"Mensa problem solving #8!"

(Note: This represents my thought process as I tried to figure this out.  I haven't edited it after going to the next attempt, other than for formatting or to correct typos.)

Bruno Velvet = 7:20 = 440 minutes
Isaac Checkers = 8:26 = 506 minutes
Damien Wellis = 9:20 = 560 minutes
Samuel Blackborough = 9:40 = 580 minutes

First thought: Roman numerals.

Bruno Velvet has two VLV (5, 50, 5), but no IXCDM..
Isaac Checkers has ICCC (1, 100, 100, 100), but no VXLDM.
Damien Wellis has DILLI (500, 1, 50, 50, 1), but no VXCM.
Samuel Blackborough has MLLC (1000, 50, 50, 10), but no IVXD.
Arthur Hastings has only I. Okay, this didn’t lead anywhere.

Next thought: Values based on positions of letters in the alphabet.

A	B	C	D	E	F	G	H	I	J	K	L	M
1	2	3	4	5	6	7	8	9	10	11	12	13
N	O	P	Q	R	S	T	U	V	W	X	Y	Z
14	15	16	17	18	19	20	21	22	23	24	25	26

Bruno Velvet = 2+18+21+14+16 (=71), 22+5+12+22+5+20 (=86) = 157
Isaac Checkers = 9+19+1+1+3 (=33), 3+8+5+3+11+5+18+19 (=72) = 105
Damien Wellis = 4+1+13+9+5+14 (=46), 23+5+12+12+9+19 (=80) = 126
Samuel Hastings = 19+1+13+21+5+12 (=71), 8+1+20+21+9+14+7+19 (=99) = 170

No luck yet...  Maybe just the initials:
B.V. = 2, 22... 22*22 − 2*22 = 440.  Interesting.
I.C. = 9, 3... 9*9 − 9*3 = 54.  Okay, the pattern doesn’t hold.

What if the vowels have a different value from the consonants?

Bruno Velvet = 2+18+21v+14+15+22+5v+12+22+5v+20 (=86) = 156 (125 consonant, 31 vowel)
Isaac Checkers = 9v+19+1v+1v+3+3+8+5v+3+11+5v+18+19 (=72) = 105 (84 consonant, 21 vowel)
Damien Wellis = 4+1v+13+9v+5v+14+23+5v+12+12+9v+19 (=80) = 126 (97 consonant, 29 vowel)
Samuel Hastings = 19+1v+13+21v+5v+12+8+1v+20+21+9v+14+7+19 = 170 (133 consonant, 37 vowel).

Let’s see:
125c + 31v = 440 or 7, 26
84c + 21v = 506 or 8, 26
97c + 29v = 560 or 9, 20
133c + 37c = 580 or 9, 40

No easy multiples here, so let’s just multiply a couple out:

(84)(97)c + (21)(97)v = (506)(97) or (8)(97), (26)(97)
(97)(84)c + (29)(84)v = (560)(84) or (9)(84), (20)(84)

  (8148c + 2037v = 49082 or 776, 2522)
-(8148c + 2436v = 47040 or 756, 1680)
-399v = 2042 or 20, 842

Alright, this seems to be too complex.  Brain teasers are supposed to have a catch that’s very simple once you see it.  So it's time to go back to the puzzle and check out the wording carefully.

Aha!  We can eliminate two of the possible answers right away: B (8 hrs. 26 min.) and C (9 hrs. 20 min.) because they are the same as two other drivers, and the question says "Each racing driver takes a different length of time to complete the race."  But that still leaves us with two possibilities: A (8 hrs. 20 min.) and B (7 hrs. 14 min.), neither of which has already been taken.

Still not getting anywhere.  Since we have four times, and four data points for each (consonants and vowels in first name and surname, aka c_f, v_f, c_s, v_s), let's try some matrix arithmetic:

Bruno Velvet has 3c_f + 2v_f + 4c_s + 2v_s = 440
Isaac Checkers has 2c_f + 3v_f + 6c_s + 2v_s = 506
Damien Wells has 3c_f + 3v_f + 3c_s + 2v_s = 560
Samuel Blackborough has 3c_f + 3v_f + 8c_s + 4v_s = 580

	cf	vf	cs	vs		time		name
	3	2	4	2		440		A1
	2	3	6	2		506		B1
	3	3	3	2		560		C1
	3	3	8	4		580		D1
	1	-1	-2	0		-66		A2 = A1 − B1
	0	5	10	2		638		B2 = B1 − 2A2
	0	6	9	2		758		C2 = C1 − 3A2
	0	6	14	4		778		D2 = D1 − 3A2
	1	-1	-2	0		-66		A2
	0	1	-1	0		120		B3 = C2 − B2
	0	0	15	2		38		C3 = C2 − 6B3
	0	0	20	4		58		D3 = D2 − 6B3
	1	0	-3	0		54		A3 = A2 + B3
	0	1	0	0.4		124		B3
	0	0	1	0.4		4		C4 = (D3−C3) ÷ 5
	0	0	0	-4		-22		D4 = D3 − 20C4
	1	0	0	1.2		66		A4 = A3 + 3C4
	0	1	0	0		121.8		B4 = B3 + C4
	0	0	1	0		1.8		C4
	0	0	0	1		5.5		D5 = D4 ÷ 5
	1	0	0	0		59.4		A5 = A4 − 1.2D5
	0	1	0	0		121.8		B5 = B4 − 0.4D5
	0	0	1	0		1.8		C5 = C4 − 0.4D5
	0	0	0	1		5.5		D5

So consonants in the first name are worth 59.4, vowels in the first name are worth 121.8, consonants in the surname are worth 1.8, and vowels in the last name are worth 5.5_.

Arthur Hastings has 4c_f + 2v_f + 6c_s + 2v_s, which is 237.6 + 243.6 + 10.8 + 11 = 503.  This means that he should complete the race in 503 minutes, which is 8 hours, 23 minutes.  Since 8 hours 23 minutes is not one of the possibilities, I can only conclude that he had a three-minute head start, so I'm choosing answer A, 8 hours 20 minutes.

And yes, I'm almost certain that I missed something so incredibly obvious that I'll slap my forehead after I find out what it is.

Well I started with #9, but then I saw your #8, could not resist.

 

Believe me or not, I knew there was an organization like that, but I only learned the name today. So thank you!

 

And here is what I think the solution should be:

 

A. Bruno Velvet - 7 hrs 20 min:

 

This is what I think the fomula is:

2 vowels pluse 5-letter word: 2+7=7 hours, last name 2 vowels plus the length of the word 6 mutiplied by 3: 2+6x3=20 minutes.

 

To Test:

 

B. Isaac Checkers - 8 hrs 26 min

 

3+5=8 hours, 2+8x3=26 mins

C. Damien Wallis - 9 hrs 20 min

 

3+6=9, 2+6x3=20

D. Samuel Blackborough - 9 hrs 40 min

 

3+6=9, 4+12x3=40

 

So, as you can see, the formula works.

 

Then the answer is:

E. Arthur Hastings -

 

2+6=8, 2+8x3=26

 

The answer is 8 hrs 26 min, which is B.

where v is number of vowels in first name, c is number of consonants in first name, n is the coefficient of v, and m is the coefficient of c
hrs=nv+mc

3m+2n=7 [given, Bruno]
3m+3n=9 [given, Damien or Samuel]
n=2 [elimination]
m=1 [substitution, subtraction]
hrs=2v+c [substitution]

where all variables are the same except last names are used instead of first names
min=nv+mc

4m+2n=20 [given, Velvet or Wallis]
6m+2n=26 [given, Checkers]
2m=6 [elimination]
m=3 [division]
n=4 [substitution]
min = 4v+2m [substitution]

Arthur v=2, c=4 [given]
hrs=4+4=8 [substitution]

Hastings v=2, c=6 [given]
min=8+12=20 [substitution]

Arthur Hastings - a) 8 hrs 20 min