Pat has $3.80 in nickels and dimes. If there are 51 coins in all how many of them are nickels?

Question

Pat has $3.80 in nickels and dimes. If there are 51 coins in all how many of them are nickels?

Answer 1

Lol I am doing these exact questions I mean exactly the same!!!

Well you need to set up equation's:

.05n+.10d=3.80  

n + d = 51

you should know the rest

Answer 2

26.  3.80 divided by 10 = 38.  If all the coin were dimes you would have 38.  38-51 = 13.  You only have nickels and dimes to work with.  So you have to have 26 nickels (1.30) and 25 dimes (2.50).

Answer 3

First, let's change nickles and dimes to more round numbers: .05 becomes 5 and .10 becomes 10; we just moved over two decimal spaces. Do the same for 3.80 and you get:

5x + 10y = 380

Now, how many total coins were there? 51. So all the nickels (x) plus all the dimes (y) equals 51:

x + y = 51

Now we can use both equations to cancel things out. When combining the two equations, the best way to cancel x out is by multiplying the bottom by -5. So:

5x + 10y = 380

-5x - 5y = -255

--------------------------

0 + 5y = 125

---------> y = 125 / 5

---------------> y = 25

Now let's plug that in to the original equation:

5x+10(25) = 380

---------> 5x + 250 = 380

--------------> 5x = 130

------------------> x = 130 / 5

----------------------> x = 26

So there are 26 nickels. Done!

Answer 4

26 nickels and 25 dimes

Answer 5

Who cares

Answer 6

26

I was by far the first to get right, so I deserve best answer:) Everyone else probably worked the simple equation off my math.

Thanks :OOOOO

Answer 7

This is a simultaneous equation with two variables.  To solve, express alegebraically, then substitute so that there is only one variable, then solve for that variable.

Let n = the number of nickels

Let d - the number of dimes

Since nickels are worth 5 cents, and dimes are worth 10 cents, we can express this problem with two equations.

5n + 10d = 380

n + d = 51

Let's solve the second equation for d.

n + d = 51

n - n + d  = 51 - n

d = 51 - n

Now substitute this in the first equation.

5n + 10d = 380 - replace d with (51-n)

5n + 10(51 - n) = 380 - multiply 10 * 51 and 10 * n

5n + 510 - 10n = 380 - add 5n + (-10n)

-5n + 510 = 380 - subtract 580 from each side)

-5n = -130 - divide each side of the equation by 5

n = 26 nickels

So there are 26 nickels (equal to $1.30) and 25 dimes (equal to $2.50).  When we add $1.30 and $2.50, we get $3.80, so the problem checks and the correct answer is 26 nickels.

Answer 8

you have 25 dimes and 26 nickels.

Answer 9

nickle=0.05 cents, dime =0.10 cents.0.1 + 0.05=0.15 cents

3.80 / 0.15=25 with remainder,

hence there are minimum 25 dime and 25 nickels,

25 x 0.10=2.50, 25.x 0.05=1.25, 2.50 + 1.25=3.75, need one more nickle to make 51 coins, 25+25+1=51 coins

3.75 + 0.05=3.80.

therefore number of nickles=25+1=26

Answer 10

25 are dimes, and 26 are nickels.

Answer 11

26 Nickles and 25 Dimes