Computer games

Let $N\,,\,M$ games were played by Teddy on first day and second day respectively. Then he won $85\%$ of $N$ games on first day and all of the $M$ games on second day.

Total number of games played = $M + N$

Total number of games won = $85\%\text{ of }N + M = \large\frac{17}{20}N + M$

Win Percentage of total games = $\displaystyle \frac{\frac{17}{20}N + M}{M + N}\times 100\%$

But net win percentage is given to be $94\%$

So, $\displaystyle \frac{\frac{17}{20}N + M}{M + N}\times 100\% = 94\%$

$\displaystyle \Rightarrow\frac{17}{20}N + M = \frac{94}{100}M + \frac{94}{100}N$

$\displaystyle\Rightarrow\frac{6}{100}M = \frac{9}{100}N$

$\displaystyle\Rightarrow 2M = 3N$

Now, Since RHS is a multiple of $3$ ,so $M$ must be a multiple of $3$ .

For total number of won games to be an integers we must have

$\displaystyle\frac{17}{20}N$ as an integer and $\displaystyle\frac{94}{100}(M + N)$ as an integer.

$\Rightarrow$ $N$ must be a multiple of $20$ and $M + N$ must be a multiple of $50$ .

So least value of $M + N$ possible is $50$ .

To get least value of $M$ we must have maximum value of $N$ as a multiple of (20) such that M is a multiple of $3$ .

Such maximum value of $N$ is $20$ with $M = 30$ .

Hence minimum number of games played by Teddy on the second day is $30$ .

• Let $W_y$ be no. of won games yesterday
• $T_y$ be no. of games played yesterday
• $W_t$ be no. of games played today

Yesterday:

$\cfrac{W_y}{T_y} = \cfrac{85}{100} \implies W_y = \cfrac{17}{20} \cdot T_y \\ \\ \\ \therefore \text{min. } T_y = \boxed{20 }$

Today:

$\begin{aligned} \cfrac{W_y + W_t}{T_y+W_t} &= \cfrac{94}{100} \\ \implies 100W_y + 100 W_t &= 94 T_y + 94 W_t \\ \implies 6 W_t &= 94 T_y - 100 W_y \\ &= 94 T_y - 85 T_y \\ &= 9 T_y \\ \\ \implies W_t &= \cfrac{3}{2} T_y \\ \implies \text{min.} W_t &= \boxed{30} \hspace{10mm} \text{ as min. } T_y = 20 \end{aligned}$