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steppenwolf111
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

If y is doubled, z is doubled too. So y becomes 2y and y^2 becomes 4y^2. Similarly, z becomes 2z and z^2 becomes 4z^2.
So 4 in numerator gets cancelled with 4 in the denominator and x remains the same.
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steppenwolf111
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

If y is doubled, z is doubled too. So y becomes 2y and y^2 becomes 4y^2. Similarly, z becomes 2z and z^2 becomes 4z^2.
So 4 in numerator gets cancelled with 4 in the denominator and x remains the same.

Unfortunatelly I still don't understand. In case 2. x varies directly with y and y varies inversely with z. The formula is as followed \(x/yz=k\) which means that when y is doubled and z is doubled x is four times bigger. In case 4 "x varies directly with y^2 and y varies directly with z" means that when z is doubled y is doubled and when y is also doubled all in all it means (2*2)^2=16
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steppenwolf111
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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