Every day, Dave eats either a sandwich or pizza for lunch. Over 42 days, Dave had pizza 3 times for every 4 times he had a sandw
ich. Over the next x days, he had pizza 3 times for every 2 times he had a sandwich. If at the end of this entire period he had pizza as many times as he had a sandwich, what is the value of x?
Alright, so every 7 days(a week) in the course of those 42 days, Dave eats 4 sandwiches and 3 pizzas. 42 / 7 = 6 weeks. So every week for 6 weeks Dave ate 4 sandwiches and 3 pizzas. This is a total of:
6 * 4 = 24 sandwiches 6 * 3 = 18 pizzas
This is what he ate in the first 42 days.
Over the next 'x' days, he eats pizza 3 times for every 2 times he eats a sandwich for lunch over the course of 5 days. He's eating 1 more pizza than sandwiches during this time period. Right now he's already eaten 24 sandwiches and 18 pizzas, this is 6 more sandwiches than pizzas. So in order to eat the same amount of pizza and sandwiches, he needs to eat 6 more pizzas. In 5 days he gains 1 pizza over sandwiches. So in 6 * 5 = 30 days he should gain 6 * 1 = 6 pizzas over sandwiches and tie the number of times he had pizza or sandwiches for lunch.
Check your work:
6 * 2 = 12 more sandwiches eaten in these 30 days 6 * 3 = 18 more pizzas eaten in these 30 days
24 + 12 = 36 sandwiches eaten in total 18 + 18 = 36 pizzas eaten in total
There are at least a couple of ways to think about this. 1) The fraction of the first 42 days that Dave ate pizza was 3/(3+4) = 3/7, so Dave ate (3/7)·42 = 18 pizzas during that time. The fraction of the next x days that Dave at pizza was 3/(3+2) = 3/5. We want the total number of pizzas to be half the total number of days. 18 + (3/5)x = (1/2)(42 + x) x(3/5 -1/2) = (1/2)·42 - 18 . . . . . subtract 1/2x + 18 x = 3/(1/10) . . . . . . . . . . . . . . . . .divide by the coeffient of x x = 30
2) Dave has 1 less pizza than sandwich in each period of 7 days, so over 42 days he has 6 fewer pizzas than sandwiches. Dave has 1 more pizza than sandwich in each period of 5 days for the next x days. In order for that to be 6 more pizzas than sandwiches (bringing the difference to zero), the value of x must be x = 6·5 = 30
