In a class of 59 pupils, there are 7 more boys than girls. How many boys are in that class? Peter spent 1/4 of his money on biscuits and 1/2 of the rest on bananas. What fraction is left?

Let $B$ be the number of boys and $G$ be the number of girls.
The total number of pupils is 59:

There are 7 more boys than girls:

There are 33 boys in the class.

The answer is:

• $\boxed{33 \text{ boys}}$

Fraction spent on biscuits = $\frac{1}{4}$
Fraction remaining = $1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4}$

Peter spent $\frac{1}{2}$ of the rest on bananas.
Fraction spent on bananas = $\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$

Total fraction spent = Fraction on biscuits + Fraction on bananas
Total fraction spent = $\frac{1}{4} + \frac{3}{8}$
To add these, find a common denominator, which is 8:

Total fraction spent = $\frac{2}{8} + \frac{3}{8} = \frac{5}{8}$

Fraction left = $1 - \text{Total fraction spent}$
Fraction left = $1 - \frac{5}{8} = \frac{8}{8} - \frac{5}{8} = \frac{3}{8}$

The answer is:

• $\boxed{\frac{3}{8}}$

This is an inverse proportion problem: as the number of students increases, the time taken to weed the farm decreases.
Let $S_1 = 21$ students and $T_1 = 4$ hours.
Let $S_2$ be the unknown number of students and $T_2 = 3\frac{1}{2}$ hours $= 3.5$ hours.

24 students can weed the farm in $3\frac{1}{2}$ hours.

The answer is:

• $\boxed{24 \text{ students}}$

Total eggs = 80
Bad eggs = 20
Good eggs = Total eggs - Bad eggs = $80 - 20 = 60$

Fraction of good eggs = $\frac{\text{Number of good eggs}}{\text{Total eggs}} = \frac{60}{80}$

Percentage of good eggs = $\frac{60}{80} \times 100\%$
Percentage of good eggs = $\frac{6}{8} \times 100\%$
Percentage of good eggs = $\frac{3}{4} \times 100\%$
Percentage of good eggs = $0.75 \times 100\%$
Percentage of good eggs = $75\%$

The answer is:

• $\boxed{75\%}$

The distance round a square plot is its perimeter.
Perimeter of a square = $4 \times \text{side}$
Given Perimeter = 36 m

Area of a square = $\text{side} \times \text{side}$
Area = $9 \text{ m} \times 9 \text{ m}$
Area = $81 \text{ m}^2$

The answer is:

• $\boxed{81 \text{ m}^2}$