Answer :
[tex]\bf slope\implies \cfrac{\stackrel{\stackrel{rise}{height}}{4}}{\stackrel{\stackrel{run}{base}}{50}}\implies \cfrac{2}{25}\ne \cfrac{1}{12}
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\textit{one could adjust the height}\qquad \cfrac{h}{50}=\cfrac{1}{12}\implies h=\cfrac{50}{12}\implies h=4\frac{1}{6}
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\textit{or one could adjust the base}\qquad \cfrac{4}{b}=\cfrac{1}{12}\implies 48=b[/tex]
Slope is defined as rise (height) over run (length) so we can express this as slope = Height/Length.
Your rule says that the maximum slope (H/L) cannot exceed 1/12. If the height is 4 this means the length must be at least 4(12) = 48.
In other words: slope = H/L = 4(1)/4(12) = 4/48 = 1/12
If L is less than 48 the the slope is Greater than 1/12 and this is not allowed. Any length longer than 48 and the slope is less than 1/12 and this is OK.
Since 50 > 48 the ramp will be acceptable
Your rule says that the maximum slope (H/L) cannot exceed 1/12. If the height is 4 this means the length must be at least 4(12) = 48.
In other words: slope = H/L = 4(1)/4(12) = 4/48 = 1/12
If L is less than 48 the the slope is Greater than 1/12 and this is not allowed. Any length longer than 48 and the slope is less than 1/12 and this is OK.
Since 50 > 48 the ramp will be acceptable