IGCSE Mathematics | Worked Solutions | 25 Questions
Section A — Recall
Questions 1–10
1.
Write $5^3$ as a product of numbers.
Answer
The index 3 means multiply the base 5 by itself 3 times:
$$5^3 = 5 \times 5 \times 5$$$5^3 = 5 \times 5 \times 5$
Note: The number being multiplied (5) is the base. The small raised number (3) is the index (or power).
2.
State the zero index rule. Give one example to support your answer.
Answer
Any number (except zero) raised to the power 0 equals 1.
$$a^0 = 1$$Example: $\quad 8^0 = 1$
$a^0 = 1$ (for any $a \neq 0$)
Exam tip: This rule applies to any base — numbers, letters, or fractions. E.g. $(100)^0 = 1$, $(x)^0 = 1$.
3.
Complete the rule for a negative index: $a^{-n} = \ldots$
Answer
A negative index means one divided by that power of the base.
$$a^{-n} = \frac{1}{a^n}$$
Remember: A negative index does NOT make the answer negative — it means the reciprocal (1 over). E.g. $2^{-3} = \dfrac{1}{8}$, not $-8$.
4.
Find the value of $4^3$.
Answer
$$4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$$
$4^3 = 64$
5.
Find the value of $6^0$.
Answer
Apply the zero index rule: any number to the power 0 equals 1.
$6^0 = 1$
6.
Find the value of $3^{-1}$.
Answer
- Apply the negative index rule: $\quad 3^{-1} = \dfrac{1}{3^1}$
- Simplify: $\quad \dfrac{1}{3^1} = \dfrac{1}{3}$
$3^{-1} = \dfrac{1}{3}$
7.
Find the value of $7^{-2}$.
Answer
- Apply the negative index rule: $\quad 7^{-2} = \dfrac{1}{7^2}$
- Calculate the denominator: $\quad 7^2 = 7 \times 7 = 49$
- Write the answer: $\quad \dfrac{1}{49}$
$7^{-2} = \dfrac{1}{49}$
8.
Write down the rule for multiplying two powers with the same base. Use the formula $a^m \times a^n = \ldots$
Answer
When multiplying powers with the same base, add the indices.
$a^m \times a^n = a^{m+n}$
9.
Write down the rule for dividing two powers with the same base. Use the formula $a^m \div a^n = \ldots$
Answer
When dividing powers with the same base, subtract the indices.
$a^m \div a^n = a^{m-n}$
10.
Write down the power of a power rule. Use the formula $(a^m)^n = \ldots$
Answer
When raising a power to another power, multiply the indices.
$(a^m)^n = a^{m \times n}$
Section B — Application
Questions 11–20
11.
Simplify $a^3 \times a^5$.
Answer
Same base ($a$), so add the indices:
$$a^3 \times a^5 = a^{3+5} = a^8$$$a^8$
12.
Simplify $x^7 \div x^3$.
Answer
Same base ($x$), so subtract the indices:
$$x^7 \div x^3 = x^{7-3} = x^4$$$x^4$
13.
Simplify $(b^2)^4$.
Answer
Power of a power — multiply the indices:
$$(b^2)^4 = b^{2 \times 4} = b^8$$$b^8$
14.
Find the value of $2^{-3} \times 2^4$.
Answer
- Same base (2), so add the indices: $$2^{-3} \times 2^4 = 2^{-3+4} = 2^1$$
- Evaluate: $\quad 2^1 = 2$
$2^{-3} \times 2^4 = 2$
15.
Find the value of $2^3 \div 2^4$.
Answer
- Same base (2), so subtract the indices: $$2^3 \div 2^4 = 2^{3-4} = 2^{-1}$$
- Apply the negative index rule: $$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$$
$2^3 \div 2^4 = \dfrac{1}{2}$
16.
Find the value of $(2^3)^2$.
Answer
- Power of a power — multiply the indices: $$(2^3)^2 = 2^{3 \times 2} = 2^6$$
- Evaluate: $\quad 2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$
$(2^3)^2 = 64$
17.
Simplify $5^3 \times 5^{-2}$ and find its value.
Answer
- Same base (5), so add the indices: $$5^3 \times 5^{-2} = 5^{3+(-2)} = 5^1$$
- Evaluate: $\quad 5^1 = 5$
$5^3 \times 5^{-2} = 5$
18.
Find the value of $10^{-2}$. Give your answer as a decimal.
Answer
- Apply the negative index rule: $$10^{-2} = \frac{1}{10^2}$$
- Calculate the denominator: $\quad 10^2 = 100$
- Convert to decimal: $\quad \dfrac{1}{100} = 0.01$
$10^{-2} = 0.01$
19.
Simplify $\dfrac{m^6}{m^2}$.
Answer
Division of same base — subtract the indices:
$$\frac{m^6}{m^2} = m^{6-2} = m^4$$$m^4$
20.
Find the value of $4^{-2}$.
Answer
- Apply the negative index rule: $$4^{-2} = \frac{1}{4^2}$$
- Calculate the denominator: $\quad 4^2 = 16$
- Write the answer: $\quad \dfrac{1}{16}$
$4^{-2} = \dfrac{1}{16}$
Section C — Challenge
Questions 21–25
21.
Simplify $3^2 \times 3^{-5} \times 3^4$. Write your answer as a single power of 3, then find its value.
Answer
- All bases are the same (3), so add all three indices: $$3^2 \times 3^{-5} \times 3^4 = 3^{2 + (-5) + 4}$$
- Calculate the combined index: $\quad 2 – 5 + 4 = 1$ $$= 3^1$$
- Evaluate: $\quad 3^1 = 3$
$3^2 \times 3^{-5} \times 3^4 = 3^1 = 3$
Tip: When multiplying more than two powers with the same base, add all the indices together in one step.
22.
(a) Write $\dfrac{1}{32}$ as a power of 2.
(b) Write $\dfrac{1}{27}$ as a power of 3.
Answer
(a)
- Recognise that $32 = 2^5$ (since $2 \times 2 \times 2 \times 2 \times 2 = 32$)
- So $\dfrac{1}{32} = \dfrac{1}{2^5}$
- Apply the negative index rule: $\quad \dfrac{1}{2^5} = 2^{-5}$
$\dfrac{1}{32} = 2^{-5}$
(b)
- Recognise that $27 = 3^3$ (since $3 \times 3 \times 3 = 27$)
- So $\dfrac{1}{27} = \dfrac{1}{3^3}$
- Apply the negative index rule: $\quad \dfrac{1}{3^3} = 3^{-3}$
$\dfrac{1}{27} = 3^{-3}$
Key step: First write the denominator as a power, then use $\dfrac{1}{a^n} = a^{-n}$.
23.
Find the value of $n$ such that $2^n = \dfrac{1}{8}$. Show your working.
Answer
- Write the right-hand side as a power of 2. Recognise that $8 = 2^3$, so: $$\frac{1}{8} = \frac{1}{2^3}$$
- Apply the negative index rule: $$\frac{1}{2^3} = 2^{-3}$$
- So the equation becomes: $$2^n = 2^{-3}$$
- The bases are equal, so the indices must be equal: $\quad n = -3$
$n = -3$
Tip: When you see $\dfrac{1}{\text{power}}$, always think of the negative index rule.
24.
(a) Simplify $(x^3)^2 \div x^4$.
(b) Find the value of your answer to part (a) when $x = 2$.
Answer
(a)
- Apply the power of a power rule to $(x^3)^2$: $$(x^3)^2 = x^{3 \times 2} = x^6$$
- Now divide — same base ($x$), subtract the indices: $$x^6 \div x^4 = x^{6-4} = x^2$$
$(x^3)^2 \div x^4 = x^2$
(b)
Substitute $x = 2$ into the simplified expression $x^2$:
$$x^2 = 2^2 = 4$$When $x = 2$: $x^2 = 4$
Strategy: Always simplify the algebra first (part a), then substitute the number (part b). This is easier than substituting before simplifying.
25.
Show that $(2^{-3} \times 2^5)^2 = 16$. Show every step of your working.
Answer
- Work inside the brackets first — same base (2), so add the indices: $$2^{-3} \times 2^5 = 2^{-3+5} = 2^2$$
- Now apply the power of a power rule to $(2^2)^2$ — multiply the indices: $$(2^2)^2 = 2^{2 \times 2} = 2^4$$
- Evaluate $2^4$: $$2^4 = 2 \times 2 \times 2 \times 2 = 16 \checkmark$$
$(2^{-3} \times 2^5)^2 = (2^2)^2 = 2^4 = 16$ ✓
Tip: This question uses all three index rules in sequence: multiply (add indices) → power of a power (multiply indices) → evaluate. Work step by step and you will not make mistakes.
