Home / iGCSE Mathematics (0580) :E1.6 Order quantities by magnitude and demonstrate familiarity with the symbols.iGCSE Style Questions Paper 4

iGCSE Mathematics (0580) :E1.6 Order quantities by magnitude and demonstrate familiarity with the symbols.iGCSE Style Questions Paper 4

Question

(a) Factorise completely.
\(3a^{2}b – ab^{2}\)

(b) Solve the inequality.
\(3x + 12 < 5x – 3\)

(c) Simplify.
\((3x^{2}y^{4})^{3}\)

(d) Solve.
\(\frac{2}{x} = \frac{6}{2 – x}\)

(e) Expand and simplify.
\((x – 2)(x + 5)(2x – 1)\)

(f) Alan invests \($200\) at a rate of \(r\%\) per year compound interest.
After 2 years the value of his investment is \($206.46\).
(i) Show that \(r^{2} + 200r – 323 = 0\).
(ii) Solve the equation \(r^{2} + 200r – 323 = 0\) to find the rate of interest.
Show all your working and give your answer correct to 2 decimal places.

▶️ Answer/Explanation
Solution

(a) Ans: \(ab(3a – b)\)

Factor out the common term \(ab\) from \(3a^{2}b – ab^{2}\).

(b) Ans: \(x > 7.5\)

Rearrange inequality: \(3x + 12 < 5x – 3 \Rightarrow 15 < 2x \Rightarrow x > 7.5\).

(c) Ans: \(27x^{6}y^{12}\)

Apply the power rule \((a^m)^n = a^{mn}\) to each term inside the parentheses.

(d) Ans: \(x = 0.5\) or \(\frac{1}{2}\)

Cross-multiply: \(2(2 – x) = 6x \Rightarrow 4 – 2x = 6x \Rightarrow x = 0.5\).

(e) Ans: \(2x^{3} + 5x^{2} – 23x + 10\)

First expand \((x – 2)(x + 5) = x^2 + 3x – 10\), then multiply by \((2x – 1)\).

(f)(i)

Using compound interest formula: \(200(1 + \frac{r}{100})^2 = 206.46\).
Expand and simplify to get \(r^2 + 200r – 323 = 0\).

(f)(ii) Ans: \(r = 1.60\%\)

Solve quadratic equation using the quadratic formula:
\(r = \frac{-200 \pm \sqrt{200^2 – 4(1)(-323)}}{2} \approx 1.60\).

Question

(a) Make \( t \) the subject of the formula \( s = k – t^{2} \).

(b) (i) Factorise \( x^{2} – 25 \).

(ii) Simplify \( \frac{x^{2} – 25}{x^{2} – 2x – 35} \).

(c) Write as a single fraction in its simplest form:

\( \frac{x – 8}{x} + \frac{3x}{x + 1} \).

(d) Find the integer values of \( n \) that satisfy the inequality:

\( 18 – 2n < 6n \leq 30 + n \).

▶️ Answer/Explanation
Solution

(a) Ans: \( t = \pm \sqrt{k – s} \)

Rearrange \( s = k – t^{2} \) to isolate \( t \). First, subtract \( s \) from both sides: \( t^{2} = k – s \). Then take the square root of both sides.

(b)(i) Ans: \( (x – 5)(x + 5) \)

This is a difference of squares, which factors into \( (x – 5)(x + 5) \).

(b)(ii) Ans: \( \frac{x – 5}{x – 7} \)

Factor numerator and denominator: \( \frac{(x-5)(x+5)}{(x-7)(x+5)} \). Cancel \( (x + 5) \) from both.

(c) Ans: \( \frac{4x^{2} – 7x – 8}{x(x + 1)} \)

Find a common denominator \( x(x + 1) \). Combine fractions and simplify the numerator.

(d) Ans: 3, 4, 5, 6

Split the inequality into two parts: \( 18 – 2n < 6n \) gives \( n > 2.25 \), and \( 6n \leq 30 + n \) gives \( n \leq 6 \). The integer solutions are \( n = 3, 4, 5, 6 \).

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