Jen and Elyse have some beads. If Jen gives 21 beads to Elyse, Elyse will have 5 times the number of beads as Jen. If Elyse gives 77 beads to Jen, Jen will have
23 as many beads as Elyse. How many beads does Jen have at first?
|
Case 1 |
Case 2 |
|
Jen |
Elyse |
Jen |
Elyse |
Before |
5 u + 21 |
25 u - 21 |
12 u - 77 |
18 u + 77 |
Change |
- 21 |
+ 21 |
+ 77 |
- 77 |
After |
1x5 = 5 u |
5x5 = 25 u |
2x6 = 12 u |
3x6 = 18 u |
The total number of beads in both cases remains unchanged. Make the total number of beads in both cases the same. LCM of 6 and 5 is 30.
Number of beads that Jen had at first is the same in both cases.
12 u - 77 = 5 u + 21
12 u - 5 u = 21 + 77
7 u = 98
1 u = 14
Number of beads that Jen had at first
= 5 u + 21
= 5 x 14 + 21
= 70 + 21
= 91
Answer(s): 91