Hi- I am new to the forum and have been frantically trying to find help with the GK Math portion of the FTCE. Ive taken it 2 written and 3 computer!! I see many in the past have had the same experience. Can anyone tell me what they finally did to pass, Did you finally pass...what did you do to prepare. Please "Pass" on any usefull info. I know the Math portion is 45 questions - 200 points - what is the lowest passing score?? Desperately seeking help! My Temporary Cert. expires in June 2007 Currently employed at Centennial Middle in Miami, Florida. I teach Language Arts and Reading :thanks:

Math is about steps and formulas. You need to find a study book that lists what concepts will be on the test. Perhaps you can get text books that students use to bring it to a tutorial level. Can you seek out such books from your staff? I dont know if you must pay each time for taking the test, but it sounds like your not investing in the right study tools. Does Kaplan publish a prep book?

Don't know your area of weakness but the is a book series called Painless Algebra, Painless Fractions, etc.. These books teach you step-by-step with easy to understand directions. You may want to look them over in a bookstore to see if they help. I have a friend that feels they are what helped her to finally understand Math.

Is there an FTCE website? Usually, you can get sample test questions and answers (from real past tests) which I personally found of most value when I was taking CBEST, MSAT, and RICA tests here in CA. Truly, the source is the best place to get released test questions and know the format of the tests, IMO. Good luck!

Florida's the odd one out, eduk8r: Florida rolls its own when it comes to teacher tests, and there isn't a nice centralized Web site as there is with Praxis or with the various NES tests. KMCTeach, do your score reports give any indication of the areas of math in which you're having difficulty? Or do you have some sense of that?

Thanks - I'll try. Conversion is another I have to work on....mostly in word problems. That is really my area of need.

I wonder if "At $1.50 per gallon it doesn't matter which car Joe drives, he will still save thousands of dollars." would be an acceptable answer?

Get a Freshman Algebra book. (I like "Algebra I by Keenan and DRessler, published by Amsco. It does a nice job of separating the word problems by type and presenting an orderly logical approach to each type.)

First things first: Test makers generally discourage the public discussion of specific actual test questions - and some of them do this rather forcefully, with consequences you'd prefer to avoid - so I'm not going to explain those questions, and I'd take it as a personal favor if you'd edit that post. Now, then: let's do perimeter. Find two paperback books. Set one of them (we'll call it book A) in front of you. Seen from above, it's a rectangle, right? Trace your finger around the outside edge: up, across, down, across, ending where you started. How far did your finger travel? Up + across + down + across. That's the perimeter of book A. Measure the length of the up edge and the across edge and the down edge and the other across edge and write those numbers down. You can measure in inches or in centimeters. Add them together, and the number that results is the perimeter. (Make sure you label the unit of measurement.) And you should find that up = down and across-left = across-right, which turns out to be true of all rectangles and a useful shortcut. Now set book B next to book A so that they meet along one edge but so that together they do NOT make a simple rectangle. You can make an L shape or a zig-zag shape or whatever. Once again, perimeter (Greek: peri 'around', meter 'measure') is all the way around the outside OF THE WHOLE SHEBANG. You use all of an edge that isn't touching another edge, but if an edge is touching another edge, you don't count that part. Does this help?

Yes, it does. Thanks for the tip. Just trying to get help...but don't want to mess up anything for anyone. Thanks again

Now the other problems you mentioned are both rate problems. A rate is a ratio, and a ratio is nothing but a fraction that shows a relationship between two numbers. If someone hands you six Hershey's kisses and you eat them all in 30 seconds, then we can make a fraction that expresses your Hershey's-kiss-eating rate: 6 kisses 30 sec which, if need be, we can restate (simply by reducing) as 1 kiss 5 sec or, if it's handier, as 12 kisses 1 minute (Notice, by the way, that once the rate is stated as a fraction, it can be dealt with using all the tools that apply to bog-standard fractions: rates can be multiplied and added and so on.) Now if you had an unlimited supply of Hershey's kisses but only 60 seconds to eat them in, how many Hershey's kisses could you eat, if you can eat 6 kisses in 30 seconds? You can solve this by (a) multiplying rate times time - and remember that multiplying a fraction by a whole number means multiplying the numerator times the whole number and, if need be, dividing by the denominator. 6 kisses x 60 sec = 12 kisses 30 sec (b) setting up a proportion, which is nothing but two fractions yoked together with an equals sign, and solving for the missing term 6 kisses = k kisses . 30 sec . . 60 sec 6 * 60 = k * 30 6 * 2 = k 12 = k (c) observing that 60 seconds is twice as long as 30 second and that therefore you should be able to eat twice as many Hershey's kisses. (Leaving aside, of course, practical considerations such as your diet and the kind of day you had: this is math, dear, not the real world.) Now suppose that you and Upsadaisy and Alice and I are sitting around talking math and we've got a big bowl of Hershey's kisses. We already know you can eat 6 kisses in 30 seconds. Suppose that in 30 seconds 'daisy can eat 4 kisses, Alice can eat 5, and I don't eat any because I'm working my way through a Trader Joe's Pound Plus Bar. Then how many kisses do we eat all together? 6 kisses + 4 kisses + 5 kisses + 0 kisses = 15 kisses . 30 sec. . . 30 sec. . . 30 sec. . . 30 sec. . . 30 sec. The denominators are all the same, so just add the numerators. If the denominators were different, you'd need to do the usual sorts of common-denominator things before you could add. In any case, you end up with a combined rate, which can then be plugged into the proportion above or into the standard rate formula d = r * t to get a result.

It's partly a matter of choosing the right manipulatives, I suppose... And thank you: this is a tack I hadn't quite taken before. Can you tell me how long it would take the four of us to eat 60 Hershey's kisses? And suppose that each Hershey's kiss is wrapped in a 3" x 3" square of foil. Can you tell me how many square inches of foil it would take to wrap 12 kisses?

okay...I think it will take use 120 seconds (2 minutes) to eat the kisses....now come on!!! How do I figure out the "foil" problem...I think it will take about 72 inches confused:

72 inches? First, how did you get the number 72? Second, plain old inches are like a string or a tape measure. Could I trouble you to tweak the label just a bit?

um 72 square inches I'm practicing algebra and the brackets throw me off...I believe you do them first like parenthesis hint "PEMDAS". The answer to the problem is given, but I don't see how they got it?? 4 + 2 (5x-4) = 6 [2-(x-)] answer 11/8? can you help please?

You didn't tell me how you got the 72, though. As to brackets, it's actually simpler than that: brackets work exactly, and I mean EXACTLY, like parentheses, so 6 [2-(x-1)] = 6(2-[x-1]) = 6- (2-(x-1)) (Brackets are used either inside or outside parentheses to try to make it easier to see which left-hand one goes with which right-hand one, that's all. Angled brackets <, > and braces {, } can also be used.) One does the innermost set of parentheses (or brackets, or angled brackets, or braces) first and then works outward. I'm a little hampered in doing your problem, however: you typed x-, but you didn't tell us minus what.

oops I have to go back to the prctice prob...I found it on the internet...algebra site....not to clear on instructions thks

Hello Cali! Okay I can't find my notes on the problem...lets say it is 1 I just want to make sure I'm completing it correctly...do I start with what's inside the brackets? 4 +2 (5x-4) = 6 [2-(x-1)] 4+2(1x) ? Also I have this practice ratio question that me and my study partner completed...could you check it for us? If Joe can type a 20 page paper in 20 minutes and Carol in 30 minutes and Sally in 45 minutes, how long will it take them to type a 20 page paper together? 20/20 + 30/20 + 45/20 = 70/20 = 3.50 hrs? Right Thanks a bunch!!

Whoa, there. Yes, you do what's in the brackets - IF YOU CAN. But your first set of brackets has (5x-4), and since you typed it that way twice, I assume that's what the book said. 5x has an x but 4 doesn't, so you can't do anything just within the brackets. Hold that thought for a second... Your second set of brackets is 6[2-(x-1)]. Here, you can do some stuff: 2-(x-1) = 2-x+1. (It will be recalled that 2-(x-1) = 2 + (-(x-1)), and (-(x-1)) = -x+1. That is, that negative distributes itself over the terms that follow.) And 2-x+1 = 1-x. Then we've got 6-(1-x), which is 6-1+x, or 5-x. So now we've got this: 4 + 2(5x-4) = 5-x Now back to side 1, for which, since we can't resolve what's in the parentheses in the parentheses, we need the distributive property: 4 + 2(5x-4) = 5-x 4 + 10x - 8 = 5-x -4 + 10x = 5-x 9x = 9 x = 1 From which we can infer that, if the book's solution of 11/8 was correct, the missing number wasn't 1. As to the rate problem, you need to label your units. The text tells you rates in pages per minute: that is, pages are on top, but ou have them on the bottom. And, really: if Sally, who takes the longest, can type the paper in 45 minutes, does it make sense that it would take the three of them longer than that?? So we have 20 pp. + 20 pp. + 20 pp. 20 min. . 30 min. . 45 min. which means it's time for a truly ugly common denominator. And I'll let you take it from there.

Lets see:You want "time" how long with it take them to type a 20 page paper together? 20/95 (divide both by 5) 4/19=4.75 minutes?? estimate

20/95? Nope, nope, nope: you can't add fractions by adding unlike denominators. You need a common denominator - the brute-force version would be 20(30)(45), but an elegant way to find the lowest common denominator is to factor the numbers 20 = 2 * 2 * 5 30 = 2 * 3 * 5 45 = 3 * 3 * 5 and then multiply one each of the factors shared by all three by each factor that isn't shared: 5 * 3 * 3 * 2 * 2 - 180