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maraudersmap
Registered User Posts: **14** New Member

Hey guys, I'm prepping for the Dec SAT and I think a daily math questions would help me improve. Can anyone help me with these questions? If you nee help too, feel free to post.

1. Kyle's lock combination consists of 3 two-digit numbers. The combination satisfies the three conditions below.

-One number is odd

-One number is a multiple of 5

-One number is the day of the month of Kyle's birthday

If each number satisfies exactly one of the conditions, which of the following could be the combination of the lock?

A. 14-20-13

B. 14-25-13

C. 15-18-16

D. 20-15-20

E. 34-30-21

I'm confused because A, B, and D all fit the conditions...

1. Kyle's lock combination consists of 3 two-digit numbers. The combination satisfies the three conditions below.

-One number is odd

-One number is a multiple of 5

-One number is the day of the month of Kyle's birthday

If each number satisfies exactly one of the conditions, which of the following could be the combination of the lock?

A. 14-20-13

B. 14-25-13

C. 15-18-16

D. 20-15-20

E. 34-30-21

I'm confused because A, B, and D all fit the conditions...

Post edited by maraudersmap on

This discussion has been closed.

## Replies to: Daily Math Questions

346MemberThe only thing you missed was the word "one." ONE number is odd. ONE number is a multiple of 5. And ONE number.. you get it.

A. 14-20-13

B. 14-25-13 - Eliminate because it has 2 odd numbers.

C. 15-18-16 - Eliminate because 15 can't satisfy two requirements.

D. 20-15-20 - Eliminate because it has two multiple of 5's.

E. 34-30-21 - Eliminate because 21 and 30 work, but 34 doesn't satisfy the birthday requirement.

A!

14New MemberHere's another one I need help with:

Let x be defined as [x]=x^2-x for all values of x. If [a]=a-2, what is the value of a?

A. 1

B. 1/2

C. 3/2

D. 6/5

E. 3

2,118Senior MemberQ: If p and q are positive integers such that pq = 50, which of the following cannot equal (p^2)q?

A: 50

B: 100

C: 200

D: 250

E: 500

2,118Senior Member[a] = a^2 - a = a - 2

a^2 - 2a + 2 = 0

a = (2 +- sqrt(4 - 8))/2, i.e. two complex solutions.

43Junior Member14New Member2,118Senior Member14New Member2,118Senior Membera^2 - a = (a-2)^2 - (a-2)

a^2 - a = a^2 - 5a + 6 (upon expanding, simplifying). a^2 cancels, leaving

-a = -5a + 6 --> a = 3/2

14New Member91Junior Member2,118Senior Member62Junior Member2,118Senior Member200 is 50*4, and 4 is not a factor of 50, so C is the answer.

3,449Senior Member