Is it 2?

fadibassil wrote:I saw this problem in MGmat i don't know if i'm allowed to past everything here. but here is the stem.

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?

the solution is to think of the factors of the white balls in the bags, so here in bag A white must be a multiple of 2 and 3 and in bag be a multiple of 4.

but when I see such a problem I start thinking of the ratios not the factors. what take away or mental note can I use to differentiate such a problem from other ratio problems. When looking at it with a hindsight they give info about the white balls but this is just analysing after the fact, and I don't feel this is transferable to other problems.

I hope I was clear with my question.

Thank you

You are supposed to past, err...post either the five answer choices if it was a PS or the two statements if it was a DS, with OA and/or the source along with the stem. It surely is a PS, since the stem reads "How many red marbles could be in bag A?". That means it doesn't have got a unique answer, and it cannot be answered without the supply of the five answer choices.

In Bag A

R:W = 1:3 and W:B = 2:3, therefore R:W:B = 2:6:9 (any doubt?). So, let's take R = 2 m, W = 6 m, and B = 9 m, in Bag A; where m is any positive integer.

In Bag B

R:W = 1:4. We can take R = n, and W = 4 n, in Bag B; where n is any positive integer.

Together, we are to find a possible value of 2 m, given that 6 m + 4 n = 30 or 3 m + 2 n = 15. This gives the following possibilities for m and n:

When m = 1, n = 6; giving 2 m = 2

When m = 3, n = 3; giving 2 m = 6

When m = 5, n = 0; giving 2 m = 10; do not take this to account anyway.

If the answer choices with you include either 2 or 6 (both together there would spoil the question), that's my answer.

Sorry for the confusion, the answer choices are

1 , 3 , 4 , 6 , 8


So it's as simple as that. I reached the 6m + 4n = 30 while solving and got stuck, thinking this is a one equation with two unknowns, and 2m + n = ? and got stuck. sometimes I don't understand how you get the idea of testing factors


So the take away from this problem is to notice the word could be as signaling testing choices, and when reaching something similar to 6m+4n=30 i should list the multiples of each to check the answer. Correct?


Thank you.