A shop sells 4 bags at 20% off for a total of 6,400 baht. What was the original price of one bag?

Step-by-step solution

1. 1

   Reduce the total to a single item first. The 6,400 baht covers $4$ bags, so the discounted price of one bag is

   $\frac{6400}{4}=1600\ \text{baht}$

   Keeping the quantity in the equation until the end only makes the algebra heavier.

2. 2

   Express the sale price as a fraction of the original. A discount of $20\%$ means the customer still pays $80\%$ of the original price $x$:

   $x-\frac{20}{100}x=\left(1-\frac{20}{100}\right)x=\tfrac{4}{5}\,x$

3. 3

   Set up and solve the equation.

   $\tfrac{4}{5}\,x=1600\quad\Longrightarrow\quad x=1600\div\tfrac{4}{5}=2000\ \text{baht}$

   $\boxed{2000\ \text{baht per bag}}$

4. 4

   Check by applying the discount forwards. $20\%$ of $2000$ is $400$, so the sale price is $2000-400=1600$ baht, and $4$ of them cost $4\times1600=6,400$ baht — exactly the amount given.

5. 5

   Avoid the classic trap. Adding $20\%$ back onto the sale price does not undo the discount: $1600\times1.20=1,920$ baht, which is short of $2000$. Percentages are taken of different bases going down (the original) and coming back up (the sale price), so the reverse operation is division by $0.80$, not multiplication by $1.20$.