Answer in detail.There were 3 times as many nickels as dimes & total of $25. How many of each coin were ther?

Question

Answer in detail.There were 3 times as many nickels as dimes & total of $25. How many of each coin were ther?

Answer 1

Let there be x dimes ; then, there are 3x nickels.

3x(5)+x(10)=2500

15x+10x=2500

25x=2500

x=100 dimes = 1000 cents = $10.00

3x=300 nickels = 1500 cents = $15.00

Total $25.00

Answer 2

Let

D = number of dimes

N = number of nickels

Equation 1 -- There were 3 times as many nickels as dimes

N = 3D

<< & total of $25.>>

0.05N + 0.10D = 25

<<How many of each coin were there? >>

From Equation 1, N = 3D and substituting this in Equation 2,

0.05(3D) + (0.1D) = 25

0.15D + 0.1D = 25

0.25D = 25

D = 100

Solving for "N" (using Equation 1)

N = 3(D) = 3(100) = 300

ANSWER:

Number of dimes = 100

Number of nickels = 300

CHECK:

0.05(300) + 0.1(100) = 15 + 10 = 25

Answer 3

you've gotta do better than that.

(3 nickels,1 dime = .25) x 100 = 25.00

300 nickels, 100 dimes

Answer 4

300 nickels = $15

100 dimes =  $10

$10+$15 =$25

Answer 5

You said you wanted detail, so here goes:

x=dimes

.1x+3*.05x=25

.1x+.15x=25

.25x=25

x=100 dimes

therefore, 300 nickels

Answer 6

n = number of nickels

d = number of dimes

n = 3d

0.05n + 0.10d = $25

Solve the system of equations using substition:

0.05(3d) + 0.10d = 25

0.15d + 0.10d = 25

0.25d = 25

d = 100

Use the value of d = 100 to find n in the first equation:

n = 3(100)

n= 300

Therefore, n = 300, d = 100

Check your answer:

0.05(300) + 0.10(100) = $25