Over the course of one day, a store owner determined that 80% of customers bought a drink and 30% of customers bought a snack. If the store sold 120 snacks that day, how many more drinks than snacks did the store sell?

200

Step-by-step explanation:

Let the total number of customers that day be "x"

30% of total customers bought a snack and that is 120.

So we can say:

30% of total (x) is 120

Note: 30% converted to decimal is 30/100 = 0.3

This can be translated to an algebraic equation as:

0.3x = 120

Now, we can solve for x, the total customers:

0.3x = 120

x = 120/0.3

x = 400

Since, 80% bought drink, we find 80% of 400:

80% = 80/100 = 0.8

0.8 * 400 = 320

The store sold 320 drinks and 120 snacks. So drinks outweigh snacks be:

320 - 120 = 200

So,

200 more drinks than snacks