+53 Dylan's dimes total $1.30 less than his quarters of which he has one-fourth as many as he has nickels, which total $0 more than his dimes. How many of each coin does he have?
+364 @ guest I think you are wrong
reason:
'Dylan's dimes total $1.30 less than his quarters "
so the quarters have to be at least 6 of them
My attempt:
So we need 2x as many nickels as dimes
so
| quarters | dimes | nickels | |||
| 6 | 2 | 4 | n | o | |
| 8 | 7 | 14 | n | o | |
| 10 | 12 | 24 | n | o | |
| 12 | 17 | 34 | n | o | |
| 14 | 22 | 44 | n | o |
| 16 | 27 | 54 | n | o |
| 18 | 32 | 64 | n | o | |
| 20 | 37 | 74 | n | o | |
| 22 | 42 | 84 | n | o |
| 24 | 47 | 94 | n | o | |
| 26 | 52 | 104 | y | e | s |
Thhis is acually a pretty bad method of doing it but if you noticed, there is a pattern
I'll let you find it
+237 Number of nickels = x ---> cost = $0.05x
Number of quarters = x/4 ---> cost = ($0.25 * x)/4
Cost of dimes = same as of nickels = $0,05x
No. of dimes * 0.1 = 0.05x
No. of dimes = (0.05x)/0.1 = 0.5x
Also cost of dime = Cost of quarters - $1.3
0.05x = (0.25x)/4 - 1.3
=> 1.3 = 0.0125x - 0.05x
=> 1.3 = 0.0125x
=> x = 1.3/0.0125 = 104
He have 104 nickels, 104/4 = 26 quarters and 1/2 * 104 = 52 dimes
