Tony has 11 more nickels than quarters. How many coins does he have if the total value of his coins is $2.65?

Answer 1

Let q be the number of quarters.
0.25*q+0.05*(q+11) = 2.65
0.30*q + 0.55 = 2.65
0.30*q = 2.10
q = 7
so there are 7 quarters and 18 nickels.

Answer 2

Just read the problem carefully, using algebraic expressions as you go. Your variable declarations and equations will come easily after that.

"Tony has 11 more nickels than quarters."
How many quarters does he have? *shrug* I don't know, either... call it "q."
How many nickels does he have? 11 more than he has quarters... and you've defined the number of quarters to be q, so what number is 11 more than that? It's "q + 11."

The total value is $2.65.
How much money are his quarters worth? $0.25 for each quarter, of which he has q. It's 0.25q.
How much money are his nickels worth? $0.05 for each nickel, and you've defined that, too. It's 0.05(q + 11).

How much money does he have in total?
Quarters + Nickels = $2.65
0.25q + 0.05(q + 11) = 2.65

You can solve this from here.

Answer 3

Try doing this without algebraic equations. Simply note that Tony has a bunch of quarter-nickel pairs, then he has 11 more nickels, for 55 cents. Take that away from $2.65 and get $2.10. A nickel-quarter pair is $0.30, so divide $2.10 by $0.30 and you will get the number of quarter-nickel pairs, and from that the total number of nickels and quarters.

Answer 4

Assuming Tony has only nickels and quarters ...

(25q) + ((11+q)(5)) = 265

25q + 55 + 5q = 265

30q + 55 = 265

30q = 265 - 55

30q = 210

q = 210/30

q = 7

7 quarters ($1.75) + 18 nickels ($0.90) = $2.65

Answer 5

Create a simple algebra equation and figure out what variables you want to use and what values you want to use for each variable.

So:
Let x = # of quarters.
Let y = # of nickels.

Because you have 11 more nickels than quarters, define one variable in terms of the other:
y = 11 + x
Create your equation:
.25x +.05y = 2.65
25x + 5y = 265
Now substitute one value in place of the variable:
25x + 5(11 + x) = 265
Solve:
25x + 55 + 5x = 265
30x = 210
x = 7
Now, substitute the value of x to find out y:
y = 18

totally, you have 25 coins: 18 nickels and 7 quarters.

hope it helped :)

Answer 6

18 nickels and 7 quarters-as answered by my grand son
he has 11 more nickels than quarters
start with 2 quarters: 2q+13n=$1.15
then try 4 quarters: 4q+15n=$1.75
then try 6 quarters; 6q+17n= $2.35
you need 30 more cents- so try 7 quarters
so 7q+18n=$2.65

Answer 7

Let q equal the number of quarters.
0.25q + 0.05(q+11) = 2.65
0.25q + 0.05q + 0.55 = 2.65
0.30q = 2.10
q = 7

There are 7 quarters and 18 nickels.

Answer 8

Let Q =number of quarters.
Let N = number of nickels.

N = 11 + Q

.25Q +.05N = 2.65
25Q + 5N = 265
25Q + 5(11 + Q) = 265
25Q + 55 + 5Q = 265
30Q = 210
Q = 7
N = 18
Q + N = 25

Answer 9

25.

7 Quarters, 18 Nickels

Answer 10

$14.88 and 2 monkeys