Question 934803
we have the following formulas
n = d + 8
p = 3n
p = 3*(d+8)
p = 3d+24
d +(d+8) +(3d+24) +q = 120
d+d+8+3d+24+q = 120
5d +32 +q = 120
5d +q = 88
.10d + .05*(d+8) + .01*(3d+24) +.25q = 5.52
.10d +.05d+.4 +.03d+.24 +.25q = 5.52
.18d +.64 +.25q = 5.52
.18d +.25q = 4.88
now we have two equations in two unknowns
5d +q = 88
.18d +.25q = 4.88
solve first equation for q
q = 88 - 5d
substitute for q in second equation
.18d +.25(88-5d) = 4.88
.18d +22-1.25d = 4.88
-1.07d +22 = 4.88
-1.07d = -17.12
d = 16
n = 16+8 = 24
p = 3*24 = 72
q = 120 - (16+24+72) = 8
therefore there are
72 pennies, 24 nickels, 16 dimes, 8 quarters
check answer
.01*72 + .05*24 + .10*16 + .25*8 = 5.52
.72 + 1.20 + 1.60 + 2.00 = 5.52
5.52 = 5.52
our answer checks :-)