Tom had three bags of rice, X, Y and Z and the mass of each bag of rice was in the ratio 5 : 2 : 4. Tom decided to transfer 10% of rice from Bag X into Bag Y and 90% of rice from Bag Z into Bag Y. Given that the mass of Bag Y was 54.9 kg in the end, how many kilograms of rice were transferred into Bag Y?
| |
X |
Y |
Z |
| Before |
5 u |
2 u |
4 u |
| Change 1 |
- 0.5 u |
+ 0.5 u |
|
| Change 2 |
|
+ 3.6 u |
- 3.6 u |
| After |
4.5 u |
6.1 u |
0.4 u |
Mass of rice transferred from Bag X into Bag Y
= 10% x 5 u
=
10100 x 5 u
= 4.5 u
Mass of rice transferred from Bag Z into Bag Y
= 90% x 4 u
=
90100 x 4 u
= 3.6 u
Mass of rice in Bag Y in the end
= 2 u + 0.5 u + 3.6 u
= 6.1 u
6.1 u = 54.9
1 u = 54.9 ÷ 6.1 = 9
Mass of rice transferred into Bag Y
= 0.5 u + 3.6 u
= 4.1 u
= 4.1 x 9
= 36.9 kg
Answer(s): 36.9 kg
