Ivan, Ahmad and Howard had a total of 115 buttons. The ratio of the number of buttons that Ahmad had to the number of buttons that Howard had was 8 : 5. After Ivan and Ahmad each gave away half of their buttons, the 3 children had 75 buttons left. How many buttons did Ahmad and Howard have at first?
| |
Ivan |
Ahmad |
Howard |
Total |
| Before |
2 p |
8 u |
5 u |
115 |
| Change |
- 1 p |
- 4 u |
|
- 40 |
| After |
1 p |
4 u |
5 u |
75 |
Total number of buttons that Ivan and Ahmad gave away
= 115 - 75
= 40
1 p + 4 u = 40 --- (1)
1 p + 4 u + 5 u = 75
1 p + 9 u = 75 --- (2)
(2) - (1)
(1 p + 9 u) - (1 p + 4 u) = 75 - 40
9 u - 4 u = 35
5 u = 35
1 u = 35 ÷ 5 = 7
Total number of buttons that Ahmad and Howard had at first
= 8 u + 5 u
= 13 u
= 13 x 7
= 91
Answer(s): 91
