Mark had three bags of flour, A, B and C and the mass of each bag of flour was in the ratio 5 : 4 : 3. Mark decided to transfer 20% of flour from Bag A into Bag B and 90% of flour from Bag C into Bag B. Given that the mass of Bag B was 15.4 kg in the end, how many kilograms of flour were transferred into Bag B?
| |
A |
B |
C |
| Before |
5 u |
4 u |
3 u |
| Change 1 |
- 1 u |
+ 1 u |
|
| Change 2 |
|
+ 2.7 u |
- 2.7 u |
| After |
4 u |
7.7 u |
0.3 u |
Mass of flour transferred from Bag A into Bag B
= 20% x 5 u
=
20100 x 5 u
= 4 u
Mass of flour transferred from Bag C into Bag B
= 90% x 3 u
=
90100 x 3 u
= 2.7 u
Mass of flour in Bag B in the end
= 4 u + 1 u + 2.7 u
= 7.7 u
7.7 u = 15.4
1 u = 15.4 ÷ 7.7 = 2
Mass of flour transferred into Bag B
= 1 u + 2.7 u
= 3.7 u
= 3.7 x 2
= 7.4 kg
Answer(s): 7.4 kg
