Your piggy bank has a total of 20 coins in it; some are dimes and some are nickels. If you have a total of $1.25, how many nickels do you have? 15 10 5 20
Okay, let's start with ten nickels.
That would be a total of 50 cents.
You still have ten coins left, and they have to be dimes.
how did you do that
i meant 15 coins not 20
And ten times ten is 100 cents, so a dollar. And a dollar plus 50 cents is 1.50, which is too much.
???
my problem on my homeowrk says this a cash register contains 15 coins in dime and nickels. the total value is 1.25. how many coins of each typer are there
Oh, okay.
Then I'll start over.
Just pick a random number of nickels to start with...let's say 7.
If you have 7 nickels, how much money is that?
35 cents
Good, and how many coins do we have left that have to be dimes?
8 coins left...
Good, and since they have to be dimes, how much money is that?
80 cents?
Sorry, my cat just walked on the keyboard and deleted a bunch of stuff. And yes, your answer is right.
And 80 + 35 is?
45
Uh, no. It's 1.15
So, not enough.
Let's try again with 8 nickels.
8 times 5 is 40. 7 times 10 is 70. 40+70 is 1.30
still too much
Yeah, what I don't understand, is with 7 nickels it's too little, and with 8 it's not enough...
ya. thanks anyway. i get it somehow. it suppose be into a equation
Oh, sorry, should I help you with an equation?
d + n = 20 -->d = 20 - n .10d + .05n = 1.25 now sub 20 - n in for d in the second equation and solve for n
Like that made sense to her.
d is dimes n is nickels
a dime is a tenth of a dollar so therefore .10
a nickel is a twentieth so therefore .05
d and n are variables, so they can be any number. if you multiply d with .10 it means for example 5 dimes times 0.10, which is .50, or 50 cents.
Do you get it @bs?
ugh, she's offline
don't you hate that....do all that explaining and she is gone
Yeah, ikr
lol, but now we both got medals! :)
yes we do :)
@texaschic101 's way is MUCH easier. A system of equations. \(\Large{d+n=20\\ .10d + .05n = 1.25}\) This is how I would set up the problem, and I am fairly confident that this is how the problem was intended to be solved.
that is how I would set it up also :)
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