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 January 8th, 2012, 04:23 PM #1 Newbie   Joined: Jan 2012 Posts: 5 Thanks: 0 Challenging maths puzzle! Give it a go!! Calling all aspirin Hey you guys! I'm a maths fanatic in my senior year of high school and I was working on an interesting maths „puzzle“ I stumbled upon today and, since I had extra time this evening, I thought it would be a good idea to post the said puzzle online just to see what you guys think of it, to compare my answers and see if there are some other approaches to solving the problems. No, I actually DID complete all of the questions in the puzzle, I'm not trying to get someone else to do the work for me, I'm just double-checking my answers and seeing if there are any different approaches to solving the puzzle than mine Okay, here goes the puzzle. It's more of an investigation really, about a trip students take, prices of the hotel… Anyway, you'll see (and appreciate I took time to write this all up): --------------------------------------- THE TRIP Students of the class of 2011 went to a travelling agency to ask about the conditions for a 4-day-long trip which they are planning. They decided on a trip which cost $1000 per person, the option when each student went to the trip individually (no group discounts). However, they do not need luxurious, single-bedded rooms the aforementioned cost covers. Furthermore, they heard that it is possible to get a cheaper price if more students went on the trip. There are 31 students in their class and all of the students would like to go, but some cannot afford it. So far, 20 students gathered$14000. A few more students consider paying for the trip if the price isn't too high, but some students are not able to pay anything at all. Finally, they wanted to know exactly what the travelling agencies can offer them. They inquired the agency which said it will review their demands and get back to them. Next day, they received an offer from the travelling agency: The price is $1120 per person, and each extra person costs$20 less. (e.g. 1 person = 1120, 2 people = 1120 + 1100 = 2220, 3 people = 1120 + 1100 + 1080 = 1300 etc. Etc.) Questions: 1. What is the price of the trip if only one person goes on it? How many students need to go on the trip for the price to equal the price of each student going individually ($1000 per person, mentioned in introduction)? 2. What is the price per student if 20 people go on the trip? What is the total price then ? 3. What is the price per student if 22 people go on the trip? What is the total price then ? 4. How many students can go on the trip with the already collected$14000? 5. How much would they pay if all 31 students would go on the trip? What would be the price per student then? 6. How many more students can go on the trip for the same price as for 31 students? What is the price per student then? 7. All of the students of the class of 2011 want to go on the trip. Suggest an option of payment which would allow all of the students of the class to go on the trip. Keep in mind the various factors we already mentioned (some students cannot pay etc.) The boss of the travelling agency, upon seeing the offer given to the students, was pleased so he asked the agent in charge of the class of 2011 to explain to him how he figured out the optimal offer. This is what he answered: „I calculated the lowest price at which 31 students can go on the trip and the agency isn't losing money. My calculations led me to a number of $15500: this was calculated by the formula = (A-Bx)x = (where A is the price, B is the discount per person), where A and B have several different values , and I chose the most optimal of those values. 8. Calculate a few different values for A and B, and comment on how they affect the price. 9. For each of the values you chose, calculate how many more students can go on the trip for a price of$15500? How does that number relate to the number 31 (number of students)? 10. Determine A and B so that less than 31 students can go on the trip for $15500? 11. Make an offer for a class of 36 students , if the total price of the trip is$13500 and if maximum 29 students can afford to pay for the trip. -------------------------------------------------- If anything sounds wrong, it will usually start to make sense as you begin to solve the question, I had the same problem. Anyway, have a go at it, and please, do not hesitate to ask me if I made something unclear or I made an error somewhere. I'll check back at this page multiple times today! Happy mathematicating! Cheers! Johnatan
January 8th, 2012, 05:25 PM   #2
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Re: Challenging maths puzzle! Give it a go!! Calling all asp

Quote:
 1. What is the price of the trip if only one person goes on it? How many students need to go on the trip for the price to equal the price of each student going individually ($1000 per person, mentioned in introduction)? a)$1120

b) $\frac{S_n}{n}=1000$

$\frac{1}{2}$$2\cdot1120+(n-1)(-20)$$=1000$

$2240-20n+20=2000$

$260=20n$

$n=13$

We could also use the fact that for each additional student, the price per student goes down $10, and it takes 12 additional students to get to$1000 per student for a total of 13 students.

With 13 students going, the price is $13000, or$1000 per person.

Quote:
 2. What is the price per student if 20 people go on the trip? What is the total price then ?
$1120-19(10)=930$

$20\cdot930=18600$

Quote:
 3. What is the price per student if 22 people go on the trip? What is the total price then ?
With 2 more than part 3, we deduct $20 from$930 to get $910. $22\cdot910=20020$ Quote:  4. How many students can go on the trip with the already collected$14000?
$\frac{n}{2}$$2\cdot1120+(n-1)(-20)$$=14000$

$2260n-20n^2=28000$

$n^2-113n+1400=0$

We take the root for which 1 < n < 32 and round down to get n = 14.

Quote:
 5. How much would they pay if all 31 students would go on the trip? What would be the price per student then?
$\frac{31}{2}$$2\cdot1120+(31-1)(-20)$$=25420$

$\frac{25420}{31}=820$

Quote:
 6. How many more students can go on the trip for the same price as for 31 students? What is the price per student then?
$2260n-20n^2=50840$

$n^2-113n+2542=0$

$(n-31)(n-82)=0$

We find that 82 students can also go for $25420. That's an additional 51 students. $\frac{25420}{82}=310$ I'll get back to the rest later if someone else doesn't finish.  January 8th, 2012, 06:47 PM #3 Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Challenging maths puzzle! Give it a go!! Calling all asp This is too much like real cost accounting work. :P  January 9th, 2012, 09:51 AM #4 Newbie Joined: Jan 2012 Posts: 5 Thanks: 0 Re: Challenging maths puzzle! Give it a go!! Calling all asp @MarkFL Real interesting stuff you got going there. My answer for question 6. was wrong an you corrected it! Thanks for your time! Please do go on with the questions, I would very much like to see how you approach the rest of the questions! Cheers! January 9th, 2012, 01:22 PM #5 Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0 Re: Challenging maths puzzle! Give it a go!! Calling all asp Quote:  Originally Posted by mobilefreak10 @MarkFL Real interesting stuff you got going there. My answer for question 6. was wrong an you corrected it! Thanks for your time! Please do go on with the questions, I would very much like to see how you approach the rest of the questions! Cheers! Is not really the "challenging maths puzzle" as promised in the title.  January 10th, 2012, 09:00 AM #6 Newbie Joined: Jan 2012 Posts: 5 Thanks: 0 Re: Challenging maths puzzle! Give it a go!! Calling all asp Please do go on with the question, MarkFL, I'd be thrilled to see your answers to the rest of the questions, especially No. 11!  January 10th, 2012, 12:15 PM #7 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,581 Thanks: 1038 Re: Challenging maths puzzle! Give it a go!! Calling all asp Le general case: n = number of students f = first student cost r = reduction Total = n(f) - r[n(n - 1) /2] Total = 31(1120) - 20(31*30/2) = 25420 Repeating what WilliamNicholasVictorLadouceur said: where's the challenge? January 10th, 2012, 12:23 PM #8 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Challenging maths puzzle! Give it a go!! Calling all asp Quote:  7. All of the students of the class of 2011 want to go on the trip. Suggest an option of payment which would allow all of the students of the class to go on the trip. Keep in mind the various factors we already mentioned (some students cannot pay etc.) Well, as we found in part 5, the cost for 31 students is$25420. If those who can pay are willing to equally cover those who cannot, then the cost to those who can pay, with n students not paying would be:

$\frac{25420}{31-n}$

Quote:
 The boss of the traveling agency, upon seeing the offer given to the students, was pleased so he asked the agent in charge of the class of 2011 to explain to him how he figured out the optimal offer. This is what he answered: "I calculated the lowest price at which 31 students can go on the trip and the agency isn't losing money. My calculations led me to a number of $15500: this was calculated by the formula = (A-Bx)x = (where A is the price, B is the discount per person), where A and B have several different values , and I chose the most optimal of those values." 8. Calculate a few different values for A and B, and comment on how they affect the price. We find that: $(1120-20\cdot31)31=15500$ $(1120-20\cdot25)25=15500$ With B and x fixed, we find the price ?P = 31?A. With A and x fixed, we find ?P = -31²?B. Quote:  9. For each of the values you chose, calculate how many more students can go on the trip for a price of$15500? How does that number relate to the number 31 (number of students)?
$Ax-Bx^2=15500$

$Bx^2-Ax+15500=0$

Taking the larger root:

$x=\frac{A+\sqrt{A^2-62000B}}{2B}$

We see that an increase in A allows more students to go, while an increase in B allows fewer students to go.

Now I'm going to let someone else finish.

 January 15th, 2012, 04:47 AM #9 Newbie   Joined: Jan 2012 Posts: 5 Thanks: 0 Re: Challenging maths puzzle! Give it a go!! Calling all asp Aww, c'mon MarkFL, I really wanted to see your offer for question 11! I came up with some crappy, half-baked offer for #11 and I was counting on you to set me straight! Pretty please Please please please it's just two more itsy bitsy questions, it must be a piece of cake for you! How much more dignity must I lose begging you to answer them?
January 15th, 2012, 05:07 AM   #10
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Re: Challenging maths puzzle! Give it a go!! Calling all asp

Quote:
 Originally Posted by mobilefreak10 How much more dignity must I lose begging you to answer them?
Dignity is worthless: does not buy groceries...

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