Contents Slow & Fast Math Cats. Interactive pre-Algebra Explore * Once upon a time, in the Math Cats' Land: * The slow cat and the fast cat, were sleeping. * Then the mice came out of their hole. The slow cat woke up and started catching mice to put in his light blue basket. The fast cat was still asleep. * A little while later, the fast cat also woke up!. He, too, began to collect mice in his own pink basket. But the fast cat caught mice faster than the slow cat did. * When did the two cats' baskets have the same number of mice? * What was that number?
(1) Choose: (2) Choose: (3) Choose:
Slow Cat's speed: Time before Fast Cat wakes up: Fast Cat's Speed:
(4) At the instant when the number of mice in each basket is equal: choose your prediction of:
(5) Your actual prediction: (6) Start the program and verify your prediction:
(7) Check your results: Calculations, and simulations with words and numbers:
(If you make a prediction, the results will be more complete)
 Results: Results simulation: Predictions: Time: Mice: Why: Slow cat Fast cat
Time rate: 1 second simulation : x seconds problem: (Optional: only for more precise results)
By default: 1 second simulation = 10 seconds in the problem
Speed of the simulation:    slower<<<<    >>>>faster
1:1 sec 1:2 sec 1:5 sec 1:10 sec 1:20 sec 1:30 sec 1:60 sec
(8) Help:
 Slow & Fast Math Cats - HELP: Here is a simple example: The slow cat puts 10 mice per minute in his basket. ......4 minutes later...... The fast cat wakes up, and puts 20 mice per minute in his basket. (The slow cat: blue line) (The fast cat: pink line) y=10*x in minutes y=20*(x-4) in minutes The 10*x and 20*x in these equations are also called: "the slope of the line". In this grid, each vertical space represents 10 mice; and each horizontal space, represents 1 minute (60 seconds) SOLUTIONS: After 8 minutes total, (4 minutes for the fast cat, since he woke after 4 minutes): The slow cat has 80 mice in his basket (8 min*10 mice/minute) The fast cat has 80 mice in his basket (4 min*20 mice/minute) If you want to solve this problem without graphing, here is one way: The Slow Cat catches mice at a rate of 10 mice/minute; the Fast Cat at a rate of 20 mice/minute. We're looking for the time at which they have the same number of mice in the 2 baskets. So we must set the number of mice in both baskets as equal: 10*x=20*(x-4) 10*x=20*x-80 10*x+80=20*x-80+80 10*x+80=20*x 10*x-10*x+80=20*x-10*x 80=10*x x=80/10 x=8 minutes The same in seconds: 10 mice/minute = 10/60 = 0.1666666 mice/second 20 mice/minute = 20/60 = 0.3333333 mice/second 4 minutes=240 seconds 0.1666666*x=0.3333333*(x-240) 0.1666666*x=0.3333333*x-80 80=0.1666666*x x=80/0.1666666 x=480 seconds = 8 minutes