Jane has a bag containing some orange and green coins. If she removes 10 orange coins, 20% of the coins in the bag are green. If she removes another 30 green coins, 10% of the coins in the bag are green. How many orange coins are in the bag?
| |
Orange coins |
Green coins |
| Before |
36 u + 10 |
9 u |
| Change 1 |
- 10 |
|
| After 1 |
4x9 =
36 u |
1x9 =
9 u |
| Change 2 |
|
- 30 |
| After 2 |
9x4 =
36 u |
1x4 =
4 u |
20% =
20100 =
1510% =
10100 =
110The number of orange coins remains unchanged at the second change when she removes another 30 green coins. Make the number of orange coins the same at the second change. LCM of 4 and 9 is 36.
Number of green coins removed
= 9 u - 4 u
= 5 u
5 u = 30
1 u = 30 ÷ 5 = 6
Number of orange coins at first
= 36 u + 10
= 36 x 6 + 10
= 216 + 10
= 226
Answer(s): 226
