There were some beads in Packet M and Packet N. After Henry transferred
25 of the beads from Packet M to Packet N, the ratio of the number of beads in Packet M to that in Packet N changed to 2 : 7. Find the ratio of the numbers of beads in Packet M to the number of beads in Packet N at first.
|
Packet M |
Packet N |
Before |
5x2 = 10 u |
17 u |
Change |
- 2x2 = - 4 u |
+ 4 u |
After |
3x2 = 6 u |
|
Comparing Packet M and Packet N in the end |
2x3 = 6 u |
7x3 = 21 u |
The number of beads in Packet M in the end is repeated. Make the number of beads in Packet M in the end the same. LCM of 3 and 2 is 6.
At first
Packet M : Packet N
10 : 17
Answer(s): 10 : 17