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My first instinct was to set up a system of equations and solve for the variables. But I quickly noticed that there were 6 unknowns (2 for each person; i.e., each collected bottles and cans). I needed 6 equations! This prospect didn't appeal to me.
So, I started a table:
"...Penny collected three times as many cans as Quincy..." => P_c = 3a and Q_c = a
"...Quincy had collected four times as many bottles as Rosa..." => Q_b = 4b and R_b = b
"...each had collected exactly the same number of items..." => Q_c + Q_b = a + 4b => Penny and Rosa also collected this many items:
P_b = (a + 4b) - 3a = 4b - 2a and R_c = (a + 4b) - b = a + 3b
So now I have all the items counted but only two unknowns!
"...the three as a group had collected as many cans as bottles..."
=>
P_c + Q_c + R_c = P_b + Q_b + R_b =>
3a + a + (a + 3b) = (4b - 2a) + 4b + b =>
5a + 3b = 9b - 2a =>
7a = 6b
At this point, the solution is obvious: a = 6, b = 7 => 42 = 42 where 42 is the least common multiple of 6 and 7. Another possible solution is a = 12, b = 14 which comes from doubling the least common multiple. If we substitute these values into 5a + 3b, we obtain 5(12) + 3(14) = 102 cans and bottles or 204 total items. However, we are told that less than 200 total items were collected, so a = 12, b = 14 is not a solution in this case.
P_c = 3(6) = 18
P_b = 4(7) - 2(6) = 16
Q_c = 6
Q_b = 4(7) = 28
R_c = 6 + 3(7) = 27
R_b = 7
So, they each collected 34 items. There were 51 cans and bottles collected, etc.
Thus our solution is a valid solution for the problem.