Some math help please... adjusting price and solving for volume

Felicia asked on Aug 19, 2018 - 3 answers

Hi Everyone, I got the below from a casebook. The way I solved it was to use an assumption for volume, say volume = 100 units. Is there a more elegant way to do so though?

One option that our client has is to give in to its customers’ demands for a price discount. However our client is concerned about the impact this will have on its profits. If we were to cut prices 5%, how much would we have to increase volumes to maintain the same amount of profits?

In this case the gross margin is 20%. So if the original price is 100, then per unit costs are 80 and profits 20. A 5% decline in price makes it 95. Costs remain at 80 and profits are now 15. To make the profits the same, you need to increase volume by one third. So cutting prices to take back market share, requires that it sell 33.3% more product simply to keep profits constant. Given its high market share levels, this is unlikely.

3 answers

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Anonymous A updated his answer on Aug 19, 2018

20% margin means C = 0.8*P.

If new price is 0.95*old price, then new margin is 0.95*P - 0.8*P=0.15*P, so your new margin is 15%.

New volume * P * 0.15 = Old volume * P * 0.2 (price is same here in both sides),

New volume / Old volume = 0.2/0.15 = 1.33. So your new profit has to be 33% higher than old.

To arrive at non-number equations, once you have calculated the solution in numbers, replace numbers with words.


replied on Aug 19, 2018
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Anonymous answer is correct. However, there is nothing wrong with using your approach of 100 units - and if it helps you, you shouldn't be worried about using it.

That said, practice algebraic equations like this (e.g. by doing GMAT problem solving) to get more used to solving problems like this in an algebraic way.

Anonymous replied on Sep 06, 2018

Hi Felicia,

Using a base price of 100 is fine to do. I have an online course that covers how to solve this, and similar problems quickly and efficiently. I have some practice problems here (check out he sections labled Inverse Proporation): .

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