Question

A new coach was appointed in the middle of a season for a football team. After he took over, the team won 80% of the 60 matches it played. But the overall success rate is only 60%. Find the minimum number of matches the team must have played that season before the new coach took over

Answer 1

Let the number of matches played by football team before the season =x matches

Number of matches played by football team when the new coach has joined =60 matches

Match won by football team when new coach has joined = 80 % of 60

$=\frac{80}{100}\times{60}=48$

Overall success rate= 60%

Now, total number of matches played by football team before and after the appointment of new coach = x + 60

So, 60% of (60 +x) $=\frac{60}{100}\times(60+x)\\\\=36 +\frac{6 x}{10}$

  But it should be an integral value.

So, when , x=10, $\frac{6 \times 10}{10}=6$

=an integer, because 36 +6=42 which is 60% of (60+10=70 matches)

So, minimum number of matches the team must have played that season before the new coach took over=10 matches

Answer 2

Answer:

20

Step-by-step explanation:

We know that out of 60 matches they won in 80% of matches

Thus total matches won is 0.8*60=48 matches

Let x be the number of matches played before the coach was hired,

We need to find the minimum number of matches played so, let the win be 48

so 48/(60+x)=60

Solving this we get x=20