Topic of the Day: Delta Math
For the most part the math questions are not as hard as they use to be. We get a lot of questions about the math portion of the interview process. On our website we have a Math quiz trainer that will walk you through the type of problems that are currently being asked on the tests. Our math quiz trainer also have problems that are part of the new test bank with their solutions.
During your interview you will be given two opportunities to show Delta how much you love math. Once during the Cog test and once during the Job knowledge test.
Cog Test Math
As part of the Cog test you 1 of 13 sections will be a math section. You will not be able to use a calculator or a paper and a pencil for this section. This section is probably one of the most difficult because of the compression of time. The type of problems you can expect are basic yet thought intensive type of math. Many of the problems deal with percentages. You will be given a short amount of time to complete these problems in your head and more than likely only complete 4-5 questions. The best technique for this section is to incorporate the answers into your scan so you can quickly derive the correct answers.
Job knowledge Test Math
During the job knowledge test you more than likely have a couple of math problems that deal with turn radius, calculating descents, fuel burns etc. The math as of the recent changes is basic and easy if you understand the formulas and understand the concept. I have received debriefs from people who have mentioned not having any math questions on their test at all. On this test you will be given the opportunity to use a paper and a pencil.
Cog Percentages Review:
How to Solve Percent Problems
A lot of percent problems turn out to be easy to solve when you give them a little thought. In many cases, just remember the connection between percents and fractions and you’re halfway home:
Solve simple percent problems
Some percents are easy to figure. Here are a few.
Finding 100% of a number: Remember that 100% means the whole thing, so 100% of any number is simply the number itself:
100% of 5 is 5
100% of 91 is 91
100% of 732 is 732
Finding 50% of a number: Remember that 50% means half, so to find 50% of a number, just divide it by 2:
50% of 20 is 10
50% of 88 is 44
Finding 25% of a number: Remember that 25% equals 1/4, so to find 25% of a number, divide it by 4:
25% of 40 is 10
25% of 88 is 22
Finding 20% of a number: Finding 20% of a number is handy if you like the service you’ve received in a restaurant, because a good tip is 20% of the check. Because 20% equals 1/5, you can find 20% of a number by dividing it by 5. But you can use an easier way:
To find 20% of a number, move the decimal point one place to the left and double the result:
20% of 80 = 8 2 = 16
20% of 300 = 30 2 = 60
20% of 41 = 4.1 2 = 8.2
Finding 10% of a number: Finding 10% of any number is the same as finding 1/10 of that number. To do this, just move the decimal point one place to the left:
10% of 30 is 3
10% of 41 is 4.1
10% of 7 is 0.7
Finding 200%, 300%, and so on of a number: Working with percents that are multiples of 100 is easy. Just drop the two 0s and multiply by the number that’s left:
200% of 7 = 2 7 = 14
300% of 10 = 3 10 = 30
1,000% of 45 = 10 45 = 450
Make tough-looking percent problems easy
Here’s a trick that makes certain tough-looking percent problems so easy that you can do them in your head. Simply move the percent sign from one number to the other and flip the order of the numbers.
Suppose someone wants you to figure out the following:
88% of 50
Finding 88% of anything isn’t an activity that anybody looks forward to. But an easy way of solving the problem is to switch it around:
88% of 50 = 50% of 88
This move is perfectly valid, and it makes the problem a lot easier. As you learned above, 50% of 88 is simply half of 88:
88% of 50 = 50% of 88 = 44
As another example, suppose you want to find
7% of 200
Again, finding 7% is tricky, but finding 200% is simple, so switch the problem around:
7% of 200 = 200% of 7
Above, you learned that to find 200% of any number, you just multiply that number by 2:
7% of 200 = 200% of 7 = 2 7 = 14
Solve more-difficult percent problems
You can solve a lot of percent problems using the tricks shown above. But what about this problem?
35% of 80 = ?
Ouch — this time, the numbers you’re working with aren’t so friendly. When the numbers in a percent problem become a little more difficult, the tricks no longer work, so you want to know how to solve all percent problems.
Here’s how to find any percent of any number:
Change the word of to a multiplication sign and the percent to a decimal.
Changing the word of to a multiplication sign is a simple example of turning words into numbers. This change turns something unfamiliar into a form that you know how to work with.
So, to find 35% of 80, you would rewrite it as:
35% of 80 = 0.35 80
Solve the problem using decimal multiplication.
So 35% of 80 is 28.
As another example, suppose you want to find 12% of 31. Again, start by changing the percent to a decimal and the word of to a multiplication sign:
12% of 31 = 0.12 31
Now you can solve the problem with decimal multiplication:
So 12% of 31 is 3.72.
Job knowledge Test Review Formulas:
CONVERTING MACH TO TAS
MACH x 6 = TAS
.8 x 6 = 480
NAUTICAL MILES PER MINUTE
60 KNOTS =1NM/per min
120 = 2
180 = 3
240 = 4
300 = 5
360 = 6
420 = 7
480 = 8
540 = 9
600 = 10
TEMP CONVERSIONS
F=([Cx2] – 10%) + 32
Example: C=10. What is F?
C=([F-32] + 10%) ÷ 2
Example: F=50. What is C?
50-32 = 18 + 2 (1.8 to be exact) = 20 ÷ 2 = 10C
60 TO 1 RULE (to figure how many radials per NM)
60 = 1 (1 Radial per NM)
60
60 = 2 (2 radials per NM)
30
60 = 3 (3 radials per NM)
20
FIGURING DISTANCE FROM THE STATION
TAS x TIME
RADIALS
Example: RMI bearing pointer falls 10 degrees in 1 minute and you are doing mach .8
480 x 1 = 48 DME from the station
10
FIGURING YOUR LEAD RADIAL
60
DME x (NM/per min -2)
Example: You are on the 15 DME arc at 240knots. When do you start your turn inbound? Note: 240 is 4 NM per/min.
60 = 4(radials per NM) x (4-2) which is 4x 2 = 8. Start your turn inbound 8 radials prior to reaching the inbound radial.
15
RADAR TILT (ESTIMATING THUNDERSTORM TOPS)
Degrees of tilt x 100 x DME + your altitude
Example: You are at 20,000 MSL and painting a cell 20 miles ahead. Your radar tilt is 5° up. How high is the cell?
5 x 100 x 20 + 20,000
5 x 100 = 500 x 20 = 10,000 + 20,000 = 30,000 MSL height of cell.
PITCH CHANGE (DESCENT ANGLE)
100’S OF FEET (in Flight levels)
DISTANCE
Example: To lose 15,000 feet in 25 nm
150 = 6° (descent angle)
25
DESCENTS
DESCENT ANGLE x 100 x Speed in NM/PER MIN
Example: 20,000 to 5,000 in 100 NM at .6 mach. How many feet per minute in the descent?
150 = 1.5 (descent angle) x 100 = 150 x 6 NM/per min = 900 FPM REMEMBER: any mach number = the same amount of miles per minute (.6 = 6 miles per minute)
100
DESCENT GRADIENT (Feet per NM)
FEET (total feet … not feet in FL’s)
NM
Exmple: To lose 15,000 feet in 25 NM.
15,000 = 600 Feet per NM
25NM
FIGURING FPM FOR A 3° GS
GROUNDSPEED ÷ 2 THEN ADD A ZERO.
Example: 200 knots on approach. what should be your FPM in descent?
200 / 2 = 100 then add a zero…. = 1000 FPM (you could also just multiply GS x 5 and get the same answer. Either rule works).
VISUAL GLIDEPATH (AGL)
DISTANCE FROM RUNWAY x 300 = the height above the airport you should be at.
Example: At 5DME from runway your should be at 1500 AGL.
2DME x 300 = 600 AGL
CALCULATING VDP
USING DME—— HAT ÷ 300 = DISTANCE FROM THRESHOLD TO PLACE VDP
USING TIME —– 10% OF THE HAT = TIME IS SECONDS TO SUBTRACT FROM THE FAF TO THE MAP.
CROSSWIND CORRECTION ANGLE
CROSSWIND IN MPH ÷ TAS IN NM/PER MIN = CORRECTION ANGLE
Example: 100 knots of crosswind. TAS = 300. What is your correction angle?
100 knot xw = 20° correction
5NM/min
HOW LONG IS THE ARC?
How long is the 15 dme arc from the 047 radial to the 011 radial?
047-011 = DIFFERENCE OF 36 RADIALS
60/15 (DME ARC) = 4 RADIALS PER NM
36 RADIALS ÷ 4 RADIALS PER NM = 9 NM
FIGURING CROSSWIND COMPONENTS
INCLUDE GUST FACTORS WHEN CALCULATING
0 degrees x 0.0
30 degrees x .5
45 degrees x .7
60 degrees x .9
90 degrees x 1.0
FIGURING HEADWIND/TAILWIND COMPONENTS
INCLUDE GUST FACTORS WHEN CALCULATING
0 degrees x 1.0
30 degrees x .9
45 degrees x .7
60 degrees x .5
90 degrees x 0.0
LOAD FACTOR AND STALL SPEED
STALL SPEED INCREASES IN PROPORTION TO THE SQUARE ROOT OF THE LOAD FACTOR
Example: Your aircraft will stall normally at 45 knots. What is the stall speed at 4 g’s?
45 x √4 = 90 knots.
ESTIMATING CLOUD BASES
THE MOIST ADIABATIC LAPSE RATE IS 2.5° PER 1,000 FEET
TAKE THE DIFFERENCE OF TEMP AND DEWPOINT THEN DIVIDE BY 2.5 = COULD BASES
Example: Temp 11 Dewpoint 6. About how high are the cloud bases?
11-6= 5 ÷ 2.5 = 2 (or 2,000 feet AGL).
BANK ANGLE FOR STD RATE TURN
(TAS ÷ 10) x 1.5
