https://file.aiquickdraw.com/mathbot/file/17268517165538jixqxq5.pdf

Let's work through the exercises step by step.

### 1. Solve the expression: 
\[
A = 10 \cdot \left( \frac{62.8}{31.4} + 1 \right)^3 - 53 \cdot 2
\]

First, calculate \( \frac{62.8}{31.4} = 2 \), so the expression inside the parentheses becomes:
\[
2 + 1 = 3
\]

Now cube 3:
\[
3^3 = 27
\]

So the expression becomes:
\[
A = 10 \cdot 27 - 53 \cdot 2 = 270 - 106 = 164
\]

Thus, \( A = 164 \).

---

### 2. Solve the expression:
\[
\frac{41}{60} - \frac{2}{5} \cdot \left( \frac{5}{3} - \frac{1}{4} \right)
\]

First, simplify \( \frac{5}{3} - \frac{1}{4} \) by finding a common denominator:
\[
\frac{5}{3} - \frac{1}{4} = \frac{20}{12} - \frac{3}{12} = \frac{17}{12}
\]

Now multiply \( \frac{2}{5} \cdot \frac{17}{12} \):
\[
\frac{2}{5} \cdot \frac{17}{12} = \frac{34}{60}
\]

Now subtract:
\[
\frac{41}{60} - \frac{34}{60} = \frac{7}{60}
\]

Thus, the result is \( \frac{7}{60} \).

---

### 3. Solve the following:

**(a)** \( 13.5551 : 0.001 \)
\[
13.5551 \div 0.001 = 13555.1
\]

**(b)** \( 0.02023 \cdot 10^5 \)
\[
0.02023 \cdot 100000 = 2023
\]

---

### 4. Find which numbers are divisible by 2, 3, and 5: 
1465, 60120, 42040, 520500

A number divisible by 2, 3, and 5 must be divisible by 30 (since \( 2 \cdot 3 \cdot 5 = 30 \)).

- **1465**: Not divisible by 2, so it's not divisible by 30.
- **60120**: Divisible by 2, 3, and 5 (divisible by 30).
- **42040**: Divisible by 2 and 5, but not by 3.
- **520500**: Divisible by 2, 3, and 5 (divisible by 30).

Thus, the numbers are **60120** and **520500**.

---

### 5. Calculate the printing costs:

**(a)** Cost of printing 80 photos:
\[
80 \cdot 0.17 = 13.60 \, \text{€}
\]

**(b)** Cost of printing 120 photos:
The first 100 photos cost \( 100 \cdot 0.17 = 17 \, \text{€} \), and the remaining 20 photos cost \( 20 \cdot 0.15 = 3 \, \text{€} \).
\[
17 + 3 = 20 \, \text{€}
\]

**(c)** Maximum number of photos with 32.5€:
The first 100 photos cost 17€. The remaining budget is:
\[
32.5 - 17 = 15.5 \, \text{€}
\]
For each additional photo, the cost is 0.15€, so:
\[
\frac{15.5}{0.15} = 103.33
\]
Thus, you can print 100 + 103 = **203 photos**.

---

### 6. Solve the pricing problem:

**(a)** After a reduction of \( \frac{1}{5} \), the new price is:
\[
150 - \frac{150}{5} = 150 - 30 = 120 \, \text{€}
\]
After adding 36€:
\[
120 + 36 = 156 \, \text{€}
\]

**(b)** The percentage increase from the original price:
\[
\frac{156 - 150}{150} \cdot 100 = \frac{6}{150} \cdot 100 = 4\%
\]

---

### 7. Maximum number of children sharing chocolates, candies, and lollipops:
You need to find the greatest common divisor (GCD) of 30, 42, and 12:
\[
\text{GCD}(30, 42, 12) = 6
\]
Thus, at most **6 children** can equally share the treats.

---

### 8. Time to fill the tank with both taps:
The first tap fills the tank in 8 hours, so it fills \( \frac{1}{8} \) of the tank per hour. The second tap empties \( \frac{1}{10} \) of the tank per hour.

The net filling rate is:
\[
\frac{1}{8} - \frac{1}{10} = \frac{5}{40} - \frac{4}{40} = \frac{1}{40}
\]
Thus, the tank will fill in 40 hours.

---

Let me know if you need further details or clarifications!

Here are some related questions:
1. How would the result change if the second tap emptied faster?
2. What if the price increase in Exercise 6 was 50€ instead of 36€?
3. How can we find divisibility for numbers other than 2, 3, and 5?
4. Can you explain the process of finding the GCD in Exercise 7 in detail?
5. How would the printing costs change with different rates for photos?

**Tip**: For mixed operations, always simplify expressions step by step, starting with parentheses and exponents.

Explore detailed solutions for math exercises including basic arithmetic operations, solving fractions, and rate of work problems. Suitable for students in Grades 7-9. Learn key concepts like divisibility rules and finding the greatest common divisor.

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