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

Question

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

a. 10
b. 14 
c. 42
d. 77
e. 87

gracie · Accepted Answer

let w=games won
p=games played
w=.58p→
w=29p/50→
w/29=p/50
only 87, a multiple of 29, will make w an integer
E

philipssonicare · Answer

There has to be a quicker way than this:

Originally
 \frac{w}{w+l} 
55/100
So the options are
11/20, 22/40, 33/60, 44/80, 55/100, 66/120, 77/140, 88/160 and it stops there as a 162 game season. (it has to be whole numbers as we are talking about games)
These numbers, modified by each of the answers, has to equal 0.58

a. 10
We can see that the smallest number of wins for 55% is 11. Eliminate

b. 14
11/20 becomes 14/23 = 0.6.... Too high. 22/40 is too many. Eliminate
 
c. 42
33/60 becomes 42/69 = 0.6..too much. So all of the others that have played less games than that would be too much too.
and all the other numbers are too high. Eliminate

d. 77
77/140 = 0.55
66/120 becomes 77/131=0.587...Pretty good
Anything smaller would be too much. 
Keep D as it's pretty close, see if E gives an exact number

e. 87
77/140 becomes 87/150=0.58
Perfect
 E

-----------

I initially tried solving it algebraically with 
0.03= \frac{w+x}{w+l+x}  -  \frac{w}{w+l}  but it didn't make things easy so had to go with the above.

philipssonicare · Answer

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%

ScottTargetTestPrep · Answer

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

MMahadevan · Answer

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 ?

GMATGuruNY · Answer

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

E

GMATGuruNY · Answer

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.

Mugdho · Answer

Great Solution!

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

Posted from my mobile device

MMahadevan · Answer

Thanks ... its been of great help

Adit_ · Answer

This is pretty straightforward.

55% of games means 11/20 of the games before winning streak.
58% of games mean 29/50 after win streak.

In that gap we see all games were won.

Now for 29/50 we only have 3 values possible.
29/50,58/100, 87/150 .

We need to find a corresponding value for 11/20 scenario.
22/40, 33/60, 44/80, 55/100, 66/120,  77/140

The bold faced represents the exact case we are looking for. One can simply write down the options and find if inclunding all wins from before to after we land up with the value in 58% column or not.

Answer:  Option E

Kinshook · Answer

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

How many wins did the team have at the end of its winning streak?

Let the games played at a certain point in the season be g and winning streak be of w games.

Team has won .55g = 55g/100 = 11g/20 games at a certain point in its season. 
Since team has won integer games, g must be a multiple of 20.

.55g + w / (g + w) = .58
.55g + w = .58g + .58w
.03g = .42w
g = 42w/3 = 14w

w = g/14; g must be a multiple of LCM(20,14) = 140

g = 140; Since 2*140 = 280 > 162
Wins at a certain point = .55g = 11*140/20 = 77 games

Winning streak = w = 140/14 = 10 games

Wins at the end of its winning streak = 77 + 10 = 87 games

IMO E

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

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

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