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?
Accepted Solution
A:
Answer:200Step-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 120Note: 30% converted to decimal is 30/100 = 0.3This can be translated to an algebraic equation as:0.3x = 120Now, we can solve for x, the total customers:0.3x = 120x = 120/0.3x = 400Since, 80% bought drink, we find 80% of 400:80% = 80/100 = 0.80.8 * 400 = 320The store sold 320 drinks and 120 snacks. So drinks outweigh snacks be:320 - 120 = 200So,200 more drinks than snacks