seconds before Frank.
B) Karl Finishes the race 0.5 seconds before Frank.
C) They finish the race at the same time.
D) Frank finishes the race 0.5 seconds before Karl.
E) Frank finishes the race 3 seconds before Karl.

According to the answer key, the answer is A, and I've tried to work it out again, but keep getting C 😭. Can someone please explain how to do this?

(edited 8 months ago)

Reply 1

mqb2766

Letting Sf being the speed of Frank and Tk being Karls time, then the first two races give
Sf = 800/(Tk+8)
Sf = 700/(Tk-4.5)
So equating and solving for the times for the third race gives A.

(edited 8 months ago)

Reply 2

Raabs_grvs

Original post by mqb2766

Letting Sf being the speed of Frank and Tk being Karls time, then the first two races give
Sf = 800/(Tk+8)
Sf = 700/(Tk-4.5)
So equating and solving for the times for the third race gives A.

Ohhh! I forgot to account for the 5 seconds at the beginning. That makes much more sense. Thanks so much 👍

(edited 8 months ago)

Reply 3

Raabs_grvs

Original post by mqb2766

Letting Sf being the speed of Frank and Tk being Karls time, then the first two races give
Sf = 800/(Tk+8)
Sf = 700/(Tk-4.5)
So equating and solving for the times for the third race gives A.

Hey, also, could you tell me how to question 24 on this paper (I can't figure out how to send this as an image sorry): https://www.admissionstesting.org/Images/687556-past-paper-october-2022-section-2.pdf

Reply 4

mqb2766

Original post by Raabs_grvs

Without working it through, Id assume "radius" 1 in both cases (youre interested in the ratio so its irrelevant what the radius is), then the area of each triangle is efffectively (again the 1/2s cancel)
sin(60) for the hexagon
sin(45) for the octagon
and you have 3 in the hexagon and 4 in the octagon ...

Reply 5

faery18!

Original post by mqb2766

Letting Sf being the speed of Frank and Tk being Karls time, then the first two races give
Sf = 800/(Tk+8)
Sf = 700/(Tk-4.5)
So equating and solving for the times for the third race gives A.