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Re: marbles [#permalink] New post 17 Apr 2009, 21:08
I have a question:

Why did we take the Number of marbles in BAG A and BAG B to be equal as K? It did not state in the question?

I both were different quantities they would be slit up as : 2x : 6x: 9 x

and BAG B as : 1y : 4y

then i took 6x + 4 y = 30 3x + 2y = 15

i took y = 1 x = 13/3 not possible
y = 2 x = 11/3 not possible
y = 3 x = 3 possible.

So i since x = 3 i got R = 6. Is my approach correct.
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Re: marbles [#permalink] New post 17 Apr 2009, 21:43
tkarthi4u wrote:
I have a question:

Why did we take the Number of marbles in BAG A and BAG B to be equal as K? It did not state in the question?

I both were different quantities they would be slit up as : 2x : 6x: 9 x

and BAG B as : 1y : 4y

then i took 6x + 4 y = 30 3x + 2y = 15

i took y = 1 x = 13/3 not possible
y = 2 x = 11/3 not possible
y = 3 x = 3 possible.

So i since x = 3 i got R = 6. Is my approach correct.


because 30 is the quantity from both bags, you can use the same variable; i.e. 6x+4x=30, x=3. then use this ratio for both bags.
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Re: marbles [#permalink] New post 27 Apr 2009, 00:00
I highly doubt that. I proceeded the same way as tkarthi. I dont think you can add two distinct ratios together.
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Ratio of Marbles [#permalink] New post 02 Jun 2009, 07:37
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
D
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Re: Ratio of Marbles [#permalink] New post 02 Jun 2009, 08:45
Combined Ratio of marbles in

Bag A \Rightarrow 2:6:9 \Rightarrow a multiple of 6. Let the while marbles = 6x

Let White marbles in Bag B = W_b

Let White marbles in Bag A = W_a

Let Red marbles in Bag A = R_a


Bag B has (30-6x), which is a multiple of 4.

If x = 1, W_b = 24, W_a = 6, \Rightarrow R_a = 2 , which is not in the solution.

If x = 2, (30-12) is not a multiple of 4

If x = 3, W_b = 12, W_a = 18, \Rightarrow R_a = 6 , which is IN the solution.

If x = 4, (30-24) is not a multiple of 4.

Thus, answer is D.
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Marbel Ratios [#permalink] New post 09 Jul 2009, 11:03
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

MGMAT/CAT1/Ratios/PS
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Re: Marbel Ratios [#permalink] New post 09 Jul 2009, 14:19
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

Solution - Combined ratio of Red:White:Blue marbles in bag A is 2:6:9.

And ratio of Red:White in Bag B = 1:4

So total white marbled in 2 bags combined = 30 (given)
=> 6x + 4x = 30
=> x = 3

Hence red marbles in Bag A = 2x = 2*3 = 6

I would go with D. Whats the OA?
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Re: Marbel Ratios [#permalink] New post 09 Jul 2009, 15:48
Answer is D.

SdRandom, there is an error in your solution.

Your ratios are right.
2/6/9 in A
1/4 in B
That makes
6x + 4y=30 This is where you have done wrong.
After it. We must look at the choices.
Red no's must be even (2/6/9 ratio)
So it may be 4, 6 or 8.
If red is 4 => whites in A becomes 12; whites in B becomes 18 (30-12) => 1/4 ratio becomes impossible thus 4 is wrong.
If red is 8 => whites in A becomes 24; whites in B becomes 6 => 1/4 ratio becomes impossible
When red is 6 => whites in A becomes 18; in B becomes 12 thus 1/4 ratio becomes possible
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Re: Marbel Ratios [#permalink] New post 17 Jul 2009, 19:28
Is my process correct?

A R:W:B =2:6:9
B R:W=1:4

Now.A+B has 30 white marbles.
Out of this 30 B has 4/5 x 30 = 24 marbles,therefore the rest 30-24=6 must be from A.:)
Hence,D.
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help please :) [#permalink] New post 28 Aug 2010, 23:39
ALSO

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

Please explain. I did not understand the explanation given. Thanks :)
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Re: help please :) [#permalink] New post 29 Aug 2010, 15:47
Already posted:
marbel-ratios-80692.html

Please search in the future.

Locked.
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Re: help please :) [#permalink] New post 29 Aug 2010, 15:49
sorry! This probably sounds stupid but I didnt know that we can search questions. Thanks!
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Re: help please :) [#permalink] New post 29 Aug 2010, 17:54
spatel121990 wrote:
ALSO

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

Please explain. I did not understand the explanation given. Thanks :)



BAG A: R to W to B is 2 to 6 to 9
BAG B: R W is 1 to 4

white in both is 30 therefore using 4+6 = 10
30/10 = 3

therefore 3 *6 = 18 white in bag a
ration is 2 to 6 therefore if 18w, 18*2/6 = 6red

D
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MGMT ratio problem [#permalink] New post 02 Aug 2011, 22:38
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.

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Re: MGMT ratio problem [#permalink] New post 02 Aug 2011, 22:59
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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, 23:14
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, 23:22
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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, 09:00
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, 13:03
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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Ratios [#permalink] New post 18 Sep 2011, 12:49
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
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Ratios   [#permalink] 18 Sep 2011, 12:49
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