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word problem with fractions
- Thread starter mleccese
- Start date
4/7 of a group of children are boys. If there are 18 more boys than girls, how many children are there altogether?
So, what work have you done? Where are you stuck?
You have two unknowns: number of boys and number of girls. Therefore, you need two equations.
Let B be the number of boys and G be the number of girls.
4/7 of a group of children are boys:
(4/7)(G + B) = B
there are 18 more boys than girls:
B – 18 = G
You now have two equations. Can you solve this system of equations?
Hello, mleccese!
Here's another approach . . .
Let \(\displaystyle N\) = the number of children.
\(\displaystyle \text{We are told that there are: }\,\frac{4}{7}N\text{ boys.}\)
. . \(\displaystyle \text{Hence, there are: }\,\frac{3}{7}N\text{ girls.}\)
\(\displaystyle \text{And we are told that:}\;\text{(boys)} \;=\; \text{(girls)} + 18\)
. . . . . . .\(\displaystyle \text{So we have: }\;\;\frac{4}{7}N \;\;=\;\;\;\frac{3}{7}N \;+\; 18\)
\(\displaystyle \text{Therefore: }\;\frac{1}{7}N \:=\:18 \quad\Rightarrow\quad \boxed{N \:=\:126}\)
Here's another approach . . .
4/7 of a group of children are boys.
If there are 18 more boys than girls, how many children are there altogether?
Let \(\displaystyle N\) = the number of children.
\(\displaystyle \text{We are told that there are: }\,\frac{4}{7}N\text{ boys.}\)
. . \(\displaystyle \text{Hence, there are: }\,\frac{3}{7}N\text{ girls.}\)
\(\displaystyle \text{And we are told that:}\;\text{(boys)} \;=\; \text{(girls)} + 18\)
. . . . . . .\(\displaystyle \text{So we have: }\;\;\frac{4}{7}N \;\;=\;\;\;\frac{3}{7}N \;+\; 18\)
\(\displaystyle \text{Therefore: }\;\frac{1}{7}N \:=\:18 \quad\Rightarrow\quad \boxed{N \:=\:126}\)