Red : Blue : Green = 2x : 3x : 7x
let y be the number of green marbles removed and A and B be the red and blue marbles added
2x + a : 3x + b : 7x -y = 4:5:9
(3x + b)/ (7x - y) = 5/9
27x + 9b = 35x - 5y
5y + 9b = 8x
b = (8x - 5y)/9
(2x + a)/(7x-y) = 4/9
18x + 9a = 28x - 4y
9a + 4y = 10x
a = (10x - 4y)/9
(2x+a)/(3x+b) = 4/5
10x + 5a = 12x + 4b
5a - 4b = 2x
so (50x - 20y)/9 - (32x - 20y)/4 = 2x
200x - 80y - 288x + 180y = 72x
100y = 160x
y = 8x/5 ,
For y to be integer x has to be an integer of type 5k ie even
minimum value of x = 5 and y = 8
Putting this a = 2 , b =0 so 2
In a bag there are 3 kind of marbles. These are red, blue and green marbles. Initially the red, blue and green marbles were in the ratio 2:3:7. Now a few green marbles are taken out and some red and blue marbles are added such that red,blue,green ratio is 4:5:9. What is the minimum possible number of red marbles that were put into the bag
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Red : Blue : Green = 2x : 3x : 7x
let y be the number of green marbles removed and A and B be the red and blue marbles added
2x + a : 3x + b : 7x -y = 4:5:9
(3x + b)/ (7x - y) = 5/9
27x + 9b = 35x - 5y
5y + 9b = 8x
b = (8x - 5y)/9
(2x + a)/(7x-y) = 4/9
18x + 9a = 28x - 4y
9a + 4y = 10x
a = (10x - 4y)/9
(2x+a)/(3x+b) = 4/5
10x + 5a = 12x + 4b
5a - 4b = 2x
so (50x - 20y)/9 - (32x - 20y)/4 = 2x
200x - 80y - 288x + 180y = 72x
100y = 160x
y = 8x/5 ,
For y to be integer x has to be an integer of type 5k ie even
minimum value of x = 5 and y = 8
Putting this a = 2 , b =0 so 2
This question is a very good question. This is a CAT level question. Chances of this coming in CUET are very slim. But still, you should practice with questions like these as they make concepts stronger.
@randominbox9 good approach. Solve other questions too. You will get good grasp of concepts
mention anyone using @username while adding comments