Three girls, Barbara, Xylia and Kylie had a total of 3900 coins. Some exchanges were made. First, Barbara gave Xylia as many coins as Xylia had. After that, Xylia gave
13 of whatever she had then to Kylie. Finally, Kylie gave
15 of whatever she had to Barbara. As a result, they each had the same number of coins. How many coins did Barbara have at first?
|
Barbara |
Xylia |
Kylie |
Total |
Before 1 |
? |
1 u |
|
3900 |
Change 1 |
- 1 u |
+ 1 u |
|
|
After 1 |
|
2 u |
|
|
Before 2 |
|
3 p |
|
3900 |
Change 2 |
|
- 1 p |
+ 1 p |
|
After 2 |
|
2 p |
|
|
Before 3 |
|
|
5 boxes |
3900 |
Change 3 |
+ 1 box |
|
- 1 box |
|
After 3 |
|
|
4 boxes |
|
After 3 |
1 group |
1 group |
1 group |
3900 |
3 groups = 3900
1 group = 3900 ÷ 3 = 1300
1 group = 4 boxes 4 boxes = 1300
1 box = 1300 ÷ 4 = 325
2 p = 1 group
1 p = 1300 ÷ 2 = 650
3 p = 3 x 650 = 1950
3 p = 2 u
2 u = 1950
1 u = 1950 ÷ 2 = 975
Number of coins that Barbara had at first
= 1 group - 1 box + 1 u
= 1300 - 325 + 975
= 1950
Answer(s): 1950