HELP ME with this question PLEAse?

Question

if i have 50 coins, all quarters and dimes, worht 9.20. how many of each do i have???

Answer 1

d = # of dimes, q = # of quarters
d+q=50
10d +25q = 920

use substitution to find the # of each...
d = 50 - q
so 10(50-q) + 25q =920
so 500-10q +25q = 920
500+15q = 920
15q = 420
q = 28
so d = 50-28 which is 22
so 28 quarters and 22 dimes
check it by 28 * .25 = 7 and 22*.10 = 2.20 and $7 +$2.20 is $9.20

Answer 2

$9.20 is 920 cents

Trial & Error method
(If you have time, using binary search method to speed up search if needed. When you give up a Dime for a Quarter, you increase the total by 15 cents)

1 Q + 49 D = 1x25 + 49x10 = 515 (you need 920)

25 Q + 25 D = 25x25 + 25x10 = 872

30 Q + 20 D = 30x25 + 20x10 = 950

28 Q + 22 D = 28x25 + 22x10 = 920

So, you need 28 quarters and 22 dimes.

Answer 3

Let x be the number of quarters you have.
Let y be the number of dimes you have.

You then know that:
x + y = 50
0.25*x + 0.10*y = 9.20

Rearranging the first equation, we get: y = 50 - x
Plugging that into the second equation, we get:
0.25*x + 0.10*(50-x) = 9.20

I'd multiply everything through by 20, just to get rid of annoying fractions.
5x + 2*(50-x) = 184

5x + 100 - 2x = 184
5x - 2x = 184 - 100
3x = 84
x = 28

So you have 28 quarters.
Since you have 50 coins total, that means you have 22 dimes.

Answer 4

Assume x is quarter and y is dime.

x+y=50
0.25x+0.1y=9.2

x=50-y

0.25(50-y)+0.1y=9.2
12.5-0.25y+0.1y=9.2
0.15y=3.3
y=22
x=50-22 = 28

Answer 5

0.25x+0.10y=9.20
x+y=50