Transfer Game - advanced version

Algebra Level 4

Four persons A A , B B , C C , and D D have a total sum of $ 1460 \$1460 . They play a game of cyclic transfer. A A goes first, and gives 20 % 20\% of what he has to B B . Next, B B gives 20 % 20\% of his new amount to C C . Next C C gives 20 % 20 \% of his new amount to player D D , and finally, D D gives 20 % 20\% of his new amount to player A A . At the end of all this, the four players check the amounts they have. They find that players A A and D D have switched their initial amounts, while players B B and C C ended up with the same amounts they started with. What was the amount that player B B had initially?


The answer is 320.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Chew-Seong Cheong
Sep 12, 2020

Let the amounts the four players have initially and after the game be a 0 a_0 , b 0 b_0 , c 0 c_0 , and d 0 d_0 ; and a 1 a_1 , b 1 b_1 , c 1 c_1 , and d 1 d_1 respectively. Then a 0 + b 0 + c 0 + d 0 = 1460 a_0 + b_0 + c_0 + d_0 = 1460 and

{ b 1 = 0.8 ( 0.2 a 0 + b 0 ) c 1 = 0.8 ( 0.04 a 0 + 0.2 b 0 + c 0 ) d 1 = 0.8 ( 0.008 a 0 + 0.04 b 0 + 0.2 c 0 + d 0 ) a 1 = 0.8 a 0 + 0.2 ( 0.008 a 0 + 0.04 b 0 + 0.2 c 0 + d 0 ) \begin{cases} b_1 & = 0.8 (0.2 a_0 + b_0) \\ c_1 & = 0.8 (0.04 a_0 + 0.2 b_0 + c_0) \\ d_1 & = 0.8 (0.008 a_0 + 0.04 b_0 + 0.2 c_0 + d_0) \\ a_1 & = 0.8a_0 + 0.2 (0.008 a_0 + 0.04 b_0 + 0.2 c_0 + d_0) \end{cases}

Since

a 1 = d 0 ⟹ 0.8016 a 0 + 0.008 b 0 + 0.04 c 0 − 0.8 d 0 = 0 . . . ( 1 ) d 1 = a 0 ⟹ − 0.9936 a 0 + 0.032 b 0 + 0.16 c 0 + 0.8 d 0 = 0 . . . ( 2 ) b 1 = c 1 ⟹ − 0.128 a 0 − 0.64 b 0 + 0.8 c 0 = 0 . . . ( 3 ) ( 1 ) + ( 2 ) : ⟹ − 0.192 a 0 + 0.04 b 0 + 0.2 c 0 = 0 . . . ( 4 ) \begin{array} {rll} a_1 = d_0 & \implies 0.8016a_0+0.008b_0+0.04c_0-0.8d_0 = 0 & ...(1) \\ d_1 = a_0 & \implies -0.9936a_0+0.032b_0+0.16c_0+0.8d_0 = 0 & ...(2) \\ b_1 = c_1 & \implies -0.128a_0-0.64b_0+0.8c_0 = 0 & ...(3) \\ (1)+(2): & \implies -0.192a_0+0.04b_0+0.2c_0 = 0 & ...(4) \end{array}

( 3 ) − 4 × ( 4 ) : 0.64 a 0 − 0.8 b 0 = 0 ⟹ b 0 = 0.8 a 0 ( 3 ) + 16 × ( 4 ) : − 0.32 a 0 + 4 b 0 = 0 ⟹ c 0 = 0.8 a 0 ( 1 ) : 0.84 a 0 − 0.8 d 0 = 0 ⟹ d 0 = 1.05 a 0 \begin{array} {rll} (3)-4\times (4): & 0.64a_0 - 0.8b_0 = 0 & \implies b_0 = 0.8a_0 \\ (3)+16\times (4): & -0.32 a_0 + 4 b_0 = 0 & \implies c_0 = 0.8a_0 \\ (1): & 0.84a_0-0.8d_0 = 0 & \implies d_0 = 1.05a_0 \end{array}

Therefore, a 0 + b 0 + c 0 + d 0 = 3.65 a 0 = 1460 ⟹ a 0 = 400 ⟹ b 0 = 0.8 a 0 = 320 a_0 + b_0 + c_0 + d_0 = 3.65a_0 = 1460 \implies a_0 = 400 \implies b_0 = 0.8 a_0 = \boxed{320} .

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