A) Find the L.C.M of 6 and 9. b) Find the smallest number when divided by either 7 or 8 the remainder is 4. 28. In a class of 80 pupils, 2/5 are girls and the rest are boys. a) What is the fraction of

$$6 = 2 \times 3$$ $$9 = 3 \times 3 = 3^2$$

$$\text{L.C.M}(6, 9) = 2^1 \times 3^2 = 2 \times 9 = 18$$

The L.C.M of 6 and 9 is $\boxed{18}$.

b) Find the smallest number when divided by either 7 or 8 the remainder is 4.

Since 7 and 8 are coprime (they share no common factors other than 1), their L.C.M is their product.

$$\text{L.C.M}(7, 8) = 7 \times 8 = 56$$

This means the number is 4 more than a multiple of the L.C.M of 7 and 8.
Smallest number = L.C.M$(7, 8) + 4$

$$= 56 + 4$$ $$= 60$$

The smallest number is $\boxed{60}$.

Question 28:
In a class of 80 pupils, $\frac{2}{5}$ are girls and the rest are boys.

a) What is the fraction of boys?

Subtract the fraction of girls from the total fraction to find the fraction of boys.

$$\text{Fraction of boys} = 1 - \text{Fraction of girls}$$ $$= 1 - \frac{2}{5}$$ $$= \frac{5}{5} - \frac{2}{5}$$ $$= \frac{3}{5}$$

The fraction of boys is $\boxed{\frac{3}{5}}$.

b) How many girls are in the class?

$$\text{Number of girls} = \frac{2}{5} \times 80$$ $$= 2 \times \frac{80}{5}$$ $$= 2 \times 16$$ $$= 32$$

There are $\boxed{32}$ girls in the class.

c) How many more boys than girls are in the class?

$$\text{Number of boys} = \text{Total pupils} - \text{Number of girls}$$ $$= 80 - 32$$ $$= 48$$

$$\text{Difference} = \text{Number of boys} - \text{Number of girls}$$ $$= 48 - 32$$ $$= 16$$

There are $\boxed{16}$ more boys than girls in the class.