PrepTest 123, Section 2, Question 13
Standard aluminum soft-drink cans do not vary in(73 more words not shown)
The conclusion of the argument(10 more words not shown)
- AThe aluminum in(8 more words not shown)
- BRecycled aluminum is(6 more words not shown)
- CAll of the(12 more words not shown)
- DNone of the(11 more words not shown)
- EAluminum soft-drink cans(12 more words not shown)
Reading the argument
Let me try some numbers. Say the old group, L, contains ten cans, each with five grams of aluminum. That's fifty grams in L. Aluminum from those old cans supplies half of the aluminum in the new group, M.
My first thought is that M must have one hundred grams: fifty from L and fifty from somewhere else. Every can contains the same amount, so one hundred grams makes twenty cans. That is twice the ten we started with. The proposed doubling looks right.
What we’re looking for
Now the question asks what assumption would make that conclusion follow. So there is a step in that calculation that needs support. I don't want to abandon the math; I want to check what lets us use it. We need an answer that establishes the exact relationship between the amount in the old group and the amount in the new group. Facts about unrelated features of aluminum won't tell us how many cans there are.
Working the answers
- A
Whether M's aluminum can be recycled again later doesn't settle how many cans are in M now. We need to justify this calculation, not decide what can happen in a future recycling process.
- B
Quality isn't the amount. Aluminum could become poorer quality and still occupy cans containing the same five grams each. The passage has already told us the amount per can stays constant. A quality comparison doesn't tell us how much aluminum moved from L to M.
- CCorrect
There it is: recycling all the old cans doesn't necessarily recover all their aluminum.
What if the process recovers only 90 percent? Our ten old cans contain fifty grams, but only forty-five grams reach M. Those forty-five grams supply half of M's aluminum, so M has ninety grams. At five grams per can, that gives us eighteen cans—not twenty. Every old can went through recycling, yet the number of new cans isn't twice the number of old ones.
C rules out that loss. All fifty grams from L make it into M and account for half its aluminum, so M has one hundred grams. The constant amount per can then gives us twenty cans. The exact doubling works once we know all the aluminum is recovered.
- D
It doesn't matter whether L's aluminum had already been recycled. It could have been recycled before and still contain the same amount now. We need to know what happens as this aluminum goes from L into M, not where it came from before L.
- E
Easier to recycle than other materials? That doesn't establish the amount recovered. Aluminum might be easier to recycle than copper and still lose some material in the process. We need the quantity needed for an exact doubling, not a comparison of how easy different materials are to recycle.
More Sufficient Assumption questions
Learn the LSAT our way
Our platform drills Sufficient Assumption questions at your level and adjusts as you improve.