BTGmoderatorLU wrote: ↑

Mon Jul 06, 2020 5:13 am

Source: Princeton Review

A baseball team’s season consists of playing 162 games. At a certain point in its season, the team has won 55 percent of the games it has played. The team then wins a certain number of consecutive games so that at the end of the winning streak, the team has won 58 percent of the games it has played. How many wins did the team have at the end of its winning streak?

A. 10
B. 14
C. 42
D. 77
E. 87

The OA is E

Say the team has so far played x games; thus, 58% of x must be an integer.

Integer = 58x/100 = 29x/50

For 29x/50 to be an integer, x must be a multiple of 50, or, x would one among 50, 100 and 150.

Let's take each value of x and calculate the no. of wins the team had at the end of its winning streak.

• x = 50 => no. of wins so far = 29x/50 = (29*50)/50 = 29; There's no option as 58.
• x = 100 => no. of wins so far = 29x/50 = (29*100)/50 = 58; There's no option as 58.
• x = 150 => no. of wins so far = 29x/50 = (29*150)/50 = 87; There's an option as 87. The correct option.

Correct answer: E

Hope this helps!

-Jay
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BTGmoderatorLU wrote: ↑

Mon Jul 06, 2020 5:13 am

Source: Princeton Review

A baseball team’s season consists of playing 162 games. At a certain point in its season, the team has won 55 percent of the games it has played. The team then wins a certain number of consecutive games so that at the end of the winning streak, the team has won 58 percent of the games it has played. How many wins did the team have at the end of its winning streak?

A. 10
B. 14
C. 42
D. 77
E. 87

The OA is E

Solution:

We can let n = the number of games they played prior to the winning streak and w = the number games they 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 won at the end of their winning streak was 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 equal 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