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ratio problem

Expert replies
by ankurmit » Fri Jul 09, 2010 3:17 am
Bag A contains red, white and blue marbles such that the red to white marble ratio is 1:3 and the white to blue marble ratio is 2:3. Bag B contains red and white marbles in the ratio of 1:4. Together, the two bags contain 30 white marbles. How many red marbles could be in bag A?

a. 1

b. 3

c. 4

d 6

e. 8
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Source: — Problem Solving |

by selango » Fri Jul 09, 2010 4:01 am
A

R:W=1:3

B

R:W=1:4

Note that white marbles in B are in multiples of 4.

So possible scenarios are A=18,B=12 or A=6,B=24

If white marble in A is 18,then red marble is 6

If white marble in A is 6,then red marble is 2

Only 6 is available in option .

Hence D
--Anand--
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by kmittal82 » Fri Jul 09, 2010 4:48 am
Plugging values is also easy in this case.

Red marbles in a (Ra) + Red Marbles in b (Rb) = 30. Let white marbles in bag b = Wb
Also, In bag B, Red:White = 1:4 which means Number of red marbles must be a multiple of 4 (you can't have a fraction of a marble).

1) If Ra = 1, Rb = 29, => Wb = 29/4 => incorrect
2) Ra = 3, Rb = 27 , => Wb = 27/4 => incorrect
3) Ra = 4, Rb = 26, => Wb = 26/4 => incorrect
4) Ra = 6, Rb = 24 => Wb = 4, possibly correct
5) Ra = 8, Rb = 22 , incorrect

Only one option satisfies the ratio, so answer should be (D)
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by Rahul@gurome » Fri Jul 09, 2010 4:50 am
Bag A: R:W = 1:3 implies R = W/3
Bag B: R:W = 1:4 implies R = W/4
If white marbles in bag A = x and white marbles in bag B = y, then x + y = 30
R = W/3 or R = x/3 implies R is an integer so x is divisible by 3. So, x = 3, 6, 9, 12, 15, 18, 21...
Similarly, R = W/4 or R = y/4 implies R is an integer so y is divisible by 4. So, y = 4, 8, 12, 16, 20, 24,...
The values that give x + y = 30 so that x is divisible by 3 and y is divisible by 4 is x = 18 and y = 12
Therefore, red marbles in bag A = 18/3 = 6

The correct answer is (D).
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