Ian had three bags of maize, V, W and X and the mass of each bag of maize was in the ratio 2 : 4 : 1. Ian decided to transfer 10% of maize from Bag V into Bag W and 60% of maize from Bag X into Bag W. Given that the mass of Bag W was 14.4 kg in the end, how many kilograms of maize were transferred into Bag W?
| |
V |
W |
X |
| Before |
2 u |
4 u |
1 u |
| Change 1 |
- 0.2 u |
+ 0.2 u |
|
| Change 2 |
|
+ 0.6 u |
- 0.6 u |
| After |
1.8 u |
4.8 u |
0.4 u |
Mass of maize transferred from Bag V into Bag W
= 10% x 2 u
=
10100 x 2 u
= 1.8 u
Mass of maize transferred from Bag X into Bag W
= 60% x 1 u
=
60100 x 1 u
= 0.6 u
Mass of maize in Bag W in the end
= 4 u + 0.2 u + 0.6 u
= 4.8 u
4.8 u = 14.4
1 u = 14.4 ÷ 4.8 = 3
Mass of maize transferred into Bag W
= 0.2 u + 0.6 u
= 0.8 u
= 0.8 x 3
= 2.4 kg
Answer(s): 2.4 kg
