Home / iGCSE Mathematics (0580) – C1.7 Indices I- Exam Style Questions Paper 1

iGCSE Mathematics (0580) - C1.7 Indices I- Exam Style Questions Paper 1- New Syllabus

IGCSE Mathematics(0580) Practice Questions Paper 1 - All Chapters

Question

Find the value of

(a) \(2^{5}\)

(b) \(6^{0}\)

Most-appropriate topic codes (Cambridge IGCSE Mathematics 0580):

• Topic C1.7 Indices I: Understand and use indices (positive, zero and negative integers) (Core)
▶️ Answer/Explanation

(a)
Calculate \(2\) raised to the power of \(5\):
\(2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32\).
✅ Answer: \(32\)

(b)
Any non-zero number raised to the power of \(0\) is \(1\).
\(6^{0} = 1\).
✅ Answer: \(1\)

Question

(a) \(k^x \times k^5 = k^{20}\)
Find the value of \(x\).

(b) Expand and simplify. \(\quad (x+5)(x-4)\)

(c) Factorise. \(\quad 21r^3 – 7r\)

Most-appropriate topic codes (Cambridge IGCSE Mathematics 0580):

• Topic C1.7 Indices I: Understand and use the rules of indices (Core)
• Topic C2.2 Algebraic manipulation: Expand products of algebraic expressions; Factorise by extracting common factors (Core)
▶️ Answer/Explanation

(a)
Using the laws of indices for multiplication, add the powers: \(x + 5 = 20\).
Solving for \(x\), we get \(x = 20 – 5 = 15\).
✅ Answer: \(15\)

(b)
Multiply out the brackets using FOIL (First, Outer, Inner, Last):
\((x \times x) + (x \times -4) + (5 \times x) + (5 \times -4) = x^2 – 4x + 5x – 20\).
Simplify by combining like terms: \(x^2 + x – 20\).
✅ Answer: \(x^2 + x – 20\)

(c)
Find the highest common factor for the numbers (\(7\)) and the letters (\(r\)).
Pull out \(7r\) from both terms: \(7r(3r^2 – 1)\).
✅ Answer: \(7r(3r^2 – 1)\)

Question

(a) Write \( \frac{1}{2 \times 2 \times 2 \times 2 \times 2} \) as a power of 2.

(b) (i) \( 3^{18} \div 3^t = 3^6 \) Find the value of t.

(ii) Simplify \( 8w^{10} \times 6w^5 \).

▶️ Answer/Explanation
Solution

Ans:

(a) \( 2^{-5} \) (The denominator is \( 2^5 \), and \( \frac{1}{2^5} = 2^{-5} \))

(b)(i) t = 12 (Using laws of indices: 18 – t = 6 → t = 12)

(b)(ii) \( 48w^{15} \) (Multiply coefficients: 8×6=48; add exponents: 10+5=15)

Question

(a) Find the value of \(137^{0}\).

(b) \(7^{12}\div 7^{P}=7^{17}\)

Find the value of P.

▶️ Answer/Explanation
Solution

(a) Ans: 1

Any non-zero number raised to the power of 0 is always 1. This is a fundamental exponent rule.

Thus, \(137^{0} = 1\).

(b) Ans: \(-5\)

Using the exponent rule \(\frac{a^m}{a^n} = a^{m-n}\), we rewrite the equation:

\(7^{12-P} = 7^{17}\).

Since the bases are equal, the exponents must be equal: \(12 – P = 17\).

Solving for \(P\) gives \(P = 12 – 17 = -5\).

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