Many a times it happens that there will approximately some error in the instruments due to negligence in measuring precisely. These approximation values with errors when used in calculations may lead to larger errors in the values. There are two ways to measure errors commonly - absolute error and relative error.The absolute error tells about how much the approximate measured value varies from true value whereas the relative error decides how incorrect a quantity is from the true value.

**Eg: **A carpenter is given a task to find the length of the showcase. Due to his negligence he takes the value as 50.32 m whereas the actual precise value is 50.324 m. In this case to measure the errors we use these formulas.

Suppose the measurement has some errors compared to true values.Relative error decides how incorrect a quantity is from a number considered to be true. Unlike absolute error where the error decides how much the measured value deviates from the true value the relative error is expressed as a percentage ratio of absolute error to the true value tells what's the error percentage?

**To calculate the relative error use the following way:**

Observe the

**true value**** (x)** and

**approximate measured value** **(x**_{o}). Then find the absolute deviation using formula

**Absolute deviation $\Delta$ x = True value - measured value = x - x**_{o}Then substitute the absolute deviation value $\Delta$ x in relative error formula given below

**Relative error = ****$\frac{\Delta\ x}{x}$**Substitute the values and get the relative error.

The relative error formula is given by

**Relative error =$\frac{Absolute\ error}{Value\ of\ thing\ to\ be\ measured}$ = $\frac{\Delta\ x}{x}$.**

In terms of percentage it is expressed as

**Relative error = $\frac{\Delta\ x}{x}$ $\times$ 100 % **Here $\Delta$ x and x are absolute error and true value of the measurement.

**Below are given some relative error examples you can go through it:** ### Solved Examples

**Question 1: **John measures the size of metal ball as 3.97 cm but the actual size of it is 4 cm. Calculate the absolute error and relative error.

** Solution: **

Given: The measured value of metal ball x_{o} = 3.97 cm

The true value of ball x = 4 cm

Absolute error $\Delta$ x = True value - Measured value = x - x_{o} = 4cm - 3.97cm = 0.03 cm

Relative error = $\frac{\Delta\ x}{x}$ = $\frac{0.03}{4}$ = 0.0075.

**Question 2: **If the approximate value of $\pi$ is 3.14. Calculate the absolute and relative errors?

** Solution: **

Given: The measured value of metal ball x_{o} = 3.14

The true value of ball x = 3.142

Absolute error $\Delta$ x = True value - Measured value = x - x_{o} = 3.142 - 3.14 = 0.002

Relative error = $\frac{\Delta\ x}{x}$ = $\frac{0.002}{3.142}$ = 0.0006.