Let $A$ be the number of marbles in bag A initially, and $B$ be the number of marbles in bag B initially.

Given:

1. $A + B = 162$
2. Celia draws $\frac{1}{4}$ of the marbles from bag A and puts them into bag B.

After this transfer, the number of marbles in each bag changes:

• Bag A has $A - \frac{1}{4}A = \frac{3}{4}A$ marbles.
• Bag B has $B + \frac{1}{4}A$ marbles.

It is also given that the number of marbles in bag A is then 30 more than that in bag B: $\frac{3}{4}A = \left(B + \frac{1}{4}A\right) + 30$

Simplify this equation: $\frac{3}{4}A = B + \frac{1}{4}A + 30$ $\frac{3}{4}A - \frac{1}{4}A = B + 30$ $\frac{2}{4}A = B + 30$ $\frac{1}{2}A = B + 30$ $B = \frac{1}{2}A - 30$

Now, substitute $B = \frac{1}{2}A - 30$ into the total number of marbles equation: $A + \left(\frac{1}{2}A - 30\right) = 162$ $A + \frac{1}{2}A - 30 = 162$ $\frac{3}{2}A - 30 = 162$ $\frac{3}{2}A = 192$ $A = 192 \times \frac{2}{3}$ $A = 128$

Using $A + B = 162$: $128 + B = 162$ $B = 162 - 128$ $B = 34$

Therefore, the original number of marbles in each bag is:

• Bag A: 128 marbles
• Bag B: 34 marbles

Would you like more details or have any questions about this solution?

Here are some related questions you might consider:

1. How would the equation change if Celia drew 1/3 of the marbles instead?
2. What if the total number of marbles was different?
3. How does this problem change if Celia adds marbles instead of transferring them?
4. What if the number of marbles in bag B was initially greater than in bag A?
5. Can you formulate a similar problem with a different constraint?

Tip: When dealing with problems involving fractions of quantities, it is often helpful to express all quantities in terms of a single variable to simplify the equations.