1. ## SAT question

Hi,

In a supermarket, Shakira bought 5 items from aisles 1 through 7, inclusive, and 7 items from aisles 4 through 10, inclusive. Which of the following could be the total number of items that Shakira bought?

I. 9
II. 10
III. 11

A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

My thoughts: the number of items bought have to lie between 5 and 12 (5 + 7).
So a good guess would be E.

CollegeBoard's explanation is too winded. I don't think I'd come up with it in 60 seconds, let alone under 60 seconds.

2. My thoughts: the number of items bought have to lie between 5 and 12 (5 + 7).
So a good guess would be E.
After some thought, I agree that this may be the way to go under time pressure, except that the minimum number is 7, not 5.

A full solution is the following. Let us define the sets of isles: I1 = {1, 2, 3}, I2 = {4, 5, 6, 7}, I3 = {8, 9, 10}. Note that these sets are pairwise disjoint, I1 together with I2 is {1, ..., 7} and I2 together with I3 is {4, ..., 10}.

Variant I: 9 - 7 = 2 items bought in I1, 5 - 2 = 3 items in I2, 7 - 3 = 4 items in I3.
Variant II: 10 - 7 = 3 items in I1, 2 items in I2, 5 items in I3.
Variant III: 4 items in I1, 1 item in I2, 6 items in I3.

3. I can't imagine why the college board would have a long-winded solution for this. The quickest way to solve this is , as you've essentially done, to find the minimum and maximum values. A moment's thought should convince you that you can obtain every value in between.

Using Emakarov's notation, you get the minimum when 5 items are bought in I2 and 2 items are bought in I3 (total 7), and the maximum when 5 items are bought in I1 and 7 items are bought in I3 (total 12).

Originally Posted by Hellbent
I don't think I'd come up with it in 60 seconds, let alone under 60 seconds.
It sounds like you are giving yourself too much time pressure here. You may be under the misconception that you need to answer every math question on the SAT. This is only true if you are consistently breaking a 700 in Math on every practice test. If this is the case, then you would want to answer this particular question in under 60 seconds (but you can still give yourself more time for the harder ones).

Unless your current ability level is 630 or higher, then you will probably lower your SAT score by attempting all questions. Perhaps at some point I'll dedicate a post to this subject.

4. Firstly, if the choices included "none", I would have chosen it. Secondly, I don't understand the question. The word inclusive seems to be a distraction. The question would read just as well without it. And how can it be inclusive if the number of items bought is less than the number of isles?

Originally Posted by emakarov
After some thought, I agree that this may be the way to go under time pressure, except that the minimum number is 7, not 5.

A full solution is the following. Let us define the sets of isles: I1 = {1, 2, 3}, I2 = {4, 5, 6, 7}, I3 = {8, 9, 10}. Note that these sets are pairwise disjoint, I1 together with I2 is {1, ..., 7} and I2 together with I3 is {4, ..., 10}.
Can't I1 or I3 sets have any number of elements up to 4? What if I1 had numbers 1 - 4 and I2 and I3 had three elements?

Originally Posted by emakarov
Variant I: 9 - 7 = 2 items bought in I1, 5 - 2 = 3 items in I2, 7 - 3 = 4 items in I3.
Variant II: 10 - 7 = 3 items in I1, 2 items in I2, 5 items in I3.
Variant III: 4 items in I1, 1 item in I2, 6 items in I3.
What do variants I, II and III represent? What does the information after each colon represent?

5. What do variants I, II and III represent?
The question lists three possibilities for the total number of items purchased:
I. 9
II. 10
III. 11
One must say which of those three variants are indeed possible.

Firstly, if the choices included "none", I would have chosen it.
Saying that none of the variants I, II and III are possible is incorrect. In fact, all three variants are possible under the conditions described in the question.

The word inclusive seems to be a distraction. The question would read just as well without it.
You may be right. The important things about these two groups of aisles (1--7 and 4--10) are: there are aisles in the first group, but not in the second; there are aisles in the second group, but not in the first; there are aisles that belong to both groups.

And how can it be inclusive if the number of items bought is less than the number of isles?
Saying the the first group contains aisles 1 through 7, inclusive, means that the group has precisely aisles #1, 2, 3, 4, 5, 6, 7. Perhaps someone may think that 1 through 7 means 1, 2, 3, 4, 5, 6, or maybe 2, 3, 4, 5, 6. To repeat, the isle numbers are not important; only the three conditions given above matter. The word "inclusive" does not mean that at least one item was bought in each aisle.

So, Shakira could have bought 2 items in aisles 1--3, 3 items in aisles 4--7 and 4 items in aisles 8--10. This makes 5 items in aisles 1--7 and 7 items in aisles 4--10, for the total of 9 items. My first post also gives the breakdown of the number of items for variants II and III.

6. Originally Posted by Hellbent
Hi,

In a supermarket, Shakira bought 5 items from aisles 1 through 7, inclusive, and 7 items from aisles 4 through 10, inclusive. Which of the following could be the total number of items that Shakira bought?

I. 9
II. 10
III. 11

A. II only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

My thoughts: the number of items bought have to lie between 5 and 12 (5 + 7).
So a good guess would be E.

CollegeBoard's explanation is too winded. I don't think I'd come up with it in 60 seconds, let alone under 60 seconds.
If there is an aisle 7, then there must be aisles 1, 2, 3, 4, 5 and 6 preceding it.
Aisles 1 through 7 "inclusive" means "including" aisles 1 and 7,
essentially referring to all of the seven aisles.

Aisles 4 through 10 "inclusive" means aisles 4, 5, 6, 7, 8, 9 and 10.

The 5 items bought in aisles 1 to 7 could conceiveably have been bought in aisles 4 to 7.

Therefore the minimum number of items bought as emakarov showed is 7,
since 7 items were bought in aisles 4 to 10.

However, the 5 items could have been bought in aisles 1 to 3.
Hence the maximum number of items bought is 5+7=12.

$\displaystyle 7\ \le\ number\;of\;items\;bought\ \le\ 12$