Tom, Brandon and Sean shared a bag of cards. Brandon took 35 as many cards as Sean. Tom took twice as many cards as the total Brandon and Sean took. After Tom had given 38 to Brandon and 18 to Sean, Brandon gave 3 to Sean. In the end, all three of them had the same number of cards. Find the difference between the number of cards that Tom and Brandon had at first.
| |
Brandon |
Sean |
Tom |
Total |
| Before |
3 u |
5 u |
16 u |
24 u |
| Change 1 |
+ 38 |
|
- 38 |
|
| Change 2 |
|
+ 18 |
- 18 |
|
| Change 3 |
- 3 |
+ 3 |
|
|
| After |
1x8 =
8 u |
1x8 =
8 u |
1x8 =
8 u |
3x8 =
24 u |
Total number of cards that Brandon and Sean had at first
= 3 u + 5 u
= 8 u
Number of cards that Tom had at first
= 2 x 8 u
= 16 u
The total number of cards remains unchanged. Make the total number of cards the same. LCM of 3 and 24 is 24.
Number of cards that Sean received from Brandon and Tom
= 8 u - 5 u
= 3 u
3 u = 18 + 3
3 u = 21
1 u = 21 ÷ 3 = 7
Difference between the number cards that Tom and Brandon had at first
= 16 u - 3 u
= 13 u
= 13 x 7
= 91
Answer(s): 91
