Lee had three bags of barley, L, M and N and the mass of each bag of barley was in the ratio 4 : 5 : 1. Lee decided to transfer 20% of barley from Bag L into Bag M and 70% of barley from Bag N into Bag M. Given that the mass of Bag M was 45.5 kg in the end, how many kilograms of barley were transferred into Bag M?
| |
L |
M |
N |
| Before |
4 u |
5 u |
1 u |
| Change 1 |
- 0.8 u |
+ 0.8 u |
|
| Change 2 |
|
+ 0.7 u |
- 0.7 u |
| After |
3.2 u |
6.5 u |
0.3 u |
Mass of barley transferred from Bag L into Bag M
= 20% x 4 u
=
20100 x 4 u
= 3.2 u
Mass of barley transferred from Bag N into Bag M
= 70% x 1 u
=
70100 x 1 u
= 0.7 u
Mass of barley in Bag M in the end
= 5 u + 0.8 u + 0.7 u
= 6.5 u
6.5 u = 45.5
1 u = 45.5 ÷ 6.5 = 7
Mass of barley transferred into Bag M
= 0.8 u + 0.7 u
= 1.5 u
= 1.5 x 7
= 10.5 kg
Answer(s): 10.5 kg
