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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? 1 3 4 6 8

Bag A: R/w = 1/3 and w/b=2/3 Bag B: R/w = 1/4

Number of White marbles in A and B is 30 . i.e A+B =30

so in Bag B Number of white marbles will be 30/5 * 4 = 24

since A+B = 30 => B= 24 then in Bag A number of white marbles will be 6

Now R/w = 1/3 => R/6 = 1/3 => R=2

But there is NO 2. what i'm doing wrong. please explain.

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? 1 3 4 6 8

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? 1 3 4 6 8

So, both 2 and 6 are possible numbers of red marbles in Bag A.

Ans: "D"

Thank you for your detail explanation. just little confusion. how did you get the red part. because i have to solve it through making equations R/w = 1/3 w/b=2/3 => R/b = 2/9 after this i don't know how to set up R:W:B I assume that you have used some short cut.
_________________

Thank you for your detail explanation. just little confusion. how did you get the red part. because i have to solve it through making equations R/w = 1/3 w/b=2/3 => R/b = 2/9 after this i don't know how to set up R:W:B I assume that you have used some short cut.

Not really;

Just take the LCM of the common term:

R:W=1:3 --------1 W:B=2:3 --------2

Here W is common; Take the LCM of 2,3=6

So, eq 1 needs to be multiplied by 2: R:W=(1:3)*2=2:6---------3

So, eq 2 needs to be multiplied by 3: W:B=(2:3)*3=6:9---------4

Now; 3 and 4 are both 6 whites; we can join them R:W:B 2:6:9
_________________

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)
_________________

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)

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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Re: MGMAT PS: Bag A contains red, white and blue marbles [#permalink]

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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.

gmatclubot

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

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