Expert: Paul Klarreich - 1/29/2009

Question
Water mixture: hot water temperature 130 degree, cold water 55 degree, desired temperature 93. mixture result: 45 gallons of 93 degree water. How many gallons of hot water and how many of cold water needed to obtain the desired quantity and temparture? Please explain procedure

Answer
Questioner:   Earnest
Category:  Advanced Math
Private:  No
 
Subject:  math water problem
Question:  Water mixture: hot water temperature 130 degree, cold water 55 degree, desired temperature 93. mixture result: 45 gallons of 93 degree water. How many gallons of hot water and how many of cold water needed to obtain the desired quantity and temparture? Please explain procedure
.............................
Hi, Earnest,

I am going to assume you cannot use two variables.  Next time, PLEASE read my instructions.

--------------------+----------+-----------+----------+
Choice of Water     | Unit val | Gals used | Product  |
--------------------+----------+-----------+----------+
Hot                 |   130    |     x     |  130x    |
--------------------+----------+-----------+----------+
Cold                |    55    |  45-x     | 55(45-x) |
--------------------+----------+-----------+----------+
Mixture             |    93    |   45      |  93*45   |
--------------------+----------+-----------+----------+
                   |          |           |          |

 
Now  130x + 55(45-x) = 93*45

You can handle the rest.  

P.S. All those percent mixture problems, investment problems, rate-time-distance problems are the same.