Liam and Albert had some marbles in the ratio of 11 : 9. Albert and Xavier had some marbles in the ratio of 8 : 1. Liam sold 156 marbles to Xavier and ended up with half the number of marbles Albert had. How many marbles did Xavier have in the end?
| Liam |
Albert |
Xavier |
| 11x8 |
9x8 |
|
| |
8x9 |
1x9 |
| 88 u |
72 u |
9 u |
The number of marbles that Albert had at first is repeated. Make the number of marbles that Albert had at first the same. LCM of 9 and 8 is 72.
| |
Liam |
Albert |
Xavier |
| Before |
88 u |
72 u |
9 u |
| Change |
- 156 |
|
+ 156 |
| After |
36 u |
72 u |
9 u + 156 |
Number of marbles that Liam had in the end
= 72 u ÷ 2
= 36 u
Number of marbles that Liam sold to Xavier
= 88 u - 36 u
= 52 u
52 u = 156
1 u = 156 ÷ 52 = 3
Number of marbles that Xavier had in the end
= 9 u + 156
= 9 x 3 + 156
= 27 + 156
= 183
Answer(s): 183
