Home / iGCSE Mathematics (0580) :E1.1 Identify and use natural numbers.iGCSE Style Questions Paper 4

iGCSE Mathematics (0580) :E1.1 Identify and use natural numbers.iGCSE Style Questions Paper 4

Question

(a) (i) Write 70 as a product of its prime factors.

(ii) Find the highest common factor (HCF) of 70 and 112.

(iii) Find the lowest common multiple (LCM) of $70x^4y^2$ and $112x^3y^5$.

(b) Simplify.

(i) \(a^{12}\) ÷ \(a^{4}\)

(ii) $\frac{5}{2b} \times \frac{bc}{20}$

(c) Solve $4 + 2x = 15$.

(d) Solve $\frac{34 + 2x}{5} = 4 – x$.

(e) $P = d + \sqrt[3]{m^2}$

(i) Find $P$ when $d = 7$ and $m = -8$.

(ii) Rearrange the formula to make $m$ the subject.

▶️ Answer/Explanation
Solution

(a)(i) $2 \times 5 \times 7$

Break down 70 into prime numbers: 70 ÷ 2 = 35, 35 ÷ 5 = 7, and 7 is prime.

(a)(ii) 14

Prime factors of 112 are $2^4 \times 7$. HCF is the product of common prime factors with lowest powers: $2 \times 7$.

(a)(iii) $560x^4y^5$

Take highest powers of all prime factors and variables from both numbers: $2^4 \times 5 \times 7 \times x^4 \times y^5$.

(b)(i) $a^8$

Subtract exponents when dividing: $a^{12-4} = a^8$.

(b)(ii) $\frac{c}{8}$

Multiply numerators and denominators: $\frac{5bc}{40b} = \frac{c}{8}$ after simplifying.

(c) $x = 5.5$

Subtract 4 from both sides: $2x = 11$. Then divide by 2: $x = 5.5$.

(d) $x = -2$

Multiply both sides by 5: $34 + 2x = 20 – 5x$. Collect like terms: $7x = -14$, so $x = -2$.

(e)(i) 11

Substitute values: $P = 7 + \sqrt[3]{(-8)^2} = 7 + \sqrt[3]{64} = 7 + 4 = 11$.

(e)(ii) $m = \pm \sqrt{(P – d)^3}$

Subtract d: $P – d = \sqrt[3]{m^2}$. Cube both sides: $(P – d)^3 = m^2$. Take square root.

Question

(a) A shop sells dress fabric for \($\)2.97 per metre.
(i) A customer buys 9 metres of this fabric. Calculate the change he receives from \($\)50.
(ii) The selling price of \($\)2.97 per metre is an increase of 8% on the cost price. Calculate the cost price.

(b) A dressmaker charges \($35\) or 2300 rupees to make a dress.
Calculate the difference in price when the exchange rate is 1 rupee=\($\)0.0153 .

Give your answer in rupees.

(c) The dressmaker measures a length of fabric as 600m, correct to the nearest 5 metres.
He cuts this into dress lengths of 9m, correct to the nearest metre.

Calculate the largest number of complete dress lengths he could cut.

▶️ Answer/Explanation
Solution

(a)(i) Ans: 23.27

Total cost = 9 × 2.97 = 26.73. Change = 50 – 26.73 = 23.27.

(a)(ii) Ans: 2.75

Let cost price be \(x\). Then 1.08x = 2.97 ⇒ x = 2.97/1.08 = 2.75.

(b) Ans: 12.40

Convert $35 to rupees: 35/0.0153 ≈ 2287.58. Difference = 2300 – 2287.58 ≈ 12.42.

(c) Ans: 70

Max fabric length = 602.5m (upper bound). Min dress length = 8.5m (lower bound). Maximum dresses = 602.5 ÷ 8.5 ≈ 70.88 ⇒ 70 complete dresses.

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