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Let's go through the questions one by one.

### **Question 1:**
- The problem is to add fractions \( \frac{3}{8} \) and \( \frac{5}{12} \).
- To do this, we need a common denominator between 8 and 12. The least common denominator (LCD) is 24.
  
  \[
  \frac{3}{8} = \frac{9}{24} \quad \text{and} \quad \frac{5}{12} = \frac{10}{24}
  \]
  
  Now, adding the fractions:

  \[
  \frac{9}{24} + \frac{10}{24} = \frac{19}{24}
  \]

So, the total amount of milk is \( \frac{19}{24} \) liters.

### **Question 2:**
- We are given the weight of a large cookie as \( \frac{5}{6} \) ounces and a small cookie as \( \frac{4}{9} \) ounces. We need to find the total weight of 2 large cookies and 1 small cookie.
  
  First, the total weight of 2 large cookies:
  \[
  2 \times \frac{5}{6} = \frac{10}{6} = \frac{5}{3}
  \]

  Now, adding the small cookie weight:

  \[
  \frac{5}{3} + \frac{4}{9}
  \]

  We need a common denominator. The least common denominator of 3 and 9 is 9:
  \[
  \frac{5}{3} = \frac{15}{9}
  \]

  Now, add the fractions:

  \[
  \frac{15}{9} + \frac{4}{9} = \frac{19}{9} = 2 \frac{1}{9}
  \]

So, the total weight is \( 2 \frac{1}{9} \) ounces.

### **Question 3:**
- We are comparing two bag sizes, \( \frac{4}{5} \) kg and \( \frac{7}{12} \) kg. To find how much more coffee is in the larger bag, we subtract the smaller from the larger:

  \[
  \frac{4}{5} - \frac{7}{12}
  \]

  The least common denominator of 5 and 12 is 60. Convert the fractions:
  \[
  \frac{4}{5} = \frac{48}{60} \quad \text{and} \quad \frac{7}{12} = \frac{35}{60}
  \]

  Subtract the fractions:
  \[
  \frac{48}{60} - \frac{35}{60} = \frac{13}{60}
  \]

So, the larger bag has \( \frac{13}{60} \) kg more coffee.

### **Question 4:**
- We are given a total of \( \frac{4}{5} \) kg of sugar, from which \( \frac{1}{10} \) kg and \( \frac{3}{20} \) kg were used. We need to subtract these amounts from the total.
  
  First, add the amounts used:
  \[
  \frac{1}{10} + \frac{3}{20}
  \]
  The least common denominator of 10 and 20 is 20:
  \[
  \frac{1}{10} = \frac{2}{20}
  \]

  Now, add the fractions:
  \[
  \frac{2}{20} + \frac{3}{20} = \frac{5}{20} = \frac{1}{4}
  \]

  Now subtract from the total sugar:
  \[
  \frac{4}{5} - \frac{1}{4}
  \]

  The least common denominator of 5 and 4 is 20:
  \[
  \frac{4}{5} = \frac{16}{20}
  \]

  Subtract the fractions:
  \[
  \frac{16}{20} - \frac{5}{20} = \frac{11}{20}
  \]

So, there are \( \frac{11}{20} \) kg of sugar left.

### **Question 5:**
- For the first cheesecake, there were 16 slices, and 3 remain. So:
  \[
  16 - 3 = 13 \, \text{slices were eaten}.
  \]

  For the second cheesecake, 4 slices were sold out of 10, so 6 slices are left.

  The total cheesecake left in the fridge is:
  \[
  3 \, \text{(from the first)} + 6 \, \text{(from the second)} = 9 \, \text{slices}.
  \]

So, there are 9 slices of cheesecake left.

---

Do you want details or have any other questions? Here are some related questions:

1. How do you find the least common denominator for two fractions?
2. What are the steps to add or subtract fractions with different denominators?
3. How do you convert an improper fraction into a mixed number?
4. How would you solve similar problems involving weights or quantities in different units?
5. How would the result change if one of the cheesecakes was cut differently?

**Tip:** Always remember to simplify your fractions after performing operations like addition or subtraction!

Solve the following math problems related to fractions, weights, and measurements.

Explore how to solve fraction problems involving the addition and subtraction of weights, measurements, and mixed numbers. Learn key concepts such as finding the least common denominator, converting improper fractions to mixed numbers, and calculating differences in weights. Suitable for students in Grades 5-7.

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