1952 exam question

Now here is the full version of the 1952 exam question we started with earlier.

  1. Use logarithms to calculate, correct to three significant figures, \[\frac{1{\cdot}152\times(3{\cdot}902)^3}{(5{\cdot}463)^2} .\]

  2. The area of a square field is \(5{\cdot}355\) hectares. Find, correct to the nearest metre, the length of a diagonal of the field.

    [1 hectare = 100 ares; 1 are = 1 square dekametre.]

    [Editorial note: 1 dekametre = 10 metres.]

UCLES O level Elementary Mathematics Alternative A: Arithmetic, Question 7, 14 July 1952 (slightly amended)

Worked exercise
Worked solution

One final twist

In this resource we have dealt only with the logarithms of numbers greater than \(1\). Can you see the difficulty with using log tables to do calculations with numbers such as \(0{\cdot}0283\) ?

First, we write the number in standard form: \[0{\cdot}0283=2{\cdot}83\times10^{-2}\] so \[\log 0{\cdot}0283=-2+\log 2{\cdot}83=-2+0{\cdot}4518 .\]

If we were to work this out as \(-2+0{\cdot}4518=-1{\cdot}5482\), we would lose the clarity that the power of \(10\) is \(-2\). So instead we write the number as \(\bar2{\cdot}4518\) (called bar notation). Arithmetic with such quantities is carried out separately on the whole number part (called the characteristic) and the decimal part (called the mantissa) as illustrated here. Use of bar notation usually makes calculations easier.

Worked example