There were black, red and silver balls in a jar. Sean put another 10 black balls and 5 silver balls in the jar, then the ratio of the number of red balls to the number of silver balls became 6 : 5. Then, he doubled the number of black balls and removed 23 silver balls. The ratio of the number of black balls to red balls became 2 : 1. He counted and found that there were a total of 92 balls left in the jar. Find the number of black balls that he had in the end.
| |
Black balls |
Red balls |
Silver balls |
Total |
| Before |
6 u - 10 |
6 u |
5 u - 5 |
|
| Change 1 |
+ 10 |
|
+ 5 |
|
After 1
|
1x6 =
6 u |
6 u |
5 u |
|
Change 2
|
+ 1x6 =
6 u |
|
- 23 |
|
After 2
|
2x6 =
12 u |
1x6 =
6 u |
5 u - 23 |
92 |
The number of red balls remains unchanged. Make the number of red balls the same. LCM of 1 and 6 is 6.
Total number of balls in the end
= 12 u + 6 u + 5 u - 23
= 23 u - 23
23 u - 23 = 92
23 u = 92 + 23
23 u = 115
1 u = 115 ÷ 23 = 5
Number of black balls in the end
= 12 u
= 12 x 5
= 60
Answer(s): 60
