Howard had three bags of rice, K, L and M and the mass of each bag of rice was in the ratio 5 : 4 : 2. Howard decided to transfer 40% of rice from Bag K into Bag L and 70% of rice from Bag M into Bag L. Given that the mass of Bag L was 66.6 kg in the end, how many kilograms of rice were transferred into Bag L?
| |
K |
L |
M |
| Before |
5 u |
4 u |
2 u |
| Change 1 |
- 2 u |
+ 2 u |
|
| Change 2 |
|
+ 1.4 u |
- 1.4 u |
| After |
3 u |
7.4 u |
0.6 u |
Mass of rice transferred from Bag K into Bag L
= 40% x 5 u
=
40100 x 5 u
= 3 u
Mass of rice transferred from Bag M into Bag L
= 70% x 2 u
=
70100 x 2 u
= 1.4 u
Mass of rice in Bag L in the end
= 4 u + 2 u + 1.4 u
= 7.4 u
7.4 u = 66.6
1 u = 66.6 ÷ 7.4 = 9
Mass of rice transferred into Bag L
= 2 u + 1.4 u
= 3.4 u
= 3.4 x 9
= 30.6 kg
Answer(s): 30.6 kg
