In the first half of the season a football team participating in the l

Question

Competition Mode Question

In the first half of the season a football team participating in the local league won 40% of its matches. It then went on to win all its matches except two in the second half of the season. If it is known that for the entire season the percentage of matches won by the team was 66.67% and that the team played double the number of matches in the second half of the season as compared to the first half of the season then how many matches did the team play in the first half of the season?

(A) 2
(B) 5
(C) 10
(D) 15
(E) 20

(A) 2
(B) 5
(C) 10
(D) 15
(E) 20

freedom128 · Answer

First half: x matches, 0.4x win
Second half: 2x matches, 2(x-1) win

Evaluate x in each choice:
(A) 2 --> No! 0.4x=0.8 win, nonsensical
 (B) 5 --> YES!  win=2+8=10, total=5+10=15, win = 2/3 of total

We stop here, because we have found the right answer.

FINAL ANSWER IS (B)

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exc4libur · Answer

first half matches = m
second half matches = n
n = 2m

wins: 
2/5m+1(n-2)=2/3(m+n)
2/5m+1(2m-2)=2/3(3m)
2m/5+2m-2=2m, 2m/5=2, 2m=10, m=5

Ans (B)

unraveled · Answer

In the first half of the season a football team participating in the local league won 40% of its matches. It then went on to win all its matches except two in the second half of the season. If it is known that for the entire season the percentage of matches won by the team was 66.67% and that the team played double the number of matches in the second half of the season as compared to the first half of the season then how many matches did the team play in the first half of the season?

(A) 2
(B) 5
(C) 10
(D) 15
(E) 20

(A) 2
(B) 5
(C) 10
(D) 15
(E) 20

Let first half matches = x
second half matches = y
y = 2x
Matches won in first half = 0.4x
Matches won in second half = y - 2

Now  \frac{0.4x + y - 2}{x + y} = \frac{66.67}{100} 
 \frac{0.4x + 2x - 2}{ x + 2x} = \frac{2}{3} 
 \frac{0.4x + 2x - 2}{ 3x} = \frac{2}{3} 
0.4x + 2x - 2 = 2x
0.4x = 2
x = 5

Answer B.

ostrick5465 · Answer

The number of matches the team played in the first half of the season is x (matches)
- The number of matches the team won in the first half of the season is 0.4x
- Because the team played double the number of matches in the second half of the season as compared to the first half of the season so the number of matches the team played in the entire season is 3x
- The number of matches the team won in the second half of the season is 2x - 2
For the entire season, the percentage of matches won by the team was 66.67%
=>  \frac{0.4x + 2x -2 }{ 3x } = \frac{2}{3}

=> x = 5 => The number of matches the team played in the first half of the season is 5 matches

=>  Choice B

madgmat2019 · Answer

No.of matches in first half=x
No.of matches in second half=2x
total matches=3x
from given total matches won =2(3x)/3=2x
so 0.4x+2x-2=2x
x=5

OA:B

Archit3110 · Answer

use plugin approach 
option C ;10
so won in first game = 4 matches
and 2nd half games = 20
so total = 30 games and won 66.67% = 20 
so won in 2nd half = 10 games but given in question that I t then went on to win all its matches except two in the second half of the season  so incorrect

option B ; 5
first half won = 5 * .4 = 2 games
and second half games= 10
so won 15*.667 = 10 games i.e lost only 2 games out of 10 hence sufficient
IMO B is correct

In the first half of the season a football team participating in the local league won 40% of its matches. It then went on to win all its matches except two in the second half of the season. If it is known that for the entire season the percentage of matches won by the team was 66.67% and that the team played double the number of matches in the second half of the season as compared to the first half of the season then how many matches did the team play in the first half of the season?

(A) 2
(B) 5
(C) 10
(D) 15
(E) 20

(A) 2
(B) 5
(C) 10
(D) 15
(E) 20

ScottTargetTestPrep · Answer

We can let the number of first half matches = n and the number of second half matches = 2n; thus:

(0.4n + 2n - 2)/(3n) = 2/3

3(0.4n + 2n - 2) = 6n

1.2n + 6n - 6 = 6n

1.2n = 6

n = 5

Answer: B

eakabuah · Answer

Let the number of games played in the first half of the season be x and the number of games played in the second half of the season be y. We are informed that y=2x. In addition, we know that the team won 0.4x matches in the first round while it won y-2 (or 2x-2) matches in the second half of the season. An overall win percentage of the team for the season is 66.67 (i.e. 200/3).

We can form the following equation for the overall percentage win:
(100*(0.4x+2x-2))/3x = 200/3
40x+200x-200=200x
40x=200
x=5.

The answer is B.

We can form the following equation for the overall percentage win:
(100*(0.4x+2x-2))/3x = 200/3
40x+200x-200=200x
40x=200
x=5.

The answer is B.

Jawad001 · Answer

In the first half,
Let Number of matches be x 
Winning = 0.40x

In the second Half,
Let Number of matches be 2x
Winning = 2/3x

2/3 of 3x = .4x + 2x -2
2x = .4x + 2x -2
.4x = 2
 x = 2/0.4
 x = 5(Answer)

ScottTargetTestPrep · Answer

We can let the number of matches in the first half of the season = n and the number of matches in the second half = 2n; thus:

(0.4n + 2n - 2)/(3n) = 2/3

3(0.4n + 2n - 2) = 6n

1.2n + 6n - 6 = 6n

1.2n = 6

n = 5

Answer: B

bumpbot · Answer

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