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.
- Explore: What do you know?
What do you want to find out?
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
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)
- Solve: SOLVE the problem.
- 5n + 4d = 413
- - 4n + 3d = 326
- n + d = 87
- 4(5n + 4d = 413)
- 20n + 16d = 1652
+ 3d = 326)
- 15d = -1630
+ 16d = 1652
20n - 15d = -1630
- Examine: How does your
plan relate to your solved problem? Is it reasonable? Do you
need to make corrections?
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:
A - 4(4n + 3d = 326), which yields -16n - 12d = -1304
B 3(5n + 4d = 413), which yields 15n + 12d = 1239
these are subtracted, we have:
- 12d = -1304
15n + 12d = 1239
- x = - 65
- so x
- I do
believe this will work!!!