
John M. answered 01/18/17
Tutor
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Engineering manager professional, proficient in all levels of Math
- This is the type of problem where you should just start jotting down every fact you can get
- Let A, B, C, D be the amount of money in each of the four envelopes, and A be the one with the largest share
- A+B+C+D =30
- Let A', B' C' and D' be the amount of money in each envelope after the money is removed from envelope A
- A'+B'+C'+D' = 30 {Eqn 1}
- B' = 2B {Eqn 2}
- C' = 2C {Eqn 3}
- D' = 2D {Eqn 4}
- The additional fact is that the amounts are the same (although in different envelopes). So, we can randomly assign these
- B' = C {Eqn 5}
- C' = D {Eqn 6}
- D' = A {Eqn 7}
- A' =B {Eqn 8}
- We now have 8 equations and 8 unknowns, so we can solve the problem uniquely. It will just require a large amount of substituting
- B+ 2B+ 2C' +D' = 30 {by substituting Eqn 8 and 6 into Eqn 1}
- B + 2B + 2(2B) + D' = 30 {by substituting first Eqn 5, then Eqn 2}
- B + 2B + 4B + 2D = 30 {substitute Eqn 4}
- B + 2B + 4B + 2C' = 30 {substitute Eqn 6}
- B + 2B + 4B + 2(2C) = 30 {substitute Eqn 3}
- B + 2B + 4B + 2(2B') = 30 {substitute Eqn 5}
- B + 2B + 4B + 2(2(2B)) = 30 {substitute Eqn 2}
- B + 2B + 4B + 8B =30
- 15B = 30
- B = 2
- Now we can solve for the other envelopes. When B = 2, C =4, D = 8 and A = 16.
- So, $14 is removed from A. Of this, $8 is placed in D, $4 in C and $2 in B. Thus, after the exchange, A' = 2, B' = 4, C' = 8 and D' = 16.