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SupersonicMax said:
I'm just happy I don't have to wear a beret and blouse my pants anymore. Everything else is just gravy!
Aircrew Selection/ACS (Merged)
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- Start date
I'm just happy I don't have to wear a beret and blouse my pants anymore. Everything else is just gravy!
Eye In The Sky
Army.ca Legend
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Pre-flight said:
This was one of the best things I left behind when I got to start going to work in the flyin' jammies. No boot bands, and some jammy TD locations and hotels...feck ya.
I'm practicing for the Aircrew Selections, mainly the Distance, Speed, and Time calculations. The main issue I am having is converting the minutes to hours mentally, as they end up in decimals usually. How can I quickly divide the minutes by 60, then use them in the DST formula quickly?
Also, did anyone here use the mental math tricks from the tecmath YouTube channel, or just use the standard age-old methods?
Also, did anyone here use the mental math tricks from the tecmath YouTube channel, or just use the standard age-old methods?
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AliTheAce said:
I’m no expert, but hopefully these ideas help a little.
1. I memorized these numbers below (chart)
2. I rarely divided by 60 minutes, but instead I tried to think in fractions.
3. For the examples I used 'easy' numbers to keep this simple. if you get a decimal I would round and estimate.
4. hopefully my math is correct and I haven't embarrassed myself
5. Guess and check is important for deciding what fraction to use
Minute(s) Occurrences per hour
1 60
2 30
3 20
4 15
5 12
6 10
10 6
12 5
15 4
30 2
60 1
For the fraction system to work you are looking for a common denominator.
Simple example
Say the AC is travelling at 200mph for 90 min. How far did it go?
90 min = 1 ½ hours. 1 hour = 200 miles. 30 min = 100 miles. 200 + 100 = 300 miles in 90 min.
example 2:
AC travels at 156mph for 75 minutes. How far did it travel?
75 min = 15 minutes x 5
15 minutes happens 4 times per hour, so
156/4= 39miles
This means that every 15 minutes the AC travels 39 miles.
15 min happens 5 times in 75min. So 39 x 5 = distance.
To make this easier, do 40 x 5 = 200, but subtract 5 (because 40 – 39 = 1 x 5 = 5)
and you get 195 miles.
example 3:
AC travels 250miles in 40 min. What is its speed?
Divide each by 10.
250 miles/10 = 25 miles. 40 min / 10 = 4 min
Therefore every 4 minutes the AC travels 25 miles.
4 minutes occurs 15 per hour, so multiple 25 x 15.
For mental math, instead of 25x15, try (25x10 = 250) + (25x5 = 125) = 375mph
** I didn't think of this. I stole the idea from a MARS officer
***** Edited several times because I'm bad at proof reading.
winnipegoo7 said:I’m no expert, but hopefully these ideas help a little.
1. I memorized these numbers below (chart)
2. I rarely divided by 60 minutes, but instead I tried to think in fractions.
3. For the examples I used 'easy' numbers to keep this simple. if you get a decimal I would round and estimate.
4. hopefully my math is correct and I haven't embarrassed myself
5. Guess and check is important for deciding what fraction to use
Minute(s) Occurrences per hour
1 60
2 30
3 20
4 15
5 12
6 10
10 6
12 5
15 4
30 2
60 1
For the fraction system to work you are looking for a common denominator.
Simple example
Say the AC is travelling at 200mph for 90 min. How far did it go?
90 min = 1 ½ hours. 1 hour = 200 miles. 30 min = 100 miles. 200 + 100 = 300 miles in 90 min.
example 2:
AC travels at 156mph for 75 minutes. How far did it travel?
75 min = 15 minutes x 5
15 minutes happens 4 times per hour, so
156/4= 39miles
This means that every 15 minutes the AC travels 39 miles.
15 min happens 5 times in 75min. So 39 x 5 = distance.
To make this easier, do 40 x 5 = 200, but subtract 5 (because 40 – 39 = 1 x 5 = 5)
and you get 195 miles.
example 3:
AC travels 250miles in 40 min. What is its speed?
Divide each by 10.
250 miles/10 = 25 miles. 40 min / 10 = 4 min
Therefore every 4 minutes the AC travels 25 miles.
4 minutes occurs 15 per hour, so multiple 25 x 15.
For mental math, instead of 25x15, try (25x10 = 250) + (25x5 = 125) = 375mph
** I didn't think of this. I stole the idea from a MARS officer
***** Edited several times because I'm bad at proof reading.
Thanks a lot, I'll try this!
Now what if the numbers aren't easily divisible by 5?
Eg: What speed covers 255 miles in 36 mins?
or
at 135 mph, how far do you travel in 16 mins?
or
What speed covers 14 miles in 42 mins?
or
At 20 mph, how far do you travel in 21 mins?
Note: These are taken straight off the speed distance time website
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I’m assuming you want to do this in your head? If you have scrap paper it’s much easier.
I hope I got these right.
1. 255 mile in 36 min
Guess and check. I tried dividing both by 6. Was too hard. Then I tried by 3.
255/3 = 85miles. 36/3 = 12 min
So 85 miles every 12 min. Multiply both by 5.
85x5= (80x5 = 400) + (5x5=25) = 425 mph.
2. This one is a bit tough
135miles in 16 min.
I would pretend it’s 136miles in 16. Then divide each by 4.
136/2= 68. 68 /2 = 34 miles
16/4=4 min (happens 15 times per hour) so do 34x15
(34x10=340) + (34x5 = 170*) = 510mph, but we know this number is high because we used 136 miles instead of 135. So we must subtract. 1 mile/4= 0.25. 0.25 x 15 = 3.75( or do 15/4. 15/2=7.5/2=3.75)
So 510-3.75= 506.25mph
(* 34x5 = 170 which is half of 340. Dividing 340/2 is easier mentally than 34x5). ... also I pretended that 340 was 34 because I find 34/2 easier than 340/2. But maybe I’m weird.
3. 14 miles in 42 min
Divide each by 7
2 miles every 6 min
Multiply by 10 because (6 x 10 =60 min)
2 x 10 = 20 mph
4.at 20mph for 21 min
Think this way 60mph = 1mile per min
So 20mph = 1/3 mile per min. So divide 21 min by 3 = 7 miles.
I recommend that you make up problems and then try to find the easiest way to solve that problem through trial and error. Do that everyday for 30 min and you’ll find it easy after a couple weeks
I hope I got these right.
1. 255 mile in 36 min
Guess and check. I tried dividing both by 6. Was too hard. Then I tried by 3.
255/3 = 85miles. 36/3 = 12 min
So 85 miles every 12 min. Multiply both by 5.
85x5= (80x5 = 400) + (5x5=25) = 425 mph.
2. This one is a bit tough
135miles in 16 min.
I would pretend it’s 136miles in 16. Then divide each by 4.
136/2= 68. 68 /2 = 34 miles
16/4=4 min (happens 15 times per hour) so do 34x15
(34x10=340) + (34x5 = 170*) = 510mph, but we know this number is high because we used 136 miles instead of 135. So we must subtract. 1 mile/4= 0.25. 0.25 x 15 = 3.75( or do 15/4. 15/2=7.5/2=3.75)
So 510-3.75= 506.25mph
(* 34x5 = 170 which is half of 340. Dividing 340/2 is easier mentally than 34x5). ... also I pretended that 340 was 34 because I find 34/2 easier than 340/2. But maybe I’m weird.
3. 14 miles in 42 min
Divide each by 7
2 miles every 6 min
Multiply by 10 because (6 x 10 =60 min)
2 x 10 = 20 mph
4.at 20mph for 21 min
Think this way 60mph = 1mile per min
So 20mph = 1/3 mile per min. So divide 21 min by 3 = 7 miles.
I recommend that you make up problems and then try to find the easiest way to solve that problem through trial and error. Do that everyday for 30 min and you’ll find it easy after a couple weeks
You have to get comfortable with using numbers that dont produce a "clean" number to work with. This will require rounding, usually to the second decimal place. Practice your mental arithmetic. There are many free sites available. The problem sets featured on the TSD website for RAF officer applicants are designed to work with relatively easy number if you can simplify the fraction. IE- You travel 138 miles in 36 minutes, what is your speed? 138 ( 60/ 36)= 138 ( 5/3) = (138/3) (5) = 46 (5) = 230 mph.
Get use to working with numbers ( in your head- no pencil and paper at ACS for the TSD section) such as 139 miles travelled in 27 min. What is speed in mph? So 139 ( 60/27) = 139 ( 20 / 9 ) = 139 ( 2.22) = 278 + 27.8 + 2.78 = 308.58 ~ 309 mph.
Round your answer to the nearest whole number.
At ACS you are going to be nervous and tired ( It is a mentally draining test) . Instead of looking for numbers and rearranging said numbers that work "nicely" get comfortable with the route execution of the calculation.
Get use to working with numbers ( in your head- no pencil and paper at ACS for the TSD section) such as 139 miles travelled in 27 min. What is speed in mph? So 139 ( 60/27) = 139 ( 20 / 9 ) = 139 ( 2.22) = 278 + 27.8 + 2.78 = 308.58 ~ 309 mph.
Round your answer to the nearest whole number.
At ACS you are going to be nervous and tired ( It is a mentally draining test) . Instead of looking for numbers and rearranging said numbers that work "nicely" get comfortable with the route execution of the calculation.
Absolutely. Practice mental arithmetic, TSD calculations and I would also recommend the SAT/ ACT portion that involves TSD. Practicing general SAT type math questions ( TSD, algebra type problem sets), in fact, are tricky and force you to really understand what is being asked. This translates well to the general math problem solving portion and the TSD portion. You can use pencil and paper for the SAT/ ACT questions since you're allotted paper and pencil for the general math problem sets at ACS.winnipegoo7 said:
Another question, haha:
What would be the most effective method of approaching something like this?
What speed covers 21 miles in 2 hours and 20 mins?
That is the question. I know that 10.5 mph covers 21 miles in 2 hours, but how would I account for the 20 minutes in there? since there are 3 occurrences of 20 minutes in one hour.
so basically: what speed covers 21 miles in 2.3 hours?
Or something like this:
At 5 mph, how far do you travel in 2 hours and 12 mins?
Or:
At 52 mph, how long does it take to travel 130 miles?
I'm mainly having trouble dividing mentally since I cannot seem to break apart the number properly, and holding all that info in my head is quite difficult. I guess I need a LOT of practice
Also, should I be using anything at all to write this or should this be completely in your head? Ie. not even typing anything onto the screen unless its your answer?
What would be the most effective method of approaching something like this?
What speed covers 21 miles in 2 hours and 20 mins?
That is the question. I know that 10.5 mph covers 21 miles in 2 hours, but how would I account for the 20 minutes in there? since there are 3 occurrences of 20 minutes in one hour.
so basically: what speed covers 21 miles in 2.3 hours?
Or something like this:
At 5 mph, how far do you travel in 2 hours and 12 mins?
Or:
At 52 mph, how long does it take to travel 130 miles?
I'm mainly having trouble dividing mentally since I cannot seem to break apart the number properly, and holding all that info in my head is quite difficult. I guess I need a LOT of practice
Also, should I be using anything at all to write this or should this be completely in your head? Ie. not even typing anything onto the screen unless its your answer?
What would be the most effective method of approaching something like this?
Realize the fraction from the minutes.
What speed covers 21 miles in 2 hours and 20 mins?
20 mins = 1/3 hour. Adding total time = 2(3/3) + 1/3 = 7/3 hours.
S= D/T = 21/7/3= 21 * (3/7) = (21/7)(3) = 9 mph
At 5 mph, how far do you travel in 2 hours and 12 mins?
12 min / 60 min/h = 1/5 hr. S=D/T, D=S(T) , T= 2(5/5)+ 1/5 = 11/5, D= 5 ( 11/5) = 11 miles
At 52 mph, how long does it take to travel 130 miles?
Divide distance by speed to get time ( ie. S= D/T, D= S(T), T= D/S )
130/ 52 = 2 hr + ( 130 - 2(52))/ 52)= 2 hr + (130-104)/52 = 2Hr + (26/52) = 2hr + 0.5 hr = 2.5 hrs.
Also, should I be using anything at all to write this or should this be completely in your head? Ie. not even typing anything onto the screen unless its your answer?
As you know, there will not be any aids ( pencil and paper) for DST calculations at ACS. Having said that, using pencil and paper in the begining will aid in understanding and seeing where you are going wrong. After time, you become "efficient" in such calculations that the mental execution becomes easier ( you know what you are calculating instead of holding numbers in your head, calculating, and being unsure if the methodology employed is correct).
Dont become discouraged, do a little everyday at least.
Roger123 said:
So I'm a little confused by what you did for this part:
130/ 52 = 2 hr + ( 130 - 2(52))/ 52)= 2 hr
Why did you multiply 52 by 2 and subtract it from 130, to get the minutes? A little unsure on how it works. I'd appreciate if you could clarify
AliTheAce said:
For this question, finding a common denominator makes the most sense and is more efficient for me. Knowing your timetables up to 19 is helpful for the SDT website:
130 miles / 52 mph <--both can be divided by 13
= 10 / 4
= 2.5
AliTheAce said:
I divided 130 by 52 and you get 2 whole divisions, hence the 2 hrs. ( 130 - 2(52)) is the remainder. 130 - 2 (52). Multiplied 52 by 2 because 52 goes into 130 twice. 130 - 2 (52)= 130- 104 = 26. 26/52 = 0.5. Total time is 2hrs + 0.5 hrs = 2.5 hrs.
I have been waiting since August for confirmation of my ASC dates.
Tentative dates, I have been told via email and in person, are early December.
I have been doing Lumosity, and Speed/Distance/Time (http://www.speeddistancetime.info/) for over a year now in preparation.
Lumosity scores are all in the 97th+ percentile. And, my Speed/Distance/Time and Fuel Calculations are basically perfect and fast.
Tentative dates, I have been told via email and in person, are early December.
I have been doing Lumosity, and Speed/Distance/Time (http://www.speeddistancetime.info/) for over a year now in preparation.
Lumosity scores are all in the 97th+ percentile. And, my Speed/Distance/Time and Fuel Calculations are basically perfect and fast.
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For me I started by doing them on paper and then moved on to doing them in my head as my understanding grew. I try to picture in my head how Id work through them on paper. Seems to be working for me. Still need to get more efficient though, I dont always solve them in the fastest (or simplest) possible way.
Ive sent in my dates for ACS waiting to hear back about tentative booking.
Ive sent in my dates for ACS waiting to hear back about tentative booking.
