Pierre had 2 coins and Liam 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 28 coins. How many ten-cent coins did they have at first?
| |
Ten-cent coins |
Fifty-cent coins |
Total |
| Before |
28 - 5 u |
1 u |
8 |
| Change |
+ 5 u |
- 1 u |
+ 4 u |
| After |
28 |
0 |
28 |
Number of coins at first
= 2 + 6
= 8
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 = 28 - 8
4 u = 20
1 u = 20 ÷ 4 = 5
Number of ten-cent coins at first
= 28 - 5 u
= 28 - 5 x 5
= 28 - 25
= 3
Answer(s): 3
