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Bunuel
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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
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­I think all the details in the given statement is not necessary. Statements alone are necessary.

Statement 1: Total 12 free throw attempts. Some could have been unsuccessful. Not Sufficient

Statement 2: \(\frac{5}{6}\) attempts are successful. Not sufficient

Combined: \(\frac{5}{6}\) of 12 attempts are successful. So the team made 10 successful points out of free throws. Sufficient­
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In my opinion this question is poorly written. The question can be How many points did the team earn from their free throws or How many free throws did the team shoot?
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We all know we have this equation of 7x + y = 45 for the score tally right! (where 7x is coming from is: 2x (2 points) + 1x (3 points) = 7x points)

Then that gives us the following table of possibilities:

7x (All 2 & 3 pointers points)Free Throw PointsTotal
42345
351045
281745
212445
143145
73845
Statement 1: It says, they got 12 free hits, it is still insufficient because the top two rows of the table become the possibilities then (so not enough to assert)

Statement 2: It says 5/6 of the Free Throws were successful, then that means Free throw point has to be a multiple of 5 and then from there we can increase the missed points to fit into the 5/6 ratio ... since only row number 2 satisfies this, then It alone is just sufficient to answer!
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There is no where written that 2 and 3 pointers can't be 0.
Your assumption to this is wrong.
Hence answer should be C

Nidzo
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
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It says "The team made twice as many 2-point shots as 3-point shots", which indicates that the 2-point and 3-point shots has to be a non-zero number!

akhil0699
There is no where written that 2 and 3 pointers can't be 0.
Your assumption to this is wrong.
Hence answer should be C

Nidzo
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
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Hello, in S2: 5/6th of free throw is successful, so shouldn't number of free throws be multiple of 6?. please explain
Nidzo
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
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