CareerCup

x := Number of traders who deal in both derivative and crude oil but not fx
y := Number of traders who deal in both fx and derivative but not crude oil
z := Number of traders who deal in both fx and crude oil but not derivative

From this info:
36 traders deal in the fx market.
32 traders deal in the derivative market.
39 traders deal in the crude oil market.
37 traders deal in only one market.
8 traders deal in all the three markets.
we can derive Number of traders who deal in two markets:
(36+32+39-8*3-37)/2=23 :=Number of traders who deal in two markets

Now we have three equations:

x=y-1 <= Number of traders who deal in both derivative and crude oil but not fx was one less than the number of traders who deal in both fx and derivative but not crude oil.
z+x-2=2y <= The sum of the number of traders who deal in both fx and crude oil but not in derivative and in both crude oil and derivative but not fx was two greater than twice the number of traders who deal in both fx and derivative but not crude oil.
x+y+z=23 <= 23 is Number of traders who deal in two markets

By solving equations we get:
z=10
So 10 traders deal in both fx and crude oil but not derivative.


Let assume that someone who has solved the first half of their collabaration question got the answer 10(same as you).
So P=(10*x/2)+floor(y/4)=52

After that we have simple linear optimization problem which i solved on python by pulp library:

from pulp import *
xTC = pulp.LpVariable("xTC", 0,cat='Integer')
xTE = pulp.LpVariable("xTE", 0,cat='Integer')
xOC = pulp.LpVariable("xOC", 0,cat='Integer')
xOE = pulp.LpVariable("xOE", 0,cat='Integer')
xMC = pulp.LpVariable("xMC", 0,cat='Integer')
xME = pulp.LpVariable("xME", 0,cat='Integer')
xLC = pulp.LpVariable("xLC", 0,cat='Integer')
xLE = pulp.LpVariable("xLE", 0,cat='Integer')
problem = pulp.LpProblem('0',pulp.LpMaximize)
problem += xLE, "Функция цели"
problem += xLC+xLE >= 50, "1"
problem += xLC+xLE <= 175, "2"
problem += xTC+xTE <= 175, "3"
problem += xTC+xTE >= 50, "4"
problem += xOC+xOE <= 175, "5"
problem += xOC+xOE >= 50, "6"
problem += xMC+xME <= 175, "7"
problem += xMC+xME >= 50, "8"
problem += xOE== xOC*0.45/0.55, "9"
problem += xTE== xTC*0.45/0.55, "10"
problem += xME== xMC*0.45/0.55, "11"
#problem += xLC == xLE, "12"
problem += xTC+xOC+xMC+xLC == 260, "13"
problem += xTC+xOC+xMC+xLC+xTE+xOE+xME+xLE == 500, "14"
problem.solve()

problem2 = pulp.LpProblem('0',pulp.LpMinimize)
problem2 += xLE, "Функция цели"
problem2 += xLC+xLE >= 50, "1"
problem2 += xLC+xLE <= 175, "2"
problem2 += xTC+xTE <= 175, "3"
problem2 += xTC+xTE >= 50, "4"
problem2 += xOC+xOE <= 175, "5"
problem2 += xOC+xOE >= 50, "6"
problem2 += xMC+xME <= 175, "7"
problem2 += xMC+xME >= 50, "8"
problem2 += xOE== xOC*0.45/0.55, "9"
problem2 += xTE== xTC*0.45/0.55, "10"
problem2 += xME== xMC*0.45/0.55, "11"
problem2 += xTC+xOC+xMC+xLC == 260, "12"
problem2 += xTC+xOC+xMC+xLC+xTE+xOE+xME+xLE == 500, "13"
problem2.solve()

print ("Результат:")
s1=[]
s2=[]

for variable in problem.variables():
    print (variable.name, "=", variable.varValue)
    s1.append(variable.varValue)
print(s1[0]/s1[1])
print ('m>n' if s1[0]/s1[1] < 0 else 'm<n')

for variable in problem2.variables():
    print (variable.name, "=", variable.varValue)
    s2.append(variable.varValue)
print(s2[0]/s2[1])
print ('m>n' if s2[0]/s2[1] < 0 else 'm<n')

x := Number of traders who deal in both derivative and crude oil but not fx
y := Number of traders who deal in both fx and derivative but not crude oil
z := Number of traders who deal in both fx and crude oil but not derivative

From this info:
36 traders deal in the fx market.
32 traders deal in the derivative market.
39 traders deal in the crude oil market.
37 traders deal in only one market.
8 traders deal in all the three markets.
we can derive Number of traders who deal in two markets:
(36+32+39-8*3-37)/2=23 :=Number of traders who deal in two markets

Now we have three equations:

x=y-1 <= Number of traders who deal in both derivative and crude oil but not fx was one less than the number of traders who deal in both fx and derivative but not crude oil.
z+x-2=2y <= The sum of the number of traders who deal in both fx and crude oil but not in derivative and in both crude oil and derivative but not fx was two greater than twice the number of traders who deal in both fx and derivative but not crude oil.
x+y+z=23 <= 23 is Number of traders who deal in two markets

By solving equations we get:
z=10
So 10 traders deal in both fx and crude oil but not derivative.


Let assume that someone who has solved the first half of their collabaration question got the answer 10(same as you).
So P=(10*x/2)+floor(y/4)=52, as long as x=10 and y=10

After that we have simple linear optimization problem which i solved on python by pulp library:

from pulp import *
xTC = pulp.LpVariable("xTC", 0,cat='Integer')
xTE = pulp.LpVariable("xTE", 0,cat='Integer')
xOC = pulp.LpVariable("xOC", 0,cat='Integer')
xOE = pulp.LpVariable("xOE", 0,cat='Integer')
xMC = pulp.LpVariable("xMC", 0,cat='Integer')
xME = pulp.LpVariable("xME", 0,cat='Integer')
xLC = pulp.LpVariable("xLC", 0,cat='Integer')
xLE = pulp.LpVariable("xLE", 0,cat='Integer')
problem = pulp.LpProblem('0',pulp.LpMaximize)
problem += xLE, "Функция цели"
problem += xLC+xLE >= 50, "1"
problem += xLC+xLE <= 175, "2"
problem += xTC+xTE <= 175, "3"
problem += xTC+xTE >= 50, "4"
problem += xOC+xOE <= 175, "5"
problem += xOC+xOE >= 50, "6"
problem += xMC+xME <= 175, "7"
problem += xMC+xME >= 50, "8"
problem += xOE== xOC*0.45/0.55, "9"
problem += xTE== xTC*0.45/0.55, "10"
problem += xME== xMC*0.45/0.55, "11"
problem += xTC+xOC+xMC+xLC == 260, "12"
problem += xTC+xOC+xMC+xLC+xTE+xOE+xME+xLE == 500, "13"
problem.solve()

problem2 = pulp.LpProblem('0',pulp.LpMinimize)
problem2 += xLE, "Функция цели"
problem2 += xLC+xLE >= 50, "1"
problem2 += xLC+xLE <= 175, "2"
problem2 += xTC+xTE <= 175, "3"
problem2 += xTC+xTE >= 50, "4"
problem2 += xOC+xOE <= 175, "5"
problem2 += xOC+xOE >= 50, "6"
problem2 += xMC+xME <= 175, "7"
problem2 += xMC+xME >= 50, "8"
problem2 += xOE== xOC*0.45/0.55, "9"
problem2 += xTE== xTC*0.45/0.55, "10"
problem2 += xME== xMC*0.45/0.55, "11"
problem2 += xTC+xOC+xMC+xLC == 260, "12"
problem2 += xTC+xOC+xMC+xLC+xTE+xOE+xME+xLE == 500, "13"
problem2.solve()

print ("Result:")
s1=[]
s2=[]

for variable in problem.variables():
    print (variable.name, "=", variable.varValue)
    s1.append(variable.varValue)
print(s1[0]/s1[1])
print ('m>n' if s1[0]/s1[1] < 0 else 'm<n')

for variable in problem2.variables():
    print (variable.name, "=", variable.varValue)
    s2.append(variable.varValue)
print(s2[0]/s2[1])
print ('m>n' if s2[0]/s2[1] < 0 else 'm<n')

which give as answer like this
Result:
xLC = 18.0
xLE = 42.0
xMC = 66.0
xME = 54.0
xOC = 88.0
xOE = 72.0
xTC = 88.0
xTE = 72.0
0.42857142857142855
m<n

P.S. i'm just 2 weeks into programming so my code is far away from ideal, it's given just to solve the question asked.

for first question:

only Fx traders + only Derivative traders + only crude oil traders = 37

fx traders = 36
derivative traders = 32
crude oil traders = 39

8 traders deal in all three.
===============================

traders deal only in Fx + another market  = 36-8 = 26
traders deal only in Derivative + another market = 32-8 = 24
traders deal only in Crude Oil + another market = 39-8 = 31


fx			derivative			crude
|				|				|
|				|				|
------------------------------------------------------
		|			|
		X+1			X
		
fx			derivative			crude
|				|				|
|				|				|
------------------------------------------------------
		|		|				|	
		Y		|				|
				---------------------------
						|
						X
						
X+Y = 2(X+1)-2
Y=X

Thus 

(X+1)+X+X = 26+24+31-37
3X+1 = 46
X = 15

Since X = Y, number of traders deal with Fx and Crude Oil is 15.