Howard had 5 times as many beads as Eric at first. Then each of them bought an equal number of new beads. In the end, Eric had 24 beads and Howard had 3 times as many beads as Eric. How many beads did both of them buy altogether?
| |
Howard |
Eric |
Difference |
| Before |
5x1 =
5 u |
1x1 =
1 u |
4x1 =
4 u |
| Change |
+ ? |
+ ? |
|
| After |
3x2 =
6 u |
1x2 =
2 u |
2x2 =
4 u |
The difference in the number of beads at first and in the end remains unchanged. Make the difference in the number of beads at first and in the end the same. LCM of 4 and 2 is 4.
Number of beads that Eric had in the end = 2 u
2 u = 24
1 u = 24 ÷ 2 = 12
Number of beads that Howard bought
= 6 u - 5 u
= 1 u
Number of beads that both of them bought altogether
= 2 x 1 u
= 2 u
= 2 x 12
= 24
Answer(s): 24
