Pierre has 10-cent coins and 20-cent coins. The total value of all his coins is $36. He has 9 less 10-cent coins than 20-cent coins. How many coins does he have altogether?
|
10-cent |
20-cent |
Total |
Number |
1 u |
1 u + 9 |
2 u + 9 |
Value |
10 |
20 |
|
Total value |
10 u |
20 u + 180 |
30 u + 180 |
$1 = 100¢
$36 = 36 x 100 = 3600¢
Pierre has 9 more 20-cent coins than 10-cent coins.
Number of 10-cent coins = 1 u
Number of 20-cent coins = 1 u + 9
Total value of 10-cent coins
= 10 x 1 u
= 10 u
Total value of 20-cent coins
= 20 x (1 u + 9)
= 20 u + 180
Total value of the coins
= 10 u + 20 u + 180
= 30 u + 180
30 u + 180 = 3600
30 u = 3600 - 180
30 u = 3420
1 u = 3420 ÷ 30 = 114
Total number of coins
= 1 u + (1 u + 9)
= 2 u + 9
= 2 x 114 + 9
= 228 + 9
= 237
Answer(s): 237