Problems from Oct 11, 2026

All 12 problems from Oct 11, 2026, across Physics, Math, Chemistry, Electrical engineering. Try each one first, then check the answer and the main steps below.

Physics

Problem 1: Around the track

A runner covers $400\ \text{m}$ in $50\ \text{s}$ at a steady pace.

What is the average speed, in metres per second?

Answer: 8 m/s

How it is solved

  1. Use speed = distance ÷ time: $400 \div 50$. ($8\ \text{m/s}$)

Problem 2: A wooden block

A $10$ cm tall block of wood (density $600$ kg/m³) floats upright in water (density $1000$ kg/m³).

How many centimetres of the block are below the surface?

Answer: 6 cm

How it is solved

  1. A floating block sinks until it pushes aside its own weight
  2. Find the fraction: $0.6$
  3. Apply it to the height: $6\ \text{cm}$

Problem 3: Small gear, big gear

A gear with $12$ teeth turns $60$ times each minute and drives a larger gear with $36$ teeth.

How many turns per minute does the large gear make?

Answer: 20

How it is solved

  1. Teeth passing a point: $720$ teeth
  2. The big gear must move the same number of teeth: $720 \div 36 = 20$ turns

Math

Problem 1: Muffins at the bakery

A bakery starts the day with $200$ muffins. By closing time, $35\%$ of them have been sold.

How many muffins are left?

Answer: 130

How it is solved

  1. Find how many were sold: $35\%$ of $200$ is $200 \times 35 \div 100$. ($70$ sold)
  2. Take that away from the start: $200 - 70$. ($130$)

Problem 2: Watering down juice

$200\ \text{mL}$ of orange syrup is $40\%$ pure. Water is added until it is only $25\%$ pure.

How many millilitres of water must be added?

Answer: 120 mL

How it is solved

  1. How much pure there is: $80\ \text{mL}$ pure
  2. Find the new total: $320\ \text{mL}$
  3. What was added: $120\ \text{mL}$

Problem 3: The corners of a square

A circle of radius $10$ cm fits exactly inside a square, touching all four sides. The square has side $20$ cm.

What is the area of the square that is not covered by the circle? Use a calculator value for $\pi$ and give the answer to the nearest whole number.

Answer: 86

How it is solved

  1. Area of the square: $400$
  2. Area of the circle: $314.16$
  3. Subtract: $85.84$, about $86$

Chemistry

Problem 1: Atoms in five units

One unit of a compound has the formula $\ce{Al2(SO4)3}$.

How many atoms are there in $5$ units?

Answer: 85

How it is solved

  1. Count the atoms in one unit: A number after a symbol multiplies that atom. A number after a bracket multiplies everything inside it. One unit has 2 Al + 3 S + 12 O. ($17$ atoms)
  2. Multiply by the number of units: $17 \times 5$. ($85$)

Problem 2: Gas volumes in a reaction

The reaction is $\ce{2H2 + O2 -> 2H2O}$, with every substance a gas at the same temperature and pressure.

How many litres of oxygen are involved when $10\ \text{L}$ of hydrogen reacts completely?

Answer: 5 L

How it is solved

  1. Gases at the same conditions react in the ratio of the coefficients
  2. Scale: $5\ \text{L}$

Problem 3: Carbon dioxide from carbon

The reaction is $\ce{C + O2 -> CO2}$. Carbon burns completely. Assume one mole of any gas takes up $22.4\ \text{L}$. Atomic masses: C $=12$.

What volume of $\ce{CO2}$ is made from $6\ \text{g}$ of $\ce{C}$?

Answer: 11.2 L

How it is solved

  1. Moles of the starting material: $0.5\ \text{mol}$
  2. Moles of gas made: $0.5\ \text{mol}$
  3. Volume: $11.2\ \text{L}$

Electrical engineering

Problem 1: Voltage across a coil

A heating coil has a resistance of $24\ \Omega$ and a current of $0.5\ \text{A}$ flows through it.

What voltage is across it, in volts?

Answer: 12 V

How it is solved

  1. Use V = I × R: $0.5 \times 24$. ($12\ \text{V}$)

Problem 2: A battery pack

A battery pack is made of $4$ cells of $1.5$ V and $2400$ mAh joined in series.

How much energy does the pack hold, in watt-hours? (Energy = voltage × amp-hours.)

Answer: 14.4

How it is solved

  1. Voltage and capacity of the pack: $6\ \text{V}$, $2.4\ \text{Ah}$
  2. Energy: $14.4\ \text{Wh}$

Problem 3: Find the hidden resistor

A $12$ V battery pushes a current of $100$ mA through two resistors in series. One is $40$ ohms.

What is the resistance of the second resistor, in ohms?

Answer: 80 ohm

How it is solved

  1. Total resistance: $120\ \Omega$
  2. Take away the known one: $80\ \Omega$