The surface area of a spill, s, is directly proportional to the time

Question

The surface area of a spill, s, is directly proportional to the time since the spill occured, t, and is inversely proportional to the square of the viscosity of the liquid spilled, v. If the time since a spill occured is tripled and the viscosity of the liquid spilled is halved, by what percentage is the surface area of the spill increased?

A. 75%
B. 900%
C. 1000%
D. 1100%
E. 1200%

A. 75%
B. 900%
C. 1000%
D. 1100%
E. 1200%

Archit3110 · Answer

given that
s= t = 1/v^2
let t=v =1
s = 1
later t = 3 and v = .5
s = 3 /.25 ; 12 
12-1/ 1 ; 1100% 
option D 
 GianKR  what is the source of this question?

GianKR · Answer

Bunuel   KarishmaB  I kindly need your help with the following:

I undestand that:
a is directly proportional to b, translates to a=kb
p is inversely proportional to q, translates to p=k*/q

But I don't quite understand why:
x is directly proportional to y and inversely proportional to z, translates to x=ky/z?

If I were to translate it, I would do as following: 
x=ky
x=k*/z

Can you please help clarify why we merged the two relationships into one equation " x=ky/z " ?

Thank you in advance!

GianKR · Answer

given that
s= t = 1/v^2
let t=v =1
s = 1
later t = 3 and v = .5
s = 3 /.25 ; 12 
12-1/ 1 ; 1100% 
option D

Hello  Archit3110  can you please help me understand why you set s= t = 1/v^2?
Isn't s supposed to = kt = k*/v^2?
Have you set the proportionality constants as 1?
And again the same question I am asking above, why s= kt/v^2?

Thank you!

Archit3110 · Answer

k is the proportionality constant which is why s = kt/v^2 ; basically k is a constant so its best to take it as 1 unless its defined in question..

Isn't s supposed to = kt = k*/v^2?  no ; because we have a given relation s is directly and inversely proportional to values... 
hope this helps..

KarishmaB · Answer

There is an unsaid constraint here:

x is directly proportional to y means x/y = constant (provided z is constant)

and x is inversely proportional to z means xz = constant (provided y is constant)

When we talk about joint variation i.e. how do all three vary with each other, both above relations should be satisfied.

i.e. xz/y = constant

Now hold z constant so you get x/y = constant

If you hold y constant, you get xz = constant

SaquibHGMATWhiz · Answer

Solution:

According to the question, s is  directly proportional to t  and  inversely proportional to the square of v  
 Thus, we can form the equation  s∝\frac{t}{v^2}  or  s=k	imes \frac{t}{v^2}  where k is a constant

So, initially  s_1=k	imes \frac{t}{v^2}  
 After changes,  s_2=k	imes \frac{3t}{(\frac{v}{2})^2}=k	imes \frac{3t}{\frac{v^2}{4}}=12	imes k	imes \frac{t}{v^2}=12	imes s_1

Percentage increase  =\frac{s_2-s_1}{s_1} 	imes 100=\frac{12s_1-s_1}{s_1} 	imes 100=1100\%

Hence the right answer is    Option D

Bunuel · Answer

NEYR0N · Answer

Bunuel , based on the image sent, I used test cases for the variables given. I found that s increases by a factor of 3, while with viscosity it increases by a factor of 2 .. However, if I multiply the factors (3*2 = 6 ) to account for total change the percentage difference is wrong.. what am I doing here that is not correct?

I am assuming halving a square is not the same as halving a non square ?

Krunaal · Answer

Firstly, there's always a proportionality constant when you equate. Here, since  s∝t , and  s∝\frac{1}{v^2}  =>  s∝\frac{t}{v^2}  and  s = k*\frac{t}{v^2} ; where k is the proportionality constant.

Secondly, when you halve the viscosity of a liquid from  v  to new viscosity,  v_n  =  \frac{v}{2} , squaring the new viscosity gives  (v_n)^2  =  \frac{v^2 }{ 4}

So the new surface area will be  s_n = k*\frac{3t}{v^2/4}  =>  k*\frac{12t}{v^2}

So, the two mistakes in your approach were ignoring proportionality constant, and taking half of the square of viscosity rather than of viscosity.

The surface area of a spill, s, is directly proportional to the time

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The surface area of a spill, s, is directly proportional to the time

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