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27 Jun 2024, 01:30
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B
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Difficulty:
95%
(hard)
Question Stats:
30% (02:53) correct
70%
(02:20)
wrong
based on 280
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There are three different ways to score in basketball. A successful shot from beyond an arc drawn on the court is worth 3 points. Shots made from inside the arc are worth 2 points. If a player is fouled by another player, the fouled player may shoot "free throws," which are worth 1 point each. A certain team scored 45 points in a game. The team made twice as many 2-point shots as 3-point shots. How many free throws did the team make?
(1) The team attempted 12 free throws.
(2) Five-sixths of the team's free-throw attempts were successful.
(1) The team attempted 12 free throws.
(2) Five-sixths of the team's free-throw attempts were successful.
27 Jun 2024, 05:34
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Let the number of one-pointers = \(x\), the number of two-pointers = \(y\) and the number of three-pointers = \(z\)
\(x+2y+3z = 45\)
We are told "The team made twice as many 2-point shots as 3-point shots". That means \(y = 2z\). Plugging that back in:
\(x+2(2z)+3z = 45\)
\(x+7z = 45\)
(1) The team attempted 12 free throws.
The team could have missed all of them, made all of them or made any number inbetween.
INSUFFICIENT
(2) Five-sixths of the team's free-throw attempts were successful.
This tells us that the number of freethrows that went in is a multiple of 5. This means that the value's unit digit will be either 0 or 5.
Looking at \(x+7z = 45\), as \(x\) will end in either 0 or 5, \(7z\) will have to end in the opposite value to ensure we have a units digit of 5.
The first number where \(7z\) will end in 0, is when \(z\) is 10 (\(7(10) = 70\). This obviously cannot hold as this is well above the 45 points scored, and the game does not have negative points. Therefore, \(7z\) needs to end in 5. The first time that occurs is when \(z = 5\) (\(7(5) = 35\)) and the next will be \(z = 15\) (\(7(15) = 105\)), which once again exceeds 45. Therefore \(z = 5\)
\(x+35 = 45\)
\(x = 10\)
SUFFICIENT
ANSWER B
\(x+2y+3z = 45\)
We are told "The team made twice as many 2-point shots as 3-point shots". That means \(y = 2z\). Plugging that back in:
\(x+2(2z)+3z = 45\)
\(x+7z = 45\)
(1) The team attempted 12 free throws.
The team could have missed all of them, made all of them or made any number inbetween.
INSUFFICIENT
(2) Five-sixths of the team's free-throw attempts were successful.
This tells us that the number of freethrows that went in is a multiple of 5. This means that the value's unit digit will be either 0 or 5.
Looking at \(x+7z = 45\), as \(x\) will end in either 0 or 5, \(7z\) will have to end in the opposite value to ensure we have a units digit of 5.
The first number where \(7z\) will end in 0, is when \(z\) is 10 (\(7(10) = 70\). This obviously cannot hold as this is well above the 45 points scored, and the game does not have negative points. Therefore, \(7z\) needs to end in 5. The first time that occurs is when \(z = 5\) (\(7(5) = 35\)) and the next will be \(z = 15\) (\(7(15) = 105\)), which once again exceeds 45. Therefore \(z = 5\)
\(x+35 = 45\)
\(x = 10\)
SUFFICIENT
ANSWER B
General Discussion
27 Jun 2024, 02:11
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2(2x)+3x+f=45
7x+f=45
S1
f<=12 (not every free throw will land a point)
x can be 5 or 6
Insufficient
S2
if 5/6 succesful but 5/6 of what ?
Insufficient
S1+2
5/6 of 12 succesful = 10 points
we also know that x is 5 so 5 3 pointers and 10 2 pointers
Hence C
7x+f=45
S1
f<=12 (not every free throw will land a point)
x can be 5 or 6
Insufficient
S2
if 5/6 succesful but 5/6 of what ?
Insufficient
S1+2
5/6 of 12 succesful = 10 points
we also know that x is 5 so 5 3 pointers and 10 2 pointers
Hence C
