kevin and randy muise have a jar containing 84 coins, all of which are either quarters or nickels. the total value of the coins in the jar is $11.80. how many of each type of coin do they have?

Answer 1

what do we know?

we know that a quarter is $0.25 and a nickel is $0.05

4x$.25 = $1

20x$0.05 = $1

and we have 84 coin

n = the number of nickels

 q = the number of quarters

 (1) n + q = 84 and

(2) 5*n + 25*q = 100*11.80 (always work in cents)

(3) 5*n + 25*q = 1180

Now substitute n from equation (1) into equation (3) and get

5*(84 - q) + 25*q = 1180

420 - 5*q + 25*q = 1180

20*q = 1180 - 420

 20*q = 760

 q = 38

Then using equation (1), plug in q

 n + 38 = 84

 n = 46

now check these values.

Is (.05*38 + .25*46 = 11.80)?

Is (2.30 + 9.50 = 11.80)?

Is (11.80 = 11.80)? Yes

Answer 2

q + n = 84

25q + 5n = 1180

q = 84 - n

25(84 - n) + 5n = 1180

2100 - 25n + 5n = 1180

-20n = -920

n = 46 nickles

d = 84 - 48

d = 38 dimes

Answer 3

Let's let q represent the NUMBER of quarters and n represent the NUMBER of nickers. Since there are a total of 84 coins , we can write:

q + n = 84

Then since each quarter is worth twenty-five cents or 0.25 and each nickel is worth five cents or 0.05, we can write:

0.25q + 0.05n = 11.80

To solve these two equations with two unknowns, use substitution.

From the first equation we can write that n= 84 -q

Substitute this expression for n into the second equation:

0.25q + 0.05(84 - q) = 11.80

Solve for q using inverses operations, combining like terms, and the distributive property:

0.25q + 4.2 - 0.05q = 11.80

0.20q + 4.2 = 11.80

0.20q + 4.2 - 4.2 = 11.80 - 4.2

0.20q = 7.60

0.20q /0.20 = 7.60/0.20

q= 38

So the number is quarters is 38 . The number of nickels will be 84 -38 or 46 nickels.

To check, 38 quarters would be worth $9.50 (0.25 * 38)

46 nickels would be worth $2.30 (0.05 * 46)

$9.50 +2.30 = $11.80