Depth · one concept from first principles · step 2 of 4

One win count sets both players' scores

A box holds $50$ cards, numbered $1$ through $50$ . Emma and Grace play a game: each round, they each draw one card from the box at the same time and compare the numbers. Whoever draws the larger number gets $5$ points, and whoever draws the smaller number gets $2$ points. They play $30$ rounds. For Emma's total score to be at least $20$ points more than Grace's, at least how many times must Emma draw the larger number? (After each round, both cards go back into the box.)

Expressions

Your answer

How to solve

Strategy Change Focus / Count the Complement — Every round Emma wins is a round Grace loses, and the other way around. So one number, Emma's wins, settles both players' scores. Write both scores with it, then set up the inequality for the gap.

1STEP 1

One count decides both

Each round has exactly one winner. If Emma wins x rounds, she loses 30 - x, and Grace is the reverse.

Emma: x wins, 30 - x losses Grace: 30 - x wins, x losses

2STEP 2

Write both scores

Wins earn 5 points and losses earn 2.

Emma = 5x + 2(30 - x) = 3x + 60, Grace = 5(30 - x) + 2x = 150 - 3x

3STEP 3

Set up the gap

Emma's score must be at least Grace's score plus 20.

3x + 60 ≥ (150 - 3x) + 20 → 6x ≥ 110 → x ≥ 55/3 = 18 1/3

4STEP 4

Round up and check

The number of wins is a whole number at least 18 1/3, so it is at least 19. Check 19 and 18.

x = 19: 117 - 93 = 24 ≥ 20, x = 18: 114 - 96 = 18 ＜ 20

Answer

19 times

With 19 wins and 11 losses, Emma scores 95 + 22 = 117 and Grace scores 55 + 38 = 93; the gap is 24. The two totals add to 210 = 30 × 7, as they must, since every round hands out 5 + 2 = 7 points.

Takeaway

When one player's win is always the other's loss, a single count tells you both scores.

One count decides both
Write both scores
Set up the gap
Round up and check

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