Respuesta :
Answer:
Second well was [tex]103 \frac{3}{6}\ m[/tex] deep below the surface.
Step-by-step explanation:
Given:
Depth of First well = [tex]46\frac{2}{3}\ m[/tex]
[tex]46\frac{2}{3}\ m[/tex] can be Rewritten as [tex]\frac{140}{3}\ m[/tex]
Depth of First well = [tex]\frac{140}{3}\ m[/tex]
Also Given:
The second well had more water, and was [tex]77\frac{5}{6}\ m[/tex] deeper than the first well.
[tex]77\frac{5}{6}\ m[/tex] can be Rewritten as [tex]\frac{467}{6}\ m[/tex]
Hence We can say that;
Depth of second well is equal to [tex]\frac{467}{6}\ m[/tex] plus Depth of First well.
framing in equation form we get;
Depth of second well = [tex]\frac{467}{6}+\frac{77}{3}[/tex]
Now the denominators are common so we can solve the numerators
now to solve the fractions we need to make the denominator common we will use L.C.M we get;
Depth of second well = [tex]\frac{467\times1}{6\times1}+\frac{77\times2}{3\times2}= \frac{467}{6}+\frac{154}{6}[/tex]
Now the denominators are common so we can solve the numerators.
Depth of second well = [tex]\frac{467+154}{6}=\frac{621}{6}\ m \ \ OR\ \ 103 \frac{3}{6}\ m[/tex]
Hence Second well was [tex]103 \frac{3}{6}\ m[/tex] deep below the surface.
