1. Problem statement. The price of a pen was 2 less in February than in March and 2 more in April than in March. The bookshop sold pens worth 4200 in February and 4500 in April. Let $x$ be the price of a pen in March. 2. (a) Expressions. (a)(i) The number of pens sold in February equals total revenue divided by price per pen, which is $\frac{4200}{x-2}$. (a)(ii) The number of pens sold in April equals $\frac{4500}{x+2}$. 3. (b) Equation from the condition. They sold 50 more pens in February than in April, so $\frac{4200}{x-2}=\frac{4500}{x+2}+50$. Solve this equation step by step. Multiply both sides by $(x-2)(x+2)$ to clear denominators obtaining $-300x+17400=50(x^2-4)$. Bring all terms to one side to get $50x^2+300x-17600=0$. Divide both sides by 50 to simplify to $x^2+6x-352=0$. Use the quadratic formula to find $x$: $x=\frac{-6\pm\sqrt{6^2-4\cdot1\cdot(-352)}}{2}$. Compute the discriminant: $6^2-4\cdot1\cdot(-352)=36+1408=1444$. So $x=\frac{-6\pm38}{2}$ which gives $x=16$ or $x=-22$. Discard $x=-22$ because price cannot be negative, thus $x=16$. Now compute number of pens sold in February: $\frac{4200}{16-2}=\frac{4200}{14}=300$. Therefore the bookshop sold 300 pens in February. 4. (c) Percentage change from February to April. Number sold in April is $\frac{4500}{16+2}=\frac{4500}{18}=250$. The change is $250-300=-50$ pens which is a fractional change of $\frac{-50}{300}=-\frac{1}{6}$. As a percentage this is $-\frac{1}{6}\times100\%= -\frac{50}{3}\%\approx -16.67\%$. Thus the number sold in April decreased by $16\frac{2}{3}\%$ compared to February. 5. Final answers. (a)(i) $\frac{4200}{x-2}$. (a)(ii) $\frac{4500}{x+2}$. (b) Number of pens sold in February is $300$. (c) Percentage change is a decrease of $16\frac{2}{3}\%\approx -16.67\%$.