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# Bag A contains red, white and blue marbles such that the red

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Kudos [?]: 62 [0], given: 42

Re: Ratios [#permalink]  18 Sep 2011, 16:10
Bag A

R/W = X/(3X) ---- 1
W/B = 2Y/((3Y) ----2

from 1 and 2 , we can see W= 3X = 2Y

Bag B
R/W = Z/4Z

X=?

the two bags contain 30 white marbles => 3X+4Z = 30

=> Z = (30-3X)/4

then by plugging in X values , we can see that only D gives an integer value for Z.

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Re: Ratios [#permalink]  19 Sep 2011, 03:18
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aeros232 wrote:
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 [highlight]red marbles could be in bag A[/highlight]? [/[color=#40FF00]color]

a 1
b 3
c 4
d 6
e 8

Bag A:
R:W = 1:3
W:B = 2:3
W is the common one here so make it equal i.e. R:W = 2:6 and W:B = 6:9 (the ratios remain the same). So R:W:B = 2:6:9
Since number of marbles has to be an integer, number of red marbles in this bag must be 2 or a multiple of 2 and number of white marbles must be 6 or a multiple of 6.

Bag B:
R:W = 1:4
The number of white marbles must be 4 or a multiple of 4.

To make 30 white marbles, you could mix white marbles from Bag A and Bag B in many ways.
BagA: 6 + BagB: 24 (No. of red marbles in BagA = 2)
BagA: 12 + BagB: 18 - Not possible because 18 is not a multiple of 4
BagA: 18 + BagB: 12 (No. of red marbles in BagA = 6)
BagA: 24 + BagB: 6 - Not possible because 6 is not a multiple of 4

No of red marbles in bag A can be both 2 and 6. Answer (D)
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Joined: 30 Sep 2009
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Kudos [?]: 12 [0], given: 66

Re: Ratios [#permalink]  21 Sep 2011, 07:52
please comment is this method wrong :

ratio of balls in bag A: red:white:blue -->2:6:9 (2x+6x+9x)
ratio of balls in bag B: red:white --->1:4 (1y+4y)

so we can derive a equation where
white balls in both bags--> 6x+4y=30 -----1 (as the total number of white balls are 30)
red balls in bag A ---> 2x=?------2

solving eqn 1 --> we get x value as 1 and 3 to get a whole number as balls cannot be in fraction.

substituting the value in 2 --> we get the value as 2 and 6 for red balls in bag A.
As 2 is not there in the answer , we select 6 (ie answer D)
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Re: Ratios [#permalink]  21 Sep 2011, 10:12
abhi398 wrote:
please comment is this method wrong :

ratio of balls in bag A: red:white:blue -->2:6:9 (2x+6x+9x)
ratio of balls in bag B: red:white --->1:4 (1y+4y)

so we can derive a equation where
white balls in both bags--> 6x+4y=30 -----1 (as the total number of white balls are 30)
red balls in bag A ---> 2x=?------2

solving eqn 1 --> we get x value as 1 and 3 to get a whole number as balls cannot be in fraction.

substituting the value in 2 --> we get the value as 2 and 6 for red balls in bag A.
As 2 is not there in the answer , we select 6 (ie answer D)

Nothing wrong with the method. Essentially, both, this one and the one above, find integral solutions of 6x + 4y = 30.
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Status: Need to read faster and get less distracted by the ticking clock!
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Joined: 19 Nov 2010
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Kudos [?]: 6 [0], given: 4

Re: MGMAT PS: Bag A contains red, white and blue marbles [#permalink]  21 Sep 2011, 11:00
D) 6

Let x be the # of Red balls in Bag A
Therefore the ratio for R/W = 1/3 = x/3x

Let y be the # of Red balls in Bag B
Therefore the ratio for R/W = 1/4 = y/4y

Now we know White (Bag A) + White (Bag B) = 30
3x + 4y = 30

So lets see what value of x solves this equation from the choices given: 1, 3, 4, 6, 8
1: y is not an integer (27/4)
3: y is not an integer (21/4)
4: y is not an integer (14/4)
6: y is an integer
8: y is not an integer

Thus 6.
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Kudos [?]: 7 [0], given: 8

Re: MGMAT PS: Bag A contains red, white and blue marbles [#permalink]  22 Sep 2011, 06:15
Used incognito1 method got 6.
Re: MGMAT PS: Bag A contains red, white and blue marbles   [#permalink] 22 Sep 2011, 06:15
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