+0

# help

0
40
3

Jim has two bags of marbles, each of which contains some red marbles and some blue marbles (and no marbles of any other color). The ratio of red marbles to blue marbles in the first bag is \$3:5.\$ The ratio of red marbles to blue marbles in the second bag is \$3:2.\$ When the two bags of marbles are mixed together, the ratio of red marbles to blue marbles is \$11:9.\$ What is the ratio of the number of marbles in the first bag to the number of marbles in the second bag?

Jul 23, 2023

#1
0

And once again, an Ao Ps hw problem.

Jul 23, 2023
#2
0

1 - First bag:

R/B =3 / 5...........................................(1)

2 - Second bag:

r / b = 3/2...........................................(2)

3 - Mixing the 2 bags:

[R + r] / [B + b] = 11/9.......................(3)

LCM(11, 9) = 99

R + r = 99............................................(4)

B +  b=81.............................................(5)

R=15,  B=25,  r=84,  b=56

Ratio of:[15 + 25] / [84 + 56] = 40 / 140 = 2 / 7 - ratio of the number of marbles in the first bag to the 2nd bag.

Jul 25, 2023
#3
0

3 + 5 = 8 sum of the ratios of the first bag

3 + 2 = 5 sum of the ratios of the 2nd bag

LCM(5, 8) = 40 - number of marbles in one of the bags.

LCM(11, 9) = 99

11 / 9 = 99 / 81

99 + 81 = 180 - number of marbles in the two bags.

180 - 40 = 140 - number of marbles in the 2nd bag

Since 140 is NOT divisible by 8, it must be the 2nd bag.

140 / 5 = 28 - the constant of the 2nd bag

[28 * 3] / [28 * 2] = 84 / 56 =140 marbles in the 2nd bag.

40 / 8 = 5 - the constant of the 1st bag

[5 * 3] / [5 * 5] = 15 / 25 = 40 marbles in the 1st bag.

40 / 140 = 2 / 7 - ratio of marbles in the 1st bag to the 2nd bag.

Jul 26, 2023