GCSE Mathematics (9-1)

Chapter 7: Review

The chapter problem, resolved

The map uses scale 1:25 0001:25\,000, and the path measures 3.23.2 cm. Its real length is

3.2×25 000=80 000 cm=800 m=0.8 km.3.2\times25\,000=80\,000\text{ cm}=800\text{ m}=0.8\text{ km}.

The walker takes 1010 minutes, which is 1/61/6 hour. The average speed is

0.8÷(1/6)=4.8 km/h.0.8\div(1/6)=4.8\text{ km/h}.

A sample recorded as 0.00420.0042 kg to 2 significant figures lies from 0.004150.00415 kg up to, but not including, 0.004250.00425 kg. In standard form, the recorded value is 4.2×10−34.2\times10^{-3} kg. These steps show how conversion, compound measures, rounding and standard form connect.

Mixed practice: choose the method first

For each question, name the method before working.

  1. Convert 2.72.7 m to mm.
  2. A cyclist travels 3636 km in 1.51.5 hours. Find the average speed.
  3. Write the interval for 5.65.6 kg rounded to the nearest 0.10.1 kg.
  4. A length is 8.28.2 cm and a width is 3.43.4 cm, each to the nearest 0.10.1 cm. Find the upper bound for the area.
  5. Work out (4×106)(3×10−2)(4\times10^6)(3\times10^{-2}) in standard form.
  6. Simplify 108−12\sqrt{108}-\sqrt{12}.
  7. A map distance is 66 cm at scale 1:50 0001:50\,000. Find the real distance in kilometres.
  8. A force of 450450 N acts on 0.15 m20.15\text{ m}^2. Find the pressure.
Answers
  1. 27002700 mm.
  2. 36÷1.5=2436\div1.5=24 km/h.
  3. 5.55≤m<5.655.55\leq m<5.65 kg.
  4. Upper bounds: 8.258.25 cm and 3.453.45 cm. Area upper bound =28.4625 cm2=28.4625\text{ cm}^2.
  5. (4×3)×106+(−2)=12×104=1.2×105(4\times3)\times10^{6+(-2)}=12\times10^4=1.2\times10^5.
  6. 63−23=436\sqrt3-2\sqrt3=4\sqrt3.
  7. 6×50 000=300 0006\times50\,000=300\,000 cm =3=3 km.
  8. 450÷0.15=3000450\div0.15=3000 Pa.

What you can now do

Your connection

Return to your example from the opening page. Which method from this chapter could help with it? Write one new connection you noticed while learning.

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