In a competition, every student from school A and from school B is a gold
medalist, a silver medalist or a bronze medalist. The number of gold medalist
from each school is the same. The ratio of the percentage of students who are gold medalist from school A to that from school B is 5:6. The ratio of the
number of silver medalists from school A to that from school B is 9:2. The
percentage of students who are silver medalists from both school is 20%. If 50%
of the students from school A are bronze medalists, what percentage of the
students from school B are gold medalists?
【Solution】
Suppose school A has 9 silver medalists. Then school B has 2, for a total of 9+2=11.
Hence the total population of the two schools is 11 ÷ 20% = 55. For gold medalists,
the numbers are the same but the percentages are in the ratio 5:6. This means that the
ratio of the populations are in the ratio 6:5. It follows that the population of school A
is 55 × 6
6 5 +
= 30 and that of school B is 55−30 = 25. Now the number of bronze
medalists from school A is 30 × 50% = 15, so that the number of gold medalists from
each school is 30−15−9=6. It follows that the percentage of students from school B
who are gold medalists is 6 ÷ 25 = 24%.
ANS: 24%