Determine the percentage of the original price paid.

A 20% reduction means the customer paid $100\% - 20\% = 80\%$ of the original price.

Let $P$ be the original price.

The correct option is C. Le43,750.

Principal ($P$) = Le 500,000
Rate ($R$) = 6% per annum
Time ($T$) = 5 years

The formula for simple interest is $I = \frac{P \times R \times T}{100}$.

The correct option is D. Le150,000.00.

Departure time: 8:30 a.m.
Arrival time: 11:50 a.m.
Duration = (11 hours 50 minutes) - (8 hours 30 minutes)
Duration = 3 hours 20 minutes.

Total time ($T$) = $3 + \frac{1}{3} \text{ hours} = \frac{9}{3} + \frac{1}{3} \text{ hours} = \frac{10}{3} \text{ hours}$.

Speed ($S$) = 66 km/hour.

The correct option is B. 220 km.

Pavement length ($L_p$) = 500 m = $500 \times 100 \text{ cm} = 50,000 \text{ cm}$.
Pavement width ($W_p$) = 150 m = $150 \times 100 \text{ cm} = 15,000 \text{ cm}$.
Area of pavement ($A_p$) = $L_p \times W_p = 50,000 \text{ cm} \times 15,000 \text{ cm} = 750,000,000 \text{ cm}^2$.

Slab length ($L_s$) = 75 cm.
Slab width ($W_s$) = 40 cm.
Area of one slab ($A_s$) = $L_s \times W_s = 75 \text{ cm} \times 40 \text{ cm} = 3,000 \text{ cm}^2$.

There seems to be a discrepancy with the options provided. Let me recheck the calculation.
$500 \times 150 = 75000 \text{ m}^2$.
$75 \text{ cm} \times 40 \text{ cm} = 0.75 \text{ m} \times 0.40 \text{ m} = 0.3 \text{ m}^2$.
Number of slabs = $\frac{75000 \text{ m}^2}{0.3 \text{ m}^2} = \frac{750000}{3} = 250000$.

Let's re-examine the options. The options are 3000, 2500, 2250, 2000. My calculated answer is 250,000. This is significantly larger than any option. It's possible there's a typo in the question or options, or I misread a unit.
Let's assume the pavement dimensions were in cm and the slabs in mm, or vice versa, but the question states m and cm.
If the pavement was 500 cm by 150 cm, then $500 \times 150 = 75000 \text{ cm}^2$.
Number of slabs = $\frac{75000 \text{ cm}^2}{3000 \text{ cm}^2} = 25$. This is too small.

Let's assume the pavement dimensions were 50 m by 15 m.
Area of pavement = $50 \text{ m} \times 15 \text{ m} = 750 \text{ m}^2$.
Area of one slab = $0.75 \text{ m} \times 0.40 \text{ m} = 0.3 \text{ m}^2$.
Number of slabs = $\frac{750 \text{ m}^2}{0.3 \text{ m}^2} = \frac{7500}{3} = 2500$.
This matches option B. It is highly probable that the pavement dimensions were intended to be 50 m by 15 m, or there is a factor of 100 error in the question's length/width. Given the multiple-choice options, I will proceed with the assumption that the pavement dimensions were 50 m by 15 m.

Revised Step 1 (Assumption: Pavement length 50 m, width 15 m to match options):
Pavement length ($L_p$) = 50 m = $5000 \text{ cm}$.
Pavement width ($W_p$) = 15 m = $1500 \text{ cm}$.
Area of pavement ($A_p$) = $L_p \times W_p = 5000 \text{ cm} \times 1500 \text{ cm} = 7,500,000 \text{ cm}^2$.

Revised Step 2 (Area of one slab remains the same):
Area of one slab ($A_s$) = $75 \text{ cm} \times 40 \text{ cm} = 3,000 \text{ cm}^2$.

Revised Step 3 (Calculate number of slabs):

The correct option is B. 2500.

The figure consists of a rectangle with two semicircles at its ends.
The perimeter is made up of:

• Two straight sides of length 100 m.
• The circumference of two semicircles, which together form one full circle. The diameter of this circle is given as 35 m.

Length of two straight sides = $2 \times 100 \text{ m} = 200 \text{ m}$.

The diameter ($d$) of the circle is 35 m.
The circumference ($C$) of a circle is $\pi d$. Using $\pi = \frac{22}{7}$.

The correct option is A. 310 m.