SOLUTION: here is what the question is.
Kristen bought her lunch for $6.85. She paid for it with nickels and quarters. If there were 53 quarters, how many of each coin type was there?
Algebra.Com
Question 876684: here is what the question is.
Kristen bought her lunch for $6.85. She paid for it with nickels and quarters. If there were 53 quarters, how many of each coin type was there?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
53 quarters = $13.25 ???
Do you mean she used 53 of them? Or they were just in the vicinity?
===========
53 nickels = 265 cents.
Each nickel replaced by a quarter add 20 cents
----
(685 - 265)/20 = 420/21 = 21 quarters
--> 21 quarters & 32 nickels
RELATED QUESTIONS
Isabelle paid for her $1.75 lunch with 87 coins. If all the coins were nickels and... (answered by greenestamps)
Isabelle paid for her $1.75 with 87 coins. If all of the coins were nickels and pennies,... (answered by fractalier)
can you help me find the formula for this problem?
Laura found $2.15 in quarters and... (answered by Edwin McCravy)
Jo bought an item for 45¢, and paid with a $1.00 bill. How many different ways could she... (answered by solver91311)
Use the five steps for problem solving to answer the following question. Please show all... (answered by ewatrrr)
Emma paid the $10.02 bill for her lunch with 234 coins consisting of pennies, nickels and (answered by ikleyn)
tom has 42 coins in nickels and quarters. He has 6 more quarters than nickels, if x=... (answered by stanbon)
I have a question on my Algebra I homework. The question is," Lucy has some coins in her... (answered by ankor@dixie-net.com)
When a child breaks open her piggy bank, she finds a total of 64 coins, consisting of... (answered by richwmiller)