Pierre has 2 more 20-cent coins than 10-cent coins. After he used 19 of his 20-cent coins, the total value of 20-cent coins is $1.10 more than the total value of 10-cent coins. How many coins did he have at first?
|
20¢ |
10¢ |
Before |
1 u + 2 |
1 u |
Change |
- 19 |
|
After |
1 u - 17 |
1 u |
|
20¢ |
10¢ |
Number |
1 u - 17 |
1 u |
Value |
20 |
10 |
Total value |
20 u - 340 |
10 u |
$1 = 100¢
$1.10 = 110¢
Number of 20¢ coins in the end
= 1 u + 2 - 19
= 1 u - 17
Total value of 20¢ coins in the end
= 20 x (1 u - 17)
= 20 u - 340
Total value of 10¢ coins in the end
= 10 x 1 u
= 10 u
After using 19 20-cent coins, the value of 20-cent coins is 110¢ more than the value of 10-cent coins. If another 110¢ is added to the 10-cent coins, the total value of 20-cent coins and 10-cent coins will be the same.
20 u - 340 = 10 u + 110
20 u - 10 u = 110 + 340
10 u = 450
1 u = 450 ÷ 10 = 45
Number of coins that Pierre had at first
= 1 u + 1 u + 2
= 2 u + 2
= (2 x 45) + 2
= 90 + 2
= 92
Answer(s): 92