Let the initial number of sweets that Ahmad, Berlinda, and Charles had be represented by $3x$, $5x$, and $8x$ respectively, since the ratio of their sweets is $3:5:8$.

After Charlie and Berlinda gave away a total of 78 sweets, the ratio of the sweets became $5:7:6$. Let’s denote the new number of sweets for Ahmad, Berlinda, and Charles as $5y$, $7y$, and $6y$ respectively.

Step 1: Set up equations based on the change in sweets

The number of sweets Ahmad had did not change, so:

3x = 5y \tag{1}

Berlinda and Charles together gave away 78 sweets, so:

$(5x - 7y) + (8x - 6y) = 78$ Simplifying this:

$(5x + 8x) - (7y + 6y) = 78$ $13x - 13y = 78$ x - y = 6 \tag{2}

Step 2: Solve the system of equations

From equation (1), we have:

$x = \frac{5y}{3}$

Substitute this into equation (2):

$\frac{5y}{3} - y = 6$ Multiply through by 3 to eliminate the fraction:

$5y - 3y = 18$ $2y = 18$ $y = 9$

Step 3: Find $x$

Substitute $y = 9$ back into equation (2):

$x - 9 = 6$ $x = 15$

Step 4: Calculate Berlinda’s initial number of sweets

Berlinda’s initial number of sweets was $5x$, so:

$5x = 5 \times 15 = 75$

Thus, Berlinda initially had 75 sweets before giving away any.

Final Answer:

Berlinda had 75 sweets before giving away.

Would you like to dive deeper into any step, or have any further questions? Here are some related problems for you:

1. If Charlie gave away 54 sweets, how many did Berlinda give away?
2. How many sweets did Ahmad have after the others gave away 78 sweets?
3. What would be the new ratio if Berlinda gave away 10 more sweets?
4. If the ratio was originally 4:6:9, how would the solution change?
5. What if Ahmad gave away 10 sweets after the total of 78 were given away?

Tip: Ratios provide a powerful way to compare quantities, but always ensure you account for the totals when they change!