The market stall's takings

Statement

A market stall sells mugs at $5$ euros each and sells $40$ per day, for takings of $(5)(40)=200$ euros. For every euro added to the price, it sells $5$ fewer mugs per day. At what prices do the daily takings become exactly $210$ euros?

Solution

Model. Let $d$ be the price increase in euros. The new price is $5+d$ and the mugs sold are $40-5d$. The takings are $(5+d)(40-5d) = 210.$ Equation. Expand: $200 - 25d + 40d - 5d^2 = 210 \implies -5d^2 + 15d + 200 = 210.$ Hence $-5d^2 + 15d - 10 = 0$, i.e. $d^2 - 3d + 2 = 0 \implies (d-1)(d-2) = 0$, so $d = 1$ or $d = 2$.

The two increases give the prices $5+1 = 6$ euros and $5+2 = 7$ euros. Check: $6\cdot 35 = 210$ and $7\cdot 30 = 210$. Correct. $\boxed{6 \text{ euro} \ \vee\ 7 \text{ euro}}$