Jeremy had 7 coins and Albert had 3 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 22 coins. How many ten-cent coins did they have at first?
| |
Ten-cent coins |
Fifty-cent coins |
Total |
| Before |
22 - 5 u |
1 u |
10 |
| Change |
+ 5 u |
- 1 u |
+ 4 u |
| After |
22 |
0 |
22 |
Number of coins at first
= 7 + 3
= 10
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 = 22 - 10
4 u = 12
1 u = 12 ÷ 4 = 3
Number of ten-cent coins at first
= 22 - 5 u
= 22 - 5 x 3
= 22 - 15
= 7
Answer(s): 7
