algebra problem"Find the number of coins and their denomination?

(0.05)n+ (0.1)d + (0.25)q= 20 where n = number of nickels , d = number of dimes and q = number of quarters

d= 10 q

n= 2(d+q) +24 - > n=2d+2q+24 -> n = 22q+24

(.05) (22q+24) + q + q/4 = 20

5 (22q+24) +100q+25q = 2000

110q +120 +125q = 2000

235q=1880

q= 1880/235 = 8

d = 10q=80

n = 22 *8 +24 = 200

check : 8 /4 +8 + 10 = 20 ...correct

Let N, D and Q be the number of nickels, dimes and quarters respectively.

Now you just need to write the appropriate equations and solve:

A sum of money amounting to $20 consists of nickels dimes, and quarters.

If you take each coin times its value, you get the total value. Let's do this in cents, to make it simpler:

5N + 10D + 25Q = 2000

There are ten times as many dimes as quarters.

Stated another way, the number of dimes is ten times the number of quarters.

D = 10Q

The number of nickels exceed twice the number of dimes and quarters by 24.

N = 2(D+Q) + 24

Let's substitute D = 10Q into the other two equations:

5N + 10(10Q) + 25Q = 2000

N = 2(10Q + Q) + 24

Simplify the second equation:

N = 2(11Q) + 24

N = 22Q + 24

Substitute this into the first equation:

5(22Q + 24) + 10(10Q) + 25Q = 2000

110Q + 120 + 100Q + 25Q = 2000

Group terms:

(110Q + 100Q + 25Q) + 120 = 2000

235Q + 120 = 2000

235Q = 2000 - 120

235Q = 1880

Q = 1880/235

Q = 8

Now solve for the other values:

D = 10Q

D = 10(8)

D = 80

N = 22Q + 24

N = 22(8) + 24

N = 176 + 24

N = 200

Answer:

200 nickels, 80 dimes, 8 quarters

Double-checking:

Total value is $10 + $8 + $2 = $20

There are ten times as many dimes as quarters (80 dimes, 8 quarters)

The number of nickels (200) exceeds twice the number of dimes and quarters (88) by 24. Two times 88 = 176. Add 24 and you get 200.