Sean had 5 coins and Ryan had 6 coins. Their coins consisted of ten-cent coins and fifty-cent coins. When all their fifty-cent coins were exchanged for ten-cent coins, they had a total of 19 coins. How many ten-cent coins did they have at first?
|
Ten-cent coins |
Fifty-cent coins |
Total |
Before |
19 - 5 u |
1 u |
11 |
Change |
+ 5 u |
- 1 u |
+ 4 u |
After |
19 |
0 |
19 |
Number of coins at first
= 5 + 6
= 11
1 fifty-cent coin can be exchanged for 5 ten-cent coins.
Increase in the number of coins after all the fifty-cent coins were exchanged for ten-cent coins
= 5 u - 1 u
= 4 u
4 u = 19 - 11
4 u = 8
1 u = 8 ÷ 4 = 2
Number of ten-cent coins at first
= 19 - 5 u
= 19 - 5 x 2
= 19 - 10
= 9
Answer(s): 9