steppenwolf111 wrote:
Hi
I am asking for help in explaining a case from the thread "variations-on-the-gmat-all-in-one-topic" BY VeritasPrepKarishma ( i cannot use hyperlinks yet)
"How will you write the joint variation expression in the following cases?
4. x varies directly with y^2 and y varies directly with z.
\(x/y^2=k\)
\(y/z=k\) which implies that \(y^2/z^2=k\)
Joint variation: \(x*z^2/y^2=k\)"
According to this logic, in the case above for instance If both y and z get doubled, x become unchanged, while it should be 16 times bigger. Please help
Most of the above is not mathematically correct.
First, if x varies directly with y^2, then x = ky^2, where k is some constant.
If y varies directly with z, then y = mz, where m is some constant. There is no reason that this constant should be equal to the "k" we used above, so we need to use a new letter (you can't use k again, as someone did in the quoted portion of your post).
If x = ky^2, and y = mz, then by substitution, x = k(mz)^2 = km^2 z^2. Since km^2 is just some new constant number, we can replace it with a single letter, say p, so
x = p * z^2
where p is some constant (equal to km^2).
I haven't followed how you arrived at your direct variation equation, but it is not correct.
Further, it doesn't make much sense to ask what happens "If both y and z get doubled" in this situation. We know y varies directly with z. Their values are related, and can't change independent of each other. Any time z is doubled, y is doubled automatically. So in the sentence "if both y and z get doubled", half of the information is redundant -- the only actual information here is that z doubles. And if z doubles, then from our variation equation x = pz^2, the value of x will quadruple.
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