Brandon, Bobby and Jeremy shared a bag of pens. Bobby took 25 as many pens as Jeremy. Brandon took twice as many pens as the total Bobby and Jeremy took. After Brandon had given 47 to Bobby and 16 to Jeremy, Bobby gave 2 to Jeremy. In the end, all three of them had the same number of pens. Find the difference between the number of pens that Brandon and Bobby had at first.
| |
Bobby |
Jeremy |
Brandon |
Total |
| Before |
2 u |
5 u |
14 u |
21 u |
| Change 1 |
+ 47 |
|
- 47 |
|
| Change 2 |
|
+ 16 |
- 16 |
|
| Change 3 |
- 2 |
+ 2 |
|
|
| After |
1x7 =
7 u |
1x7 =
7 u |
1x7 =
7 u |
3x7 =
21 u |
Total number of pens that Bobby and Jeremy had at first
= 2 u + 5 u
= 7 u
Number of pens that Brandon had at first
= 2 x 7 u
= 14 u
The total number of pens remains unchanged. Make the total number of pens the same. LCM of 3 and 21 is 21.
Number of pens that Jeremy received from Bobby and Brandon
= 7 u - 5 u
= 2 u
2 u = 16 + 2
2 u = 18
1 u = 18 ÷ 2 = 9
Difference between the number pens that Brandon and Bobby had at first
= 14 u - 2 u
= 12 u
= 12 x 9
= 108
Answer(s): 108
