Container S and Container T were filled completely with flour. The total mass of
16 of the flour in Container T and
111 of the flour in Container S was 540 g. If
311 of the flour in Container S was poured out, the total mass of the flour in both containers became 3.44 kg. How much flour was in
- Container S in grams?
- Container T in grams?
(a)
Let the mass of the flour in Container T be T.
Let the mass of the flour in Container S be S.
16 T +
111 S = 540 --- (1)
Fraction of the flour left in Container S after
311 of it was poured out
= 1 -
311 =
811 1 kg = 1000 g
3.44 kg = 3440 g
1 T +
811 S = 3440
1 T = 3440 -
811 S --- (2)
Make T the same.
(1)
x6 66 T +
611 S = 3240
1 T +
611 S = 3240
1 T = 3240 -
611 S --- (3)
(3) = (2)
3240 -
611 S = 3440 -
811 X
811 S -
611 S = 3440 - 3240
211 S = 200
111 S = 200 ÷ 2 = 100
1111 S = 11 x 100 = 1100
1 S = 1100
Mass of Container S = 1100 g
(b)
From (1)
16 T + 100 = 540
16 T = 540 - 100 = 440
66 T = 6 x 440 = 2640
1 T = 2640
Mass of Container T = 2640 g
Answer(s): (a) 1100 g; (b) 2640 g