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1,001 questions about the ASTB (post your scores & ask your questions here!)

GreenLantern330

Active Member
Well, IMHO that is a bit high. I believe somewhere around a 50 is closer to the average. Remember you are seeing a very small sampling on here and over at usnavyocs.com. Also, many people do not post their scores if they are noticing many people getting much higher.

You're absolutely right. And I meant to say average selected scores (at least for those selected that are members of AW). Thanks for the correction.
 

fattestfoot

In it for the naked volleyball
Not sure about the Marines, but for Navy last year the average SNA selectee was 60 7/6/7 (or something close to that; positive about the OAR part though). NFO was slightly lower, and the average OCS selectee was around 48-50 I believe.
 

GreenLantern330

Active Member
I was just going over the Marine Gouge for the math section and had some questions as to how to solve some problems. If anyone could please explain how to solve these problems, I'd greatly appreciate it!

9) Perimeter of a triangle is 180 feet the width is 1/2 the length, what is the length?
Correct answer was 60. No clue how to get that.

16) The average temperature for a 4 day period was 73. If the temperature for the first three days was 75, 74, and 70, what was the temperature of the 4th day?
Correct answer was 73. When I solved, I got 72.

24) A square tarp is X feet on a side. If 4 feet of another tarp is sewn across one end so that the tarp will cover a fighting position, then the area of the tarp after the addition is made, in square feet, must be
Correct answer was x^2 + 4x. When I tried to solve, I got X^2 + 4.

25) It will take 4 hours, 32 minutes, 30 seconds to dig one part of a mortar pit. Another part will take 3 hours time, how long will the second part take?
Correct answer was 3 hours, 24 mins, 22.5 secs. I'm not even sure how to go about this one.
 

Lucy

Member
Just to be a little more specific for the averages. 60 for SNA, 56-58 for NFO and around a 52 for selects. 48-50 is the average for test takers, not selects. Breaking that 50 is important.

16) Average of four days is 73. 73 times 4 days is 292. 292-75=217. 217-74=143. 143-70=73. Thus 73 is the fourth day.

Not good enough to explain the others.
 

CUPike11

Still avoiding work as much as possible....
None
Contributor
Here's #9:

Perimeter of a Triangle is 180.
Width is half the length.
What is the length?

The perimeter of a triangle is found by the summation of all the sides.
So P = L1 + L2 + L3 or P = 3*L

*For sake of argument we're assuming the the sides are the same distance. The length is a solid line drawn from the top vertex to the midpoint of the base (width).

So we know from the problem that W = 1/2 * L or as I simplify, W = L/2.

Looking at our original equation of P = 3*L, we need to substitute for L. So rearrange the width eqn we just made. Multiply both sides by 2 and you go from
W = L/2 to L = 2W.

Substitute in the length eqn now into the Perimeter equation and you end up with:

180 = 3 * (2W) ----> 180 = 6W.

Solve for W and you get Width (W) = 30.

Plug that into your length eqn: L = 2W ----> L = 2*30

L = 60.

I'm working on the other ones too, but hopefully that made sense.
 

CUPike11

Still avoiding work as much as possible....
None
Contributor
For your last problem, its going to require you to do some long division. There's no way around, so hope your division is up to par.

First make sure copied the problem right. I believe the problem says that the second pit is dug in three quarters the time of the first one, NOT 3 hours. If i'm mistaken, then disregard this entire post.

First translate 4 hours 32 minutes and 30 seconds (04:32:30) into minutes.

You should get 272.5 minutes. If you're unsure how to get this, PM me and I'll further explain it.

Then multiply 3/4 or .75 by 272.5.

272.5 divided by 4 will get you 68.125 (yes that exact)

Then multiply that by 3, and you get 204.375.

First 204 minutes doesn't round to a complete hour, so you get the closest, which is 3 hours and then the rest is the minutes.

So you should have 3 hours and 24 minutes so far.

There's only one answer with the 3 hours and 24 minutes so I'd just go with that one, but if you need further convincing.

.375 * 60 will give you 22.5 seconds.

Hence, your answer is 3 hours 24 minutes and 22.5 seconds to dig the second mortar pit.
 

VNE

Member
9) Perimeter of a triangle is 180 feet the width is 1/2 the length, what is the length?
Correct answer was 60. No clue how to get that.



Checked my Marine Gouge and discovered the problem. Here is the question it reads.....

"Perimeter of a rectangle is 180 feet the width is 1/2 of the length, what is the length?"
 

GreenLantern330

Active Member
For your last problem, its going to require you to do some long division. There's no way around, so hope your division is up to par.

First make sure copied the problem right. I believe the problem says that the second pit is dug in three quarters the time of the first one, NOT 3 hours. If i'm mistaken, then disregard this entire post.

First translate 4 hours 32 minutes and 30 seconds (04:32:30) into minutes.

You should get 272.5 minutes. If you're unsure how to get this, PM me and I'll further explain it.

Then multiply 3/4 or .75 by 272.5.

272.5 divided by 4 will get you 68.125 (yes that exact)

Then multiply that by 3, and you get 204.375.

First 204 minutes doesn't round to a complete hour, so you get the closest, which is 3 hours and then the rest is the minutes.

So you should have 3 hours and 24 minutes so far.

There's only one answer with the 3 hours and 24 minutes so I'd just go with that one, but if you need further convincing.

.375 * 60 will give you 22.5 seconds.

Hence, your answer is 3 hours 24 minutes and 22.5 seconds to dig the second mortar pit.

Thank you so much for the help! I understand it now.
 

GreenLantern330

Active Member
9) Perimeter of a triangle is 180 feet the width is 1/2 the length, what is the length?
Correct answer was 60. No clue how to get that.



Checked my Marine Gouge and discovered the problem. Here is the question it reads.....

"Perimeter of a rectangle is 180 feet the width is 1/2 of the length, what is the length?"

You're absolutely right. Thanks for pointing that out.. obviously my reading comprehension isn't up to par haha
 

USSRyan

New Member
I am going to take the ASTB one Tuesday and still studying as much as I can. I have this question bothering me and I like to get it out of the way, so here it is...

Am I allowed to use a calculator during the Math section and if so do I have to bring my own?
Anything else I need to bring to the testing appointment?

thanks!
 

marathonman

New Member
Alright, I have a question for someone that might know...

I'm taking the ASTB for the 3rd and final time. My recruiter told me to retake after the first one when I got a 56 on the OAR, then it went down to 47. Right now I plan on applying for SWO and Intel. Instead of stretching myself over ASTB material, is it really necessary to take the portions that aren't involved in the OAR? Will it hurt my application if I don't take the last 3 portions of the test even though I'm not applying for anything related to aviation? Any recommendations would be great.
 

Lucy

Member
The whole ASTB is required to apply for Intel.

Section-
040202. AVIATION SELECTION TEST BATTERY​
a. Testing Overview. The Aviation Selection Test Battery
(ASTB) was developed by statistical methods to predict the
success of students in aviation officer training programs....
(2) The ASTB should be given in its entirety to all
applicants who indicate an interest in aviation and intelligence
programs....
(3) The Officer Aptitude Rating (OAR) portion of the
test, the MST, RCT and MCT sections, may be given to individuals
who are not interested/not qualified for aviation and
intelligence programs. This score is applicable only to various
non-aviation designators as specified in the Program
Authorizations, however, Intelligence Program applicants are
required to take the entire test.

In Chapter 4 of the recruiting manual.
http://www.cnrc.navy.mil/publications/Directives/1131%202D_CHAPTER%204_CH3.pdf


 
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