UCAT: Decision Making

[mars_02]

I might be really dumb or the question is wrong... Another DM question,


I do not understand why B is an incorrect answer. The answers say 'none of the schools that offer childcare offer counselling and consultation & training', but clearly there is an overlapping region for the square, parallelogram and rectangle making B correct?

I also do not understand, in the solution's explanation of why D is incorrect, how they got 14 schools as the number of schools that offer childcare to children under 4. Isn't it 7+4+5+6+3+2=27?

Thanks so much for the help, I don't know what I'm missing I think I've been looking at this question for the past 30 minutes lol!

 

[2xq]

I might be really dumb or the question is wrong... Another DM question,
View attachment 4469
View attachment 4470
I do not understand why B is an incorrect answer. The answers say 'none of the schools that offer childcare offer counselling and consultation & training', but clearly there is an overlapping region for the square, parallelogram and rectangle making B correct?

I also do not understand, in the solution's explanation of why D is incorrect, how they got 14 schools as the number of schools that offer childcare to children under 4. Isn't it 7+4+5+6+3+2=27?

Thanks so much for the help, I don't know what I'm missing I think I've been looking at this question for the past 30 minutes lol!

Rectangle is offering kids over 4 NOT under. This means the whole square other than the 2 is out of play. HOWEVER the 2 at the top isn't in the parallelogram, thus B is wrong.

 

[mars_02]

Rectangle is offering kids over 4 NOT under. This means the whole square other than the 2 is out of play. HOWEVER the 2 at the top isn't in the parallelogram, thus B is wrong.

Hi, thanks for the answer, I should've read the question more carefully, my apologies. What do you think about their explanation for D? They say that 14 schools offer childcare to children under 4, but how do we know that? All we know is that 14 schools do not offer childcare for children older than four years old, but I don't think that is equivalent to 14 schools offer childcare for children under 4. What if those 14 schools do not provide childcare at all? Thanks..

 

[JL538]

Hi, thanks for the answer, I should've read the question more carefully, my apologies. What do you think about their explanation for D? They say that 14 schools offer childcare to children under 4, but how do we know that? All we know is that 14 schools do not offer childcare for children older than four years old, but I don't think that is equivalent to 14 schools offer childcare for children under 4. What if those 14 schools do not provide childcare at all? Thanks..

Im pretty sure in decision making (from what I can remember, been a long time lol ) that if its outside the shape, then the opposite is true. So you assume that all of the numbers added up outside of the rectange (which is kids>4) is kids<4. This is an assumption obviously but thats just how DM works

 

[mars_02]

Ahh I see, will have to remember that, thank you so much as always for the answer JL538! Really appreciate it

 

[2xq]

Im pretty sure in decision making (from what I can remember, been a long time lol ) that if its outside the shape, then the opposite is true. So you assume that all of the numbers added up outside of the rectange (which is kids>4) is kids<4. This is an assumption obviously but thats just how DM works

I mean we did kind of assume that 4 year olds don't exist

 

[cocodreams]

Could someone please clarify why the answer would be C? I chose C intuitively using process of elimination, but I don't quite understand it
Thanks!

 

[JL538]

View attachment 4488
Could someone please clarify why the answer would be C? I chose C intuitively using process of elimination, but I don't quite understand it
Thanks!

p is the probability of going at a speed faster than 100m per minute. So this is any speed faster than 1.666... m/sec

Since Todd doesn't tire, and this is 1 individual event (1000m race, not 10 x 100m races), this means Todd's speed will be constant throughout the whole race no matter how long it is, so could be like 2 million metres and it would still be the same speed. Therefore, it only takes the p chance to occur/trigger once for the statement to be true.

If it were 10 x 100 m races, you would do p^10 to find out the probability.

Idk if that makes sense lol

 

[mars_02]

View attachment 4488
Could someone please clarify why the answer would be C? I chose C intuitively using process of elimination, but I don't quite understand it
Thanks!

Hi, where did you get this question from? Just curious.. Adding to JL538's answer, since 0 <= p <= 1, 0 <= 10p <= 10 so I immediately had alarm bells ringing that the answer was going to be a No.

 

[JL538]

Hi, where did you get this question from? Just curious.. Adding to JL538's answer, since 0 <= p <= 1, 0 <= 10p <= 10 so I immediately had alarm bells ringing that the answer was going to be a No.

Just FYI no discussion of paid services so I don't think cocodreams can reveal unless it is an official UCAT question

 

[jinjja21]

Hi guys, I'm struggling to understand the solution to this question


The answer is yes, because all numbers of have equal chances of being rolled.

And here is the solution:


I don't really understand how the chance of not rolling a four (ie. the probability of rolling the rest of the five numbers) become 1-x.
- My understanding is that the probability of rolling a four is x/6 and the probability of rolling any other dice other than four is 1/6 (which is definitely wrong but I can't understand why this would be incorrect)

Cheers.

 

[ucatboy]

Hi guys, I'm struggling to understand the solution to this question
View attachment 4522

The answer is yes, because all numbers of have equal chances of being rolled.

And here is the solution:
View attachment 4521

I don't really understand how the chance of not rolling a four (ie. the probability of rolling the rest of the five numbers) become 1-x.
- My understanding is that the probability of rolling a four is x/6 and the probability of rolling any other dice other than four is 1/6 (which is definitely wrong but I can't understand why this would be incorrect)

Cheers.

The possibilities of all events must add up to 1 or 100% (whatever you prefer). The question tells us that the probability of rolling a four is X. Now we don't know what X could be, it could be 0.45, it could be 0.579, but that's why we call it X: an unknown.

Everything in life either happens or doesn't happen i.e. you either win the lottery or you don't, there's no inbetween. It doesn't matter that the two sides are unequal (e.g. 0.001% chance of winning the lottery and 99.999% of losing), the combined probabilities of something both happening and NOT happening MUST add up to 100%, or 1. In the same vein, the probability of rolling a four and NOT rolling a four must ALSO add up to 1. And since not rolling a four = rolling any other number, then the probability of that happening is 1 (all possibilities) - X (rolling a four).

We are also told that for the remaining five sides of the die, they are equally likely to be landed on. So we have to split that (1-X) into five equal portions, or (1-X)/5, which is the answer. Hence, the probability of rolling a 6 (one of the equal five sides), is (1-X)/5.

In summary:

• The probability of rolling a four is X.
• The probability of rolling anything other than a four is 1 - X.
  • The probabilities of rolling 1, 2, 3, 5 or 6 are (1 - X)/5 each.

I think your problem is that you fixated too much on the 6 sides of the die (hence putting 6 as a denominator in your answer). In reality, this die has been modified so much that the probability of rolling a certain number could be anything. It's generally a bad idea to put the number of possibilities as the denominator unless you're certain they're equally likely to happen.

 

[fm8]

Hey does anyone know of a really quick method to answering DM questions like these: the list of 5 friends, food/drink type... x did not order coke etc..
I currently use the table method but is there a faster way?

 

[nb]

Hey does anyone know of a really quick method to answering DM questions like these: the list of 5 friends, food/drink type... x did not order coke etc..
I currently use the table method but is there a faster way?

Something that helped me was getting the obvious statements out of the way first, so you can eliminate some of the multiple choice options, putting you in a better position to answer the question. From what I remember, there are always a few ones which are just obvious and you can make it out to be false in your head. You will also be more likely to 'guestimate' the answer if you're running out of time by doing this. I personally never used tables and stuff as it's just too time consuming for me.

 

[cocodreams]

p is the probability of going at a speed faster than 100m per minute. So this is any speed faster than 1.666... m/sec

Since Todd doesn't tire, and this is 1 individual event (1000m race, not 10 x 100m races), this means Todd's speed will be constant throughout the whole race no matter how long it is, so could be like 2 million metres and it would still be the same speed. Therefore, it only takes the p chance to occur/trigger once for the statement to be true.

If it were 10 x 100 m races, you would do p^10 to find out the probability.

Idk if that makes sense lol

Sorry this is so late, I just saw the solution! But yeah looking back at it and reading your solution, it makes much more sense now! Thanks

 

[fm8]

Hey does anyone know of a really quick method to answering DM questions like these: the list of 5 friends, food/drink type... x did not order coke etc..
I currently use the table method but is there a faster way?

 

[mars_02]

Hey does anyone know of a really quick method to answering DM questions like these: the list of 5 friends, food/drink type... x did not order coke etc..
I currently use the table method but is there a faster way?

Hey, the official name for that question type is logical puzzles. (From the question tutorial on ucat.edu.au)

They're also known as Einstein's riddle - [os] Einstein's five-houses riddle

I did some research on the most efficient way to solve them, and the table method was the only one I could find, like how they do it here [os] Solution to Einstein's five-houses riddle

But if someone knows a better way please share it with us!

 

[fm8]

Hey
So I understand that if all A's are B's, and you get a statement saying some A's are B's - then that statement is wrong.
but what if it says some B's are A's?
I just want to confirm that this statement would be a correct

 

[dotwingz]

I forget where exactly, but users on this site emailed pearson and they confirmed that the definitions are

“All” is 100%
“Some” is 0 < amount < 100
“At least some” is 0 < amount<= 100
"Most" = 50% < amount < 100%

or if you understand set notation

All = 100%
Some = (0, 100%)
At least some = (0, 100%]
Most = (50%, 100%)

 

[mars_02]

if all A is B, then wouldn't some A is B be true?
Similarly, if all A is B, wouldn't some B is A also be true?



I'm terrible at DM, so I'm most likely wrong. If someone knows an answer, please let us know!