Little correction: it should be Y + B = 13 and R + B = 7.

A certain number of marbles are to be removed from a box containing only solid-colored red, yellow, and blue marbles. How many more yellow marbles than red marbles are in the box before any are removed?

We need to find Y - R.

(1) To guarantee that a red marble is removed, the smallest number of marbles that must be removed from the box is 14.

This implies that Y + B = 13 because only if there are total of 13 blue and yellow marbles, removing 14 marbles guarantees getting a red one: you should pick all blue and yellow and the next one for sure will be red. But knowing that B + Y = 13 is not enough to get Y - R. Not sufficient.

(2) To guarantee that a yellow marble is removed, the smallest number of marbles that must be removed from the box is 8.

The same ways as above this means that R + B = 7. Not sufficient.

(1)+(2) Subtract R + B = 7 from Y + B = 13: (Y + B) - (R + B) = 13 - 7 --> Y - R = 6. Sufficient.

Answer: C.
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Let there be R,Y and B red,yellow and blue balls be there initially. We need to find out Y-R = ?

St 1 basically says that B+Y=14 ( many combinations possible )
B : Y
1:13
2:12
3:11
4:10....

Not sufficient

St 2 says R+B=8

R: B
1:7
2:6
3:5
4:4
5:3...

Combining we see that Blue balls cannot be more than 7, So if

B=7, R=1 and Y=7 Y-R=6
B=6,R=2, and Y=8 then 8-2=6

If B=4,R=4, then Y=10 Y-R=6..

For all values Y-R=6...

Ans is C
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Hope it helps.
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A certain number of marbles are to be removed from a box containing only solid-colored red, yellow, and blue marbles. How many more yellow marbles than red marbles are in the box before any are removed?

(1) To guarantee that a red marble is removed, the smallest number of marbles that must be removed from the box is 14.

Y + B = 14

(2) To guarantee that a yellow marble is removed, the smallest number of marbles that must be removed from the box is 8.

B + R = 8

Combined-
Y - R = 6
C.

\(? = Y - R\,\,\,\,\,\left( {{\text{yellow,}}\,{\text{red and}}\,\,{\text{blue}}\,\,{\text{marbles}}\,\,{\text{only}}} \right)\)

\(\left( 1 \right)\,\,Y + B = 13\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {Y,R,B} \right) = \left( {12,1,1} \right)\,\,\,\, \Rightarrow \,\,\,? = 11 \hfill \cr
\,{\rm{Take}}\,\,\left( {Y,R,B} \right) = \left( {12,2,1} \right)\,\,\,\, \Rightarrow \,\,\,? = 10 \hfill \cr} \right.\)

\(\left( 2 \right)\,\,R + B = 7\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {Y,R,B} \right) = \left( {12,6,1} \right)\,\,\,\, \Rightarrow \,\,\,? = 6 \hfill \cr
\,{\rm{Take}}\,\,\left( {Y,R,B} \right) = \left( {13,6,1} \right)\,\,\,\, \Rightarrow \,\,\,? = 7 \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,Y + B = 13 \hfill \cr
\,R + B = 7 \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,\,\,? = Y - R = 13 - 7\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.\)


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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it's an interesting exercise reviewing the error log. when I could not figure out how to evaluate both statements, I blindly assumed the answer is E, which is wrong. my rationale now that led to the right answer.

when we pick marbles from a box, as long as there is at least one marble of a color, there is a chance of picking up that marble. the only way we can guarantee a marble of a specific color being picked up is when all other colors have already been picked.

(1) if a minimum of 14 marbles have to be picked to guarantee a red marble, then yellow + blue = 14
(2) if a minimum of 8 marbles have to be picked to guarantee a yellow marble, then blue + red = 8.

since the question is asking for how many more yellow than red, you can find the answer by subtracting the equations (1) and (2) which gives yellow - red = 6. there are six more yellow marbles than red.

Answer is C.

cheers.

I have have done the following based on my understanding of the Question

From Statement 1: R+Y+B=14
R=1, as per the statement
Eq 1: So, Y+B=13. Therefore not sufficient
From Statement 2: R+Y+B=8
Y=1, as per statement 2
Eq 2: So, R+B=7. Therefore not sufficient

From Both (1) & (2) R+Y+B=14
From (2) R+B=7
So, Y=7
So, from Eq 1 & Eq 2: R+Y+2B=20
We know Y=7 & R+B=7

So, 7+7+B=20
B= 6
Finally, R=1