Ahmad had 5 coins and David had 4 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 29 coins. How many ten-cent coins did they have at first?
|
Ten-cent coins |
Fifty-cent coins |
Total |
Before |
29 - 5 u |
1 u |
9 |
Change |
+ 5 u |
- 1 u |
+ 4 u |
After |
29 |
0 |
29 |
Number of coins at first
= 5 + 4
= 9
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 = 29 - 9
4 u = 20
1 u = 20 ÷ 4 = 5
Number of ten-cent coins at first
= 29 - 5 u
= 29 - 5 x 5
= 29 - 25
= 4
Answer(s): 4