I turn 30 years old tomorrow. 30 f’ing years old. CLING TO EVERY SECOND OF YOUR YOUTH. SQUEEZE EVERY OUNCE OF JOY OUT OF IT LIKE JUICE FROM YOUR FAVORITE FRUIT. SOMEDAY YOU WILL BE OLD LIKE ME.

The prize for the challenge: Free access to the PWN the SAT Math Guide Beta.

Week 1: Served customers numbered a through b
Week 2: Served customers numbered c through d
Week 3: Served customers numbered e through f
Week 4: Served customers numbered g through h
Week 5: Served customers numbered i through j

The proprietor of a deli is trying to project how many customers he usually has on Mondays, so that he can order enough roast beef, but not order too much because nobody likes rotten roast beef. At the deli, customers take numbers before they are served, so he plans to collect data over the next 5 Mondays in the format of the list above, and is looking for an expression for average number of customers that he can plug numbers into as he collects them. What is the expression for the average number of customers (in terms of a through j) that are served in the deli on those 5 Mondays?

Good luck, children. I’ll post the solution Monday.

UPDATE: Congrats, Serrilius. I hope you enjoy the book.

Solution below the cut.

As Collin so deftly pointed out in the comments, a little plugging in does wonders here. If, for example, the deli serves customers numbered 35 through 39, how many customers were served? 35, 36, 37, 38, 39. 39-35 = 4, but 5 customers were served. So when you want to know how many customers passed through on a day when numbers a through b were served, it’s b – a + 1.

So then, to find the average, you find the sum, and then divide by 5.

[(b a + 1) + (dc + 1) + (fe + 1) + (hg + 1) + (ji + 1)]/5
= (-a + bc + de + fg + hi + j + 5)/5
= (-a + b – c + d – e + f – g + h – i + j)/5 + 1

Total number of customers divided by number of days to get average per day.  ((b-a) + (d-c) + (f-e) +(h-g) + (j-i)) / 5.  ?

Collin Zeng says:

I’ll give you guys a hint since I already won a Weekend Challenge …

Collin Zeng says:

StaceyHL’s answer is incorrect because she forgot that one must add 1 when including two endpoints. She is, however, precariously close to the right answer :o.

For example: how many numbers are there from 0-9, inclusive?

9 – 0 = 9 ==> 9 + 1 = 10 numbers from 0 to 9 inclusive. (You are including both endpoints).

Therefore, j – i + 1 = number of customers in Week 5.

Or let’s say j = 5 and i = 1

5 – 1 = 4. But 1, 2, 3, 4, 5 … there are 5 numbers from 1 to 5 inclusive!?!?!?!

Thus, it is 5 – 1 + 1, not just 5 – 1.

Well, let me be the first to wish you a HAPPY BIRTHDAY —  and then to assure you, it’s not so bad over here on the other side of the hill.

Excellent weekend challenge, but I’m going to pass today.  I’m suffering from “decision fatigue.”

Serrilius says:

((b-a+1)+(d-c+1)+(f-e+1)+(h-g+1)+(j-i+1))/5
=
(b+d+f+h-a-c-e-g+5)/5
=
((b+d+f+h-a-c-e-g)/5 + 1

Sam Fu says:

Oops, this is Serrilius; apparently I was supposed to use my Google account. I hate shedding my anonymity, but I think this shouldn’t be a problem.

Sam Fu says:

Thanks, and happy belated birthday! My SAT is coming up soon, so this is perfect 🙂

help! says:

Do I have to type the working out of the problem??
I’m not sure but this is what I did:

Monday 1 = (b-a+1)/7
M 2 = (d-c+1)/7
M3 = (f-e +1)/7
M4 = (h-g+1)/7
M5 = (j-i+1)/7

then add all and divide by 5 that’s how I got the above answer…hope its right..

Collin Zeng says:

Serrilius/Sam Fu is correct. It’s over, children :).

Happy birthday to you Mike :)!