Answer:
number of nickels = 58
dimes = 231
quarters = 19
Explanation:
Let x,y,z represent the number of nickels, dimes and quarters that it contains.
The total number of coins in the box is 308
x+y+z = 308 .....1
The number of dimes is three times the number of nickels and quarters together;
y = 3(x+z) .......2
Nickel = $0.05
Dime = $0.10
Quarters = $0.25
the box contains 30 dollars and 75 cents
0.05x + 0.10y + 0.25z = 30.75 ......3
Let y = 3(x+z)
0.05x + 0.30(x+z) + 0.25z = 30.75
0.35x + 0.55z = 30.75
x + (0.55/0.35)z = 30.75/0.35 ........4
From equation 1
x+y+z = 308
Substituting equation 2
x + 3(x+z) + z = 308
4x + 4z = 308
x+z = 77 .......5
Subtract equation 5 from 4
(0.55/0.35 - 1)z = 30.75/0.35 - 77
z = 19
from equation 5
x = 77-z = 77-19 = 58
x = 58
From equation 2
y = 3(x+z) = 3(77) = 231
y = 231
number of nickels x = 58
dimes y = 231
quarters z = 19