A baseball team’s season consists of playing 162 games.

Bunuel, has there been analysis of what makes some quant questions particularly difficult?
gracie shows that this question can be very simple and not abstract, yet, at the time of writing answer distribution is
A 24%
B 18%
C 18%
D 24%
E 18%

We can let n = the number of games they have played right before the winning streak and w = the number games they have won during the winning streak. We can create the following equation:

(0.55n + w)/(n + w) = 0.58

0.55n + w = 0.58n + 0.58w

0.42w = 0.03n

42w = 3n

14w = n

From the above, we see that n is a multiple of 14. Furthermore, since 0.55n = 11/20 x n = 11n/20 is a whole number, we see that n is a multiple of 20. A multiple of 20 that is also a multiple of 14 is their LCM, which is 140 (the next common multiple is 280, but that is greater than 182 already). Therefore, n must be 140 and w must be 10. So the number of games they have won at the end of their winning streak is 0.55 x 140 + 10 = 77 + 10 = 87.

Alternate Solution:

We know that the number of games played and the number of games won at the end of the winning streak both need to be whole numbers. If we let n be the total number of games played at the end of the winning streak, then 0.58n, or 58n/100 = 29n/50 must be a whole number. Notice that 29 and 50 have no common factors and 29n/50 is the number of games won at the end of the winning streak; therefore, the number of games won at the end of the winning streak must be a multiple of 29. Among the choices, only 87 = 29 x 3 is a multiple of 29; therefore, it is the correct answer.

Answer: E

But they must have given 162 matches for some reasons right .. should we or should we not assume the team played all the 162 matches ?

Percentage of games won at the end of the winning streak = 58% = 58/100 = 29/50
Since 29/50 of these games are won -- and the season consists of 162 games -- the total number of games at the end of the winning streak must be a MULTIPLE OF 50 LESS THAN 162.

The following approach is called ALLIGATION.

Winning percentage for the first part of the season = 55%
Winning percentage for the winning streak = 100%
Winning percentage for the MIXTURE of the first part and the winning streak = 58%

Let F = the first part and S = the streak

Step 1: Plot the 3 percentages on a number line, with the percentages for F and S on the ends and the percentage for the mixture in the middle.
F 55--------58--------100 S

Step 2: Calculate the distances between the percentages.
F 55---3---58---42---100 S

Step 3: Determine the ratio in the mixture.
The ratio of F to S is equal to the RECIPROCAL of the distances in red.
F:S= 42:3 = 14:1

The sum of the parts of the resulting ratio = 14+1 = 15.
Implication:
The total number of games at the end of the winning streak must be not only a multiple of 50 less than 162 but also a multiple of 15.
Only 150 is viable.

Since 29/50 of these 150 games are won, we get:
29/50 * 150 = 87

If you consult my solution above, you'll note that the total number of games at the end of the winning streak must be both a multiple of 50 and a multiple of 15.
Since the LCM of 50 and 15 = 150, the total number of games at the end of the winning streak must be a multiple of 150:
150, 300, 450, 600...
The season is constrained to 162 games so that only 150 is a viable option for the total number of games at the end of the winning streak.

Great Solution!

In your second solution, shouldn’t n be the multiple of 50? instead of 29 ?

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Thanks ... its been of great help

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