If all the numbers referred to are the same number, we can use n to represent it. Then, "the ratio of a number and 4 times that number" is n/(4n) = 1/4. "The ratio of 5 times the number and 95 more than the number" is (5n)/(n+95)
The equation for the problem can be written ...
1/4 = (5n)/(95+n)
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If you're a purist, you might leave the factors of "a number" in the first raio and write it ...
n/(4n) = (5n)/(95+n)
Solving this gets you to a quadratic form that has the extraneous solution n=0.
