Game coins in a board game are placed at locations A, B, C, D, and E,

Question

Game coins in a board game are placed at locations A, B, C, D, and E, in the ratio 1:2:3:4:5, respectively. To win the game, a player must acquire at least 50 percent of the coins in each of three or more of the five locations. What minimum percent of the total coins must a player acquire to win the game?

(A) 10%
(B) 20%
(C) 30%
(D) 40%
(E) 50%

(A) 10%
(B) 20%
(C) 30%
(D) 40%
(E) 50%

houston1980 · Answer

1+2+3+4+5 = 15. Choose a convenient number to represent the total. Choose 30.

Therefore location A has 2 coins, B has 4 coins, C has 6 coins, D has 8 coins, and E has 10 coins. 2+4+6+8+10 = 30.

50% of minimum of locations A, B and C = 50% (2+4+6) = 50%(12) = 6

6 is 20% of 30.

6 (20% of total) is minimumof the total coins must a player acquire to win the game?

Hence   B = 20%    is the answer.

Hero8888 · Answer

The answer will change depending on the number of coins and odd/even.

e.g Number of coins 1,2,3,4,5
To win you have to pick min: 1 in A, 1 in B, 2 in C . Total 4 of 15 = 26.6%, wich is closer to Answer C (30%) rather than Answer B (20%).

In each case the number will be in some range. I think that the question should be modified, saying "the number of coins in each slot is even)

houston1980 · Answer

Hello Hero8888,

the question is "What minimum percent of the total coins must a player acquire to win the game?". The question expect us to think of when the  minimum  is possible. The  minimum  will be possible when the coins at each location are  even  in number.

Hero8888 · Answer

Hi houston1980, Thank you for explanation. But the Q askes " ... must a player acquire to win the game'', which has nothing to do with "find possible min". Since there are no conditions for even/odd of coins, both cases should be implied e.g. any DS question

el2010 · Answer

To start, we know we have to have at least 50% of the coins in 3 of the 5 locations. Easiest way to do this is use the ratio given to us (no manipulation is required) and take half of 1, 2, and 3 (first 3 of the 5 ratios). This gives us 1/2, 1, and 3/2, which totals to 3. We know the sum of the ratios equals 15 (1+2+...+5); therefore, 3/15 can be reduced to 1/5, which equates to 20%. Answer choice B.

Hero8888 · Answer

We can not have the half of a coin, as if we couldn't have the half of a human. When you have 3 people/coins, at least 50% of them is 2, but not 1.5. When you have 4 coins/people, indeed you can pick 2 of them to get "at least 50%". I know how to calculate all this stuff, my comment was on the fact that % will be changing due to odd/even factor x of the ratio 1x, 3x, 5x.

zishu912 · Answer

Let's assume we are having 15 (1+2+3+4+5) coins on the board,

To win we need 7.5 coins that is 50 percent of the total coins

Maximum coin on the board is in location E, Hence E needs to acquire 3 more to win the game

Therefore the minimum percent of the total coin, to win the game = 3/15 * 100 = 20%

Note--> Hit kudos if it helps

ScottTargetTestPrep · Answer

We can let the total coins = x + 2x + 3x + 4x + 5x = 15x.

Taking 50% of the smallest number of coins we have:

0.5(x + 2x + 3x) = 0.5x + x + 1.5x = 3x

So the minimum percent of the total coins a player must acquire to win the game is 3x/15x = 1/5 = 0.2 = 20%.

Answer: B

Game coins in a board game are placed at locations A, B, C, D, and E,

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Game coins in a board game are placed at locations A, B, C, D, and E,

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