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

I just don't seem to understand multiple ratios such as in this questions. Please help

I just did it in mgmat1; I used picking method pls see my way.
Bag A: Ra/Wa = 1/3,
Bag B: Rb/Wb = 1/4
and Wa+Wb = 30
hidden information is that each color must be positive integer =>
- Wb must be divisible by 4 to make Rb be an integer
- Wa must be divisible by 3 to make Ra be an integer

Pick Wb = 4 => wa will be 26 => not ok becausee 26 is not divisible by 3
Pick Wb = 8 => wa will be 22 => not ok becausee 22 is not divisible by 3
Pick Wb = 12 => wa will be 18 => ok becausee 18 is divisible by 3
Pick Wb = 16 => wa will be 14 => not ok becausee 14 is not divisible by 3
Pick Wb = 20 => wa will be 10 => not ok becausee 10 is not divisible by 3
Pick Wb = 24 => wa will be 6 => ok becausee 6 is divisible by 3

Now because the answer has only case wb12 => wa will be 18 =>Ra =6

It doable within 2 min

Bag A: has

R: W: B :: 2: 6: 9

Bag b:

R: w: :: 1: 4

white = 30 so

(6/17)*X + (4/5)* Y = 30

X = 17 , Y = 40 , satisfies the equation
X= 17*3 , Y = 15 , also satisfies the equation and both results in different results.....

Please explain where i am going wrong?

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?

bag 1
R:W = 1:3 SO.... W = 3R
W:B = 2:3 SO.... B = 3W/2 = 9R/2
SO R:W:B = 1:3:9/2 = 2:6:9


BAG 2
R:W = 1:4

now look at the two ratios B1:W1 = 6:9


B2:W2 = 1:4


we have w1+w2 = 30

so we can achieve this sum only by B1:W1 = 12:18
AND B2:W2 = 3:12

SO THAT THE SUM ABOVE OF W2 AND W1 IS 30.....

WHEN W1 IS 18 R1 = 4

That gives us (c)

Bag 1 contains 2X (red) + 6X (white) + 9X (blue)
Bag 2 contains 1Y (red) + 4Y (white)

6X + 4Y = 30
3X + 2Y = 15

here both X and Y are positive integers (number of balls are positive integers)
we get following possible values of (X,Y) -- (1,6), (3,3)

hence red marbles in Bag A could be 2, 6.

6 is Option D, (2 is not mentioned as an option)

Here's how I keep track of these "separate but related" ratios: Line them up!

Red, White and Blue are mentioned, so on paper I write:
R : W : B

Underneath, I write each given ratio on a separate line, vertically lining up the numbers with the right letter:
R : W : B
1 : 3 : _
_ : 2 : 3

Finally, multiply the rows, sort-of. The number of white marbles must be an integer (no partial marbles), and a multiple of 2 and 3 (because the number Red and Blue marbles must be integers, too). Thus, multiply the first row by 2 and the second row by 3.
R : W : B
1 : 3 : _ (times 2)
_ : 2 : 3 (times 3)
2 : 6 : 9 (resulting combined ratio)

Note that a final check is a good idea:
2:6 reduces to 1:3, which was given.
6:9 reduces to 2:3, which was given.

Bag A therefore has R:W:B = 2x:6x:9x, where x is an integer. This gives us enough to eliminate the odd answers: the number of Red marbles in Bag A must be even!

In Bag B, R:W = 1y:4y, where y is an integer. You SHOULD NOT combine this ratio with the one for Bag A above, as they are truly, physically separate (i.e. different bags). The number of W in Bag A must be a multiple of 6, and the number of W in Bag B must be a multiple of 4. This does NOT mean that the number of W in the combined bags must be 24. (in fact, we are told that it isn't!)

At this point I would just plug numbers to see how the 30 white marbles could be distributed between the two bags:

0W in Bag B-->30 W in Bag A-->30 is the 5th multiple of 6, so x = 5 and that means 2*5 Red marbles in Bag A.
4W in Bag B-->26 W in Bag A-->26 is not a multiple of 6, so not possible.
8W in Bag B-->22 W in Bag A-->22 is not a multiple of 6, so not possible.
12W in Bag B-->18 W in Bag A-->18 is the 3rd multiple of 6, so x = 3 and that means 2*3 Red marbles in Bag A.
etc.

10 is a possible answer, but the only answer listed is 6 Red marbles in Bag A.

_________________
Emily Sledge
Instructor
ManhattanGMAT

I used the following method. Please advice me if it was pure luck:

A
R:W = 1:3 --> Red balls = 1/4 (partial / total)
W:B = 2:3 --> Red balls = 0/5

B
R:W = 1:4 --> Red balls = 1/5

Then I have (1/4) + (1/5) + (0/5) = 30

The only way I can have 30 is saying that I have 5/20 + 1/5 + 0/5 --> 6 red balls!

Is this a consistent method?

Thank you!

hi --

this is definitely pure luck if it yields the correct answer -- you're treating the two conditions given in part (a) as if they were two totally separate collections of marbles.

please see emily's post above (she posts as "esledge") for the correct way to combine multiple ratios involving the same information.

Can't we just use the unknown multiplier here?

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

Bag B
R : W
1x : 4x

The total number of white marbles in both bags is 30, so we have:

6x + 4x = 30
x = 3 (our unknown multiplier)

So the number of red marbles in Bag A would be 2(3) = 6

What am I missing with this simplistic approach?

Major error is that you are assuming the total number of marbles to be the same in both cases. It works here but might not work in other cases.

Let total marbles in Bag A=x
So,
Bag A
R : W : B
2x : 6x : 9x


Let total marbles in Bag B=y
So,
Bag B
R : W
1y : 4y

So 6x+4y=30
or 3x+2y=15-----only 2 solutions possible. x =1 and y=6 or x=3 and y=3

So number of red marbles in A could either be 2 or 6

I understand. Thanks gokul_nair1984!

Great! Thanks Gokul.

Not sure if this thread is still active, but I just came across this question and read Emily's solution - she mentioned at the end that 10 is also a possible answer. I don't really understand why? I would think that 18 and 12 are the only numbers which would work since it's clearly mentioned in the prompt that Bag B does contain 4 parts of white marbles. To hypothesize Bag B has 0 white marbles while Bag A has all 30 marbles seems to change the question a little.

Just wondering if my interpretation of the question is accurate. I imagine that details like these might come up DS questions...whether there is only one answer or multiple solutions. Though I haven't done enough to see such questions, lol.

Hope I'm not bouncing this off a wall :)

Hey Sarah, I agree with you and I think it's just a small error on Emily's part. You are right; for Bag A to have 10 red marbles it would also have 30 white marbles. Thus Bag B would have no marbles, yet the problem clearly states the ratio of marbles in Bag B. Great catch!

_________________
Jamie Nelson
ManhattanGMAT Instructor

BEST WAY TO DO THIS:

Bag A - Red:White - 1:3
White:Blue - 2:3

The intention here is to find what is common in this single bag. It is White marbles BUT the ratio (always understand ratio as ="in relation to") in which the white is present is different toward red and blue marbles. Now, first try to make the relation equitable i.e look for what can be multiplied to 3 in the 1:3 ratio(red,white) and to 2 in the 2:3 ratio (white,blue) to make them comparable. Yes, we must multiply the 1:3 ratio (red,white) by (2) in order to make it into a 2:3 and the (red,blue) 2:3 by (3).

Final ratios are red,white - 2:6
and white:blue - 6:9

NOTE: the white is present in the same ratio (6 in both)

Now, Bag B -> Red:White = 1:4 (Again white is present and is common in both bags)
So one must multiple the WHITES in both bags to equate their relationship however, NOTE- the question says there are 30 white marbles in all.

NOTICE: if one was to multiply the whites in bag A(i.e 6) by 3 and the whites in bag B(i.e 4) by 3 as well, overall we would reach the target number i.e 30 (i.e 6X3 +4X3)= 30 (Remember this multiplication is only helping us reach 30 as a ratio NOT as a number)
Since, we are multiplying Bag A whites by (3), the rest of the marbles in Bag A also must be multiplied by (3) to maintain consistency in the relationship within the Bag i.e in Bag A-> Reds= 2X(3) and Blues = 9X(3). Now FINALLY we have met all the criteria and we have 6 REDS ANS.

Reach out for simple approaches with a proper STORY (hehe) on my e-mail id. !