Supper with Skiing



A ski resort offers two package plans. Plan A gives one person four nights lodging and three dinners for $326. Plan B gives one person five nights lodging and four dinners for $413. Assuming that the costs per night for lodging and dinner are the same for both plans, how much does one dinner cost?

(Source: Mathematics Teaching in the Middle School, Mar-Apr 1996).

I would use the Four Step Plan for Problem Solving here.

We know that Plan A costs $326 for:

4 nights lodging and 3 dinners > 4n + 3d = 326

(n represents the number of nights, while d represents the cost of the dinner)

We also know that Plan B costs $413 for:

5 nights lodging and 4 dinners > 5n + 4d = 413


We want to find out the cost of 1 night of lodging (hoping and assuming they are the same) and 1 dinner

  • Plan: Estimate your answer. Make a list of data that pertains to the problem and see if it would be reasonable. If the problem does not make sense, try to solve a simpler problem...or try solving it in another way!!
  • First, we need to subtract the two equations in order to show the cost of 1 night plus the cost of 1 dinner, which totals $87: (see subtraction #1 under Solve)
    Second, we need to isolate the d in order to solve for the cost of 1 dinner. To do this, we need to multiply both sides of Plan B's equation by 4 (this will eventually get rid of n and leave d all alone...see the multiplication #2 under Solve)
    Third, we need to continue the isolation of d by multiplying both sides of Plan A's equation by -5:(see the multiplication #3 under Solve)
    Finally, we need to subtract the two equations we now have, leaving only d, the cost of the dinner!!! Let's hope it works!! (see the subtraction #4 under Solve)


    5n + 4d = 413
    - 4n + 3d = 326
    n + d = 87
    4(5n + 4d = 413)
    20n + 16d = 1652



    -5(4n + 3d = 326)

    -20n - 15d = -1630



    20n + 16d = 1652

    - 20n - 15d = -1630

    d = 22

    $22 for the cost of dinner is quite reasonable, especially since this is a ski resort!!! :-) Also, re-solving the equations using $22 as the price of dinner in each Plan, gave the cost of lodging as $65, respectively. We may also isolate n in order to solve for the cost of lodging by multiplying Plan A's equation by - 4 and Plan B's equation by 3...we would have the following:
    Plan A - 4(4n + 3d = 326), which yields -16n - 12d = -1304
    Plan B 3(5n + 4d = 413), which yields 15n + 12d = 1239
    Once these are subtracted, we have:
    -16n - 12d = -1304
    15n + 12d = 1239
    - x = - 65
    so x = 65
    I do believe this will work!!!