Question

There are 120 boys and 114 girls in class X of a school. Principal of the school decided as a policy matter to have a maximum number of mixed sections, each section has to accommodate an equal number of boys and equal number of girls. What is the maximum number of such sections?

Answer 1

Hint: To solve this type of problem we will assign an alphabet for consideration and then write the equations based on the given content. By doing mathematical operations and by using the ratios concept we will arrive at the final solution.

Complete step-by-step answer:
Let 'x' be the total number of sections.
'y' is the number of boys in a section.
'z' is the number of girls in a section.
As there 120 boys we can write,
\[120=x\times y\]. . . . . . . . . . . . . . . . . . . . . . . . . (a)
As there are 114 girls we can write,
\[114=x\times z\]. . . . . . . . . . . . . . . . . . . . . . . . . . . (b)
Dividing both the equations (a) and (b) we get,
\[\dfrac{120}{114}=\dfrac{x\times y}{x\times z}\]
\[\dfrac{20}{19}=\dfrac{y}{z}\] . . . . . . . . . . . . . . . . . . . . . . . . . (c)
From (c) we can say that y=20 and z=19.
Now we get,
\[x=\dfrac{120}{20}=6\]
Thus there are a total of 6 sections.

Note: In the above we have used boys' values and got the number of sections. We can use the value of the number of girls and get the number of sections. Be careful while solving. This is a direct question with little tricky.