Answer:
There are 100 boys and 300 girls in School A and there are 300 boys and 500 girls in School B .
Explanation:
The ratio of the boys in school A and the boys in school B is 1:3
Let the ratio be x
No. of boys in School A = x
No. of boys in School B = 3x
We are given that The number of boys in school B is 200 higher than the number of boys in school A.
So,
![x+200=3x](https://img.qammunity.org/2021/formulas/mathematics/high-school/lqnnwx0c976w1ov58pkfr1gqpoe9k62382.png)
![200=3x-x](https://img.qammunity.org/2021/formulas/mathematics/high-school/tups3s7l7cr2jwpa311so7027r4f89hrhi.png)
![200=2x](https://img.qammunity.org/2021/formulas/mathematics/high-school/fm4j3p2f1tw8q8fm2lqk5kq6pqpm2hwf5c.png)
![(200)/(2)=x](https://img.qammunity.org/2021/formulas/mathematics/high-school/29dj2o6gm7e9u4br3z36zuysol5mozaemw.png)
100=x
So, No. of boys in School A = x = 100
No. of boys in School B = 3x =3(100)=300
We are given that the ratio of the girls in school A and the girls in school B is 3:5.
Let the ratio be y
So, No. of girls in School A = 3y
No. of girls in School B = 5y
So, Total No. of students in School A= No. of boys in school A + No. of girls in school A=100+3y
Total No. of students in School B= No. of boys in school B + No. of girls in school B=300+5y
We are given that The number of pupils in school A is equal to half the number of pupils in school B.
So,
![100+3y=(1)/(2)(300+5y)](https://img.qammunity.org/2021/formulas/mathematics/high-school/qd8wgjqwqynxvb1y63higfmjvfebxsb46v.png)
![200+6y=300+5y](https://img.qammunity.org/2021/formulas/mathematics/high-school/44ampidsbd0uthptfysjdp31wam5jrt13h.png)
![y=100](https://img.qammunity.org/2021/formulas/mathematics/high-school/2gdztzfdjgl2dcn0k60cc05eemmb95zf3q.png)
No. of girls in School A = 3y=3(100)=300
No. of girls in School B = 5y=5(100)=500
Hence there are 100 boys and 300 girls in School A and there are 300 boys and 500 girls in School B .