CPhill

Well...the question is a little unclear....I took "both" to mean the sum...but maybe the question  should be "each".....

No matter....each will have 91  and  both will have 182 total

We want to solve this

61 + 10D  = 82 + 3D     where D is the number of days we are looking for

subtract 3D, 61 from  both sides

7D  = 21    divide both sides  by 7

D  = 3  =   number of days it will take to have the same number of friends

And the number of friends both them will have  is just twice the number of friends that either will have  =  2 [  82 + 3*3]   =  2 [ 82 + 9 ]  =   2 [ 91  ] =    182

Very nice, hectictar.....!!!

We have  C(5,2)   = 10 pairs

They are 

1,2    1,3    1,4   1,5

2,3    2,4    2,5

3,4    3,5    

4,5

The pairs in red have sums < their products

So..the probability is   6 / 10    =  3 / 5

We have that

Δ ADB  ~ Δ AEC      and

Δ ADB  ~ Δ PEB

So

ΔAEC  ~   Δ PEB

AE / EC  = PE / EB

AE / [CP + EP ]    = PE / EB

AE  / [ 10 + 20 ]  = 20 / 30

AE / 30  = 20 / 30  ⇒   AE   = 20

The diagonals bisect each other...therefore......the point where the diagonals intersect  must be the mid-point of these two coordinates  =

  ( [ 1 + 11 ] / 2,  [ -5 + 7 ] / 2 )  =  ( 12 / 2, 2 / 2 )  =  (6,  1 )

(6 , -5)  reflected across the line y = x  ....   reverse the coordinates ⇒  (-5, 6) 

Call the side of the square  = x

Notice that  

 Δ ABE  ~  Δ  EGF

 Δ  EGF ~  Δ  FCD

So

 Δ  ABE ~  Δ FCD 

So

AB / BE   = FC / CD

28 / x  = x / 58   cross-multiply

x^2  = 58 * 28

x^2  = 1624  

So...the area of the square is 1624 units^2

The volume  of the mixture filling the original pan is

9 * 13 * 1   =   117 in^3

1. 5 times this volume   =   175.5 in^3

So

175.5  =  10 * 15   *  depth

175.5 = 150 * depth     divide both sides  by 150

175.5 / 150  = depth  = 1.17 in

The perimeter  =   2 [ W + L ] .....where  W and  L   are the dimensions of the fencing

So we have

250  = 2 [ W + L ]     divide both sides by 2

125  = W + L        subtract L  from both sides

125 - L   = W

So...the area  A   = 

A  = W * L       substitute  125 - L  for W.....and we require that

W* L  >   1000

( 125 - L) * L  > 1000 

125L - L^2  > 1000

Sub  x for L   and see the  graph here :

https://www.desmos.com/calculator/jzrsiji5qi

The possible  lengths, L,  of the fence  =   8.59 ft < L < 116.41 ft