Question 692965
y varies jointly as m and the square of n and inversly as p 
y=15 when m=2, n=1, and p=6. Find y when m=3, n=4 and p=10

<pre>
For all proportion problems, start with this:

Varying           "directly" or product of "jointlys" or 1 if none 
quantity  = k · ----------------------------------------------------------
                inversely variable or product of "inverselys" or 1 if none

In this problem the varying quantity is y.
the "jointlys" are m and n².  We have one inversely, p.  So we have m·n² 
on top and p on the bottom:

y = k·{{{m*n^2/p}}}

>>...y=15 when m=2, n=1, and p=6....<<

Substitute these values:

15 = k·{{{2*1^2/6}}}

15 = k·{{{2*1/6}}}

15 = k·{{{2/6}}}

15 = k·{{{1/3}}}

Multiply both sides by 3

45 = k

Now substitute 45 for k in the first equation:

y = 45·{{{m*n^2/p}}}

>>...Find y when m=3, n=4 and p=10...<<

Substitute those values

y = 45·{{{3*4^2/10}}}

y = 45·{{{3*16/10}}}

y = 216

Edwin</pre>