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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
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in the joint vacation. are you sure there is no typo in adding k^2 VeritasKarishma
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k represents a constant, any constant. k^2 will be a constant too. Basically, its value does not change when x, y or z change. It is like (1/2) in this relation: Area of triangle = (1/2) * b*h If b or h or area change, (1/2) stays as it is.
Now whether the value of that constant is 1/2 or 1/4 or 200, it doesn't matter.
Replace k by "Constant" and you might feel better about it.
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
Show more
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.
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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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
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
Show more
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.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.