Mark had 7 coins and Pierre had 12 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 27 coins. How many ten-cent coins did they have at first?
|
Ten-cent coins |
Fifty-cent coins |
Total |
Before |
27 - 5 u |
1 u |
19 |
Change |
+ 5 u |
- 1 u |
+ 4 u |
After |
27 |
0 |
27 |
Number of coins at first
= 7 + 12
= 19
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 = 27 - 19
4 u = 8
1 u = 8 ÷ 4 = 2
Number of ten-cent coins at first
= 27 - 5 u
= 27 - 5 x 2
= 27 - 10
= 17
Answer(s): 17