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# In the first M games of a team's season, the ratio of the

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In the first M games of a team's season, the ratio of the  [#permalink]

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Updated on: 21 May 2013, 07:38
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65% (03:01) correct 35% (02:50) wrong based on 207 sessions

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In the first M games of a team's season, the ratio of the team's wins to its losses was 1:2. In the subsequent N games, the ratio of the team´s wins to losses was 1:3. If M:N = 4:5, what was the ratio of the team's wins to its losses for all M+N games?

A. 7:18
B. 9:23
C. 11:27
D. 23:54
E. 31:77

Originally posted by josemarioamaya on 21 May 2013, 07:10.
Last edited by josemarioamaya on 21 May 2013, 07:38, edited 2 times in total.
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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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21 May 2013, 07:19
5
josemarioamaya wrote:
In the first M games of a team´s season, the ratio of the team´s wins to its losses was 1:2. In the subsequent N games, the ratio of the team´s wins to losses was 1:3. If M:N = 4:5, what was the ratio of the team´s wins to its losses for all M+N games?

a) 7:18
b) 9:23
c) 11:27
d) 23:54
e) 37:77

The choice [E], I feel should be 31:77. **Request you to look into the source for the value.

As for the solution, if M is the games played in the first half,
Total wins = M/3
Total Losses = 2*M/3

Similarly for N games, with the ratio 1:3
Total wins = N/4
Total Losses = 3N/4

Hence when combined, total games become M+N
Total wins = M/3 + N/4
Total losses = 2M/3 + 3N/4

Now with the above, we know M:N = 4:5 i.e. M = 4N/5. Substituting the values,
Total wins = 31N/60
Total losses = 77N/60

Hence the ratio is 31:77. Please correct me if I am wrong!

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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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21 May 2013, 07:34
josemarioamaya wrote:
In the first M games of a team´s season, the ratio of the team´s wins to its losses was 1:2. In the subsequent N games, the ratio of the team´s wins to losses was 1:3. If M:N = 4:5, what was the ratio of the team´s wins to its losses for all M+N games?

a) 7:18
b) 9:23
c) 11:27
d) 23:54
e) 37:77

Are you sure that the option E is 37:77. I'm getting 31:77.
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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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21 May 2013, 07:46
I corrected the answer choice e it is 31:77
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Posts: 611
Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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21 May 2013, 11:08
2
1
josemarioamaya wrote:
In the first M games of a team´s season, the ratio of the team´s wins to its losses was 1:2. In the subsequent N games, the ratio of the team´s wins to losses was 1:3. If M:N = 4:5, what was the ratio of the team´s wins to its losses for all M+N games?

a) 7:18
b) 9:23
c) 11:27
d) 23:54
e) 31:77

As the final ratio is being asked, we can consider the value of M:N as 48:60.

First M(48) games, Wins = 16;Losses = 32

Succesive N(60) games, Wins = 15;Losses = 45

Total Wins:Total Losses = (16+15)/(32+45) = 31/77
E.
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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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21 May 2013, 11:10
vinaymimani wrote:
josemarioamaya wrote:
In the first M games of a team´s season, the ratio of the team´s wins to its losses was 1:2. In the subsequent N games, the ratio of the team´s wins to losses was 1:3. If M:N = 4:5, what was the ratio of the team´s wins to its losses for all M+N games?

a) 7:18
b) 9:23
c) 11:27
d) 23:54
e) 31:77

As the final ratio is being asked, we can consider the value of M:N as 48:60.

First M(48) games, Wins = 16;Losses = 32

Succesive N(60) games, Wins = 15;Losses = 45

Total Wins:Total Losses = (16+15)/(32+45) = 31/77
E.

Why cant we consider 12:15 and end up with the ratio 9:23 which is option (B)
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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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21 May 2013, 11:15
avinashrao9 wrote:
vinaymimani wrote:
josemarioamaya wrote:
In the first M games of a team´s season, the ratio of the team´s wins to its losses was 1:2. In the subsequent N games, the ratio of the team´s wins to losses was 1:3. If M:N = 4:5, what was the ratio of the team´s wins to its losses for all M+N games?

Why cant we consider 12:15 and end up with the ratio 9:23 which is option (B)

You will never got 9:23 for 12:15 .You will still get 31:77.
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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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21 May 2013, 11:23
I considered the ratio wrong..
Though 12:15 satisfies 4:5 condition, it does not work out for the number of matches mentioned. It has to be 48:60..
Thanks
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Joined: 15 May 2013
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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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17 May 2014, 07:31
How about 12:20, its a perfect fit for 4:5 - M:N, an also all the other ratios.. and with this result I get 9:23.. really confused why this is the wrong answer..

Any help from the experts will be of great help!

avinashrao9 wrote:
I considered the ratio wrong..
Though 12:15 satisfies 4:5 condition, it does not work out for the number of matches mentioned. It has to be 48:60..
Thanks

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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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17 May 2014, 07:41
pkhats wrote:
How about 12:20, its a perfect fit for 4:5 - M:N, an also all the other ratios.. and with this result I get 9:23.. really confused why this is the wrong answer..

Any help from the experts will be of great help!

avinashrao9 wrote:
I considered the ratio wrong..
Though 12:15 satisfies 4:5 condition, it does not work out for the number of matches mentioned. It has to be 48:60..
Thanks

4:5 = 8:10 = 12:15 = 16:20 = ... but not 12:20.
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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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17 May 2014, 09:00
2
2
pkhats wrote:
How about 12:20, its a perfect fit for 4:5 - M:N, an also all the other ratios.. and with this result I get 9:23.. really confused why this is the wrong answer..

Any help from the experts will be of great help!

avinashrao9 wrote:
I considered the ratio wrong..
Though 12:15 satisfies 4:5 condition, it does not work out for the number of matches mentioned. It has to be 48:60..
Thanks

12:20 is 3:5 and not 4:5 as required.
We know that the ratio of in first M games in 1:2 which means, we have denominator as 3 if we want to find the numbers of wins and losses.
In second case, the ratio of next N games is 1:3, making denominator 4.

Now M = 4x = 4 * 12x = 48 x (We are choosing 12 as it is LCM of 3 & 4)
N = 5x = 5* 12x = 60 x
Hence, we are taking the ratio as 48:60 and not any other ratio. Hope it helps!!!
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Joined: 15 May 2013
Posts: 15
GPA: 3.3
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Re: In the first M games of a team´s season, the ratio of the te  [#permalink]

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17 May 2014, 10:12
Oh!

Thanks both, Silly me!
These silly mistakes are gonna make me cry one day - Hope not! :p

mittalg wrote:
pkhats wrote:
How about 12:20, its a perfect fit for 4:5 - M:N, an also all the other ratios.. and with this result I get 9:23.. really confused why this is the wrong answer..

Any help from the experts will be of great help!

avinashrao9 wrote:
I considered the ratio wrong..
Though 12:15 satisfies 4:5 condition, it does not work out for the number of matches mentioned. It has to be 48:60..
Thanks

12:20 is 3:5 and not 4:5 as required.
We know that the ratio of in first M games in 1:2 which means, we have denominator as 3 if we want to find the numbers of wins and losses.
In second case, the ratio of next N games is 1:3, making denominator 4.

Now M = 4x = 4 * 12x = 48 x (We are choosing 12 as it is LCM of 3 & 4)
N = 5x = 5* 12x = 60 x
Hence, we are taking the ratio as 48:60 and not any other ratio. Hope it helps!!!

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In the first M games of a team's season, the ratio of the  [#permalink]

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18 Aug 2016, 17:14
M = 4/9 of total games
N = 5/9 of total games

Wins out of M games = 1/3
Losses out of M games = 2/3

Wins out of N games = 1/4
Losses out of N games = 3/4

At this point, we should look at the denominators and come up with the least common multiple.
LCM (9,3,4) = 108. This gives us the total number of games.

M = (4/9)*108 = 48
N = (5/9)*108 = 60

Wins out of M games = (1/3)*48 = 16
Losses out of M games = (2/3)*48 = 32

Wins out of N games = (1/4)*60 = 15
Losses out of N games = (3/4)*60 = 45

Total wins = 16 + 15 = 31
Total losses = 32 + 45 = 77

Desired ratio = 31/77

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Re: In the first M games of a team's season, the ratio of the  [#permalink]

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26 Jul 2018, 08:31
$$\frac{Total Wins}{Total Losses}$$ = $$\frac{(16+15)}{(32+45)} = \frac{31}{77}$$

Option E is the correct answer
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In the first M games of a team's season, the ratio of the  [#permalink]

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26 Jul 2018, 10:39
josemarioamaya wrote:
In the first M games of a team's season, the ratio of the team's wins to its losses was 1:2. In the subsequent N games, the ratio of the team´s wins to losses was 1:3. If M:N = 4:5, what was the ratio of the team's wins to its losses for all M+N games?

A. 7:18
B. 9:23
C. 11:27
D. 23:54
E. 31:77

let M=3x
let N=4y
3x/4y=4/5→
x/y=16/15
3*16=M=48 games: 16 wins; 32 losses
4*15=N=60 games: 15 wins; 45 losses
(16+15)/(32+45)=31/77
E
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Re: In the first M games of a team's season, the ratio of the  [#permalink]

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30 Jul 2018, 10:47
josemarioamaya wrote:
In the first M games of a team's season, the ratio of the team's wins to its losses was 1:2. In the subsequent N games, the ratio of the team´s wins to losses was 1:3. If M:N = 4:5, what was the ratio of the team's wins to its losses for all M+N games?

A. 7:18
B. 9:23
C. 11:27
D. 23:54
E. 31:77

Using the given ratio M:N of 4:5, we can let M = 4(12) = 48 and N = 5(12) = 60. So in the first 48 games, the team has ⅓(48) = 16 wins and ⅔(48) = 32 losses, and in the subsequent 60 games, the team has ¼(60) = 15 wins and ¾(60) = 45 losses. So the team’s win-loss ratio is (16 + 15)/(32 + 45) = 31/77.

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