Question

Un elev citeste o carte de 480 file citind in fiecare zi acelasi numar de pagini. Daca ar citi cu 16 pagini mai mult in fiecare zi, ar termina cartea cu 5 zile mai devreme. In cate zile citeste cartea?

Answer 1

   
Notatii:
p = numarul de pagini citite in fiecare zi
z = numarul de zile in care termina decitit cartea

Scriem sistemul de ecuatii:


[tex]\displaystyle \\ \frac{480}{p} =z \\ \\ \frac{480}{p+16} =z-5 \\\\ \text{- - - - -}\\ p\times z = 480 \\ (p+16)(z-5) = 480\\ \text{- - - - -}\\ p\times z = 480\\ p\times z + 16 z - 5p - 80 = 480\\ \text{- - - - -}\\ p\times z = 480\\ 480 + 16 z - 5p - 80 = 480\\ \text{- - - - -}\\ p\times z = 480\\ 16 z - 5p = 480+80 -480\\ \text{- - - - -}\\ p\times z = 480\\ 16 z - 5p = 80 ~~~~~\text{Rezulta substitutia: } \boxed{z=\frac{80 + 5p}{16} } \\ \text{- - - - -} [/tex]


[tex]\displaystyle \\ p \times \frac{80 + 5p}{16} = 480 \\ \\ p(80 + 5p)= 480 \times 16 \\ 5p^2 +80p= 7680 ~~~|: 5 \\ p^2 + 16p = 1536 \\ p^2 + 16p - 1536 = 0\\ \text{Fiind o ecuatie de gradul 2, vom obtine 2 solutii.}\\ \text{Ecuatia de gradul 2 se rezolva cu formula:} [/tex]


$\displaystyle\\ p_{12}=\frac{-b \pm\sqrt{b^2-4ac}}{2a}=\frac{-16\pm\sqrt{16^2-4\times 1 \times(-1536)}}{2\times1}=\\\\ =\frac{-16\pm\sqrt{256+6144}}{2}=\frac{-16\pm\sqrt{6400} }{2} = \frac{-16\pm80}{2}=-8\pm40\\\\ p_1=-8+40=\boxed{32~\text{de pagini pe zi}}\\\\ p_2=-8-40=-48~\text{Solutie eliminata deoarece }-48\notin N\\\\ \text{Ne intoarcem la substitutie:}\\\\ z=\frac{80+5p}{16}=\frac{80+5\times32}{16}=\frac{80+160}{16}= \frac{240}{16}= \boxed{15~\text{zile}}$


⇒A citit  32 de pagini pe zi timp de 15 zile.  
Verificare:
32 × 15 = 480

(32 + 16)(15 - 5) = 48 × 10 = 480