Hey dudes I see everyone is saying the mechanical/aviation, math sections on form 2 are basically identical to the marine study guide. So what about form 1 is it completely different or what? I'm just curious because I'll be taking form 1, I hope it is'nt harder than #2. Thanks for the input!
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ASTB Prep Question
- Thread starter Yukon
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Hey Ghost or anybody else that can answer my question: Can you give me some tips on how to solve #28,42,47 on the marine prep. I can't seem to find any clues in any of the books I've got. #28- this is the one about 1stLt and Captains and how many total company grade officers there are between them all. #42- this is the one on the grid with coordinates (1,2)(4,6) it asks how long line XY is. Lastly, #47 is the one that asks you to find the area of an equilateral triangle all sides measure 6. Hey thanks for any help you can give me I really appreciate it...Pete
the second question is just finding the hypotenus of a triangle. Find the length of the two sides. For the bottom side do 4-1=3. For the vertical side do 6-2=4. Then square each of them. 3^2=9 and 4^2=16. Then add the together. 16+9=25. then take the square root of that SQT 25 = 5. They give this question a lot where the sides are 3 and 4. Standard 3,4,5 triangle.
The other way is to draw an imaginary line down the center of the triangle (dividing it equally). Then take one part (a right triangle). The base length will be 3 (half of 6). The hypotenuse will be 6 (given) and the height is unknown. You can find it using the same formula for question #2 only this time you have to find the side length. So..... you have the equation 3^2 + x^2 = 6^2. Then you get 9 + x^2 = 36 then x^2 = 27 and finally x = 5.19. Then just use the standard formula for a right triangle, which is 1/2 base * height. So 1/2 * 3 * 5.19 = 7.794. Then double it since you originally had two of those triangle. So 7.794 + 7.794 = 15.588. This way is longer, but nothing to memorize.
on the other question, I don't remember the numbers but C=captain, L=1st lui and X=2nd lui they give you 3 equations where L+X=20 C+X=30 L+C=10 There are multiple ways to work this, since they want the complete total of all the easiest way is to add up all the C's, L's, and X's which is 2L+2C+2X=60 (20+30+10 = 60) then you can take the 2 out of each variable. 2(L+C+X)=60
L+C+X=60/2=30
L+C+X=60/2=30
Which version of the ASTB is harder? Is version 1 harder than 2, and which one most resembles the notorious Marine ASTB study prep? It's been 7 months since I've taken it, so I think they can give me the same version if they so desire. I think I took version 2.
While I'm on the ASTB....on the Marine ASTB study prep, how in the hell do you work #47 on the math section? I'm stumped. Any help would be appreciated.
While I'm on the ASTB....on the Marine ASTB study prep, how in the hell do you work #47 on the math section? I'm stumped. Any help would be appreciated.
kenjamin17
Registered User
To add to what tali said in the 2nd post about #47 I don't think you get a calculator so finding x in x^2=27 would be tough. When you get x = SQRT(27) it helps to notice that 27 is divisible by 9 (3*9=27). So you can break it down into SQRT(9)*SQRT(3), which equals 3*SQRT(3).
Actually the easiest way to do #47 is to use this simple formula for an equalateral triangle which is side squared times the square root of 3 divided by 4. By using this formula you will get 36 times the square root of 3 divided by 4 and that breaks down to 9 times the square root of 3 which is the answer. That exact question is on the 2nd version of the ASTB.
kenjamin17
Registered User
poor Tali, Kent is trying to steal her formula, hehe.
] I guess I should have kept my posts shorter [
] Maybe no one wanted to read the whole thing. hehe- Status