Struggling with our UMAT questions? See how you went with these solutions.
As of the 24th of September 2017, UMAT has now been replaced by the University Clinical Aptitude Test – UCAT.
To find out more about the change to UCAT, please read our post: UMAT replaced by UCAT for 2019.
To learn more about UCAT and to find sample questions, please read our Beginner’s Guide to UCAT.
1. This is a relatively simple question. Since the values on dice A and B are equal, Bonita goes first. As B+D=2+6=8, Adam rolls and obtains C=4. Since A+C=2+4=6, both players lose (c).
2. In this game, A = 1, 3 or 5 and B = 2, 4 or 6. From the last piece of information, we know that Bonita wins the game and that this occurs only after C is rolled. Hence we must consider the left part of the flowchart and so Adam goes first (a). Furthermore, since Adam goes first, B cannot be 6 as this would mean that Bonita would go first (b). Given that A = 3 and we know that Adam goes first, this means that the only value that B can take is 2 as it must be an even number less than 3. Hence, given that Bonita wins and B+D = 6, it follows that D = 6 (d). As Adam goes first, then A = 3 or 5 and it is possible for Adam to win if C = 4, since A+C = 3+4 = 7. Thus C must not necessarily be odd (c).
3. This question requires careful examination of all the possible combinations of dice that both players may have. It is important to note that we do not know the order in which the respective dice that Adam and Bonita possess are rolled. The following table shows the possible combinations of dice that each player may have, in the order in which they are possibly rolled:
Adam | Bonita | ||
2 then 5 | 2 then 6 | 3 then 4 | 4 then 4 |
5 then 2 | 6 then 2 | 4 then 3 | 4 then 4 |
This means that there are a total of 16 possible outcomes in this new game.
Suppose that Adam rolled a 2 first and Bonita rolled a 3 first. According to the rules in the flowchart, Bonita must roll her second die and obtains a 4, regardless of what the value of Adam’s second die was. Hence Bonita wins at Stop 3 (1/16). Additionally, if Bonita rolled a 4 first, then Bonita rolls again and obtains either a 4 or a 3 (2/16). There are 2 cases in which Bonita rolls a 4 first and rolls a 4 again. Hence Bonita wins at Stop 3 four times out of 16.
Continuing this logic with the other possibilities, the following list summarises the overall results from the game:
Hence, both players have an equal chance of winning (c).