Q13 - Standard aluminum soft-drink cans

[dan]

13. (C)
Question Type: Assumption
What is the author assuming in making the conclusion that M contains twice as many cans as L?

Well, suppose for the sake of argument that Group L contained only 1 aluminum can and that can had 10 units of aluminum, but that only 5 units were recovered in the recycling process (and the other 5 were lost). The passage tells us that all the recycled aluminum was used to create cans in Group M, so all 5 recovered units were used to create M. According to the passage, 5 represents half of M’s aluminum, so M must have 10 units in total. Remember, every standard aluminum can has the same amount of aluminum (according to the passage). Cans in L and cans in M are standard cans. If each can in L has 10 units of aluminum, then each can in M has 10 units of aluminum. Thus, if M has 10 units total, M is comprised of exactly one can! Can we conclude that L originally had half as many cans as M? No! In this case, they both have exactly 1 can.

In order for this argument to work, the author must assume that no aluminum is lost in the recycling process. If no aluminum is lost, then all 10 units from L are used to make M, and 10 represents half of its total. So, M would have 20 units total, which would make two cans of 10 each (twice as many as L). Answer (C) correctly expresses this assumption.

(A) is out of scope. This argument focuses on taking L’s cans and creating M’s cans. We don’t care what happens after.
(B) is out of scope. The quality of the aluminum is irrelevant
(D) is out of scope. This argument focuses on taking L’s cans and creating M’s cans. We don’t care what happens before.
(E) is out of scope. Can made from other materials are irrelevant.


#officialexplanation

[kiwistory]

Hi! I just have a quick question about this problem..

After reading your solution, I'm still bit confused.

The problem begins with "Standard aluminum soft-drink cans do not vary in the amount of aluminium that they contain",therefore, no matter how much recycled standard aluminum soft-drink cans were used, wouldn't can M always contain the same amount of aluminum as can L? (since they do not vary in the amt of al that they contain)?

Sorry if this sounded confusing, I'm just thrown off by the first premise. Thanks!

[dan]

Excellent point! This is a fact I overlooked in my original solution. I've edited the original -- take a look and let us know if it makes more sense. The idea is exactly the same (the argument assumes that all the aluminum is recovered in the recycling process), but now the numbers used in the example adhere to the constraints in the passage.

Thanks again for pointing this out.

dan

[kiwistory]

Makes much more sense now! Thank you so much for the quick reply. I have recently bought both of your prep books (LG/LR) and I am really surprised that not many people mentions this book online. It has been tremendously helpful in both of those subjects! (I also really love the "how to read like a debater" section, though I wish there were some more drills regarding outlining the prompts)

I hope you guys won't mind me dropping in every now and then with a new question, I really love this forum : ).

[dan]

Thanks for the kind words. We're glad you're finding the books and forum useful.

We'll look forward to seeing you in here!

dan

[jewels0602]

I'm surprised more people didn't have an issue with this one. I spent far too much time on this and still got the question wrong

I had it narrowed down to C and D... and just picked D after essentially giving up.

But your explanation was perfect- thank you!

There are so many moving components

[asafezrati]

This question feels very open.

Here's a possible option that can lead to a different conclusion, even if we assume C -
Group L contains one can, and it is being recycled to 2 M cans. But these two M cans are recycled in turn to two more M cans. This gives us group L=1, group M=4.
The cans in each group don't have to exist at the same time to be a part of the same group, so the quantity doesn't have to be connected to the material used.

Am I missing something, or should I join LSAC and show them how to do the job properly?

Thanks

[maryadkins]

But then the 4 M cans would only have 25% of L aluminum each. We are told they have 50%.

So say you have 10 L cans. These are recycled wholly into a group of M cans, such that in the new group of M cans, 50% is from L. That means we have 20 M cans, each with a former L can composing half of it (on average).

But say these were then recycled into a new group of M cans under your hypothetical, such that we had twice as many cans as before—40 M cans, not 20. Well then each can in the new group would contain only 25% of a former L can on average, because 10 L cans as part of 40 M cans amounts to 25%.

Since that's not the case in this question, the hypo doesn't contradict it. Hope this helps.

[asafezrati]

But then the 4 M cans would only have 25% of L aluminum each. We are told they have 50%.

So say you have 10 L cans. These are recycled wholly into a group of M cans, such that in the new group of M cans, 50% is from L. That means we have 20 M cans, each with a former L can composing half of it (on average).

But say these were then recycled into a new group of M cans under your hypothetical, such that we had twice as many cans as before—40 M cans, not 20. Well then each can in the new group would contain only 25% of a former L can on average, because 10 L cans as part of 40 M cans amounts to 25%.

Since that's not the case in this question, the hypo doesn't contradict it. Hope this helps.

I'll try to illustrate it:

On Monday I had 10 L cans.

Tuesday - I used them all up, recycled them completely into a chunk of aluminum, and added about the same amount of aluminum to create 20 M cans. Now I've got 20 M cans, each containing 50% aluminum originated in the L cans.

Wednesday - I've used all the M cans I had, and again recycled them completely into a chunk of aluminum, still containing 50% that originated in the L cans. I've made 20 new M cans, each still containing 50% of the aluminum originated in the L cans.

Now, in the history of my cans possession, there were 40 M cans in total, even though 20 of them perished while making the new ones (each and every one of them had 50% of L originated aluminum), and 10 L cans.

I guess my problem was that I went too far while the argument was meant to be simpler. I just couldn't help to think about the possibility of the M can group as an open group - still being manufactured from used M cans, thus not being limited in the total amount of M cans produced even if the material source was clearly defined.

[tommywallach]

Does it make sense now, or do you want a further followup from Mary?

-t

[hnadgauda]

I am confused by this question and the answer given above. Is there a different explanation for why the answer is C and not D or A?

[ohthatpatrick]

Here's a quick metaphor:

I have a plate of rice. Half it came from Benny, who dumped his entire plate of rice onto my plate. Since I have 100 grains of rice (half of them, 50, from Benny), it follows that I have twice as many grains of rice as Benny had.

POSSIBLE OBJECTION:
What if Benny is sloppy at pouring rice from his plate and some of the grains of rice fell off?

In that case, 50 grains of rice landed on my plate from Benny, but maybe another 20 grains bounced off the plate and hit the floor.

This ruins the author's math, because this means that Benny had 70 grains on his plate.

The author is assuming that number of grains of rice I got from Benny is THE SAME as the number of grains of rice that Benny had on his plate when he dumped it ... the author is assuming that "nothing was lost in transition".

If you are wondering why (A) and (D) are wrong, you might not fully understand what Sufficient Assumption's task is. You have to 'mathematically' derive the conclusion.

For (D), we already know that L is standard aluminum, and they already told us that standard doesn't vary in the amount of aluminum contained. So (D) isn't patching up any holes in our knowledge. We already know how to judge L's aluminum content.

(A) isn't giving us anything related to proving the quantitative comparison made in the conclusion: "M has twice as many cans as L".

(A) is about what could happen NEXT, to the aluminum in M.

Our job is only to examine the transfer of aluminum from L to M and see if we can prove that there were half as many cans in L as there are in M.