#color(blue)("Expansion of Taimur's solution")#

Let original count of girls be #g#

Let original count of boys be #b#

Using ratio in fractional format

#("girls")/("boys")->g/b->5/4#

#color(green)(5/4color(red)(xx1) " "-=" "5/4color(red)(xx3/3)" "=" "15/12#

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If we were to add 3 more boys then the numbers would become:

#("girls")/("boys")->5/4 ->15/(12+3) = 15/15=1/1 larr" new ratio as required"#

#color(purple)("As the final ration of "1/1" was achieved the previous ratio of "15/12" is correct")#

So we have: 15 girls and 12 boys as the original count

#color(white)(.)#

#color(white)(.)#

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#color(blue)("Solved using substitution")#

#g/b=5/4#.....................Equation(1) #larr" 1st ratio"#

#g/(b+3)=1/1#.................Equation(2) #larr" 2nd ratio"#

Consider Equation(2)

#=>g=b+3#..................Equation(3)

Substitute for #g# in Equation(2) using Equation(3)

#g/b=5/4" "->" "(b+3)/b=5/4#

#=>4b+12=5b#

#b=12#

By substitution of #b# in Equation(1) givies

#g/b=5/4" "->" "g/12=5/4#

#g=(5xxcancel(12)^3)/(cancel(4)^1) = 15#