Răspuns: Ai rezolvarea mai jos
Explicație pas cu pas:
Salutare!
Notam cu x lungimea totala a traseului
(A)
[tex]\circledast \underline{\bf Prima ~I ~zi}:~\bf \dfrac{1}{4}\cdot x=\boxed{\bf\dfrac{x}{4}}[/tex]
[tex]\bf Rest~ dupa~prima~zi: x-\dfrac{x}{4}= \dfrac{4x-x}{4}= \boxed{\bf \dfrac{3x}{4}}[/tex]
[tex]\circledast \underline{\bf A~doua~zi}:\bf \dfrac{1}{3}\cdot\dfrac{3x}{4} +200=\dfrac{1}{\not3}\cdot\dfrac{\not3x}{4} +200=\boxed{\bf\dfrac{x}{4} +200}[/tex]
[tex]\bf Rest~dupa~a-II-a~zi: \dfrac{3x}{4} -\bigg(\dfrac{x}{4} +200\bigg)=\dfrac{3x-x}{4}-200=\boxed{\bf\dfrac{x}{2}-200}[/tex]
[tex]\circledast \underline{\bf A ~a-III-a~zi}:\bf \dfrac{13}{25}\cdot\bigg(\dfrac{x}{2}-200\bigg)=\dfrac{13x}{50}- \dfrac{13\cdot\not 200}{\not25}=\boxed{\bf \dfrac{13x}{50}-104}[/tex]
[tex]\bf Rest~dupa~a-III-a~zi: \!{^{^{^{^{^{^{\displaystyle 25)}}}}}}}\!\!\!\dfrac{x}{2} -200-\bigg(\dfrac{13x}{50} -104\bigg)=[/tex]
[tex]\bf =\dfrac{25x-13x}{50} -200 +104=\boxed{\bf \dfrac{12x}{50}-96}[/tex]
Restul din a treia zi înseamnă cat a mers in ultima zi (adica in a patra zi).
Iar in ultima zi a mers 120
[tex]\bf \dfrac{12x}{50}-96 = 120~~\bigg |\cdot 50[/tex]
[tex]\bf 12x-4800 = 6000[/tex]
[tex]\bf 12x = 6000+4800[/tex]
[tex]\bf 12x = 10800~~\bigg|:12[/tex]
[tex]\boxed{\boxed{ \bf x = 900~km}}[/tex]
(B)
[tex]\circledast~\underline{\bf Prima ~I ~zi}:~\bf \dfrac{900}{4}=\boxed{\bf 225~km}[/tex]
[tex]\circledast~ \underline{\bf A~doua~zi}:\bf \dfrac{900}{4} +200=225+200=\boxed{\bf 425~km}[/tex]
[tex]\circledast~ \underline{\bf A ~a-III-a~zi}:\bf \dfrac{13\cdot 900}{50}-104=234-104=\boxed{\bf130~km}[/tex]
Verificare:
225 km + 425 km + 130 km + 120 km = 900 km (adevarat)
PS: Daca esti pe telefon te rog sa glisezi spre dreapta pentru a vedea rezolvarea completa.
==pav38==
