Julian had three bags of oats, C, D and E and the mass of each bag of oats was in the ratio 3 : 5 : 4. Julian decided to transfer 10% of oats from Bag C into Bag D and 80% of oats from Bag E into Bag D. Given that the mass of Bag D was 59.5 kg in the end, how many kilograms of oats were transferred into Bag D?
| |
C |
D |
E |
| Before |
3 u |
5 u |
4 u |
| Change 1 |
- 0.3 u |
+ 0.3 u |
|
| Change 2 |
|
+ 3.2 u |
- 3.2 u |
| After |
2.7 u |
8.5 u |
0.8 u |
Mass of oats transferred from Bag C into Bag D
= 10% x 3 u
=
10100 x 3 u
= 2.7 u
Mass of oats transferred from Bag E into Bag D
= 80% x 4 u
=
80100 x 4 u
= 3.2 u
Mass of oats in Bag D in the end
= 5 u + 0.3 u + 3.2 u
= 8.5 u
8.5 u = 59.5
1 u = 59.5 ÷ 8.5 = 7
Mass of oats transferred into Bag D
= 0.3 u + 3.2 u
= 3.5 u
= 3.5 x 7
= 24.5 kg
Answer(s): 24.5 kg
