Transfer Game - Percentages

Algebra Level 4

Four persons A A , B B , C C , and D D are playing 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. Player A A finds that he has lost 10 % 10 \% of what he had initially. Players B B and C C ended up with the same amounts they started with. What was the percentage of the gain that player D D made compared to his initial amount?

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The answer is 33.33.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Chew-Seong Cheong
Sep 14, 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

{ 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 = 0.9 a 0 ⟹ − 0.0984 a 0 + 0.008 b 0 + 0.04 c 0 − 0.8 d 0 = 0 . . . ( 1 ) b 1 = b 0 ⟹ 0.16 a 0 − 0.2 b 0 = 0 . . . ( 2 ) c 1 = c 0 ⟹ 0.032 a 0 + 0.16 b 0 − 0.2 c 0 = 0 . . . ( 3 ) \begin{array} {lll} a_1 = 0.9a_0 & \implies -0.0984a_0+0.008b_0+0.04c_0-0.8d_0 = 0 & ...(1) \\ b_1 = b_0 & \implies 0.16a_0 -0.2b_0 = 0 & ...(2) \\ c_1 = c_0 & \implies 0.032a_0 + 0.16b_0 -0.2c_0 = 0 & ...(3) \end{array}

( 2 ) : b 0 = 0.16 a 0 0.2 ⟹ b 0 = 0.8 a 0 ( 3 ) : c 0 = ( 0.032 + 0.8 × 0.16 ) a 0 0.2 ⟹ c 0 = 0.8 a 0 ( 1 ) : d 0 = ( − 0.0984 + 0.8 ( 0.008 + 0.04 ) ) a 0 0.8 ⟹ d 0 = 0.3 a 0 \begin{array} {lll} (2): & b_0 = \dfrac {0.16a_0}{0.2} & \implies b_0 = 0.8a_0 \\ (3): & c_0 = \dfrac {(0.032+0.8 \times 0.16)a_0}{0.2} & \implies c_0 = 0.8a_0 \\ (1): & d_0 = \dfrac {(-0.0984+0.8(0.008+0.04))a_0}{0.8} & \implies d_0 = 0.3a_0 \end{array}

Then

d 1 = 0.8 ( 0.008 a 0 + 0.04 b 0 + 0.2 c 0 + d 0 ) = ( 0.0064 + 0.0256 + 0.128 + 0.24 ) a 0 = 0.4 a 0 \begin{aligned} d_1 & = 0.8 (0.008 a_0 + 0.04 b_0 + 0.2 c_0 + d_0) \\ & = (0.0064+0.0256+0.128+0.24)a_0 \\ & = 0.4a_0 \end{aligned}

Therefore d 1 − d 0 d 0 = 0.4 a 0 − 0.3 a 0 0.3 a 0 = 1 3 ≈ 33.3 % \dfrac {d_1-d_0}{d_0} = \dfrac {0.4a_0-0.3a_0}{0.3a_0} = \dfrac 13 \approx \boxed{33.3}\% .

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