Jean has a bag containing some green and gold buttons. If she removes 3 green buttons, 40% of the buttons in the bag are gold. If she removes another 55 gold buttons, 20% of the buttons in the bag are gold. How many green buttons are in the bag?
| |
Green buttons |
Gold buttons |
| Before |
12 u + 3 |
8 u |
| Change 1 |
- 3 |
|
| After 1 |
3x4 =
12 u |
2x4 =
8 u |
| Change 2 |
|
- 55 |
| After 2 |
4x3 =
12 u |
1x3 =
3 u |
40% =
40100 =
2520% =
20100 =
15The number of green buttons remains unchanged at the second change when she removes another 55 gold buttons. Make the number of green buttons the same at the second change. LCM of 3 and 4 is 12.
Number of gold buttons removed
= 8 u - 3 u
= 5 u
5 u = 55
1 u = 55 ÷ 5 = 11
Number of green buttons at first
= 12 u + 3
= 12 x 11 + 3
= 132 + 3
= 135
Answer(s): 135
