Chapter 6.4, Problem 35HP

To determine

To find: Smallest and largest diameters that are acceptable.

Answer to Problem 35HP

Largest diameter = $l=255$ millimeters.

Smallest diameter = $s=245$ millimeters.

Explanation of Solution

Given information:

Actual value of diameter $=250$ millimeters.

Given percent of error of no more than $2\%$ .

Percent error $=\frac{A}{O}\times 100$

Where

  $A=$ Absolute value.

  $O=$ Actual value.

From given information

Actual value of diameter $=250$ millimeters.

Given percent of error of no more than $2\%$ .

As we need to find the smallest and largest possible values of diameters.

It means that the diameter will decrease by $2\%$ or increase by $2\%$ .

So,

Smallest possible value of diameter is when percent of error $2\%$ decrease.

Let us assume smallest diameter be $\text{'}s\text{'}$ .

Absolute value = calculated value $-$ Actual value.

  $\Rightarrow A=s-250$

Therefore,

Percent error $=\frac{s-250}{250}\times 100$

  $\begin{array}{l}\Rightarrow -2=\frac{s-250}{250}\times 100\\ \Rightarrow -2=\frac{100s-25000}{250}\\ \Rightarrow 100s-25000=250\times \left(-2\right)\\ \Rightarrow 100s-25000=-500\\ \Rightarrow 100s=25000-500\\ \Rightarrow 100s=24500\\ \Rightarrow s=\frac{24500}{100}\\ \Rightarrow s=245\end{array}$ .

Therefore

Smallest diameter = $s=245$ millimeters.

So,

largest possible value of diameter is when percent of error $2\%$ increase.

Let us assume largest diameter be $\text{'}l\text{'}$ .

Absolute value = calculated value $-$ Actual value.

  $\Rightarrow A=l-250$

Therefore,

Percent error $=\frac{l-250}{250}\times 100$

  $\begin{array}{l}\Rightarrow 2=\frac{l-250}{250}\times 100\\ \Rightarrow 2=\frac{100l-25000}{250}\\ \Rightarrow 100l-25000=250\times 2\\ \Rightarrow 100l-25000=500\\ \Rightarrow 100l=25000+500\\ \Rightarrow 100l=25500\\ \Rightarrow l=\frac{25500}{100}\\ \Rightarrow l=255\end{array}$ .

Therefore

Largest diameter = $l=255$ millimeters.