Flynn had three bags of oats, S, T and U and the mass of each bag of oats was in the ratio 5 : 4 : 5. Flynn decided to transfer 20% of oats from Bag S into Bag T and 80% of oats from Bag U into Bag T. Given that the mass of Bag T was 81 kg in the end, how many kilograms of oats were transferred into Bag T?
| |
S |
T |
U |
| Before |
5 u |
4 u |
5 u |
| Change 1 |
- 1 u |
+ 1 u |
|
| Change 2 |
|
+ 4 u |
- 4 u |
| After |
4 u |
9 u |
1 u |
Mass of oats transferred from Bag S into Bag T
= 20% x 5 u
=
20100 x 5 u
= 4 u
Mass of oats transferred from Bag U into Bag T
= 80% x 5 u
=
80100 x 5 u
= 4 u
Mass of oats in Bag T in the end
= 4 u + 1 u + 4 u
= 9 u
9 u = 81
1 u = 81 ÷ 9 = 9
Mass of oats transferred into Bag T
= 1 u + 4 u
= 5 u
= 5 x 9
= 45 kg
Answer(s): 45 kg
