A drawer contains \(10\) socks and \(5\) of them are red. When three socks are drawn at random, what is the probability that all three of them are red?
a) \(\frac{3}{4}\)
b) \(\frac{2}{13}\)
c) \(\frac{5}{37}\)
d) \(\frac{1}{2}\)
e) \(\frac{1}{12}\)
Answer: e)
2. System of Equations
If \(x\) and \(y\) are positive numbers that satisfy
\[\begin{cases}x^{2}-y^{2}=24\\ x^{2}+y^{2}=26\end{cases}\]
then \(x+y\) is
a) 3
b) 4
c) 5
d) 6
e) none of these
Answer: d)
3. Tangent Circles
The larger circle of radius 9 centered at \(C\) is tangent to the smaller circle of radius 4 centered at \(D\). If the circles are tangent to the line \(l\) at \(A\) and \(B\), then what is the area of the triangle \(\Delta ABC\)?
Diagram for Question 3
a) 36
b) 47
c) 54
d) 67
e) none of these
Answer: c)
4. Trigonometry
If \(\theta\) is a solution of the equation
\[\cos(50^{\circ})+\cos(70^{\circ})=\cos\theta^{\circ}\]
then \(\cos(9\theta^{\circ})\) is
a) 0
b) \(\frac{1}{2}\)
c) \(\frac{1+\sqrt{3}}{2}\)
d) \(\frac{\sqrt{3}}{2}\)
e) 1
Answer: a)
5. Greedy Algorithm
There are four boxes full of \$10, \$20, \$50, \$100 bills. You can choose your boxes and take 2, 3, 4, and 5 bills from the chosen boxes respectively. What is the largest amount of money you can get?
a) \$480
b) \$560
c) \$780
d) \$850
e) \$920
Answer: c)
6. Logarithms
If \(x\) and \(y\) are positive numbers satisfying
\[\log_{3}(x^{2}y^{2})=2\log_{3}y+6\]
then \(x\) is
a) 1
b) 3
c) 9
d) 27
e) cannot be determined
Answer: d)
7. Triangle Geometry
Suppose that \(\triangle ABC\) is a triangle, that \(M\) is the midpoint of \(\overline{AC}\), and the segments \(\overline{AM}\), \(\overline{MC}\), \(\overline{MB}\) and \(\overline{AB}\) all have length 1. Find the area of \(\triangle ABC\).
Diagram for Question 7
a) \(\frac{1}{2}\)
b) \(\frac{\sqrt{2}}{2}\)
c) \(\frac{3}{2}\)
d) \(\frac{\sqrt{3}}{2}\)
e) none of these
Answer: d)
8. Trigonometric Equation
If \(\sin x=2\cos x\), then what is \((\sin x)(\cos x)\)?
a) \(\frac{1}{2}\)
b) \(\frac{2}{3}\)
c) \(\frac{2}{5}\)
d) \(\frac{6}{7}\)
e) \(\frac{3}{11}\)
Answer: c)
9. Polynomial Roots
If the sum of two of the roots of \(x^{3}-ax^{2}+bx-c=0\) is zero, then \(ab-c\) is
a) \(-2\)
b) \(-1\)
c) \(0\)
d) \(1\)
e) \(2\)
Answer: c)
10. Magic Square
Suppose that the 25 squares below are filled with the integers 1 through 25 such a way that the sums of integers in each row and column are the same. What is the common sum?
a) 65
b) 72
c) 83
d) 96
e) none of these
Answer: a)
11. Remainder Theorem
The remainder when \(6^{30}+8^{30}\) is divisible by 49 is
a) 0
b) 2
c) 11
d) 12
e) 13
Answer: b)
12. Polynomial Degree
If a polynomial \(p(x)\) of degree \(\geq 1\) satisfies
\[p(p(x))=5p(x^{3})\]
then the degree of \(p(x)\) must be
a) 2
b) 3
c) 5
d) 7
e) none of these
Answer: b)
13. Sum of Powers Remainder
What is the remainder when \(2^{0} + 2^{1} + 2^{2} + \cdots + 2^{99}\) is divided by 9?
a) 0
b) 1
c) 3
d) 6
e) none of these
Answer: d)
14. Integer Solutions
What is the number of positive integer solutions \((x, y)\) of the equation
\[x^{2} + y^{2} = 3x + 3y + 4 - 2xy?\]
a) 0
b) 1
c) 2
d) 3
e) none of these
Answer: d)
15. Isosceles Triangle Problem
Consider the triangles in the following diagram: Suppose that \(\triangle ABC\) is isosceles with \(AB = AC\) and \(\triangle ABD\) is isosceles with \(AD = BD\). Suppose that \(AD = 9\) and \(BC = 4\). Then \(AB\) is
Diagram for Question 15
a) 6
b) \(6\sqrt{2}\)
c) 8
d) \(8\sqrt{2}\)
e) 9
Answer: a)
16. Area of Shaded Region
Find the area of the shaded region formed by a large quarter circle of radius 10 and two smaller semicircles as shown below.
Diagram for Question 16
a) \(25\pi\)
b) \(50\pi\)
c) \(75\pi\)
d) \(100\pi\)
e) \(100\pi - 200\)
Answer: e)
17. Staircase Problem
In how many ways can you walk to a stairway with \(7\) steps if you can take one or two steps. (For example, you can walk up a stairway with \(3\) steps in three different ways: i) three \(1\) steps ii) \(1\) step and then \(2\) steps and iii) \(2\) steps and then \(1\) step.)
a) 15
b) 18
c) 21
d) 27
e) 29
Answer: c)
18. Reflection Geometry
Two points \(A\) and \(B\) are vertically \(3\) and \(5\) ft away from the line \(l\), respectively, and they are horizontally \(10\) ft apart. When a point \(P\) moves along the line \(l\), what is the smallest value of \(AP+PB\).
Diagram for Question 18
a) \(\sqrt{97}\)
b) \(2\sqrt{41}\)
c) \(3\sqrt{43}\)
d) \(4\sqrt{45}\)
e) \(\sqrt{185}\)
Answer: b)
19. Equation with Reciprocal
The sum of the solutions to
\[2\left(x+\frac{1}{x}\right)^{2}+27=21\left(x+\frac{1}{x}\right)\]
is
a) \(\frac{2}{3}\)
b) \(-\frac{5}{7}\)
c) \(\frac{7}{6}\)
d) \(\frac{21}{2}\)
e) \(\frac{29}{5}\)
Answer: d)
20. Dog Running Problem
A student walks home from her school at a speed of \(\frac{3}{2}\) miles per hour. She was greeted by her dog that runs 8 miles per hour when she was 2 miles away from home. The dog goes back home and then comes back to her and it keeps doing it until she arrives home. What is the total distance the dog runs?
a) 10
b) \(\frac{29}{3}\)
c) \(\frac{32}{3}\)
d) \(\frac{33}{2}\)
e) \(\frac{52}{5}\)
Answer: c)
Why alpha particles are more ionising but less penetrating
A clear, exam-friendly note for IGCSE (includes the mark-scheme explanation and the deeper physics).
Beta particle = electron. Charge: −1 (or +1 for positron). Mass: very small.
Gamma ray = electromagnetic radiation (no charge, no rest mass).
Why alpha particles are more ionising
Alpha particles cause more ionisation because they are heavy and carry a +2 charge. These properties make them interact strongly with atoms, knocking out many electrons per unit distance travelled. In many radioactive decays alphas also have larger kinetic energy than betas, which further increases energy transfer per collision.
Why alpha particles are less penetrating
Because alphas lose energy rapidly through intense ionisation, they cannot travel far. A sheet of paper or the outer dead layer of skin will stop them. Betas are lighter and interact less strongly, so they penetrate further than alphas but less than gammas.
Quick kinetic energy check (short)
Kinetic energy: E_k = ½ m v².
Alphas have a very large mass; even if their speed is ≈ 1/20 of a beta, the product m v² can be larger for alphas. Thus alphas often have more KE than betas in real decays — this supports Cambridge’s simplified phrasing that alphas are more ionising because they have more kinetic energy.
Exam tip (IGCSE)
If the mark scheme gives the statement linking ionising power to kinetic energy, choose the option that matches the mark scheme. In your own answer you can add that mass and charge are the fundamental reasons for high ionisation by alpha particles.
Summary table
Radiation
Mass / charge
Ionising power
Penetration
Alpha
Large mass, +2
High
Low (stopped by paper/skin)
Beta
Very small mass, −1/+1
Lower than alpha
Medium (stopped by thin metal)
Gamma
No mass, no charge
Lowest
Highest (requires thick lead/concrete)
How Smoke/Fire alarms work
In ionization smoke alarms, alpha particles from a small amount of americium-241 are used.
Here’s how it works:
The americium-241 gives off alpha radiation, which ionizes the air molecules in a small chamber inside the alarm.
This creates a flow of ions (positive and negative charges), allowing a small electric current to pass between two electrodes.
When smoke enters the chamber, it attaches to the ions and reduces their mobility, which disrupts the current.
The alarm senses this drop in current and triggers the warning sound.
In short:
In smoke alarms, alpha particles are used to ionize the air, allowing current to flow in the circuit.
Derived by taking natural logs of the exponential decay law.
Worked example
Suppose N0=200 counts, N=50 counts after t = 9 hours.
\(\dfrac{N_0}{N}=4\) so \(\ln(4)=1.386...\). Then \(T_{1/2}=\dfrac{9\times0.6931}{1.386}=\dfrac{6.238}{1.386}=4.5\) hours.
5. Decay constant and exponential form
\(\lambda = \dfrac{\ln 2}{T_{1/2}},\quad N = N_0 e^{-\lambda t}\)
Use this when the question uses the exponential constant \(\lambda\) instead of half-life.
Quick step-by-step pattern for solving problems
Write down what you know: N, N0, t, or T1/2.
Choose the formula above that contains the unknown.
If needed, take natural logs to solve for the unknown; show units.
Check special cases: if t = T1/2, remaining fraction = 1/2; if t = 0, remaining = N0.
Worked example (full)
Problem: A 160 g sample decays to 10 g in 30 days. Find the half-life.
Use formula: T_{1/2} = t*ln2 / ln(N0/N)
Substitute: t = 30, N0 = 160, N = 10
N0/N = 16, ln(16) = 2.7726
T_{1/2} = 30 * 0.6931 / 2.7726 = 20.79 / 2.7726 = 7.5 days
Answer: T_{1/2} = 7.5 days
In smoke alarms, alpha particles are used to ionize the air, allowing current to flow in the circuit.