Represented 8020

The race was divided into four stages. The length of Stages 1 and 4 was 160 km. The length of the second stage represented a third of the total length of the race. The length of the 4th stage was twice as long as the third stage.

What percentage of the total length of the race was the sum of the lengths of the 3rd and 4th Stages. ?

Correct answer:

p =  45 %

Step-by-step explanation:

$$\begin{gathered} a=160 \\ d=160 \\ b=(a+c+d)/3 \\ d=2\cdot c \\ a=160 \\ d=160 \\ a-3b+c+d=0 \\ 2c-d=0 \\ Row3-Row1\to Row3 \\ a=160 \\ d=160 \\ -3b+c+d=-160 \\ 2c-d=0 \\ Pivot:Row2\leftrightarrow Row3 \\ a=160 \\ -3b+c+d=-160 \\ d=160 \\ 2c-d=0 \\ Pivot:Row3\leftrightarrow Row4 \\ a=160 \\ -3b+c+d=-160 \\ 2c-d=0 \\ d=160 \\ d=\frac{160}{1}=160 \\ c=\frac{0+d}{2}=\frac{0+160}{2}=80 \\ b=\frac{-160-c-d}{-3}=\frac{-160-80-160}{-3}=133.33333333 \\ a=\frac{160}{1}=160 \\ a=160 \\ b=\frac{400}{3}\doteq 133.333333 \\ c=80 \\ d=160 \\ p=100\cdot \frac{c+d}{a+b+c+d}=100\cdot \frac{80+160}{160+133.3333+80+160}=45\% \end{gathered}$$

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