Find the Factor for Change in y Due to x & z

[kee23]

Homework Statement

Given that y is directly proportional to the cube of x and inversly proportional to the square of z, if x decreases by 23.3% and z increases by 45.3%, by what factor does y change?

Homework Equations

Couldn't make a equation but I tried like this:
y=K/x^3+1/z^2

The Attempt at a Solution

I couldn't get the question and where to start adding variables into the equation.
I think the equation is wrong to me though. Someone help?
(am I in the right thread for this question?)
Also my English is sort of weak so please make statement simple as possible. :)

 

[rock.freak667]

If c is proportional to a, and proportional to b, then

c∝a and c∝b ⇒ c∝ab


Try remaking your equation now.

 

[Mark44]

Start with y = Kx³/z².

 

[Rohit Mallya]

Is the answer 21.37%?

 

[kee23]

umm so I did it as y=Kx³/z²
and so I got the value : y=K12649/2052.09.
So do I replace the "y" with -23.3% and +45.3% then solve for "K"?

 

[kee23]

Is the answer 21.37%?

umm I think so because the answer they left was in decimals so real answer is 0.214

I don't what they're saying from the question "if x decreases by 23.3% and z increases by 45.3%, by what factor does y change?"

 

[Rohit Mallya]

umm I think so because the answer they left was in decimals so real answer is 0.214

I don't what they're saying from the question "if x decreases by 23.3% and z increases by 45.3%, by what factor does y change?"

x decreases BY 23.3% means x=100-23.3=76.6% So you have to sub. x=76.6. But if they had given decreases to 23.3% then x=23.3
z=145.5

substituting this in the equation with k=1 we get x=21.37 in the answer which you have mentioned they have divided it by 100...

 

[kee23]

hmmm I don't know how to get rid of "k" when I have y=K76.7³/145.5²

 

[Mark44]

Look at the ratio of the new y value to the old y value. In the ratio, the K's cancel, as do the x's and z's.

BTW, 100-23.3$\neq$76.6.

 

[Rohit Mallya]

Oops...Sorry Sir...
its 76.7...

 

[Rohit Mallya]

yeah...Mark Sir is correct...

Eqn(1) y=kx³/y²

Eqn(2) y'=k(76.7)³/(145.5)²

Eqn(2)/Eqn(1)=> y'=21.37y

 

[kee23]

I think I got it. Thanks for help people :D