Charlie had three bags of red beans, S, T and U and the mass of each bag of red beans was in the ratio 5 : 4 : 5. Charlie decided to transfer 10% of red beans from Bag S into Bag T and 90% of red beans from Bag U into Bag T. Given that the mass of Bag T was 36 kg in the end, how many kilograms of red beans were transferred into Bag T?
|
S |
T |
U |
Before |
5 u |
4 u |
5 u |
Change 1 |
- 0.5 u |
+ 0.5 u |
|
Change 2 |
|
+ 4.5 u |
- 4.5 u |
After |
4.5 u |
9 u |
0.5 u |
Mass of red beans transferred from Bag S into Bag T
= 10% x 5 u
=
10100 x 5 u
= 4.5 u
Mass of red beans transferred from Bag U into Bag T
= 90% x 5 u
=
90100 x 5 u
= 4.5 u
Mass of red beans in Bag T in the end
= 4 u + 0.5 u + 4.5 u
= 9 u
9 u = 36
1 u = 36 ÷ 9 = 4
Mass of red beans transferred into Bag T
= 0.5 u + 4.5 u
= 5 u
= 5 x 4
= 20 kg
Answer(s): 20 kg