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

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In a game of Numberball, teams can score in increments of 3, 4, or 7  [#permalink]

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New post Updated on: 22 Jul 2018, 18:03
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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

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

Originally posted by Newman2019 on 22 Jul 2018, 13:52.
Last edited by chetan2u on 22 Jul 2018, 18:03, edited 1 time in total.
Added the topic name and formatted the question
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In a game of Numberball, teams can score in increments of 3, 4, or 7  [#permalink]

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New post 22 Jul 2018, 17:42
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1
Newman2019 wrote:
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
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Re: In a game of Numberball, teams can score in increments of 3, 4, or 7  [#permalink]

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New post 23 Jul 2018, 00:56
1
Newman2019 wrote:
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?


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 1591 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 the game of Numberball, teams can score in increments of 3, 4, or 7  [#permalink]

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New post 12 Aug 2018, 15:54
Hello Everyone! This my first post. This question came up on my last practice exam and I am having trouble wrapping my head around it. It is 700-800 level, take a shot and explain please! :cool:

In the 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 seven-point baskets, then what is the range of the possible number of four-point baskets made by the winning team?

A. 0 - 2
B. 0 - 4
C. 0 - 9
D. 1 - 4
E. 1 - 9
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Re: In a game of Numberball, teams can score in increments of 3, 4, or 7  [#permalink]

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New post 12 Aug 2018, 23:23
1
onmywaytoboston wrote:
Hello Everyone! This my first post. This question came up on my last practice exam and I am having trouble wrapping my head around it. It is 700-800 level, take a shot and explain please! :cool:

In the 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 seven-point baskets, then what is the range of the possible number of four-point baskets made by the winning team?

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


Please check the discussion above.
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Re: In a game of Numberball, teams can score in increments of 3, 4, or 7 &nbs [#permalink] 12 Aug 2018, 23:23
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