Logical Reasoning Question - II

[Qwertyuiop[]]

I need help to solve this problem, wasn't able to because I don't have an idea how to go about solving it. I think i need help better understanding the question. Can't make sense of the options given.

 

[Dr.Peterson]

I need help to solve this problem, wasn't able to because I don't have an idea how to go about solving it. I think i need help better understanding the question. Can't make sense of the options given.

Just think about one of the five statements at a time, and consider whether it is necessarily true. If you can think of a situation in which the facts you are given are true, but the statement is not, then cross it off. Hopefully only one will be left!

I myself would start by looking at the given facts, and trying out a few arrangements, to get a feel for what this is about. For example, here is one possible arrangement:

x​

x​

x​

x​

xxxxxxxxxxxxxx​

Here is another:

xxx​

xxx​

xxx​

xxx​

xxxxxx​

In each case, there are 18 x's on 5 lines, with at least one x on each.

We see that another way to look at it is to list 5 integers, each greater than zero, whose sum is 18:

1+1+1+1+14​

3+3+3+3+6​

Now (A) claims that any such arrangement that includes the number 6 will contain at least one other number that is 3 or more. Is that always true? It's true of my second example; could you make one in which it isn't true?

Tell us your thoughts on (A), then move on to (B). (If in any case you can't think of a counterexample, move on to the next: Maybe that's the answer, but you need to check them all.)

 

[Qwertyuiop[]]

For A) I did something like this , just let me know what is wrong here. If it says "If there is a floor with six offices, there is another one with at least three" Then I can have more then 3 offices in a floor like 9 for example but as long as there is 1 office in each remaining floors.

Is this a possible arrangement ?
xxxxxx
xxxxxxxxx
x
x
x
OR
xxxxxx
xxxxxx
x
x
x


There are lot of ways to arrange this so I think A) seems correct and in fact A) is the answer.

In my first try, I rejected A) because it seemed incorrect to me because i made an arrangement like this :
xxxxxx
xxx
x
x
x

That was the first arrangement that came to my mind. One floor with six offices and another with three and the remaining ones will have just one each but it does not add up to 15?? It adds up to 12 so I rejected choice A) . Why the arrangement above can't be one ? I did the same thing with all the options and ended up rejecting all 5 of them lol . Then i decided to post this question here because i thought i don't understand the question. I would simply write down the arrangement and see if it adds up to 15, none of them did so thought they were all wrong.
Similarly , here is the arrangement i tried for B)
It says "If there is floor with two offices, there is another with with at least five "
xx
xxxxx
x
x
x
This one only adds up to 10 don't know what i am doing wrong i just made the arrangement like it says in the statement ?

 

[Dr.Peterson]

For A) I did something like this , just let me know what is wrong here. If it says "If there is a floor with six offices, there is another one with at least three" Then I can have more then 3 offices in a floor like 9 for example but as long as there is 1 office in each remaining floors.

Is this a possible arrangement ?
xxxxxx
xxxxxxxxx
x
x
x
OR
xxxxxx
xxxxxx
x
x
x


There are lot of ways to arrange this so I think A) seems correct and in fact A) is the answer.

The first of these examples fits the statement. The second does not.

The fact that you can make many cases in which it is true is not in itself evidence that it is always true. But the fact that you presumably were not able to find a counterexample suggests that it may be the answer. But this is why I said to keep looking at others. If you end up with two statements that seem like they are always true, you can go back and look more closely at those, to PROVE that one is always true; if all but one are definitely not always true, then you have your answer.

In my first try, I rejected A) because it seemed incorrect to me because i made an arrangement like this :
xxxxxx
xxx
x
x
x

That was the first arrangement that came to my mind. One floor with six offices and another with three and the remaining ones will have just one each but it does not add up to 15?? It adds up to 12 so I rejected choice A) . Why the arrangement above can't be one ? I did the same thing with all the options and ended up rejecting all 5 of them lol . Then i decided to post this question here because i thought i don't understand the question. I would simply write down the arrangement and see if it adds up to 15, none of them did so thought they were all wrong.

This is not a valid arrangement, because the total is not 18 as required. So you can't consider it. But that means that it contributes nothing to your reasoning; it is not a counterexample. (When you say 15, is that a typo? There is no 15 in the problem.)

Similarly , here is the arrangement i tried for B)
It says "If there is floor with two offices, there is another with with at least five "
xx
xxxxx
x
x
x
This one only adds up to 10 don't know what i am doing wrong i just made the arrangement like it says in the statement ?

ONLY CONSIDER EXAMPLES THAT MEET THE BASIC REQUIREMENTS: 18 total, 5 floors, at least one on each floor.

You are not trying to make something that fails to meet those requirements; you are trying to make something that fits those requirements, and the conditions of the statement you are checking, but not its conclusion.

Your thinking for B should involve trying to make a counterexample that fits those requirements and has a floor with (exactly) 2 offices, but has no floor with 5 or more. So put 2 on one floor, as you did, and then try to arrange the remaining 16 among the other 4 floors without having as many as 5 on any one. You will find that you can do that:

xx​

xxxx​

xxxx​

xxxx​

xxxx​

Here there are 18 offices, at least one on each floor, exactly 2 on one floor, but never more than 4 on any (because 16/4 = 4).

So that is a counterexample, and we know that B is false.

Does that make things clear? Try the remaining 3 statements, and see if you can show all of them to be false.

Again, the form of the problem is, "Given P, is it true that Q implies R?" A counterexample is a case in which P and Q are both true, but R is false. Cases in which P or Q is not true don't count.

 

[Steven G]

Say floor 1 has 6 offices.
Now there are 5-1 = 4 floors to distribute 18-6=12 offices.
Let's see if there must be a floor with at least three offices. SUPPOSE OTHERWISE. If we put 2 offices on each of the remaining 4 floors that will account for 8 offices.
Continue.

 

[Qwertyuiop[]]

The first of these examples fits the statement. The second does not.

The fact that you can make many cases in which it is true is not in itself evidence that it is always true. But the fact that you presumably were not able to find a counterexample suggests that it may be the answer. But this is why I said to keep looking at others. If you end up with two statements that seem like they are always true, you can go back and look more closely at those, to PROVE that one is always true; if all but one are definitely not always true, then you have your answer.


This is not a valid arrangement, because the total is not 18 as required. So you can't consider it. But that means that it contributes nothing to your reasoning; it is not a counterexample. (When you say 15, is that a typo? There is no 15 in the problem.)


ONLY CONSIDER EXAMPLES THAT MEET THE BASIC REQUIREMENTS: 18 total, 5 floors, at least one on each floor.

You are not trying to make something that fails to meet those requirements; you are trying to make something that fits those requirements, and the conditions of the statement you are checking, but not its conclusion.

Your thinking for B should involve trying to make a counterexample that fits those requirements and has a floor with (exactly) 2 offices, but has no floor with 5 or more. So put 2 on one floor, as you did, and then try to arrange the remaining 16 among the other 4 floors without having as many as 5 on any one. You will find that you can do that:

xx​

xxxx​

xxxx​

xxxx​

xxxx​

Here there are 18 offices, at least one on each floor, exactly 2 on one floor, but never more than 4 on any (because 16/4 = 4).

So that is a counterexample, and we know that B is false.

Does that make things clear? Try the remaining 3 statements, and see if you can show all of them to be false.

Again, the form of the problem is, "Given P, is it true that Q implies R?" A counterexample is a case in which P and Q are both true, but R is false. Cases in which P or Q is not true don't count.

"(When you say 15, is that a typo? There is no 15 in the problem.)"
Sorry that should be 18 yes! Ok i see , i have to come up with an arrangement that fits the conditions but makes the conclusion false. Like for B, as you said, we can come up with an arrangement where there is no floor with 5 offices given a floor that has 2 offices. We can get 18 offices in total , with at least one office on each and there is none that has 5 or more. Great, understood. i will try C D and E now.
I actually didn't know what is a counterexample before you explained it and how to create one, it's really the first time I've come across logic questions that's why they're kicking my a** right now.

 

[Steven G]

Look carefully at A

 

[Qwertyuiop[]]

Say floor 1 has 6 offices.
Now there are 5-1 = 4 floors to distribute 18-6=12 offices.
Let's see if there must be a floor with at least three offices. SUPPOSE OTHERWISE. If we put 2 offices on each of the remaining 4 floors that will account for 8 offices.
Continue.

Like this?
xxxxxx
xx
xx
xx
xx
That does not add up to 18. So we must have at least one floor that will have 3 or more offices to get 18 offices in total.

 

[Dr.Peterson]

Correct. This is a proof that A is in fact correct, as opposed to what we'd done before, just not finding reason to think it's wrong.

I've been holding off on this, so you wouldn't do this yet, and instead can learn all you can about counterexamples from the other cases. But when you can do this, it's a better approach. (Sometimes it's harder to make an actual proof than to eliminate options; both are useful skills.)

 

[Steven G]

I'm sorry that I gave my hint earlier than you wished.

 

[Dr.Peterson]

I'm sorry that I gave my hint earlier than you wished.

No, it's a good one, and I might have neglected to get around to it. And if, as I hope, we're dealing with a motivated student, it won't prevent continuing with the other pieces of learning that are needed.

 

[Qwertyuiop[]]

Hi, I did the last 4 options left, B C D and E.
B says: If there is a floor with two offices, there is another one with at least five
I made this arrangement as a counterexample to disprove that if we have two offices, we can get 18 offices in total without having a floor with 5 offices or more.

xx
xxxx
xxxx
xxxx
xxxx
No floor has 5 offices or more.

Now C says : There are certainly at least two floors with at most three offices each
I came up with this:
xx
xxxx
xxxx
xxxx
xxxx

We need a minimum of 2 floors that have no more than 3 offices. In this arrangement, there is only 1 floor with less than 3 offices. Statements C and D seems different than other statements which have this pattern if p then q. This one requires both part of the statement to be true right?

D : There are certainly two floors with at least three offices each

xxxxxxxxxx
xx
xx
xx
xx
There is only one floor here (instead of two) that has more than 3 offices rest of them have less than three.

E : If there is a floor with only one office, there is another one with at least seven

E was the easiest of find a counterexample for:

x
xxx
xxxx
xxxxx
xxxxx

There is a floor with one office in the arrangement but there is none with seven offices or more.
So are my arrangements correct and are these good counterexamples?

 

[Dr.Peterson]

Everything looks good (assuming I'm not reading it too quickly). I just want to comment on one:

Now C says : There are certainly at least two floors with at most three offices each
I came up with this:
xx
xxxx
xxxx
xxxx
xxxx

We need a minimum of 2 floors that have no more than 3 offices. In this arrangement, there is only 1 floor with less than 3 offices. Statements C and D seems different than other statements which have this pattern if p then q. This one requires both part of the statement to be true right?

That is the one I expected to cause most difficulty, because the statement is most convoluted; I myself find it helpful to restate "at least" and "at most" as "no less than" and "no more than", so we want an example in which it is not true that no less than 2 floors have no more than 3 offices. For this not to be true, there must be fewer than 2 floors with 3 or fewer offices on each. So what you did, putting 2 offices on only 1 floor, and distributing the remaining 16 offices evenly among 4 floors (so that none of them have too many) is just right. The tricky part is to be clear on what the negation of a complex statement is, and you got it.

Good job.

 

[Steven G]

B says: If there is a floor with two offices, there is another one with at least five
Stop looking at examples (in my opinion)
One floor has 2 offices. So there are 18-2=16 offices to put into 5-1=4 floors
Immediately you should see that 4 offices can be on each floor. So there does not need to be a floor with at least 5 offices.

C says : There are certainly at least two floors with at most three offices each
Let's try to make this not work.
Is it possible that no floor has at most 3 offices? That means that all floors have at least 4 offices. This can't happen since 4*5>18
Can exactly one floor have at most 3 offices? If one floor has 2 offices then there are 16 offices to put onto 4 floors. You can put 4 offices on each floor and all is good. So there are certainly at least two floors with at most three offices each is not true.

This is the way I would handle these type of problems.

 

[JeffM]

There are frequently many ways to do problems. The discussion about counter-examples is a great general way to do all kinds of logic problems, but frequently there are short cuts.

By the way, it is not clear to me from the wording of the problem that NECESSARILY only one statement is correct?

Here simple division will help you.

If there is a floor with 6 offices, then there are 12 offices left and 4 floors left. Thus, the average number of offices per floor is 3. Can we average 3 per floor if every floor has fewer than 3? So how do you feel about A?

If there is a floor with 2 offices, then there are 16 offices and 4 floors left. Thus, the average number of offices per floor is 4. Can we average 4 if every floor has at most 4? So how do you feel about B?

If you take care to understand the statements, all of these require nothing more than simple division to solve.

 

[Qwertyuiop[]]

There are frequently many ways to do problems. The discussion about counter-examples is a great general way to do all kinds of logic problems, but frequently there are short cuts.

By the way, it is not clear to me from the wording of the problem that NECESSARILY only one statement is correct?

Here simple division will help you.

If there is a floor with 6 offices, then there are 12 offices left and 4 floors left. Thus, the average number of offices per floor is 3. Can we average 3 per floor if every floor has fewer than 3? So how do you feel about A?

If there is a floor with 2 offices, then there are 16 offices and 4 floors left. Thus, the average number of offices per floor is 4. Can we average 4 if every floor has at most 4? So how do you feel about B?

If you take care to understand the statements, all of these require nothing more than simple division to solve.

hmm yes , this of course is an easier way to solve but in the logic questions, knowing counter examples method is very important so I wanted to learn this. This method can probably cut the time in half.

"By the way, it is not clear to me from the wording of the problem that NECESSARILY only one statement is correct?"- I think they are saying that only one is true regardless of how you distribute the offices in the floors, it says this in brackets. Some of them might be true for only certain arrangements of offices on the floors imo.

 

[JeffM]

hmm yes , this of course is an easier way to solve but in the logic questions, knowing counter examples method is very important so I wanted to learn this. This method can probably cut the time in half.

"By the way, it is not clear to me from the wording of the problem that NECESSARILY only one statement is correct?"- I think they are saying that only one is true regardless of how you distribute the offices in the floors, it says this in brackets. Some of them might be true for only certain arrangements of offices on the floors imo.

They say "which of these statements is always true," a wording that grammatically implies only one is true, but I hate to rely on a grammatical nuance in a math problem.