Anthony, Bobby and Carol had some money. Bobby gave $190 to Carol. He then received 3/13 of Anthony's money. Carol then received 20% of Anthony's money. Finally, 1/7 of Anthony's money was given to Bobby Each of them had $1200 in the end. How much did Bobby and Carol have altogether at first?

Question

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Answer 1

Let B be the initial amount of money Bobby had and C be the initial amount of money Carol had.
After receiving $190 from Bobby, Carol has C + $190.
After receiving 3/13 of Anthony's money, Bobby has B + (3/13)A.
After receiving 20% of Anthony's money, Carol has C + $190 + (0.2)A.
After receiving 1/7 of Anthony's money, Bobby has B + (3/13)A + (1/7)A.
Since each of them had $1200 in the end, we can write the following equations:
B + (3/13)A + (1/7)A = $1200,
C + $190 + (0.2)A = $1200.
We can subtract $190 from both sides of the second equation to get:
C + (0.2)A = $1010.
We can then rewrite the equations as:
B + (3/13)A + (1/7)A = $1200,
C + (0.2)A = $1010.
Since we have two equations and two unknowns, we can solve this system of equations.
To eliminate the denominators in the first equation, we can multiply it by 91 (13*7):
91B + 21A + 13A = 91*$1200,
91B + 34A = 91*$1200,
91B + 34A = $109200.
We can then solve the system of equations:
91B + 34A = $109200,
C + (0.2)A = $1010.
Now, we can try different values of A and B until we find a solution that satisfies both equations. Let's start with A = 1000 and B = 200. Plugging these values into the first equation:
91(200) + 34(1000) = $109200,
18200 + 34000 = $109200,
52200 = $109200.
The first equation is true.
Plugging A = 1000 into the second equation, we get:
C + (0.2)(1000) = $1010,
C + 200 = $1010,
C = $810.
Thus, Bobby and Carol had a total of B + C = 200 + 810 = $<<200+810=1010>>1010 at first. Answer: \boxed{1010}.