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

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MGMT ratio problem [#permalink] New post 02 Aug 2011, 21:38
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.

regards
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Re: MGMT ratio problem [#permalink] New post 02 Aug 2011, 21:59
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shrive555 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 red marbles could be in bag A?
1
3
4
6
8


Bag A:

R:W:B|R:W:B
1x:3x:-|2x:6x:-
-:2x:3x|-:6x:9x

Bag A:
R:W:B
2x:6x:9x

Bag B:
R:W
1y:4y

If we just consider white marbles now:
6x+4y=30

y=(30-6x)/4

So,
x=1; y=6
x=2; y=Non integer. Ignore
x=3; y=3
x=4; y=Non integer
x=5; y=0(Not possible)

Thus, x can be either 1 or 3.
If x=1;

Bag A:
R:W:B
2x:6x:9x
R=2*1=2

If x=3;
R=2*3=6

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

Ans: "D"
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Re: MGMT ratio problem [#permalink] New post 02 Aug 2011, 22:14
fluke wrote:
shrive555 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 red marbles could be in bag A?
1
3
4
6
8


Bag A:

R:W:B|R:W:B
1x:3x:-|2x:6x:-
-:2x:3x|-:6x:9x

Bag A:
R:W:B
2x:6x:9x


Bag B:
R:W
1y:4y

If we just consider white marbles now:
6x+4y=30

y=(30-6x)/4

So,
x=1; y=6
x=2; y=Non integer. Ignore
x=3; y=3
x=4; y=Non integer
x=5; y=0(Not possible)

Thus, x can be either 1 or 3.
If x=1;

Bag A:
R:W:B
2x:6x:9x
R=2*1=2

If x=3;
R=2*3=6

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. :roll:
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Re: MGMT ratio problem [#permalink] New post 02 Aug 2011, 22:22
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shrive555 wrote:
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. :roll:


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
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Re: MGMT ratio problem [#permalink] New post 03 Aug 2011, 08:00
you guys are very much right 2 / 6 are the options possible
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Re: MGMT ratio problem [#permalink] New post 03 Aug 2011, 12:03
good question .. Bag A R:W:B = 2x: 6x: 9x and Bag B R:W = y:4y

6x + 4y = 30 => 3x + 2y = 15 ..

the only +ve integers that satisfy the above is 1, 6 and 3, 3 ..

red marbles in Bag A = 2x = 2 or 6 => answer is D
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Re: Ratios [#permalink] New post 18 Sep 2011, 15:10
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.

Answer is D.
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Re: Ratios [#permalink] New post 19 Sep 2011, 02:18
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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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Re: Ratios [#permalink] New post 21 Sep 2011, 06:52
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] New post 21 Sep 2011, 09:12
Expert's post
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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Re: MGMAT PS: Bag A contains red, white and blue marbles [#permalink] New post 21 Sep 2011, 10:00
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.
Re: MGMAT PS: Bag A contains red, white and blue marbles   [#permalink] 21 Sep 2011, 10:00
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