Question

A. A group of twelve art students are visiting a local art museum for a field trip. The total cost of admission for the students is $125. What is the cost of admission for each student?
b. The school van can carry twelve passengers at a time. What is the least number of trips the van must make in order to bring 125 passengers to the same location?

c. Charlotte and her mother baked 125 cookies to give as Christmas gifts to their neighbors. If they plan to give a dozen cookies to each neighbor, how many neighbors will receive a gift?

d. Nicholas and Elaine are planning to serve cheesecake for dessert at their wedding and have purchased twelve cheesecakes. If the cheesecakes are divided evenly among the 125 wedding guests, how much cheesecake will each guest receive?

Answer 1

Answer:

a) $10.42

b) 11 trips

c) 10 cookies

d) 12/125ths of a cheesecake (or 9.6%)

Step-by-step explanation:

a)  

Let x = the cost of admission for each student

Let y = the total cost of admission

⇒  y = 12x  

⇒  125 = 12x  

⇒  x = 125 ÷ 12  

⇒  x = 10.42

So the cost of admission for each student = $10.42

b)

Let x = the number of trips

Let y = the total number of passengers

⇒  y = 12x

⇒  125 = 12x  

⇒  x = 125 ÷ 12  

⇒  x = 10.42

Therefore, the van must make 11 trips in order to bring 125 passengers to the same location.  (If it took 10 trips, then it would only bring 120 passengers, hence why we have to round make 11 trips to bring the last 5 passengers).

c)

Let x = the number of cookies each neighbor will receive

Let y = the total number of baked cookies

⇒  y = 12x

⇒  125 = 12x  

⇒  x = 125 ÷ 12  

⇒  x = 10.42

Therefore, each neighbor will receive 10 cookies (we have to round it down, as if each neighbor received 11 cookies, we would need 132 in total).

d)

Let x = the number of cheesecakes purchased

Let y = the amount of cheescake each guest will receive

⇒  y = 125x

⇒  12 = 125x  

⇒  x = 12 ÷ 125

⇒  x = 12/125 = 0.096 = 9.6%

Therefore, each guest will receive 12/125ths of a cheesecake (or 9.6%).