Mixed Variation Anyone

knight247 wrote:x varies inversely as y^2, when z is constant and x varies directly a square root of z, when y is constant. x= 16 when y=2 and z=4. Find x when y=4 and z=25

One method:

The given information implies the equation x=k√z/y^2, where k is a non-zero constant. Plug in the given values to solve for the constant, k: 16=k√4/2^2, so k=32. So, x=32√25/4^2=10

Ans: [spoiler]x=10[/spoiler]

x varies inversely as y^2, when z is constant, implies x=z/y^2
x varies directly as square root of z, when y is constant, implies x=y*sqroot(z)
y*sqroot(z)=z/y^2
square both sides to obtain y^2*z=z^2/y^4, y^2=z/y^4, z=y^6

x=z/y^2 or x=y^6/y^2=y^4

plz check: x= 16 when y=2 and z=4, 16=2^4, hence when y=4, x=4^4=256

@GmatMathPro, why in your solution x varies directly with sqroot(z) when y isn't const but variable (inverse var.)?

pemdas wrote:x varies inversely as y^2, when z is constant, implies x=z/y^2

pemdas,
why should it be x=z/y^2 and not x = 1/(z*(y^2)) or x=sqroot(z)/y^2 ?

PFB my version(same as Pete's)

x varies inversely as y^2, when z is constant and x varies directly a square root of z, when y is constant. x= 16 when y=2 and z=4. Find x when y=4 and z=25

x varies inversely as y^2, when z is constant
x = K*f(z)/y^2 Where K is a constant and f(z) is a function of z

x varies directly a square root of z
x = M*f(y)*(√z) Where K is a constant and f(y) is a function of y

if both y and z vary, then
x = O * (√z)/y^2 Where O is a constant(f(y) = y^2 and f(z)=√z)

16 = O * 2/4 => O =32

if y = 4 and z =25
x = 32 * (√z)/y^2
[spoiler]x = 32 * (5)/16 = 10[/spoiler]

it says z is constant not thought 'K', so place it in the numerator
in another case it says y is constant

moreover, when you put x=1/(z*(y^2)) you are having x inversely variable with z too

neelgandham wrote:

pemdas wrote:x varies inversely as y^2, when z is constant, implies x=z/y^2

pemdas,
why should it be x=z/y^2 and not x = 1/(z*(y^2)) or x=sqroot(z)/y^2 ?

My bad, I should have stopped at 'why should it be x=z/y^2 and not x=sqroot(z)/y^2 ?'. Do you still stand by your solution ?

pemdas wrote:x varies inversely as y^2, when z is constant, implies x=z/y^2
x varies directly as square root of z, when y is constant, implies x=y*sqroot(z)
y*sqroot(z)=z/y^2
square both sides to obtain y^2*z=z^2/y^4, y^2=z/y^4, z=y^6

x=z/y^2 or x=y^6/y^2=y^4

plz check: x= 16 when y=2 and z=4, 16=2^4, hence when y=4, x=4^4=256

@GmatMathPro, why in your solution x varies directly with sqroot(z) when y isn't const but variable (inverse var.)?

The given information describes a variable, x, that is influenced by two variables, y and z. Thus, it makes sense that the equation for x should be defined in terms of both variables. When they say "x varies inversely as y^2 when z is constant", they don't mean that z is the proportionality constant, they mean that x and y^2 are inversely proportional if the value of the variable z is held constant.

A simple real world example: From Wikipedia: "Boyle's law describes the inversely proportional relationship between the absolute pressure and volume of a gas, if the temperature is kept constant within a closed system." If pressure and volume are inversely proportional, that means that if you double the pressure, the volume gets cut in half, but if the temperature is all over the place, that won't be true, hence the specification that temperature is held constant. Same idea here.

GmatMathPro, thanks for intro with Boyle's however this question specifies different conditions
two functions are given with three values, two are variable and one is const.

you have made all three variables in one function

GmatMathPro wrote:When they say "x varies inversely as y^2 when z is constant", they don't mean that z is the proportionality constant, they mean that x and y^2 are inversely proportional if the value of the variable z is held constant.

please explain how the value of a variable can he held constant? The very inherent nature of variability is to be not constant.

pemdas wrote:GmatMathPro, thanks for intro with Boyle's however this question specifies different conditions
two functions are given with three values, two are variable and one is const.

you have made all three variables in one function

But doesn't it bother you at all that none of your data even comes close to satisfying your original equations?

You say x=z/y^2. The given data is x=16, y=2, z=4. Plugging this into your equation, you get 16=4/2^2 or 16=1. Then at the end you say when y=4 and z=25, then x=256. Plugging this in to this equation gives you 256=25/16.

Your other equation is x=y√z. Plugging in the first set of data gives 16=2√4 or 16=4. Plugging in your second set of data gives 256=4√25 or 256=20.

Meanwhile, my equation, x=32√z/y^2, for x=16, y=2, z=4 gives 16=32√4/2^2 or 16=16. My second set of data is x=10, y=4, z=25. Plugging in to my equation gives 10=32√25/4^2 or 10=10.

Also, it doesn't say that z is a constant, it says "when z is constant", and it doesn't say that y is a constant, it says "when y is constant". That is, y and z are variables, but certain relationships are true when each of these variables is held constant. For instance, let's say z is held constant at some specific value, then in my equation, x=32√z/y^2, 32√z will always be the same number, because we are holding z constant. Let's call that number A. So A=32√z, and the equation becomes x=A/y^2. In words, when is z held constant, x varies inversely with y^2. Now, when y is held constant at some specific value, 32/y^2 will always be the same number. Let's call it B. So the equation becomes x=B√z. In words, when y is held constant, x varies directly as the square root of z.

pemdas wrote: please explain how the value of a variable can he held constant? The very inherent nature of variability is to be not constant.

You can hold a variable constant if you assign it a value and only consider cases where it is equal to exactly that value. In this case, my equation is x=32√z/y^2. So I can say, "okay, let's see what happens if y is always 2." In that case the equation reduces to x=8√z, and I can explore what kind of relationship x and z has if y is held at this (or any) constant value.

cleared. Thanks Pete!