1 nickle = 5 cents

1 dime = 10 cents

1 quarter = 25 cents

let

n = the number of nickels, n>0

d = the number of dimes, d>0

q = the number of quarters, q>0

She received $1.60 = 160 cents in quarters, dimes, and nickels

5n + 10d + 25q = 150 divide both sides by 5

n + 2d + 5q = 30

She received at least 1 of each coin.

n>=1

d>=1

q>=1

There are more dimes than nickels.

d>n

There is an even number of quarters

2|q

There are as many quarters and nickels to get her as there are dimes:

q + n = d

plug q + n = d into n + 2d + 5q = 30

n + 2(q + n) + 5q = 30

3n + 7q = 30

since 2|q follows 2|n

let q = 2 by solving for n we have

3n + 7*2 = 30

3n = 30 - 14

3n = 16

n = 16/3

follows q is not 2 since 16/3 is not integer

for q = 4

3n + 7*4 = 30

3n = 30 - 28

n = 2/3

follows q is not 4 since 2/3 is not integer

for q = 6

3n + 7*6 = 30

3n = - 12

but n > 0 thus this problem has no solutions.