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primary 6 | Maths
| Fractions
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Date Posted: 3 years ago
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Let x and y be the number of black and white beads, respectively in the box initially.number of black beads was 2/5 of number of white beads
=> x=2/5y -----(i)
After 12 black and 44 white beads were
taken out of the box then remaining.black beads are x - 12 and remaining white beads are y - 44 in the box
we are given now (x-12) = 2/3 of (y-44)
=> x-12 = 2/3 (y-44)
=> x=2/3y-88/3+12 ----(ii)
from (i) and (ii) => 2/5y = 2/3y + -88+36/3
=>(2/5-2/3)y=-52/3 => 2(3-5)/15y = -52/3
=> -4/15y = -52/3 => y = 13 * 5 = 65
then x = 2/5y = 2/5 * 13 * 5 = 26 now In the box
black remaining are x-12 = 26 - 12 = 14
white remaining are y-44 = 65 - 44 = 21
So there are 14 + 21 = [35] beads left in the box
=> x=2/5y -----(i)
After 12 black and 44 white beads were
taken out of the box then remaining.black beads are x - 12 and remaining white beads are y - 44 in the box
we are given now (x-12) = 2/3 of (y-44)
=> x-12 = 2/3 (y-44)
=> x=2/3y-88/3+12 ----(ii)
from (i) and (ii) => 2/5y = 2/3y + -88+36/3
=>(2/5-2/3)y=-52/3 => 2(3-5)/15y = -52/3
=> -4/15y = -52/3 => y = 13 * 5 = 65
then x = 2/5y = 2/5 * 13 * 5 = 26 now In the box
black remaining are x-12 = 26 - 12 = 14
white remaining are y-44 = 65 - 44 = 21
So there are 14 + 21 = [35] beads left in the box
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