Michael has 6 more 50-cent coins than 10-cent coins. After he used 7 of his 50-cent coins, the total value of 50-cent coins is $3.90 more than the total value of 10-cent coins. How many coins did he have at first?
| |
50¢ |
10¢ |
| Before |
1 u + 6 |
1 u |
| Change |
- 7 |
|
| After |
1 u - 1 |
1 u |
| |
50¢ |
10¢ |
| Number |
1 u - 1 |
1 u |
| Value |
50 |
10 |
| Total value |
50 u - 50 |
10 u |
$1 = 100¢
$3.90 = 390¢
Number of 50¢ coins in the end
= 1 u + 6 - 7
= 1 u - 1
Total value of 50¢ coins in the end
= 50 x (1 u - 1)
= 50 u - 50
Total value of 10¢ coins in the end
= 10 x 1 u
= 10 u
After using 7 50-cent coins, the value of 50-cent coins is 390¢ more than the value of 10-cent coins. If another 390¢ is added to the 10-cent coins, the total value of 50-cent coins and 10-cent coins will be the same.
50 u - 50 = 10 u + 390
50 u - 10 u = 390 + 50
40 u = 440
1 u = 440 ÷ 40 = 11
Number of coins that Michael had at first
= 1 u + 1 u + 6
= 2 u + 6
= (2 x 11) + 6
= 22 + 6
= 28
Answer(s): 28
