Question 1201676
<pre>
If s is inversely proportional to the square root of t , than if s=28 and t=64  , find: 
a) S when t=81
b) t when s=60 

28=k/√64  and therefore k=224

a) I have no difficulty in letter a  (s=224/√81=24.9) but ..

b)  60=224/√t
     t=√(224/60)=1.93
But the answer on my book is  13.94  (letter b ) 
How is letter b  worked out ?

s is inversely proportional to the square root of t, so that gives us: {{{matrix(1,3, s, "=", k/sqrt(t))}}}, so "28 = k/√64" is CORRECT,
and therefore k = 224 (8 * 28)

a) I have no difficulty in letter a  (s=224/√81=24.9) but ..

b) t when s=60 

b)  60 = 224/√t
     t = √(224/60)=1.93 <font color = red><font size = 4><b><==== This is where you're WRONG!</font></font></b>. I think you got a little confused.
You seem to have taken the square root of {{{224/60}}} instead of SQUARING it! You need to APPLY the INVERSE
operation here, and the INVERSE of taking the square root is squaring!

In other words, {{{matrix(1,3, 60, "=", 224/sqrt(t))}}}
              {{{matrix(1,3, 60sqrt(t), "=", 224)}}} ------- Cross-multiplying
                {{{matrix(1,3, sqrt(t), "=", 224/60)}}}
             {{{matrix(1,3, (sqrt(t))^2, "=", (224/60)^2)}}} ---- Squaring BOTH sides
                {{{highlight_green(matrix(1,10, t, "=", 13.938, "=", 13.94, "(rounded", to, 2, decimal, "places)"))}}}  

NEVER, EVER, EVER, EVER round prematurely like the other person did! Most times when you do, your answer will be
different from the correct answer! Furthermore, he's still WRONG, since {{{matrix(1,3, 3.73^2 <> "13.93,", nor, 13.94)}}}</pre>