testing header
Math Goodies is a free math help portal for students, teachers, and parents.
Free Math
Newsletter
 
 
Interactive Math Goodies Software

Buy Math Goodies Software
testing left nav
Math Forums @ Math Goodies
Math Forums @ Math Goodies
Home | Profile | Register | Active Topics | Members | Search | FAQ
 All Forums
 Homework Help Forums
 Standardized Test Preparation Help
 odd, even expressions

Note: You must be registered in order to post a reply.
To register, click here. Registration is FREE!

Screensize:
UserName:
Password:
Format Mode:
Format: BoldItalicizedUnderlineStrikethrough Align LeftCenteredAlign Right Horizontal Rule Insert HyperlinkInsert EmailInsert Image Insert CodeInsert QuoteInsert List Insert Special Characters Insert Smilie
   
Message:
* HTML is OFF
* Forum Code is ON

Math Symbols
Squared [squared] Cubed [cubed] Square Root [sqrt] Cube Root [cbrt] Pi [pi]
Alpha [alpha] Beta [beta] Gamma [gamma] Theta [theta] Angle [angle]
Degrees [degrees] Times [times] Divide [divide] Less Than or Equal To [less-than] Greater Than or Equal To [greater-than]
Plus Minus [plus-minus] Integral [integral] Sum [sum] Sub 1 [sub-1] Sub 2 [sub-2]
Element Of [element-of] Union [union] Intersect [intersect] Subset [subset] Empty Set [empty-set]

  Check here to subscribe to this topic.
 
   

T O P I C    R E V I E W
mingshum Posted - 02/24/2009 : 11:39:08
What is the fastest way to solve this SAT prep problem?

If x, y, and z are positive integers such that the value of x+y is even and the value of (x + y) + x + z is odd, which of the following must be true?

A) x is odd
B) x is even
C) If z is even, then x is odd.
D) If z is even, then xy is even.
E) xy is even.

I eliminated choices A,B and E.

Answer is C.
2   L A T E S T    R E P L I E S    (Newest First)
mingshum Posted - 02/24/2009 : 13:42:08
very clear explanation...thanks
Haven Posted - 02/24/2009 : 12:56:58
On way of looking at this problem is to substitute letters for number values "O" for odd and "E" for even.

The problem tells you that the value of x + y = a even number. So that means that x and y must both be odd or x and y must both be even.

The problem also tells you that (x + y) is squared. Since (x + y) is an even number it can be reasoned that (x + y) will be an even number also.

Now with this information you can write an equation substituting possible number values with letters

(x + y)+ x + z = odd

even + odd + z = odd In this case z must be even when x is odd

or

even + even + z = odd In this case z must be odd when x is even

I hope that this explanation was of some help to you.

Math Forums @ Math Goodies © 2000-2004 Snitz Communications Go To Top Of Page
This page was generated in 0.04 seconds. Snitz Forums 2000
testing footer
About Us | Contact Us | Advertise with Us | Facebook | Blog | Recommend This Page




Copyright © 1998-2014 Mrs. Glosser's Math Goodies. All Rights Reserved.

A Hotchalk/Glam Partner Site - Last Modified 22 Oct 2014