There were blue, white and grey balls in a container. Sean put another 3 blue balls and 8 grey balls in the container, then the ratio of the number of white balls to the number of grey balls became 4 : 3. Then, he doubled the number of blue balls and removed 18 grey balls. The ratio of the number of blue balls to white balls became 2 : 1. He counted and found that there were a total of 117 balls left in the container. Find the number of blue balls that he had at first.
| |
Blue balls |
White balls |
Grey balls |
Total |
| Before |
4 u - 3 |
4 u |
3 u - 8 |
|
| Change 1 |
+ 3 |
|
+ 8 |
|
After 1
|
1x4 =
4 u |
4 u |
3 u |
|
Change 2
|
+ 1x4 =
4 u |
|
- 18 |
|
After 2
|
2x4 =
8 u |
1x4 =
4 u |
3 u - 18 |
117 |
The number of white balls remains unchanged. Make the number of white balls the same. LCM of 1 and 4 is 4.
Total number of balls in the end
= 8 u + 4 u + 3 u - 18
= 15 u - 18
15 u - 18 = 117
15 u = 117 + 18
15 u = 135
1 u = 135 ÷ 15 = 9
Number of blue balls at first
= 4 u - 3
= 4 x 9 - 3
= 36 - 3
= 33
Answer(s): 33
