SOLUTION: Two airline passengers have baggage totaling 190lbs. A passenger is charged a fixed number of dollars per pound if his or her baggage weight is over a certain maximum value. One pa

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Two airline passengers have baggage totaling 190lbs. A passenger is charged a fixed number of dollars per pound if his or her baggage weight is over a certain maximum value. One pa      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1014295: Two airline passengers have baggage totaling 190lbs. A passenger is charged a fixed number of dollars per pound if his or her baggage weight is over a certain maximum value. One passenger is charged $8. The other passenger is charged $2. If all the baggage belonged to one passenger, that person would have been charged $24. Find the charge per pound for extra weight and the maximum weight allowed per person without penalty.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = pounds of extra weight for
passenger paying $8
Let +b+ = pounds of extra weight for
passenger paying $2
Let +c+ = pounds of extra weight for
passenger paying $24
---------------------------------
+8%2Fa+ = cost/pound of extra weight
+2%2Fb+ = cost/pound of extra weight
+24%2Fc+ = cost/pound of extra weight
----------------------------------
(1) +190+-+%28+a+%2B+b+%29+=+2%2A%28+190+-+c+%29+ ( I plugged in real numbers and came up with this )
(2) +8%2Fa++=+2%2Fb+
(3) +2%2Fb+=+24%2Fc+
---------------
This is 3 equations and 3 unknowns, so it's solvable
----------------
(1) +190+-+a+-+b+=+380+-+2c+
(1) +-a+-+b+%2B+2c+=+190+
------------------------
(2) +8b+=+2a+
(2) +a+=+4b+
and
(3) +2c+=+24b+
(3) +c+=+12b+
-----------------
Substitute (2) and (3) into (1)
(1) +-a+-+b+%2B+2c+=+190+
(1) +-4b+-+b+%2B+2%2A12b+=+190+
(1) +24b+-+5b+=+190+
(1) +19b+=+190+
(1) +b+=+10+
and
(2) +a+=+4b+
(2) +a+=+40+
and
(3) +c+=+12b+
(3) +c+=120+
-----------------
+8%2Fa+ = cost/pound of extra weight
+2%2Fb+ = cost/pound of extra weight
+24%2Fc+ = cost/pound of extra weight
+8%2F40+=+1%2F5+
+2%2F10+=+1%2F5+
+24%2F120+=+1%2F5+
$.20 = cost/pound of extra weight
-------------------------------
+190+-+c+=+190+-+120+
+190+-+120+=+70+
Maximum weight allowed without penalty:
70 pounds
----------------
check:
+70+%2B+a+=+70+%2B+40+=+110+
and
+70+%2B+b+=+70+%2B+10+=+80+
and
+110+%2B+80+=+190+
OK
You might want another opinion on this
-I'm not 100% sure
-hope I got it!