Asher had three bags of maize, D, E and F and the mass of each bag of maize was in the ratio 5 : 3 : 1. Asher decided to transfer 10% of maize from Bag D into Bag E and 60% of maize from Bag F into Bag E. Given that the mass of Bag E was 28.7 kg in the end, how many kilograms of maize were transferred into Bag E?
| |
D |
E |
F |
| Before |
5 u |
3 u |
1 u |
| Change 1 |
- 0.5 u |
+ 0.5 u |
|
| Change 2 |
|
+ 0.6 u |
- 0.6 u |
| After |
4.5 u |
4.1 u |
0.4 u |
Mass of maize transferred from Bag D into Bag E
= 10% x 5 u
=
10100 x 5 u
= 4.5 u
Mass of maize transferred from Bag F into Bag E
= 60% x 1 u
=
60100 x 1 u
= 0.6 u
Mass of maize in Bag E in the end
= 3 u + 0.5 u + 0.6 u
= 4.1 u
4.1 u = 28.7
1 u = 28.7 ÷ 4.1 = 7
Mass of maize transferred into Bag E
= 0.5 u + 0.6 u
= 1.1 u
= 1.1 x 7
= 7.7 kg
Answer(s): 7.7 kg
