Theo bài ra ta có : \(\frac{23+n}{40+n}=\frac{3}{4}\)
\(\Rightarrow4.\left(23+n\right)=3.\left(40+n\right)\)
\(\Rightarrow92+4n=120+3n\)
\(\Rightarrow4n-3n=120-92\)
\(\Rightarrow n=28\)
Vậy số n cần tìm là 28
=>( 23+a)/(40+a)=3/4
=>4*(23+a)=3*(40+a)
=> 92+4a=120+3a
=> a=28
=.số N đó là 28
\(\frac{23+n}{40+n}=\frac{3}{4}\)
\(=>4.\left(23+n\right)=3.\left(40+n\right)\)
\(=>92n+4n=120+3n=4n-3n=120-92=n=28\)
~Study well~ :)
Ta có : \(\frac{23+n}{40+n}=\frac{3}{4}\)
\(\Leftrightarrow4\left(23+n\right)=3\left(40+n\right)\)
\(\Leftrightarrow92+4n=120+3n\)
\(\Leftrightarrow4n-3n=120-92\)
\(\Leftrightarrow n=28\)