# If Theresa tests G on the second day, then which one of the following must be true?

on November 22 at 03:52AM

Setup

Hello I am really stuck on this game setup, can u please setup? Thanks

Irina on November 22 at 05:16AM

@suzanne,
'
Happy to help. The game requires us to determine the bicycle-tester pairs for each of the days. We have four riders - R S T Y - tasting four bicycles - F G H J. Each rider tests only one bicycle per day and all bicycles are tested every day, meaning we have four tests total per day. Each rider tests a different bicycle on a second day, in other words no rider tests the same bicycle on both days.

I am going to use bikes as a base and what we are looking for in any scenario is all four riders in each row - day 1 & day 2, but for each of the bikes riders must be different for different days.

1. __ __ __ __
2. __ __ __ __
F G H J

The following rules apply:
(1) R cannot test F, Y cannot test J

1. __ __ __ __
2. __ __ __ __
F G H J
~R. ~Y

This means that we only have 3 potential riders for F - STY and J - RST - and we must definitely pick two of them since we cannot have the same rider on both days.

(2) T must be one of the testers for H.

1. __ __ __ __
2. __ __ __ __
F G H J
~R. T. ~Y

(3) The bicycle that Y tests on the first day has to be tested by S on the second day.
We know that Y cannot test J and H must be tested by T on one the days, hence Y cannot test H either. Thus, the bicycle tested by Y & S must be either F or G.

1. __ __ __ __
2. __ __ __ __
F G H J
~R. T.~Y
/YS /YS

The question asks us if T tests G on the second day, which of the following must be true?
Let's diagram this scenario:

1. __ __ __ __
2. __ T __ __
F G H J

We can infer that if T tests G on the second day, T must test H on the first day since T must test H on one of the days per rule (2) and a rider can only test one bicycle per day. Since we only have two potential spots for YS combination per our initial setup, and one of them is taken, we can conclude that F must be tested by Y on the first day and S on the second day.

We still have to assign two riders to day one - R & S -and two riders to day two - R & Y. Since Y cannot test J, Y must test H on the second day, and the remaining rider R must test J on the second day. Since a rider cannot test the same bike on both days, we can conclude that S must test J on the first day and R must test G on the first day.
1. Y R T S
2. S. T. Y R
F G H J

Thus, we can conclude that (E) is the only answer choice that must be true and since we have determined the complete assignment of riders to bikes, all the other answer choices must be false.

Let me know if this makes sense and if you have any other questions.