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Going in to the last game of his basketball season, Adrian had average

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Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 07 Jul 2018, 09:51
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Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 07 Jul 2018, 10:35
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Bunuel wrote:
Going in to the last game of his basketball season, Adrian had averaged 24 points per game. In his last game, he scored 50 points, bringing his average to 26 points per game for the season. How many games did Adrian play that season?

A. 12
B. 13
C. 14
D. 15
E. 16


Sum of games played before last game = S
Number of games played before final game= n
S/n = 24 or S=24n

After going to his last game we have
(S+50)/(n+1)=26
Substituting S=24n we get
24n+50=26n+26
2n=24
n=12
Since we denoted "n" as the number of games before final game his total number of games played for the whole season would be - n+1=12+1=13.
Answer B.
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Re: Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 07 Jul 2018, 10:50
Bunuel wrote:
Going in to the last game of his basketball season, Adrian had averaged 24 points per game. In his last game, he scored 50 points, bringing his average to 26 points per game for the season. How many games did Adrian play that season?

A. 12
B. 13
C. 14
D. 15
E. 16


Let the no of games played by Adrian be x before the last game.

Given Adrian had averaged 24 points per game.
So, the total points earned=24x
New average=26 when he played the last game
Total points scored=24x+50
So, \(\frac{24x+50}{x+1}=26\)
Or, 24x+50=26x+26
Or, 2x=24
Or, x=12
Total no of games played= no of games played by Adrian be x [u]before the last game+1=x+1=12+1=13
Ans. (B)
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Re: Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 07 Jul 2018, 10:57
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Bunuel wrote:
Going in to the last game of his basketball season, Adrian had averaged 24 points per game. In his last game, he scored 50 points, bringing his average to 26 points per game for the season. How many games did Adrian play that season?

A. 12
B. 13
C. 14
D. 15
E. 16



Let there be x matches

three ways

1) logical....
Increase in avg of 2 for x games means 2x points
50 points are 24 points more than the new average..
so 2x=24.....x=12
and the 13th is the one in which 50 points were scored
ans 13

2) Algebraic

24 avg for x games .. total points = 24x
new total = 24x+50
new number of games = x+1
avg = \(\frac{24x+50}{x+1}=26................24x+50=26x+26..................2x=24........x=12\)
so games = 12+1=13

3) weighted average method

x games worth 24 and 1 game worth 50
average 26
so \(\frac{x}{1}=\frac{(50-26)}{(26-24)}\frac{24}{2}=12\)
so x+1=12+1=13

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Re: Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 07 Jul 2018, 11:08
Bunuel wrote:
Going in to the last game of his basketball season, Adrian had averaged 24 points per game. In his last game, he scored 50 points, bringing his average to 26 points per game for the season. How many games did Adrian play that season?

A. 12
B. 13
C. 14
D. 15
E. 16


\(24G + 50 = (G +1)26\)

Or, \(24G + 50 = 26G + 26\)

Or, \(2G = 24\)

Or, \(G = 12\)

So, Total Number of games played that season is 12 + 1 = 13 games,Answer must be (B)
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Re: Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 07 Jul 2018, 12:31
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Increase in average by 2 points from 24 to 26 points. Hence a 2 point increase above average in every game.

The last game contributed (50 - 24) = 26 points to increase the average to 26.

Hence total # of games played = 26/2 = 13

Answer B.



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Re: Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 17 Sep 2018, 09:40
Bunuel wrote:
Going in to the last game of his basketball season, Adrian had averaged 24 points per game. In his last game, he scored 50 points, bringing his average to 26 points per game for the season. How many games did Adrian play that season?

A. 12
B. 13
C. 14
D. 15
E. 16


The homogeneity nature of the average makes this problem trivial:

\(? = N\)

\(\sum\nolimits_{N - 1} = \,\,\,24\left( {N - 1} \right)\)

\(\sum\nolimits_N = \,\,\,26\,N\)

\(26N = 24\left( {N - 1} \right) + 50\,\,\,\,\,\, \Rightarrow \,\,\,\,2N = 26\,\,\,\,\, \Rightarrow \,\,\,\,? = N = 13\)


This solution follows the notations and rationale taught in the GMATH method.

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Re: Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 17 Sep 2018, 10:59
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Bunuel wrote:
Going in to the last game of his basketball season, Adrian had averaged 24 points per game. In his last game, he scored 50 points, bringing his average to 26 points per game for the season. How many games did Adrian play that season?

A. 12
B. 13
C. 14
D. 15
E. 16


Let G = total number of games that Adrian played in the ENTIRE season

Going in to the last game of his basketball season, Adrian had averaged 24 points per game.
At this point, Adrian has played G-1 games
So, we can write: (total number of points in G-1 games)/(G-1) = 24
Multiply both sides of the equation by (G-1) to get: total number of points in G-1 games = 24(G-1)
Expand to get: total number of points in G-1 games = 24G - 24

In his last game, he scored 50 points...
We already know that total number of points in G-1 games = 24G - 24
So, TOTAL number of points for all G games = 24G - 24 + 50
Simplify to get: TOTAL number of points for all G games = 24G + 26

...bringing his average to 26 points per game for the season.
We can write: (total number of points in all G games)/G = 26
So, (24G + 26)/G = 26
Multiply both sides by G to get: 24G + 26 = 26G
Subtract 24G from both sides: 26 = 2G
Solve: G = 13

Answer: B

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Re: Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 18 Sep 2018, 18:52
Bunuel wrote:
Going in to the last game of his basketball season, Adrian had averaged 24 points per game. In his last game, he scored 50 points, bringing his average to 26 points per game for the season. How many games did Adrian play that season?

A. 12
B. 13
C. 14
D. 15
E. 16


We can let x = the number of games played in the season; thus, we have:

(24(x - 1) + 50)/x = 26

24x - 24 + 50 = 26x

26 = 2x

13 = x

Answer: B
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Re: Going in to the last game of his basketball season, Adrian had average  [#permalink]

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New post 21 Sep 2018, 22:26
Before the last game, Adrian has an average score of 24 points per game. That average can be represented as:

\(\frac{Total Points (S)}{Number of Games(N)}=24\)

Solving for \(S\), we have \(S= 24N\).

For the last game, he scored 50 points, bringing his average to 26 points per game for the season, which can be represented as:

\(\frac{(24N+50)}{(N+1)}=26\)

Solving for \(N\), we have:

\(24N+50=26(N+1)\)
\(24N+50=26N+26\)
\(50-26=26N-24N\)
\(24=2N\)
\(N=12\)

Thus, with the first 12 games and 1 additional game, there were 13 total games. The final answer is .
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Re: Going in to the last game of his basketball season, Adrian had average   [#permalink] 21 Sep 2018, 22:26
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