One team won $ 19 $ out of every $ 20 $ games played, and a second team won $ 7 $ out every $ 8 $ games played. Which team has a higher percentage of wins?
Answer
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Hint: First we will see about the percentage, which is the number that is expressed or the ratio that is expressed as the fraction of $ 100 $ .
To calculate the percentage, we need to find the number of parts and total number given and divide both values, and multiply with hundred.
Formula used: $ Per = \dfrac{{parts}}{{whole}} \times 100\% $ where the number of parts divides the number of whole things given into the times of hundred percent.
Complete step by step answer:
Since there are two games played. One team has won the games in nineteen out of the twenty games where it has been played every time.
And the second team has won seven games out of the eight games each time they play.
Here we need to find the higher percentage of win, like calculate the percentage of win from team one as well as from the team two, then compare them both and find which has more percentage.
Starting with the term one is won nineteen out of the twenty games each time they play,
Hence by the use of the percentage formula, total events are twenty, and parts of the events are nineteen thus we get, $ Per = \dfrac{{parts}}{{whole}} \times 100\% \Rightarrow \dfrac{{19}}{{20}} \times 100\% $ .
Further solving we get; team one percentage is $ 95\% $
Similarly, team two won $ 7 $ out every $ 8 $ game played.
From the percentage formula we get, $ Per = \dfrac{{parts}}{{whole}} \times 100\% \Rightarrow \dfrac{7}{8} \times 100\% $ and further solving we get, the second-team percentage is $ 87.5\% $ .
Therefore, the first team has a higher percentage of the winning opportunities.
Note: we are also able to solve this problem using a simple method, which is if the denominator value is higher while combining the two values, the best possibility of getting the higher percentage is the denominator low values part.
Like twenty is the team one denominator and eight is the team two denominators; while seeing this we get the higher percentage is team two. Thus, the resultant.
To calculate the percentage, we need to find the number of parts and total number given and divide both values, and multiply with hundred.
Formula used: $ Per = \dfrac{{parts}}{{whole}} \times 100\% $ where the number of parts divides the number of whole things given into the times of hundred percent.
Complete step by step answer:
Since there are two games played. One team has won the games in nineteen out of the twenty games where it has been played every time.
And the second team has won seven games out of the eight games each time they play.
Here we need to find the higher percentage of win, like calculate the percentage of win from team one as well as from the team two, then compare them both and find which has more percentage.
Starting with the term one is won nineteen out of the twenty games each time they play,
Hence by the use of the percentage formula, total events are twenty, and parts of the events are nineteen thus we get, $ Per = \dfrac{{parts}}{{whole}} \times 100\% \Rightarrow \dfrac{{19}}{{20}} \times 100\% $ .
Further solving we get; team one percentage is $ 95\% $
Similarly, team two won $ 7 $ out every $ 8 $ game played.
From the percentage formula we get, $ Per = \dfrac{{parts}}{{whole}} \times 100\% \Rightarrow \dfrac{7}{8} \times 100\% $ and further solving we get, the second-team percentage is $ 87.5\% $ .
Therefore, the first team has a higher percentage of the winning opportunities.
Note: we are also able to solve this problem using a simple method, which is if the denominator value is higher while combining the two values, the best possibility of getting the higher percentage is the denominator low values part.
Like twenty is the team one denominator and eight is the team two denominators; while seeing this we get the higher percentage is team two. Thus, the resultant.
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