In Linguistics 101, the ratio of the number of juniors to the number of seniors is 3:2. When six more juniors join the class, and one senior drops the class, the ratio of the number of juniors to the number of seniors becomes 2:3. How many students are in the class after these changes?
I can tell you right now that increasing the number
of juniors will not invert the ratio. But I'll go through
the motions for you anyway.
J 3
–– = ––
S 2
J + 6 2
–––– = –––
S – 1 3
Cross multiply both 2J = 3S (eq 1)
3J + 18 = 2S – 2 (eq 2)
Get either J or S in terms of the
other. It doesn't matter which, but
J would be simpler. From eq 1: J = 3S/2
Substitute this in eq 2 (3) • (3S/2) + 18 = 2S – 2
9S/2 + 18 = 2S – 2
Multiply both sides by 2 9S + 36 = 4S – 4
Subtract 4S from both sides 5S + 36 = –4
Subtract 36 from both sides 5S = –40
S = –8
You cannot have a negative number of seniors
so something is wrong somewhere. The error
is in the initial proposition. This problem has ...... no solution.
In a different problem that I solved, the solution depended on accepting
the concept of negative money. Well, we *sort of* do have such a thing
as that, it's called debt. But I can't rationalize negative students. LOL.
.
