Let's assume the following:

• Hema initially bought $x$ chocolates.
• Shireen initially bought $y$ sweets.

Step 1: Analyzing Hema's situation

• Hema gives half of her chocolates to Shireen, so she is left with $\frac{x}{2}$ chocolates.
• Shireen gives half of her sweets to Hema, so Hema receives $\frac{y}{2}$ sweets.
• Hema eats 10 sweets, so the number of sweets Hema has after eating is $\frac{y}{2} - 10$.
• The number of chocolates Hema has doesn't change, so she still has $\frac{x}{2}$ chocolates.

According to the problem, after all this, the ratio of sweets to chocolates that Hema has is 6:5. Therefore, we can write the equation:

$\frac{\frac{y}{2} - 10}{\frac{x}{2}} = \frac{6}{5}$

Multiplying both sides by $\frac{x}{2}$ to eliminate the denominator:

$\frac{y}{2} - 10 = \frac{6}{5} \times \frac{x}{2}$

Simplifying:

$\frac{y}{2} - 10 = \frac{3x}{5}$

Step 2: Analyzing Shireen's situation

• Shireen receives half of Hema's chocolates, so she gets $\frac{x}{2}$ chocolates.
• Shireen eats 4 chocolates, so she is left with $\frac{x}{2} - 4$ chocolates.
• Shireen gives half of her sweets to Hema, so she is left with $\frac{y}{2}$ sweets.

The problem states that the ratio of sweets to chocolates that Shireen has is 11:3. Therefore, we can write the equation:

$\frac{\frac{y}{2}}{\frac{x}{2} - 4} = \frac{11}{3}$

Multiplying both sides by $\frac{x}{2} - 4$:

$\frac{y}{2} = \frac{11}{3} \times \left( \frac{x}{2} - 4 \right)$

Simplifying:

$\frac{y}{2} = \frac{11}{6} \times (x - 8)$

Step 3: Solving the system of equations

We now have two equations:

1. $\frac{y}{2} - 10 = \frac{3x}{5}$
2. $\frac{y}{2} = \frac{11}{6} \times (x - 8)$

Let's solve these equations step by step.

From equation (1):

$\frac{y}{2} = \frac{3x}{5} + 10$

Substituting this into equation (2):

$\frac{3x}{5} + 10 = \frac{11}{6} \times (x - 8)$

Multiplying both sides by 30 to eliminate the denominators:

$6 \times (3x + 50) = 55 \times (x - 8)$

Expanding both sides:

$18x + 300 = 55x - 440$

Bringing like terms together:

$300 + 440 = 55x - 18x$

$740 = 37x$

Solving for $x$:

$x = \frac{740}{37} = 20$

Step 4: Finding the number of sweets $y$

Substitute $x = 20$ into the equation $\frac{y}{2} = \frac{3x}{5} + 10$:

$\frac{y}{2} = \frac{3 \times 20}{5} + 10 = 12 + 10 = 22$

So,

$y = 44$

Conclusion:

Hema initially bought 20 chocolates.

Would you like more details or have any questions?

Here are 5 relative questions that extend this problem:

1. What would happen if Hema bought 30 chocolates instead?
2. How would the solution change if Shireen ate 6 chocolates instead of 4?
3. What if the ratio of sweets to chocolates Hema had was 7:5 instead of 6:5?
4. How does changing the number of sweets Shireen buys affect the final ratios?
5. What if Shireen gave 1/3 of her sweets to Hema instead of half?

Tip: When dealing with ratios, always express both quantities in the same form (like fractions) to simplify solving equations.