Quadratic equation problem

Question

Nino sells on average three cell phones more than Will. At the end of the day, they both receive $35.00$ together for the day's sales. The store manager finds that: If Will sold at price W the number of cell phones Nino sold, he would receive $24.00$. Nino would receive $12.50$ if he sold at price N the quantity Will sold. At the end of the day, the guys got together and discovered that the equation that calculates the number of cell phones sold by each one is: (Answer:$x^2-20x+75=0$)

I try $n = w+3\\ nN+wW = 35(I)\\ nW = 24(II) \\ wN=12,5(III)\\ nN+wW=35 \implies (w+3)N+(n-3)W = 35\\ wN+3N+nW-3W = 35 \implies 12,5+3(N-W)+24=35\\ \therefore \boxed{W-N =0,5}$

but I can't finish.

Answer 1

Let $A$ be the (average) number of cell phones sold each day by Will, and $B$ the average number of cell phones sold each day by Nino. They tell you that $B=A+3$, and let $W$ and $N$ be the sale price of each cell phone owned by Will and Nino respectively

Then they also tell you that $W⋅A+N⋅B=35$, $W⋅B=24$, and that $N⋅A=12.5$. You are asked to find the value of $A$ (or that of $B$). From the above you have to:

$WA+NB=WA+N(A+3)=(W+N)A+3N=35,N=\frac{12,5}{A},W=\frac{24}{A+3}$ Now substituting in the first equation the value of $N$ and $W$ is left

$[\frac{12,5(A+3)+24A}{A(3+A)}].A+(\frac{3⋅12,5}{A})=35⟺12,5(A+3)A+24A^2+(3+A)3⋅12,5=35A(3+A)⟺24A^2+(3+A)37,5=22,5A(3+A)⟺1,5A^2−30A+112,5=0⟺3A^2−60A+225=0⟺A^2−20A+75=0$

(Solution by Masacroso)

Answer 2

Your result $ \ W - N \ = \ 0.5 \ $ is useful, but your choice of labeling for the numbers of phones sold complicated matters. The "reference" is the number of phones $ \ x \ $ that Will sold: Nino then sold $ \ x + 3 \ $ phones. Your equations (II) and (III) would then be $ \ W \ = \ \frac{24}{x + 3} \ $ and $ \ N \ = \ \frac{12.5}{x} \ \ . \ $ From all this, we have $$ \frac{24}{x + 3} \ - \ \frac{12.5}{x} \ \ = \ \ 0.5 \ \ \Rightarrow \ \ 24·x \ - \ 12.5·(x+3) \ \ = \ \ 0.5·x·(x+3) $$ [after multiplying the equation through by $ \ x·(x + 3) \ \ ] $ $$ \Rightarrow \ \ (24 \ - \ 12.5)·x \ - \ 37.5 \ \ = \ \ 0.5·x^2 \ + \ 1.5·x $$ [re-arranging terms and multiplying through by $ \ 2 \ \ ] $ $$ \Rightarrow \ \ x^2 \ + \ 2·(1.5 \ - \ 24 \ + \ 12.5)·x \ + \ 2·37.5 \ \ = \ \ 0 $$

$$ \Rightarrow \ \ x^2 \ - \ 20·x \ + \ 75 \ \ = \ \ (x \ - \ 5)·(x \ - \ 15) \ \ = \ \ 0 \ \ . $$

So your approach does work, but you needed to settle on a single variable as a "reference" for how many phones each person sold.

The solutions are then

• Will sold $ \ 5 \ $ at $ \ \frac{24}{8} \ = \ 3 \ $ (currency units) each and Nino sold $ \ 8 \ $ at $ \ \frac{12.5}{5} \ = \ 2.5 \ $ each (the prices differing by $ \frac12 \ ) \ \ , \ $ bringing in a total of $ \ 15 \ + \ 20 \ = \ 35 \ \ ; $ or

• Will sold $ \ 15 \ $ at $ \ \frac{24}{18} \ = \ \frac43 \ $ each and Nino sold $ \ 18 \ $ at $ \ \frac{12.5}{15} \ = \ \frac56 \ $ each (the prices again differ by $ \ \frac36 \ = \ \frac12 \ ) \ , \ $ for a total of $ \ \frac{60}{3} \ + \ \frac{90}{6} \ = \ 20 + 15 \ = \ 35 \ \ . $