Henry had 10 coins and Paul had 7 coins. Their coins consisted of ten-cent coins and twenty-cent coins. When all their twenty-cent coins were exchanged for ten-cent coins, they had a total of 21 coins. How many ten-cent coins did they have at first?
|
Ten-cent coins |
Twenty-cent coins |
Total |
Before |
21 - 2 u |
1 u |
17 |
Change |
+ 2 u |
- 1 u |
+ 1 u |
After |
21 |
0 |
21 |
Number of coins at first
= 10 + 7
= 17
1 twenty-cent coin can be exchanged for 2 ten-cent coins.
Increase in the number of coins after all the twenty-cent coins were exchanged for ten-cent coins
= 2 u - 1 u
= 1 u
1 u = 21 - 17 = 4
Number of ten-cent coins at first
= 21 - 2 u
= 21 - 2 x 4
= 21 - 8
= 13
Answer(s): 13