Irene, Diana and Xylia had a total of 68 marbles. The ratio of Diana's marbles to Xylia's marbles was at first 6 : 7. After Irene and Diana each gave half of their marbles away, the 3 children had 41 marbles left. How many marbles did Irene and Xylia have at first?
| |
Irene |
Diana |
Xylia |
Total |
| Before |
2 p |
6 u |
7 u |
68 |
| Change |
- 1 p |
- 3 u |
|
- 27 |
| After |
1 p |
3 u |
7 u |
41 |
Number of marbles that Irene and Diana gave away
= 68 - 41
= 27
1 p + 3 u = 27
1 p = 27 - 3 u --- (1)
1 p + 3 u + 7 u = 41
1 p + 10 u = 41
1 p = 41 - 10 u --- (2)
(1) = (2)
27 - 3 u = 41 - 10 u
10 u - 3 u = 41 - 27
7 u = 14
1 u = 14 ÷ 7 = 2
From (1)
1 p = 27 - 3 u
1 p = 27 - 3 x 2
1 p = 27 - 6 = 21
Number of marbles that Irene and Xylia had at first
= 2 p + 7 u
= 2 x 21 + 7 x 2
= 42 + 14
= 56
Answer(s): 56
