Question 350387: Please help.I have not been able to figure out 3 homework coin problems:
1... $14.55 in quarters, dimes and nickels.There are 3 more than twice as many dimes as nickels and 3 less than 3 times as many quarters as nickels.How many of each coin.
I'm stuck at: 2d+3+3Q-3+n= 14.55 I know it is not right
2....$8.80 in quarters, dimes and nickels. there are 2 more than 5times as many nickels as quarters & 4 less than twice as many dimes as quarters How many of each coin.
3.....$6.10 in q,d,n. There are 5 less dimes than quarters and 7 less nickels than dimes.How many of each coin.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You do all of these the same way.
You combine the relationships that describe the number of objects with the relationships that describe the value of the objects.
nickels: 
dimes: 
quarters: 
Twice as many dimes as nickels:  (so, for example if you had 5 nickels, you would have twice as many, or 10 dimes) and three more than that is:  . Then for the quarters you have 
Messing about with decimal fraction coefficients is rather untidy, so I always like to change the amount of money from dollars and cents to just cents. You have $14.55, so I just make that 1455 cents.
Now we know that nickels are worth 5 cents, so nickels are worth  cents. Likewise dimes are worth  cents. But since we know that  , we can say the dimes in this collection are worth ) cents. Using the same sort of logic, our quarters are worth ) cents.
Adding it all up:
\ +\ 25(3n\ -\ 3)\ =\ 1455)
All you have to do is solve for
The other two problems are done the same way.
John
My calculator said it, I believe it, that settles it
|
|
|
