Help solve this algebra word problem?

The total = $20, therefore,

The number of quarters x .25 +

the number of dimes x .1 +

the number of nickels x .05 = $20

.25q + .1d + .05n = $20

The number of dimes = 10 times the number of quarters

d = 10q

The number of nickels is = the number of dimes + the number of quarters + 24

n = d + q + 24

let q = 10

by substitution:

d = 100

n = 100 + 10 + 24 = 134

if so, .25 * 10 + .1 * 100 + .05 * 134 = $2.50 + $10 + $6.70 = $19.20 = not enough quarters

let q = 11

by substitution

d = 110

n = 110 + 11 + 24 = 145

if so, .25 x 11 + .1 x 110 + .05 x 145 = $2.75 + $11 + $7.25 = $21 = too many quarters

For lower numbers of q, both d and n decrease, and so does the total.

For higher numbers of q, both d and q increase, and so does the total.

This means there is no correct answer, and something is, indeed, wrong with the problem statement

Just for fun, I wrote a little computer program, and here's its output for d = 10q and n = d + q + 24, where q ranges from 1 to 24:

1 quarter + 10 dimes + 35 nickles = $3.00

2 quarters + 20 dimes + 46 nickles = $4.80

3 quarters + 30 dimes + 57 nickles = $6.60

4 quarters + 40 dimes + 68 nickles = $8.40

5 quarters + 50 dimes + 79 nickles = $10.20

6 quarters + 60 dimes + 90 nickles = $12.00

7 quarters + 70 dimes + 101 nickles = $13.80

8 quarters + 80 dimes + 112 nickles = $15.60

9 quarters + 90 dimes + 123 nickles = $17.40

10 quarters + 100 dimes + 134 nickles = $19.20

11 quarters + 110 dimes + 145 nickles = $21.00

12 quarters + 120 dimes + 156 nickles = $22.80

13 quarters + 130 dimes + 167 nickles = $24.60

14 quarters + 140 dimes + 178 nickles = $26.40

15 quarters + 150 dimes + 189 nickles = $28.20

16 quarters + 160 dimes + 200 nickles = $30.00

17 quarters + 170 dimes + 211 nickles = $31.80

18 quarters + 180 dimes + 222 nickles = $33.60

19 quarters + 190 dimes + 233 nickles = $35.40

20 quarters + 200 dimes + 244 nickles = $37.20

21 quarters + 210 dimes + 255 nickles = $39.00

22 quarters + 220 dimes + 266 nickles = $40.80

23 quarters + 230 dimes + 277 nickles = $42.60

24 quarters + 240 dimes + 288 nickles = $44.40

If one were to choose a "sensible" dollar amount for such an algebra problem, one might choose a number such as $300 (166 quarters + 1660 dimes + 1850 nickels.) This would probably prevent the student from doing the problem in his head, but as you could guess, it would be trivial on a computer.

Use a table.

coin ---------- amount --------- worth of each ----- total worth

quarters -------- x ------------------- .25 ----------------- .25x

dimes -------- 10x ------------------- .10 ----------------- .10 * 10x

nickels ------- x + 10x + 24 ------- .05 ----------------- .05(11x + 24)

.25x + .10 * 10x + .05(11x + 24) = 20

25x + 100x + 55x + 120 = 2000

180x = 1880

Since 1880 is not divisible by 180 I suspect you made a mistake somewhere. Check your work, fix it, then make the necessary changes in the above work to get your answer.

your 2 numbers are: x and x+35. to discover the numbers, plug them into the formula. which might become: 2(x+35) - x=87 (multiply 2 with each and each huge style in the parenthesis) 2x + 70 -x=87 (subtract 70 from the two factors) x=17 So your 2 numbers are 17 and fifty two(have been given fifty two from 35+17) desire that this facilitates.