Michael had 7 coins and Mark had 7 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 30 coins. How many ten-cent coins did they have at first?
|
Ten-cent coins |
Fifty-cent coins |
Total |
Before |
30 - 5 u |
1 u |
14 |
Change |
+ 5 u |
- 1 u |
+ 4 u |
After |
30 |
0 |
30 |
Number of coins at first
= 7 + 7
= 14
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 = 30 - 14
4 u = 16
1 u = 16 ÷ 4 = 4
Number of ten-cent coins at first
= 30 - 5 u
= 30 - 5 x 4
= 30 - 20
= 10
Answer(s): 10