Jenson had three bags of maize, D, E and F and the mass of each bag of maize was in the ratio 4 : 4 : 3. Jenson decided to transfer 40% of maize from Bag D into Bag E and 90% of maize from Bag F into Bag E. Given that the mass of Bag E was 74.7 kg in the end, how many kilograms of maize were transferred into Bag E?
| |
D |
E |
F |
| Before |
4 u |
4 u |
3 u |
| Change 1 |
- 1.6 u |
+ 1.6 u |
|
| Change 2 |
|
+ 2.7 u |
- 2.7 u |
| After |
2.4 u |
8.3 u |
0.3 u |
Mass of maize transferred from Bag D into Bag E
= 40% x 4 u
=
40100 x 4 u
= 2.4 u
Mass of maize transferred from Bag F into Bag E
= 90% x 3 u
=
90100 x 3 u
= 2.7 u
Mass of maize in Bag E in the end
= 4 u + 1.6 u + 2.7 u
= 8.3 u
8.3 u = 74.7
1 u = 74.7 ÷ 8.3 = 9
Mass of maize transferred into Bag E
= 1.6 u + 2.7 u
= 4.3 u
= 4.3 x 9
= 38.7 kg
Answer(s): 38.7 kg
