Complete the table: Items, Price of each, Total expenses. 5kg of meat 3000frw/kg, 5kg of rice 6600frw, litre of honey 4050frw. If he had 45000frw and save the rest, how much did he save?

Multiply the numerator and denominator by 1000 to remove decimals:

The result is $\boxed{80}$.

b)

The product is $\boxed{46,500,632}$.

Question 31:

a) Complete the table:

Price of meat $= 3000 \text{ Frw/kg}$.
Total expenses for meat $= 5 \text{ kg} \times 3000 \text{ Frw/kg} = 15000 \text{ Frw}$.
So, (i) is $\boxed{15000 \text{ Frw}}$.

Total expenses for rice $= 6600 \text{ Frw}$.
Price of rice per kg $= \frac{6600 \text{ Frw}}{5 \text{ kg}} = 1320 \text{ Frw/kg}$.
So, (ii) is $\boxed{1320 \text{ Frw/kg}}$.

Price of honey $= 4050 \text{ Frw/litre}$.
Total expenses for honey $= 18225 \text{ Frw}$.
Quantity of honey $= \frac{18225 \text{ Frw}}{4050 \text{ Frw/litre}} = 4.5 \text{ litres}$.
So, (iii) is $\boxed{4.5 \text{ litres}}$.

Total expenses $= (\text{Meat expenses}) + (\text{Rice expenses}) + (\text{Honey expenses})$
Total expenses $= 15000 \text{ Frw} + 6600 \text{ Frw} + 18225 \text{ Frw} = 39825 \text{ Frw}$.
So, (iv) is $\boxed{39825 \text{ Frw}}$.

b) If he had 45000 Frw and saved the rest, how much did he save?

Amount saved $= \text{Money had} - \text{Total expenses}$
Amount saved $= 45000 \text{ Frw} - 39825 \text{ Frw} = 5175 \text{ Frw}$.
He saved $\boxed{5175 \text{ Frw}}$.

Question 32:

a) Find the sum of all prime numbers between 0 and 10.

Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves.
The prime numbers between 0 and 10 are $2, 3, 5, 7$.

The sum is $\boxed{17}$.

b) Use quick calculation: $235 \times 99$

The result is $\boxed{23265}$.

Question 33:

a) If 7 kg of rice is equivalent to 6,300 Frw. How much are 12 kg of the same rice?

Price per kg $= \frac{6300 \text{ Frw}}{7 \text{ kg}} = 900 \text{ Frw/kg}$.

Cost of 12 kg $= 12 \text{ kg} \times 900 \text{ Frw/kg} = 10800 \text{ Frw}$.
12 kg of rice costs $\boxed{10800 \text{ Frw}}$.

b) Find the sum of LCM and GCF of 12, 15 and 36.

The GCF is the product of the lowest powers of all common prime factors. The only common prime factor is 3, and its lowest power is $3^1$.

The LCM is the product of the highest powers of all prime factors present in any of the numbers.
Highest power of 2 is $2^2$.
Highest power of 3 is $3^2$.
Highest power of 5 is $5^1$.

The sum of the LCM and GCF is $\boxed{183}$.

Question 34:
Calculate the area of a right-angled triangle whose base is 3 cm and height of 7 cm.

The area of the triangle is $\boxed{10.5 \text{ cm}^2}$.