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Archite
Registered User Posts: **118** Junior Member

Here is a picture of a math problem from a practice test I just took: [url]http://s1244.****/albums/gg580/agbond2013/?action=view¤t=fbae955f.jpg&evt=user_media_share[/url]

I've never seen the arch symbol before... What does it mean? The book explains the answer, but the answer is no help when I don't understand the context of the problem.

I've never seen the arch symbol before... What does it mean? The book explains the answer, but the answer is no help when I don't understand the context of the problem.

Post edited by Archite on

This discussion has been closed.

## Replies to: Math Q

369MemberA ∪ B means the set of those elements which are either in A, or in B, or in both

A ∩ B means the set that contains all those elements that A and B have in common

List of mathematical symbols - Wikipedia, the free encyclopedia

932MemberI think you should have learned some basics in middle school.

2,529Super Moderator@Archite -

You'll never see these symbols on the SAT.

So the question is - as reiterated on CC gazillion times - why bother with the non-CB questions?

932Member2,529Super ModeratorIt has been pointed out so many times on the CC that in your SAT practice you should stick with the CB authentic resources (there are plenty of them for the SAT I; SAT Subject tests is a different matter).

SAT questions from other than the CB sources are often ambiguous, outside of the SAT scope, overly difficult, and so on - which means, among other things, that creators of those questions are either not aware of exactly what is covered and what is not on the SAT or don't give a hoot about it.

1,058Senior Member261Junior Member109Junior MemberNo. E (3/14)

Method 1 (works if you don't know that much about probability)

A = {1, 2, 4, 8}

B = {1, 3, 6, 8, 9}

C = A U B = {1, 2, 3, 4, 6, 8, 9}

D = A intersection B = {1, 8}

So here are the possible combinations, where the first number c is from C and the second number d is from D

1&1, 1&8, 2&1, 2&8, 3&1, 3&8, 4&1, 4&8, 6&1, 6&8, 8&1, 8&8, 9&1, 9&8 (14 combinations)

Of the above combinations, only 1&1, 3&1, and 9&1 yield an odd product. (3 combinations)

So the probability of choosing c and d so that cd is odd is 3/14.

Method 2 (how I prefer to solve the problem)

For the product of two integers to be odd, both numbers must be odd.

The probability of picking an odd number for C is 3/7

The probability of picking an odd number for D is 1/2

The probability of picking an odd number for C as well as for D is 3/7 * 1/2 = 3/14