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

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08 Sep 2010, 19:56

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Question Stats:

69% (02:09) correct 31% (02:38) wrong based on 247 sessions

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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

The explanation given is little confusing. Can anyone suggest a easy method to approach such types of problems?

We are told that bag B contains red and white marbles in the ration 1:4. This implies that WB, the number of white marbles in bag B, must be a multiple of 4.

What can we say about WA, the number of white marbles in bag A? We are given two ratios involving the white marbles in bag A. The fact that the ratio of red to white marbles in bag A is 1:3 implies that WA must be a multiple of 3. The fact that the ratio of white to blue marbles in bag A is 2:3 implies that WA must be a multiple of 2. Since WA is both a multiple of 2 and a multiple of 3, it must be a multiple of 6.

We are told that WA + WB = 30. We have already figured out that WA must be a multiple of 6 and that WB must be a multiple of 4. So all we need to do now is to test each candidate value of WA (i.e. 6, 12, 18, and 24) to see whether, when plugged into WA + WB = 30, it yields a value for WB that is a multiple of 4. It turns out that WA = 6 and WA = 18 are the only values that meet this criterion.

Recall that the ratio of red to white marbles in bag A is 1:3. If there are 6 white marbles in bag A, there are 2 red marbles. If there are 18 white marbles in bag A, there are 6 red marbles. Thus, the number of red marbles in bag A is either 2 or 6. Only one answer choice matches either of these numbers.

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

Bag A: R/W=1/3=2/6; W/B=2/3=6/9; So R/W/B=2/6/9 --> # of marbles in bag A would be \(2x\), \(6x\), and \(9x\), for some positive integer multiple \(x\), where \(6x\) corresponds to the # of white marbles and \(2x\) corresponds to the # of red marbles (so # of red marbles must be a multiple of 2, so answers A and B are out at this stage);

Bag B: R/W=1/4 --> # of marbles in bag B would be \(y\) and \(4y\), for some positive integer multiple \(y\), where \(4y\) corresponds to the # of white marbles;

Given: \(6x+4y=30\) --> \(3x+2y=15\) --> there are two positive integer solutions for this equation: \(x=3\) and \(y=3\) --> in this case # of red marbles equals to \(2x=6\); Or: \(x=1\) and \(y=6\) --> in this case # of red marbles equals to \(2x=2\);

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

Guys - can someone please help by explaining this problem?

Responding to a pm:

We need the relation between number of red, white and total number of marbles.

We don't know what that relation in bag A. Let's find it out first. Red:White = 1:3 = 2:6 White:Blue = 2:3 = 6:9 Red:White:Blue = 2:6:9 In bag A, if you have 17 marbles, 2 will be Red, 6 will be White and 9 will be Blue. If you have 34 marbles, 4 will be Red, 12 will be White and 18 will be Blue etc No of white marbles could be 6/12/18/24... etc No of red marbles could be 2/4/6/8/... etc

In bag B, Red:White = 1:4 If you have 5 marbles, 1 will be Red, 4 will be White. If you have 10 marbles, 2 will be Red, 8 will be White. etc No of white marbles could be 4/8/12/16... etc No of red marbles could be 1/2/3/4... etc

Total number of white marbles = 30 There are two ways in which we could have obtained 30. White marbles in (Bag A, Bag B) = (6, 24) or (18, 12)

If Bag A has 6 white marbles, it will have 2 red marbles. If Bag A has 18 white marbles, it will have 6 red marbles.

Re: Bag A contains red, white and blue marbles such that the red [#permalink]

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13 Nov 2014, 15:30

The ratio for bag A is 2:6:9 for red to white to blue marbles. You can get this number by multiplying the ration of red to white marbles by two to get 2:6, then multiplying the white to blue ration by three to get 6:9 and you can put them all together to get 2:6:9

The ratio of bag B equals 1:4 of red to white marbles and 0 blue marbles.

The problem also tells us that together the two bags contain 30 white marbles. This means that 6x+4y=30. There are two different combinations that work for this. x and y could both be 3. That would mean the total number of red marbles in bag A would be 6, which is D.

You could also have y=6 and x=1 and the total number of red marbles in bag A would be 2, but that is not an answer choice, so you would have to keep looking.
_________________

Re: Bag A contains red, white and blue marbles such that the red [#permalink]

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30 Aug 2016, 12:01

Bunuel wrote:

vigneshpandi 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. 1 2. 3 3. 4 4. 6 5. 8

Bag A: R/W=1/3=2/6; W/B=2/3=6/9; So R/W/B=2/6/9 --> # of marbles in bag A would be \(2x\), \(6x\), and \(9x\), for some positive integer multiple \(x\), where \(6x\) corresponds to the # of white marbles and \(2x\) corresponds to the # of red marbles (so # of red marbles must be a multiple of 2, so answers A and B are out at this stage);

Bag B: R/W=1/4 --> # of marbles in bag B would be \(y\) and \(4y\), for some positive integer multiple \(y\), where \(4y\) corresponds to the # of white marbles;

Given: \(6x+4y=30\) --> \(3x+2y=15\) --> there are two positive integer solutions for this equation: \(x=3\) and \(y=3\) --> in this case # of red marbles equals to \(2x=6\); Or: \(x=1\) and \(y=6\) --> in this case # of red marbles equals to \(2x=2\);

Only 6 is in the answer choices.

Answer: D.

How did you solve that equation to get those values? (I have marked the same in red in your answer) Could not understand it..
_________________

"The fool didn't know it was impossible, so he did it."

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

Bag A: R/W=1/3=2/6; W/B=2/3=6/9; So R/W/B=2/6/9 --> # of marbles in bag A would be \(2x\), \(6x\), and \(9x\), for some positive integer multiple \(x\), where \(6x\) corresponds to the # of white marbles and \(2x\) corresponds to the # of red marbles (so # of red marbles must be a multiple of 2, so answers A and B are out at this stage);

Bag B: R/W=1/4 --> # of marbles in bag B would be \(y\) and \(4y\), for some positive integer multiple \(y\), where \(4y\) corresponds to the # of white marbles;

Given: \(6x+4y=30\) --> \(3x+2y=15\) --> there are two positive integer solutions for this equation: \(x=3\) and \(y=3\) --> in this case # of red marbles equals to \(2x=6\); Or: \(x=1\) and \(y=6\) --> in this case # of red marbles equals to \(2x=2\);

Only 6 is in the answer choices.

Answer: D.

How did you solve that equation to get those values? (I have marked the same in red in your answer) Could not understand it..

_____________ By trial and error.
_________________

Re: Bag A contains red, white and blue marbles such that the red [#permalink]

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22 Oct 2017, 06:41

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