The following is a set of four GMAT Integrated Reasoning MSR questions. There are two information cards in this question. You are allowed to use a calculator. For the set of four, give yourself 10 minutes.
1. Player A initially draws three-of-a-kind and chooses not to discard at all on the discard round. Player B initially draws three-of-a-kind, discards the remaining two cards. For the following final card combinations of Player B, tell whether Player A outscores Player B.
2. Suppose player J draws two pair and chooses not to discard any cards. Suppose player K draws three-of-a-kind and choose not to discard any cards. Which of the following discard choices and final card combination would have a higher point value than player J’s hand but not than player K’s hand?
(A) discard 1 card, three of a kind
(B) discard 2 cards, flush
(C) discard 2 cards, straight
(D) discard 3 cards, full house
(E) discard 4 cards, full house
3. Player G received the following cards on the initial five-card draw: 2 of Clubs, 5 of Diamonds, 8 of Hearts, 8 of Spades, and King of Diamonds. Player G will choose to discard exactly two cards, the 2 of Clubs and 5 of Diamonds. Over all possible hands that could result, what is the range of the possible point values of Player G’s final hand?
4. Suppose player S draws a straight on the first five-card draw and choose not to discard any cards. Player T initially draws the following five cards: 4 of Diamonds, 4 of Hearts, 6 of Hearts, 7 of Hearts, 7 of Clubs. Which of Player T’s discard choices and final results will outscore Player S?
1. Three-of-a-kind = A outscores B
Full House = A does not outscore B
Four-of-a-kind = A does not outscore B
4. first row = T does not outscore S
second row = T outscores S
third row = T outscores S
1) Player A draw three-of-a-kind (worth 60 points), and does not discard, so A’s hand is worth 60 points. In all three cases, B discards two cards, for a reducing factor of 1/6.
(a) B’s final hand = three-of-a-kind, which is (60)*(1/6) = 10 points. A outscores B.
(b) B’s final hand = Full House, which is (720)*(1/6) = 120 points. A does not outscore B.
(c) B’s final hand = Four-of-a-kind, which is (3600)*(1/6) = 600 points. A does not outscore B.
It wasn’t relevant in any of these scenarios, but notice the exact wording — if A and B were tied, the answer would be “A does not outscore B.”
2) Player J, with two pair and no discards, has 36 points. Player K, with three-of-a-kind and no discards, has 60 points. We want a combination worth more than 30 points but less than 60 points.
(A) discard 1 card, three of a kind = (60)*(1/2) = 30. Less than 36, no good.
(B) discard 2 cards, flush = (540)*(1/6) = 90. Higher than 60, no good.
(C) discard 2 cards, straight = (300)*(1/6) = 50. This could work.
(D) discard 3 cards, full house = (720)*(1/10) = 72. Higher than 60, no good.
(E) discard 4 cards, full house = (720)*(1/60) = 12. Less than 36, no good.
(C) is the only hand & discard combination that is in the required range.
3) The five original cards in Player G’s hand are: 2 of Clubs, 5 of Diamonds, 8 of Hearts, 8 of Spades, and King of Diamonds. Player G discards the 2 of Clubs and 5 of Diamonds, leaving two 8’s and a king. We know that Player G’s reducing factor for this hand will be 1/6.
The highest possible hand would be four-of-a-kind, in the unlikely scenario that G picked up the other two 8’s. That would result in (3600)*(1/6) = 600 points, the maximum.
The lowest possible hand would be if G picks up garbage and just has the two 8’s. That would result in (12)*(1/6) = 2 points, the minimum.
The range of any set is the max minus the min. Range = 600 – 2 = 598.
Answer = (B)
4) Player S’s hand is worth 300 points. That’s fixed. Now, we have to compare T to that 300 point total. discard 4 of Diamonds & 7 of Clubs, winds up with a flush
Scenario #1: discard 4 of Diamonds & 7 of Clubs, winds up with a flush. Flush is worth 540, times the reducing factor of 1/6, for a point value of 540*(1/6) = 90, which is less than S. Player T does not outscore player S.
Scenario #2: discard 6 of Hearts, winds up with a full house. Full house is worth 720, times reducing factor of 1/2, for a point value of 720*(1/2) = 360, which beats S. T outscores S.
Scenario #3: discards 4 of Diamonds & 4 of Hearts & 6 of Hearts, winds up with four-of-a-kind. Four-of-a-kind is worth 3600, times a reducing factor of (1/10), for a point value of 3600*(1/10) = 360, which beats S. T outscores S.
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