Asked by Bonnie
                Two greyhounds are chasing a rabbit that has a 90 rabbit-jump start. While the rabbit jumps 10, the greyhound jumps 7, but the length of two jumps of the greyhound equals the length of 5 rabbit-jumps. In how many jumps do the greyhounds catch the rabbit?
            
            
        Answers
                    Answered by
            mathhelper
            
    define the following:
number of rabbit jumps --- r
number of greyhound jumps ---- g
length of rabbit jump ---- x
length of greyhound jump --- y
we are told: 10r = 7g
r = 7g/10
also : 2y = 5x
y = 5x/2
distance covered by rabbit = rx + 90x
distance covered by greyhound = gy = g(5x/2)
rx + 90x = gy
(7g/10)(x) + 90x = g(5x/2)
times 10
7gx + 900x = 25gx
18gx = 900x
18g = 900
g = 50 , then r = 35
So the greyhounds must make 50 jumps
it appears that the length of the jumps, that is the x and the y, are independent of the above, notice the x's cancelled out.
suppose we let x = 10, then y =5(10)/2 = 25
distance covered by rabbit = rx + 90x = 35*10 + 90*10 = 1250
distance covered by greyhound = gy = 50(25) = 1250
pick any other x so that both x and y are integers, and it will work
unusual question, never seen one like it.
    
number of rabbit jumps --- r
number of greyhound jumps ---- g
length of rabbit jump ---- x
length of greyhound jump --- y
we are told: 10r = 7g
r = 7g/10
also : 2y = 5x
y = 5x/2
distance covered by rabbit = rx + 90x
distance covered by greyhound = gy = g(5x/2)
rx + 90x = gy
(7g/10)(x) + 90x = g(5x/2)
times 10
7gx + 900x = 25gx
18gx = 900x
18g = 900
g = 50 , then r = 35
So the greyhounds must make 50 jumps
it appears that the length of the jumps, that is the x and the y, are independent of the above, notice the x's cancelled out.
suppose we let x = 10, then y =5(10)/2 = 25
distance covered by rabbit = rx + 90x = 35*10 + 90*10 = 1250
distance covered by greyhound = gy = 50(25) = 1250
pick any other x so that both x and y are integers, and it will work
unusual question, never seen one like it.
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