Q16 - All etching tools are either

[smiller]

What does the Question Stem tell us?
This is a Sufficient Assumption question. We're looking for an assumption that makes the argument completely valid.

Break down the Stimulus:
Bbirdwell did great job of explaining how to diagram this argument. It might also be helpful to examine the incorrect answer choices. Let’s review:

Premises:
etch tool —> pin OR bladed
bladed < some > for engraving
bladed < some > ~for engraving
pin —> for engraving

Conclusion:
etch tool for engraving > etch tool ~for engraving

There are a few important facts to note about this argument. We know that there is at least one of each type of tool mentioned in the premises: pin-tipped used for engraving, bladed used for engraving, and bladed not used for engraving. The conclusion is specifically comparing the number of etching tools that are used for engraving to the number that are not used for engraving.

Any prephrase?
For the conclusion to be true, we have to make an assumption about the number of each type:

(pin + bladed for engraving) > (bladed ~for engraving)

The correct answer choice could state this in several different ways.

If
pin > bladed ~for engraving
or
bladed for engraving > bladed ~for engraving

either one alone would be enough to guarantee the conclusion.

Correct answer:
Choice (B) is correct.

Answer choice analysis:
A) This would allow us to conclude that all engraving tools are either pin-tipped or bladed etching tools. It doesn't tell us anything about the number of etching tools that are used for engraving vs. the number that aren't.

B) This is the correct answer. If the total number of pin-tipped etching tools is equal to the total number of bladed etching tools, but at least one of those bladed etching tools is used for engraving, the number of number of pin-tipped tools must be greater than the number of bladed tools not used for engraving.

C) At best, this tells us that pin-tipped for engraving and bladed for engraving are completely separate tools. It doesn't tell us about the number of tools that are for engraving compared to the number that are not for engraving.

D) This might seem to help at first glance, but it doesn't tell us anything about pin-tipped tools, so it doesn't guarantee the truth of the conclusion.

E) The premises, taken together, already tell us this.

Takeaway/Pattern: The premises of this argument only describe different types of engraving tools, while the conclusion makes a specific comparison involving the number of each type. We need additional information about the number of each type in order to draw that conclusion.

#officialexplanation

[linzhu2]

Hi, my question on this etching tools question is whether or not it is possible to find the answer to this question by diagramming?

I can diagram the premises ok, and conclusion too (I think.):

ET -> PT or B
B some Eng
B some ~Eng
Conclusion: ET most Eng

Given that this is a sufficient assumption question, can the missing condition be represented as a conditional relationship? I was able to get the answer by thinking about it intuitively, but am very curious about whether or not there is an easier way to solve it that involves diagramming a missing conditional relationship. Thanks!

[yoohoo081]

Hi, my question on this etching tools question is whether or not it is possible to find the answer to this question by diagramming?

I can diagram the premises ok, and conclusion too (I think.):

ET -> PT or B
B some Eng
B some ~Eng
Conclusion: ET most Eng

I actually drew diagram as well to see what I needed to assume since I'm more visual, but I think we just needed to follow the logical behind this.

For example: let's just say i had 50 bladed and 10 pin-tipped. Out of the 50 bladed, 40 are not engraved and 10 is engraved. That means only 20 are engraved (all pin-tipped are engraved-10 and 10 of bladed are engraved which equals 20) and 40 are NOT engraved.

So, for the conclusion to follow more engrave than not engrave, there needs to be equal number of pin-tipped to bladed.

To show how it would work, let's assume that we have 10 for both the pin-tipped and bladed. So that means even if 9 bladed are NOT engraved, 1 bladed is engraved which leads to the given cconclusion that there are more engraved (10 pin-tipped which are all engraved plus 1 bladed engraved)

Just remember some is at least 1, so no matter how big the numbers vary between bladed engraved and not engraved, the minimum requirement for SOME engraved will mmake engraved bigger than not engraved only if we assume B is correct.

I apologize if this explanation is confusing but this is how I got to the right answer.

[sranksonly]

Hi, my question on this etching tools question is whether or not it is possible to find the answer to this question by diagramming?

I can diagram the premises ok, and conclusion too (I think.):

ET -> PT or B
B some Eng
B some ~Eng
Conclusion: ET most Eng

I actually drew diagram as well to see what I needed to assume since I'm more visual, but I think we just needed to follow the logical behind this.

For example: let's just say i had 50 bladed and 10 pin-tipped. Out of the 50 bladed, 40 are not engraved and 10 is engraved. That means only 20 are engraved (all pin-tipped are engraved-10 and 10 of bladed are engraved which equals 20) and 40 are NOT engraved.

So, for the conclusion to follow more engrave than not engrave, there needs to be equal number of pin-tipped to bladed.

To show how it would work, let's assume that we have 10 for both the pin-tipped and bladed. So that means even if 9 bladed are NOT engraved, 1 bladed is engraved which leads to the given cconclusion that there are more engraved (10 pin-tipped which are all engraved plus 1 bladed engraved)

Just remember some is at least 1, so no matter how big the numbers vary between bladed engraved and not engraved, the minimum requirement for SOME engraved will mmake engraved bigger than not engraved only if we assume B is correct.

I apologize if this explanation is confusing but this is how I got to the right answer.

Why can't some are not engraved be 10 and engraved are 40.
because some is aleast one could be anything above 1...

[yoohoo081]

sranksonly is correct, some are not engraved be 10 and engraved are 40, scenario is possible.

I just used the numbers that I did the show easily that it's a necessary assumption that there needs to be equal number for pin-tripped and bladed in order to come to the conclusion given in the paragraph.

[bbirdwell]

Hey guys, great discussion! Maybe I can help out; I'll try to answer the original question regarding how to "diagram a missing conditional."

I'd think about it this way:

Etching Tool --> Pin or Blade
Pin --> Engraving
Blade some Engraving
Blade some ~Engraving

Conclusion:
ET+Engraving > ET+~Engraving
In other words:
Pin + BladeEngraved > B~Engraved;
P + (BE) > (B~E)

Isolating the conclusion in this way is very helpful for me. it's difficult to diagram a missing conditional since the conclusion isn't really conditional -- it's just a statement about how many there are.

So, when going to the choices, I know I'm looking for something that, given the information above, GUARANTEES that i will arrive at the conclusion above.

Answer choice (B) does this by saying P=B. ALL P are E, and therefore it doesn't matter how many B has. As long as B = 1 or more (and we know it does), the total number of E will be greater than the total ~E.

Do you see? Let's say there are 100 PE. If there are 100 B, that means at worst: 1 BE and 99B~E. And the total E is greater. At best, 99BE and 1B~E, in which case the total E is even greater still.

Hope that helps!

[PepitoH243]

The explanation is harder than the exercise.

[JessicaX194]

Hey guys! Thanks for all the responses. My question is that if it’s possible that there are pin-tipped etching tools can be used for both engraving and not engraving. Can someone explain this for me?

Thank you!