Seth had three bags of barley, X, Y and Z and the mass of each bag of barley was in the ratio 1 : 1 : 4. Seth decided to transfer 30% of barley from Bag X into Bag Y and 90% of barley from Bag Z into Bag Y. Given that the mass of Bag Y was 39.2 kg in the end, how many kilograms of barley were transferred into Bag Y?
| |
X |
Y |
Z |
| Before |
1 u |
1 u |
4 u |
| Change 1 |
- 0.3 u |
+ 0.3 u |
|
| Change 2 |
|
+ 3.6 u |
- 3.6 u |
| After |
0.7 u |
4.9 u |
0.4 u |
Mass of barley transferred from Bag X into Bag Y
= 30% x 1 u
=
30100 x 1 u
= 0.7 u
Mass of barley transferred from Bag Z into Bag Y
= 90% x 4 u
=
90100 x 4 u
= 3.6 u
Mass of barley in Bag Y in the end
= 1 u + 0.3 u + 3.6 u
= 4.9 u
4.9 u = 39.2
1 u = 39.2 ÷ 4.9 = 8
Mass of barley transferred into Bag Y
= 0.3 u + 3.6 u
= 3.9 u
= 3.9 x 8
= 31.2 kg
Answer(s): 31.2 kg
