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, $A = 164$.

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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 $\frac{7}{60}$.

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3. Solve the following:

(a) $13.5551 : 0.001$ $13.5551 \div 0.001 = 13555.1$

(b) $0.02023 \cdot 10^5$ $0.02023 \cdot 100000 = 2023$

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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.

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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.

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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: $\frac{156 - 150}{150} \cdot 100 = \frac{6}{150} \cdot 100 = 4\%$

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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.

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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.

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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.