Michael had three bags of maize, S, T and U and the mass of each bag of maize was in the ratio 5 : 1 : 3. Michael decided to transfer 40% of maize from Bag S into Bag T and 90% of maize from Bag U into Bag T. Given that the mass of Bag T was 22.8 kg in the end, how many kilograms of maize were transferred into Bag T?
| |
S |
T |
U |
| Before |
5 u |
1 u |
3 u |
| Change 1 |
- 2 u |
+ 2 u |
|
| Change 2 |
|
+ 2.7 u |
- 2.7 u |
| After |
3 u |
5.7 u |
0.3 u |
Mass of maize transferred from Bag S into Bag T
= 40% x 5 u
=
40100 x 5 u
= 3 u
Mass of maize transferred from Bag U into Bag T
= 90% x 3 u
=
90100 x 3 u
= 2.7 u
Mass of maize in Bag T in the end
= 1 u + 2 u + 2.7 u
= 5.7 u
5.7 u = 22.8
1 u = 22.8 ÷ 5.7 = 4
Mass of maize transferred into Bag T
= 2 u + 2.7 u
= 4.7 u
= 4.7 x 4
= 18.8 kg
Answer(s): 18.8 kg
