Challenging math problem

Discus: SAT/ACT Tests and Test Preparation: July 2004 Archive: Challenging math problem
 By Etsrep78328 (Etsrep78328) on Monday, July 12, 2004 - 04:34 pm: Edit

Here is a challenging question -- not so much an SAT I or II math question.

To monitor the temperature of an auto cooling system, you intend to use a meter that reads from 0 to 100. You devise a new temperature scale based on the approximate melting and boiling points of a typical antifreeze solution (-45 degrees Celsius and 115 degrees Celsius). You wish these points to correspond to 0 degrees "A" and 100 degrees "A", respectively.

1. Derive an expression for converting between "A" and Celsius.
2. Derive an expression for converting between degrees Fahrenheit and "A." Recall that the temperature in fahrenheit is given by the formula F = 1.8C + 32.
3. At what temperature would your thermometer and a Celsius thermometer give the same numerical reading?
4. Your thermometer reads 86 degrees "A." What is the temperature in Celsius and Fahrenheit?
5. What is a temperature of 45 Celsius in degrees "A?"

 By Al0 (Al0) on Saturday, July 17, 2004 - 11:34 pm: Edit

1. We have the data points (A,C): (0,-45) and (100,115). We can plug these into point slope form:
(C-C0)=(A1-A0)/(C1-C0)*(C-C0)
(A-0)=(100-0)/(115-(-45))*(C-(-45))
A=5/8*(C+45)
and conversely:
C=8/5*A-45
2. We have F=9/5*C+32 and C=8/5*A-45
substituting the latter into the former:
F=9/5(8/5*A-45)+32
F=72/25*A-49
and conversely"
A=25/72*(F+49)
To check: @C=115 (A=100),
F=1.8*115+32=207+32=239
F=72/25*100-49=288-49=239
F=F, so our formula works.
3. We have C=8/5*A-45 and C=A, so
C=8/5*C-45
-3/5*C=-45
C=75, this checks as well.
4. A=5/8*(C+45)
A=5/8*(45+45)
A=5/8*90=56.25