Question

Two mixed schools have 90 and 120 children respectively. In the first
60% and in the second 50% of the children are boys. What % of the
children in the two schools are boys?​

Answer 1

Step-by-step explanation:

54.28%is the correct answer

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Answer 2

Given: Two mixed schools have 90 and 120 children respectively. In the first

60% and in the second 50% of the children are boys.

To find: The % of the children in the two schools are boys.

Solution:

According to the question, 60% of the students in the first school are boys and there are 90 boys in the first school. So, the number of boys in the first school is

$\frac{60}{100} * 90 = 54$

Also, 50% of the students in the second school are boys and there are 120 boys in the second school. So, the number of boys in the second school is

$\frac{50}{100} * 120 = 60$

Now, the total number of boys in both the schools and the total number of children in both the schools is calculated as follows.

$60 + 54 = 114$

$90 + 120 = 210$

Thus, the percentage of boys in both schools is

$\frac{114}{210} * 100 = 54.28$

Therefore, 54.28% of the children in the two schools are boys.