Min and Fanny had 860 coins altogether. They decided to play a game. In the first round, Fanny lost 63 coins to Min. In the second round, Min lost 33 coins to Fanny. At the end of the two rounds, Fanny had thrice as many coins as Min.
- How many coins did Min have in the end?
- How many coins did Fanny have at the beginning of the game?
| |
Min |
Fanny |
Total |
| Before |
1 u + 33 - 63 |
3 u - 33 + 63 |
860 |
| Round 1 |
+ 63 |
- 63 |
|
| Round 2 |
- 33 |
+ 33 |
|
| After |
1 u |
3 u |
860 |
(a)
Total number of coins
= 1 u + 3 u
= 4 u
4 u = 860
1 u = 860 ÷ 4 = 215
Number of coins that Min had in the end
= 1 u
= 215
(b)
Working backwards.
Number of coins that Fanny had at the end of round 2
= 3 u
= 3 x 215
= 645
Number of coins that Fanny had at the end of round 1
= 645 - 33
= 612
Number of coins that Fanny had at the beginning of the game
= 612 + 63
= 675
Answer(s): (a) 215; (b) 675
