In 81 games, Rocky won 51 by KO. Assuming Rocky wins all future matches by K.O, how many more matches must he complete so his K.O wins and matches

[eddy2017]

Check response #14 - for typo

I did. Found that and rectified it 20 minutes ago when you mentioned the typo

I was hoping we would help Eddy to derive this

 

[eddy2017]

Check pls 14cand let me know it there's still a typo there

 

[Deleted member 4993]

Check pls 14cand let me know it there's still a typo there

AFAICT......................you do not have any typo any more in response #14.

 

[eddy2017]

Correct, but do you understand why we have to add x to both the numerator and denomiator?

Well it is a ratio that we need to reach , right?. So we we have to the same unknown in both numerator and denominator. That is all I can say.

 

[BigBeachBanana]

Well it is a ratio that we need to reach , right?. So we we have to the same unknown in both numerator and denominator. That is all I can say.

[math]\text{The win rate} =\frac{\text{number won matches}}{\text{total matches competed}}[/math]Initially, we are given [imath]\text{The win rate} =\frac{51}{81}[/imath]. We want to know how many more matches Rocky needs to participate in in order to have a win rate of 2/3.
The total # match he will compete in = 81 + x
Next, the question told us that he will win all of his upcoming matches. Thereby, we have
Number won matches = 51 + x.
Does this make sense?

 

[eddy2017]

Yes, it does make a world of sense. It is much like the win rate formula
to calculate winning percentage where we divide wins by games played. Thanks.

 

[Steven G]

Right. That is it. I am using a text editor. It is easy to type in math.
Yes, my wife saw that cross multiplication was a possibility and equating both sides. She is also studying a bit. and we see that 3 is a common multiple to both 51 and 81.

Yes, 3 is a multiple of 51 and 81. Does this help you with solving the problem.

 

[eddy2017]

Good question, Dr Khan. At first sight we noticed that 3 was a multiple of 51 as I am always looking out for multiples to reduce fractions before we perform them as I have learned to do here. so ,on second thought, we though we could reduce the equation a little bit, but we tried and the result was not the same. So, the answer is no. it was not helpful, anyway we had to distribute the 2/3 when cross-multiplying so we saw there was nothing to it. if there is something that could have been done we missed it.