Howard and Ethan had some beads in the ratio of 12 : 7. Ethan and Oscar had some beads in the ratio of 10 : 1. Howard sold 765 beads to Oscar and ended up with half the number of beads Ethan had. How many beads did Oscar have in the end?
| Howard |
Ethan |
Oscar |
| 12x10 |
7x10 |
|
| |
10x7 |
1x7 |
| 120 u |
70 u |
7 u |
The number of beads that Ethan had at first is repeated. Make the number of beads that Ethan had at first the same. LCM of 7 and 10 is 70.
| |
Howard |
Ethan |
Oscar |
| Before |
120 u |
70 u |
7 u |
| Change |
- 765 |
|
+ 765 |
| After |
35 u |
70 u |
7 u + 765 |
Number of beads that Howard had in the end
= 70 u ÷ 2
= 35 u
Number of beads that Howard sold to Oscar
= 120 u - 35 u
= 85 u
85 u = 765
1 u = 765 ÷ 85 = 9
Number of beads that Oscar had in the end
= 7 u + 765
= 7 x 9 + 765
= 63 + 765
= 828
Answer(s): 828
