In a game of Numberball, teams can score in increments of 3, 4, or 7

Majority of points must be scored with 7 pointers. and on your second graph number of 7 pointers is 6 and number of 3 pointers is 9. How is 9 bigger tham 6.

Oh wait, does the problem mean that not the number of 7 pointers, but actually the total score by 7 pointers was the biggest? If this is the logic then ok, but the problem is written in ambigous language, which needs very very close attention

Posted from my mobile device

(Step 1). Start with the constraint that the majority of the TOTAL POINTS must be 7 pointers

(39 + 34) = 73 total points

50% of these points ———> 36.5

Thus, if we have 5 of the (7point) baskets, we wouldn’t have a majority of the total points.


Thus, we need a MIN of 6 of the (7point) baskets by the two teams together


Winning team has 39

3a + 4b + 7c = 39


Losing team has 34

3x + 4y + 7z = 34


(Step 2) since the MIN number of 4 point baskets is either None or 1, first let’s see if 0 (4point) baskets is possible


We need to fulfill the condition that at least 6 scores must be (7point) scores

The most the losing team can have is 4 of these (z = 4)

7z = 7(4) = 28 ———> leaves us with 34 - 28 = 6, a multiple of 3

y = 2 ———> 3y = 3(2) = 6

Thus, we can have the losing team score 4 of the 6 minimum (7 point) scores that we need


Winning team would thus need to contribute at least 2 of these (7 point) scores. We can try that first.

3a + 4b + 7(2) = 39

3a + 4b = 25

Can we have b = 0? No, 25 is not a multiple of 3


Next, what if c = 3 of the (7 point baskets)

39 - 7(3) = 39 - 21 = 18 points would remain


3a + 4b = 18

B CAN equal 0, because 18 is a multiple of 3

Final tally:

Winning team: 3 of the (7point) scores and 6 of the (3point) scores. ———> total = 21 + 18 = 39

Losing team: 4 of the (7point) scores and 2 of the (3point) scores ———> total = 28 + 6 = 34

And the majority of the total points scored by both sides is from (7point) scores

21 from winning team and 28 from losing team = 49 points ————> which is greater than 50% of the 73 total points scored

Eliminate D and E

(Step 3) can the winning team score 9 of the (4point) scores such that C is the correct answer?

From above, we already know that the winning team must score at minimum 2 of the (7point) scores

3a + 4b + 7(2) = 39

3a + 4b = 25

We can not make a = 0 since 25 is not a multiple of 4

a = 1? 22 remains

a = 2? 19 remains

a = 3? 16 remains ——-> which IS divisible by 4

In which case we can have 4 of the scores be (4points)

Thus we can eliminate answer A.

If we can not find a scenario in which 9 4 point baskets are scored, then the answer must be B

Since the winning team must, at the very least, cover 2 of the (7point) scores, that means there is only 25 points to distribute among (3point) and (4point) scores

3a + 4b = 25

In such a case, if b = 9, we would be way over the allowable points scored by the winning team. Therefore, 9 can NOT be the MAX

Since we found 4 worked, the answer must be B

B

0 of the (4point) scores is possible by the winning team

4 of the (4point) scores is possible by the subbing team

These are the MIN and MAX

Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.