Container Y and Container Z were filled completely with flour. The total mass of
16 of the flour in Container Z and
111 of the flour in Container Y was 510 g. If
211 of the flour in Container Y was poured out, the total mass of the flour in both containers became 3.42 kg. How much flour was in
- Container Y in grams?
- Container Z in grams?
(a)
Let the mass of the flour in Container Z be Z.
Let the mass of the flour in Container Y be Y.
16 Z +
111 Y = 510 --- (1)
Fraction of the flour left in Container Y after
211 of it was poured out
= 1 -
211 =
911 1 kg = 1000 g
3.42 kg = 3420 g
1 Z +
911 Y = 3420
1 Z = 3420 -
911 Y --- (2)
Make Z the same.
(1)
x6 66 Z +
611 Y = 3060
1 Z +
611 Y = 3060
1 Z = 3060 -
611 Y --- (3)
(3) = (2)
3060 -
611 Y = 3420 -
911 X
911 Y -
611 Y = 3420 - 3060
311 Y = 360
111 Y = 360 ÷ 3 = 120
1111 Y = 11 x 120 = 1320
1 Y = 1320
Mass of Container Y = 1320 g
(b)
From (1)
16 Z + 120 = 510
16 Z = 510 - 120 = 390
66 Z = 6 x 390 = 2340
1 Z = 2340
Mass of Container Z = 2340 g
Answer(s): (a) 1320 g; (b) 2340 g