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805+ Level|   Algebra|   Min-Max Problems|      
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Newman2019
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In a game of Numberball, teams can score in increments of 3, 4, or 7 Updated on: 22 Jul 2018, 19:03
Originally posted by Newman2019 on 22 Jul 2018, 14:52.
Last edited by chetan2u on 22 Jul 2018, 19:03, edited 1 time in total.
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In a game of Numberball, teams can score in increments of 3, 4, or 7 points at a time. If the final score of a game was 39-34, and the majority of the total points were scored by way of 7 point baskets, then what is the range of the possible number of 4 point baskets made by the winning team?"

A) 0-2
B)0-4
C)0-9
D)1-4
E) 1-9

Show Spoiler
How can you use algebra to solve this problem without working backwards by plugging in the answer choices?
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B
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In a game of Numberball, teams can score in increments of 3, 4, or 7 23 Jul 2018, 01:56
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Newman2019
In a game of Numberball, teams can score in increments of 3, 4, or 7 points at a time. If the final score of a game was 39-34, and the majority of the total points were scored by way of 7 point baskets, then what is the range of the possible number of 4 point baskets made by the winning team?"

A) 0-2
B)0-4
C)0-9
D)1-4
E) 1-9

Show Spoiler
How can you use algebra to solve this problem without working backwards by plugging in the answer choices?
Show more

We are told that the majority of points are scored by 7 point baskets.
For this to happen the total points scored by 7 points baskets need to be greater than \(\frac{39+34}{2}\) or \(36.5\)
This is only possible for a minimum of six 7 point baskets or 42 points(shot by 7-point baskets)

Maximum 4 point baskets by winning team
We need a minimum of 7 point baskets. This will enable the winning team to have the maximum
number of 4 point baskets. The maximum 7 point baskets the losing team can have is 4(giving it
28 of the total 34 points). Now, we have 14 points(two 7 point baskets) of the 39 points. For the
remaining 25 points, we can have 3x + 4y = 25. Testing values - y = 1,x = 7 | y = 4,x = 3.

Minimum 4 point baskets by winning team
We need a maximum of 7 point baskets. This will enable the winning team to have the minimum
number of 4 point baskets. The minimum 7 point baskets the losing team can have is 2(giving it
14 of the total 34 points). Now, we can have 28 points(four 7 point baskets) of the 39 points. For
the remaining 7 points, we have one 3-point basket and one 4-point basket.

Let's try three 7-point baskets for both the team. That way 21 points of 39 points for the winning
team will be via 7-point baskets. The remaining 18 points can be scored by six 3-point baskets,
making the total number of 4-point baskets ZERO(which is the minimum 4-point baskets)

Attachment:
Point.JPG
Point.JPG [ 33.99 KiB | Viewed 12830 times ]

Therefore, the total range of the 4-point baskets possible made by the winning team is 0-4(Option B)
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In a game of Numberball, teams can score in increments of 3, 4, or 7 22 Jul 2018, 18:42
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Newman2019
Hi Everyone,

"In a game of Numberball, teams can score in increments of 3, 4, or 7 points at a time. If the final score of a game was 39-34, and the majority of the total points were scored by way of 7 point baskets, then what is the range of the possible number of 4 point baskets made by the winning team?"

A) 0-2
B)0-4
C)0-9
D)1-4
E) 1-9

How can you use algebra to solve this problem without working backwards by plugging in the answer choices?
Show more


Majority of (39+34=73) are scored from 7 pointers..
So >73/2 ~>36.5 are scored by 7 pointers...
42 is next greater multiple of 7, so AT LEAST 42 out of 73 are through 7 points..

Let's find the range of winning team ...
1) Max points by 4...
Now the team could have scored at the max 28, which is 4*7 and remaining 6 could be 2*3 pointers..
34=7*4+3*2
So remaining points from 7-pointers that is 42-28=14 must be from the winning team
7x+4y+3z=39.....
We have to maximize y...
If we take X as minimum possible that is 2, 14+4y+3z=39
4y+3z=25.......
\(y=\frac{25-3z}{4}\)
What is the least value of z such that 25-3z is multiple of 4...
It is 3, so y=(25-3*3)/4=16/4=4

2) min points
39..
7x+4y+3z=39
Now max value of 7 is 5..
7*5+4y+3z=39......4y+3z=4, so y is 1
But if we take 7 pointers as 4
7*4+4y+3z=39.....4y+3z=39-28=11, so y=2 and z=1
But we are looking for min value of y
7*3+4y+3z=39.....4y+3z=18..so y can be 0 when z is 6

So range is 0 to 4

B
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In a game of Numberball, teams can score in increments of 3, 4, or 7 03 Oct 2021, 09:52
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pushpitkc
Newman2019
In a game of Numberball, teams can score in increments of 3, 4, or 7 points at a time. If the final score of a game was 39-34, and the majority of the total points were scored by way of 7 point baskets, then what is the range of the possible number of 4 point baskets made by the winning team?"

A) 0-2
B)0-4
C)0-9
D)1-4
E) 1-9

Show Spoiler
How can you use algebra to solve this problem without working backwards by plugging in the answer choices?

We are told that the majority of points are scored by 7 point baskets.
For this to happen the total points scored by 7 points baskets need to be greater than \(\frac{39+34}{2}\) or \(36.5\)
This is only possible for a minimum of six 7 point baskets or 42 points(shot by 7-point baskets)

Maximum 4 point baskets by winning team
We need a minimum of 7 point baskets. This will enable the winning team to have the maximum
number of 4 point baskets. The maximum 7 point baskets the losing team can have is 4(giving it
28 of the total 34 points). Now, we have 14 points(two 7 point baskets) of the 39 points. For the
remaining 25 points, we can have 3x + 4y = 25. Testing values - y = 1,x = 7 | y = 4,x = 3.

Minimum 4 point baskets by winning team
We need a maximum of 7 point baskets. This will enable the winning team to have the minimum
number of 4 point baskets. The minimum 7 point baskets the losing team can have is 2(giving it
14 of the total 34 points). Now, we can have 28 points(four 7 point baskets) of the 39 points. For
the remaining 7 points, we have one 3-point basket and one 4-point basket.

Let's try three 7-point baskets for both the team. That way 21 points of 39 points for the winning
team will be via 7-point baskets. The remaining 18 points can be scored by six 3-point baskets,
making the total number of 4-point baskets ZERO(which is the minimum 4-point baskets)

Attachment:
Point.JPG

Therefore, the total range of the 4-point baskets possible made by the winning team is 0-4(Option B)
Show more
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 :(

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In a game of Numberball, teams can score in increments of 3, 4, or 7 06 Oct 2021, 21:13
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(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

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In a game of Numberball, teams can score in increments of 3, 4, or 7 04 Dec 2024, 08:49
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"majority of the total points were scored by way of 7 point baskets". By both teams? By a single team?

No way this comes on the actual exam. Too lengthy and ambiguous.
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In a game of Numberball, teams can score in increments of 3, 4, or 7 23 Dec 2024, 19:12
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chetan2u
Newman2019
Hi Everyone,

"In a game of Numberball, teams can score in increments of 3, 4, or 7 points at a time. If the final score of a game was 39-34, and the majority of the total points were scored by way of 7 point baskets, then what is the range of the possible number of 4 point baskets made by the winning team?"

A) 0-2
B)0-4
C)0-9
D)1-4
E) 1-9

How can you use algebra to solve this problem without working backwards by plugging in the answer choices?


Majority of (39+34=73) are scored from 7 pointers..
So >73/2 ~>36.5 are scored by 7 pointers...
42 is next greater multiple of 7, so AT LEAST 42 out of 73 are through 7 points..

Let's find the range of winning team ...
1) Max points by 4...
Now the team could have scored at the max 28, which is 4*7 and remaining 6 could be 2*3 pointers..
34=7*4+3*2
So remaining points from 7-pointers that is 42-28=14 must be from the winning team
7x+4y+3z=39.....
We have to maximize y...
If we take X as minimum possible that is 2, 14+4y+3z=39
4y+3z=25.......
\(y=\frac{25-3z}{4}\)
What is the least value of z such that 25-3z is multiple of 4...
It is 3, so y=(25-3*3)/4=16/4=4

2) min points
39..
7x+4y+3z=39
Now max value of 7 is 5..
7*5+4y+3z=39......4y+3z=4, so y is 1
But if we take 7 pointers as 4
7*4+4y+3z=39.....4y+3z=39-28=11, so y=2 and z=1
But we are looking for min value of y
7*3+4y+3z=39.....4y+3z=18..so y can be 0 when z is 6

So range is 0 to 4

B
Show more
When majority is by 7 pointers. Cant it be the case that 7:4:3 are in the ratio 40:30:30 percents
Here it may work, but it doesnt always mean more than half rght?
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