There were brown, green and red balls in a jar. Sean put another 9 brown balls and 3 red balls in the jar, then the ratio of the number of green balls to the number of red balls became 6 : 5. Then, he doubled the number of brown balls and removed 25 red balls. The ratio of the number of brown balls to green balls became 2 : 1. He counted and found that there were a total of 67 balls left in the jar. Find the number of brown balls that he had at first.
| |
Brown balls |
Green balls |
Red balls |
Total |
| Before |
6 u - 9 |
6 u |
5 u - 3 |
|
| Change 1 |
+ 9 |
|
+ 3 |
|
After 1
|
1x6 =
6 u |
6 u |
5 u |
|
Change 2
|
+ 1x6 =
6 u |
|
- 25 |
|
After 2
|
2x6 =
12 u |
1x6 =
6 u |
5 u - 25 |
67 |
The number of green balls remains unchanged. Make the number of green balls the same. LCM of 1 and 6 is 6.
Total number of balls in the end
= 12 u + 6 u + 5 u - 25
= 23 u - 25
23 u - 25 = 67
23 u = 67 + 25
23 u = 92
1 u = 92 ÷ 23 = 4
Number of brown balls at first
= 6 u - 9
= 6 x 4 - 9
= 24 - 9
= 15
Answer(s): 15
