Albert had three bags of barley, C, D and E and the mass of each bag of barley was in the ratio 1 : 3 : 4. Albert decided to transfer 10% of barley from Bag C into Bag D and 60% of barley from Bag E into Bag D. Given that the mass of Bag D was 22 kg in the end, how many kilograms of barley were transferred into Bag D?
| |
C |
D |
E |
| Before |
1 u |
3 u |
4 u |
| Change 1 |
- 0.1 u |
+ 0.1 u |
|
| Change 2 |
|
+ 2.4 u |
- 2.4 u |
| After |
0.9 u |
5.5 u |
1.6 u |
Mass of barley transferred from Bag C into Bag D
= 10% x 1 u
=
10100 x 1 u
= 0.9 u
Mass of barley transferred from Bag E into Bag D
= 60% x 4 u
=
60100 x 4 u
= 2.4 u
Mass of barley in Bag D in the end
= 3 u + 0.1 u + 2.4 u
= 5.5 u
5.5 u = 22
1 u = 22 ÷ 5.5 = 4
Mass of barley transferred into Bag D
= 0.1 u + 2.4 u
= 2.5 u
= 2.5 x 4
= 10 kg
Answer(s): 10 kg
