There were blue, white and silver balls in a box. Sean put another 6 blue balls and 5 silver balls in the box, then the ratio of the number of white balls to the number of silver balls became 6 : 5. Then, he doubled the number of blue balls and removed 17 silver 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 52 balls left in the box. Find the number of blue balls that he had in the end.
| |
Blue balls |
White balls |
Silver balls |
Total |
| Before |
6 u - 6 |
6 u |
5 u - 5 |
|
| Change 1 |
+ 6 |
|
+ 5 |
|
After 1
|
1x6 =
6 u |
6 u |
5 u |
|
Change 2
|
+ 1x6 =
6 u |
|
- 17 |
|
After 2
|
2x6 =
12 u |
1x6 =
6 u |
5 u - 17 |
52 |
The number of white balls remains unchanged. Make the number of white balls the same. LCM of 1 and 6 is 6.
Total number of balls in the end
= 12 u + 6 u + 5 u - 17
= 23 u - 17
23 u - 17 = 52
23 u = 52 + 17
23 u = 69
1 u = 69 ÷ 23 = 3
Number of blue balls in the end
= 12 u
= 12 x 3
= 36
Answer(s): 36
