Paul had three bags of oats, A, B and C and the mass of each bag of oats was in the ratio 1 : 2 : 5. Paul decided to transfer 40% of oats from Bag A into Bag B and 60% of oats from Bag C into Bag B. Given that the mass of Bag B was 54 kg in the end, how many kilograms of oats were transferred into Bag B?
| |
A |
B |
C |
| Before |
1 u |
2 u |
5 u |
| Change 1 |
- 0.4 u |
+ 0.4 u |
|
| Change 2 |
|
+ 3 u |
- 3 u |
| After |
0.6 u |
5.4 u |
2 u |
Mass of oats transferred from Bag A into Bag B
= 40% x 1 u
=
40100 x 1 u
= 0.6 u
Mass of oats transferred from Bag C into Bag B
= 60% x 5 u
=
60100 x 5 u
= 3 u
Mass of oats in Bag B in the end
= 2 u + 0.4 u + 3 u
= 5.4 u
5.4 u = 54
1 u = 54 ÷ 5.4 = 10
Mass of oats transferred into Bag B
= 0.4 u + 3 u
= 3.4 u
= 3.4 x 10
= 34 kg
Answer(s): 34 kg
