Let the initial number of sweets that Maria, Shanta, and Farida had be represented by the ratio 10:5:3. Let $x$ be the common factor, so their respective numbers of sweets are:

• Maria: $10x$
• Shanta: $5x$
• Farida: $3x$

Maria gave away 68 sweets, splitting them between Shanta and Farida such that all three girls ended up with the same number of sweets. Let's assume that after the transfer, each girl has $y$ sweets.

Thus, after the transfer:

• Maria has $10x - 68$ sweets.
• Shanta has $5x + k$ sweets, where $k$ is part of the 68 sweets Maria gave.
• Farida has $3x + (68 - k)$ sweets.

Since all three have the same number of sweets after the exchange, we can set up the following equations:

$10x - 68 = 5x + k = 3x + (68 - k) = y$

First, equate $10x - 68$ and $5x + k$:

$10x - 68 = 5x + k \implies k = 5x - 68 \quad \text{(Equation 1)}$

Now, equate $10x - 68$ and $3x + (68 - k)$:

$10x - 68 = 3x + (68 - k)$

Substitute the value of $k$ from Equation 1 into this equation:

$10x - 68 = 3x + (68 - (5x - 68))$

Simplify the right-hand side:

$10x - 68 = 3x + 68 - 5x + 68$ $10x - 68 = -2x + 136$

Now, solve for $x$:

$10x + 2x = 136 + 68$ $12x = 204$ $x = \frac{204}{12} = 17$

Now that we know $x = 17$, we can calculate the initial number of sweets for each:

• Maria: $10x = 10(17) = 170$
• Shanta: $5x = 5(17) = 85$
• Farida: $3x = 3(17) = 51$

Thus, the total number of sweets they initially had is:

$170 + 85 + 51 = 306$

Therefore, the total number of sweets the girls had is 306.

Would you like further details or explanations on any step?

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Here are 5 related questions for further exploration:

1. How many sweets did Maria give to Shanta and Farida individually?
2. If Maria gave more sweets to Farida than Shanta, how would the problem change?
3. What would happen if the ratio was different, say 8:6:4?
4. Can you express the total number of sweets as a function of $x$?
5. How would the solution change if Maria had given away 80 sweets instead of 68?

Tip: When dealing with ratios, assigning variables to represent the parts in terms of a common factor simplifies the problem.