Pierre had three bags of oats, N, P and Q and the mass of each bag of oats was in the ratio 1 : 1 : 4. Pierre decided to transfer 20% of oats from Bag N into Bag P and 70% of oats from Bag Q into Bag P. Given that the mass of Bag P was 8 kg in the end, how many kilograms of oats were transferred into Bag P?
| |
N |
P |
Q |
| Before |
1 u |
1 u |
4 u |
| Change 1 |
- 0.2 u |
+ 0.2 u |
|
| Change 2 |
|
+ 2.8 u |
- 2.8 u |
| After |
0.8 u |
4 u |
1.2 u |
Mass of oats transferred from Bag N into Bag P
= 20% x 1 u
=
20100 x 1 u
= 0.8 u
Mass of oats transferred from Bag Q into Bag P
= 70% x 4 u
=
70100 x 4 u
= 2.8 u
Mass of oats in Bag P in the end
= 1 u + 0.2 u + 2.8 u
= 4 u
4 u = 8
1 u = 8 ÷ 4 = 2
Mass of oats transferred into Bag P
= 0.2 u + 2.8 u
= 3 u
= 3 x 2
= 6 kg
Answer(s): 6 kg
