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Bag A contains red, white and blue marbles such that the red [#permalink]
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29 Jun 2007, 15:38
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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? A. 1 B. 3 C. 4 D. 6 E. 8 OPEN DISCUSSION OF THIS QUESTION IS HERE: bagacontainsredwhiteandbluemarblessuchthattheredto127477.html
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Joined: 28 Dec 2006
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is there a strategy for this....i solved by brute force.
# white marbles in A + # white marbles in B = 30
# white marbles in A is div 3
# white marbles in B is div 4
Bag A..................Bag B
30 w, 10 r...........0 w, 0 r
18 w, 6 r...........12 w, 3 r
6 red marbles in A works



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Even I went by brute force ;
Its only when red marbles in Bag A are 6, the # of White marbles in A is 18 and thus # of white marbles in B is 12 which is Div by 4. For rest of other %s # of white marbles in B is not div by 4. hence 6.



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Re: PS: Marble Ratios [#permalink]
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01 Feb 2008, 18:30
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# of Red marbles in Bag A can be either 2 or 6. No 2 in the choices, so 6. D. Explanation Bag A: R:W:B = 2:6:9 Bag B R:W = 1:4 6X + 4Y = 30 i.e 3X + 2Y = 15 X has to be odd to make an odd sum from the eq. X = 1 , Y = 6 OR X = 3, Y = 3 So R can be 2X i.e 2 or 6.
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Re: PS: Marble Ratios [#permalink]
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01 Feb 2008, 19:15
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white(a)+white(b) = 30
white(a)=3*red(a) white(b)=4*red(b)
3*red(a) + 4*red(b) = 30
now just see what number from the answer choices work, i.e. plug in the answer choices for red(a) and see if red(b) comes out to an integer value. Lets try a, 1:
3*1 + 4*red(b)=30 =>red(b) = 27/4 , which is not an integer and so cant be the answer.
The only value that works is 6.



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Re: Marbles Bag A and Bag B [#permalink]
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24 Feb 2008, 17:42
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Manbehindthecurtain 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?
This was a MGMAT CAT question. Can you tell me how to solve it with algebra?
1 3 4 6 8 (D) 6 I'd rather use basic ratio arithmetic for a problem like this. I imagine algebra will be a little tedious. Bag A: R:W = 1:3; and W:B = 2:3 R:W:B = 2:6:9  (I) Bag B: R:W = 1:4 Given: there are 30 white marbles. If we are to split this between the two bags, the split amounts must be multiples of 6 and 4, respectively. That is, [number of white marbles in Bag A, number of white marbles in Bag B] could be either [6, 24] or [18, 12]. Substitute both values of 6 and 18 white marbles in ratio (I) above and you'll see that in Bag A, you can have either 2 or 6 red marbles. Hence (D).
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Re: Ratio question [#permalink]
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01 Jul 2008, 11:48
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Bag A = 2x (Red), 6x (White) and 9x (Blue) Bag B = x(Red), 4x (white)
6x + 4x = 30 x= 3 Red in Bag A = 2x = 6



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Re: Ratio question [#permalink]
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01 Jul 2008, 23:53
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i think u have assumed that both the bags contain equal no balls ( x is the total no of balls in the bag). The euation should be 6x+4y=30



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Re: Ratio question [#permalink]
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02 Jul 2008, 00:21
rajesh04 wrote: From Manhattan CAT  3
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)1 2)3 3)4 4)6 5)8
Sorry wrong forum!! Number of reds in A # number of red in B. So can not assume Xa=Xb (here I call Y) 6x+4y =30 ==> x=3, y=3 (Ra=6) or x =1, y=6 (Ra=2) D is best answer.



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Re: Zumit PS 025 [#permalink]
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22 Sep 2008, 03:25
Is it 6 red balls in A?
If there are 6 red balls in A , then there will be 18 while balls in A , leaving 12 (30  12) white balls for B .
Now we can divide 12 by 4 .. so i will go for this answer.
for all other option we will not get white balls values ,so that it will be divide by 4.. (keeping we have 1:4ratio in bag b)
Thanks Vishal Shah



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Re: Zumit PS 025 [#permalink]
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22 Sep 2008, 06:24
dancinggeometry 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 W1 + W2 = 30 Where W1 is divisible by 6 and w2 is divisible by 4 6 12 18 24 4 8 12 16 20 24 28 only 18 and 12 can be correct values. W1=18 if W1=18, Red=6 (1:3) D



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Re: Zumit PS 025 [#permalink]
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22 Sep 2008, 11:06
If x is number of white marbles in A and y is number of white marbles in B then
x + y = 30.
We need to find out x/3 + y/4 or (4x+3y)/12 = (x + 90)/12
In order for the above expression to be integer, (x+90) should be a multiple of 12.....this is possible for x=6, 18, ........
or x/3 = 2, 6, .....
Since 2 is not an answer choice.....6 should be the answer.



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Re: marbles [#permalink]
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29 Mar 2009, 09:59
Its 6 R:W:B wise A will have ratio of> 2:6:9 and B> 1:4 Total White marbles =30=10(k) so red marbel wise , A will have 6. 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? ==
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Re: marbles [#permalink]
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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]
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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: Ratio of Marbles [#permalink]
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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 (306x), 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, (3012) 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, (3024) is not a multiple of 4.
Thus, answer is D.



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Re: Marbel Ratios [#permalink]
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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]
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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 (3012) => 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]
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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 3024=6 must be from A. Hence,D.
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Re: help please :) [#permalink]
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29 Aug 2010, 17:54
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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




Re: help please :)
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