Both of the previous explainations seem to be copied from one another, I frankly think this question should be approached like this.

The word "same number of red and blue marbles" are added back means then the ratio of red and blue marbles would be 1:1 hence A is the right option.

Is this the correct approach Carcass?

Step-by-Step Explanation

1. Algebraic Approach

Let the initial number of red marbles be 3 x and blue marbles be 5 x , where x is a positive integer.

Let k be the positive integer representing the number of marbles added to each color.

Quantity A: The new ratio of red to blue marbles:


\(\text { New Ratio }=\frac{3 x+k}{5 x+k}\)


Quantity B: The original ratio:


\(\text { Original Ratio }=\frac{3}{5}\)




To compare Quantity \(\mathrm{A}\left(\frac{3 x+k}{5 x+k}\right)\) and Quantity \(\mathrm{B}\left(\frac{3}{5}\right)\) , cross-multiply:



\( Quantity A } \stackrel{?}{\longleftrightarrow} \text { Quantity B }\)
\(5(3 x+k) \stackrel{?}{\longleftrightarrow} 3(5 x+k)\)
\(15 x+5 k \stackrel{?}{\longleftrightarrow} 15 x+3 k\)

Step-by-Step Explanation

1. Algebraic Approach

Let the initial number of red marbles be $3 x$ and blue marbles be 5 x , where x is a positive integer.

Let k be the positive integer representing the number of marbles added to each color.

Quantity A: The new ratio of red to blue marbles:


\(\text { New Ratio }=\frac{3 x+k}{5 x+k}\)


Quantity B: The original ratio:


\(\text { Original Ratio }=\frac{3}{5}\)




To compare Quantity \(\mathrm{A}\left(\frac{3 x+k}{5 x+k}\right) \) and Quantity \(\mathrm{B}\left(\frac{3}{5}\right) \), cross-multiply:


\(Quantity A } \stackrel{?}{\longleftrightarrow} \text { Quantity B } \)

\(5(3 x+k) \stackrel{?}{\longleftrightarrow} 3(5 x+k) \)
\(15 x+5 k \stackrel{?}{\longleftrightarrow} 15 x+3 k\)

Subtract $15 x$ from both sides:

$$
5 k \stackrel{?}{\longleftrightarrow} 3 k
$$

Since k>0 (marbles were added), 5 k>3 k . Therefore, Quantity A is strictly greater than Quantity B.

2. Numerical Example (Testing Numbers)
Initial State: Suppose there are 3 red and 5 blue marbles (Ratio \( =\frac{3}{5}=0.60\) ).
Add Marbles: Add 1 red and 1 blue marble ( k=1 ).

New red marbles =3+1=4
New blue marbles =5+1=6

New Ratio (Quantity A): \(\frac{4}{6}=\frac{2}{3} \approx 0.67 \)
Quantity B: \(\frac{3}{5}=0.60\)


Since 0.67 > 0.60, Quantity A > Quantity B.