# How do I solve this Inverse proportion question?

I am currently doing maths at GCSE Level. I was revising, then I came across a question that I became stuck on:

Brian's band is playing at a concert in a hall. The loudness of a band varies inversely as the square of the distance from the band. Brian measures the normal loudness of his band as 100 Db at a distance of $5$m. He has to stop playing if the loudness is above 85Db or more at a distance of $5.4$m. Does the band have to stop playing?

I understand I have to use the formula $y = \frac{k}{x^2}$ I do not know how to apply it though.

Would I do $Db = \frac{k}{(distance)^2}$?

I would appreciate any hints, but no answers please.

Assumption: There is no rule for normal decibels that for a sound ten times more powerful than $x$Db, only $10$Db is added.

## 2 Answers

I will use different symbols for distance and loudness $d,L$

$$k= L d^2,\, L=\dfrac{k} {d^2}$$

To evaluate $k$ the data is given:

$$k= 100 \, 5^2 = 2500$$

So no one can be seated at a distance less than

$$d_{min}= \sqrt{\dfrac{2500}{85}} = 5.42 \,m$$

The band need not stop playing but ensure that first row chair distance must be more than than 5.4 m

• I believe that $k$ = 2500 ; we have got $100 = \frac{k}{25}$, which means that $k = 100 \times 25 = 2500$ – VortexYT Feb 22 '17 at 14:52
• Yes, a simple error, thank you for pointing it out – Narasimham Feb 22 '17 at 15:05

$l = k/d^2$

$100 = k/5^2$

$100 = k/25$

$100$ x $25$ = $2500 = k$

$k = 2500$

$l = 2500/5.4^2$

$l = 2500/29.16$

$l = 85.73$

$85.73 > 85$ therefore the band has to stop playing

side note: '^2' means squared

• Welcome to MSE. Please format your answer using MathJax. – user507623 Jan 10 '18 at 20:38