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4t ESO · Maths challenges Worksheet 30 Sep 2026

3rd ESO revision

Seven exercises on percentages, volumes and areas of composite shapes to prepare for the test on 5 October. Each exercise has its solution folded away.

Download the worksheet (PDF, in Catalan) Download the solutions (PDF, in Catalan)

Do the following exercises to prepare for the test on 5 October.

Each part has its solution folded away: open the ▶ once you have tried it.

Exercise 1

In a school of $480$ students, $25\ \%$ play a team sport, $30\ \%$ play an individual sport and $15\ \%$ do both.

a) How many students play a team sport?
$$480 \cdot 0{,}25 = 120$$

$120$ students.

b) How many play an individual sport?
$$480 \cdot 0{,}30 = 144$$

$144$ students.

c) How many do both?
$$480 \cdot 0{,}15 = 72$$

$72$ students.

d) What percentage of the school does at least one of the two kinds of sport?

Those who do both must not be counted twice:

$$25\ \% + 30\ \% - 15\ \% = 40\ \%$$
e) How many students is that?
$$480 \cdot 0{,}40 = 192$$

$192$ students. Check: $120 + 144 - 72 = 192$.

Exercise 2

A library has $1\,200$ books. $35\ \%$ are novels. Of these novels, $40\ \%$ were originally written in Catalan.

a) How many novels are there?
$$1\,200 \cdot 0{,}35 = 420$$

$420$ novels.

b) How many were originally in Catalan?
$$420 \cdot 0{,}40 = 168$$

$168$ novels.

c) What percentage of all the books are these Catalan novels?
$$\dfrac{168}{1\,200} = 0{,}14$$

That is, $14\ \%$. It also comes out directly: $0{,}35 \cdot 0{,}40 = 0{,}14$.

Exercise 3

A tank is made of a cylinder with radius $4$ m and height $10$ m with, on top, a hemisphere of the same radius. Inside there is a solid cylindrical structure with radius $1{,}5$ m and height $8$ m that takes up space. Work out the real capacity of the tank in m³. Remember that $V_{\text{cylinder}} = \pi r^2 h$ and $V_{\text{sphere}} = \frac{4}{3}\pi r^3$.

r = 4 mr = 1,5 mh = 10 mh = 8 m
See the solution

Cylinder:

$$\pi \cdot 4^2 \cdot 10 = 160\pi$$

Hemisphere (half a sphere):

$$\dfrac{1}{2} \cdot \dfrac{4}{3}\pi \cdot 4^3 = \dfrac{128}{3}\pi$$

Inner structure, subtracted because it takes up space:

$$\pi \cdot 1{,}5^2 \cdot 8 = 18\pi$$

Capacity:

$$160\pi + \dfrac{128}{3}\pi - 18\pi = \dfrac{554}{3}\pi \approx 579{,}85 \text{ m}^3$$

With $\pi = 3{,}14$. Using the calculator's $\pi$ key gives $580{,}12\ \text{m}^3$.

Exercise 4

A metal part is a cylinder with outer radius $6$ cm and height $15$ cm. A cylindrical hole of radius $2$ cm is drilled right through its centre. Use $\pi = 3{,}14$.

r = 6 cmr = 2 cmh = 15 cm
a) Work out the volume of metal that is left.
$$V = \pi \cdot 6^2 \cdot 15 - \pi \cdot 2^2 \cdot 15 = \pi \cdot 15 \cdot (36 - 4) = 480\pi = 1\,507{,}2 \text{ cm}^3$$
b) Work out how many cm³ of metal were removed when drilling the hole.
$$V_{\text{hole}} = \pi \cdot 2^2 \cdot 15 = 60\pi = 188{,}4 \text{ cm}^3$$

Exercise 5

A track is shaped like a rectangle $30$ m long and $16$ m wide, with a semicircle on each short side. Inside there is an area of the same shape, with a $30 \times 10$ m rectangle and two semicircles of diameter $10$ m. Work out the area of the track, that is, the region between the outer and the inner shape.

30 m16 m10 m
See the solution

Outer shape: the $30 \times 16$ rectangle plus two semicircles of diameter $16$ (radius $8$), which together make a circle:

$$30 \cdot 16 + \pi \cdot 8^2 = 480 + 64\pi$$

Inner shape: the $30 \times 10$ rectangle plus a circle of radius $5$:

$$30 \cdot 10 + \pi \cdot 5^2 = 300 + 25\pi$$

Track:

$$(480 + 64\pi) - (300 + 25\pi) = 180 + 39\pi \approx 180 + 122{,}46 = 302{,}46 \text{ m}^2$$

With $\pi = 3{,}14$. Using the $\pi$ key gives $302{,}52\ \text{m}^2$.

Exercise 6

A square plate has sides of $20$ cm. In the centre there are two concentric circles: the large one has radius $9$ cm and the small one radius $5$ cm. The shaded region is the part of the square outside the large circle, together with the inside of the small circle. Work out the total shaded area.

9 cm5 cm20 cm
See the solution

Square:

$$20^2 = 400$$

Large circle and small circle:

$$\pi \cdot 9^2 = 81\pi \qquad \pi \cdot 5^2 = 25\pi$$

Shaded: the square minus the large circle, plus the small circle:

$$400 - 81\pi + 25\pi = 400 - 56\pi \approx 400 - 175{,}84 = 224{,}16 \text{ cm}^2$$

With $\pi = 3{,}14$.

Exercise 7

A square has sides of $12$ cm. From each of the four corners a quarter circle of radius $6$ cm is drawn inwards. Work out the shaded area, that is, the part of the square not covered by the four quarter circles. Use $\pi = 3{,}14$.

6 cm12 cm
See the solution

The four quarter circles of radius $6$ cm add up to a whole circle:

$$A_{\text{circles}} = \pi \cdot 6^2 = 3{,}14 \cdot 36 = 113{,}04 \text{ cm}^2$$

Area of the square and shaded region:

$$12^2 = 144 \qquad 144 - 113{,}04 = 30{,}96 \text{ cm}^2$$