1. **State the problem:** A group of $x$ people agreed to contribute equally to buy books worth 1200. 5 people pulled out, so the remaining $x-5$ people contributed an extra 10 each. This extra contribution allowed them to buy books worth 200 more, i.e., 1400. 2. **Define variables and expressions:** - Original contribution per person: $\frac{1200}{x}$ - New contribution per person: $\frac{1400}{x-5}$ 3. **Set up the equation:** The new contribution per person is the original plus 10: $$\frac{1400}{x-5} = \frac{1200}{x} + 10$$ 4. **Solve the equation:** Multiply both sides by $x(x-5)$ to clear denominators: $$1400x = 1200(x-5) + 10x(x-5)$$ Expand terms: $$1400x = 1200x - 6000 + 10x^2 - 50x$$ Bring all terms to one side: $$0 = 1200x - 6000 + 10x^2 - 50x - 1400x$$ Simplify: $$0 = 10x^2 - 250x - 6000$$ Divide entire equation by 10: $$0 = x^2 - 25x - 600$$ 5. **Factor or use quadratic formula:** Using quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-25$, $c=-600$: $$x = \frac{25 \pm \sqrt{(-25)^2 - 4(1)(-600)}}{2} = \frac{25 \pm \sqrt{625 + 2400}}{2} = \frac{25 \pm \sqrt{3025}}{2}$$ $$\sqrt{3025} = 55$$ So, $$x = \frac{25 \pm 55}{2}$$ 6. **Find possible values:** - $x = \frac{25 + 55}{2} = \frac{80}{2} = 40$ - $x = \frac{25 - 55}{2} = \frac{-30}{2} = -15$ (not possible since number of people cannot be negative) 7. **Final answer:** The original number of people is $\boxed{40}$.