Billy had three bags of maize, S, T and U and the mass of each bag of maize was in the ratio 1 : 5 : 3. Billy decided to transfer 10% of maize from Bag S into Bag T and 70% of maize from Bag U into Bag T. Given that the mass of Bag T was 57.6 kg in the end, how many kilograms of maize were transferred into Bag T?
| |
S |
T |
U |
| Before |
1 u |
5 u |
3 u |
| Change 1 |
- 0.1 u |
+ 0.1 u |
|
| Change 2 |
|
+ 2.1 u |
- 2.1 u |
| After |
0.9 u |
7.2 u |
0.9 u |
Mass of maize transferred from Bag S into Bag T
= 10% x 1 u
=
10100 x 1 u
= 0.9 u
Mass of maize transferred from Bag U into Bag T
= 70% x 3 u
=
70100 x 3 u
= 2.1 u
Mass of maize in Bag T in the end
= 5 u + 0.1 u + 2.1 u
= 7.2 u
7.2 u = 57.6
1 u = 57.6 ÷ 7.2 = 8
Mass of maize transferred into Bag T
= 0.1 u + 2.1 u
= 2.2 u
= 2.2 x 8
= 17.6 kg
Answer(s): 17.6 kg
