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By selling two tables for 1200 each, a man gains 10% on one and loses 10% on the other. What is his gain/loss percent in the whole transaction?

I calculated cost price for table one by equating selling price to 1.1 cost price.and got the cost price as 1,091(approx) and for table 2 I equated selling price to .9 cost price which comes out to be 1,333 (approx). Total cost price becomes 2024. There is a loss of 24. Now if I calculate loss on cost price the answer is 0.99 which is close to the answer given in the book which 1% loss.

Is there some other easier way to solve this?

Curious
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  • That is the correct approach. By the way you mistyped $ 2024$. Did you want to say $ 2424$? – Math Lover Dec 17 '20 at 07:44
  • and $1%$ answer in the book is just rounding off. – Math Lover Dec 17 '20 at 07:45
  • This is probably beyond your level, but if you were forced to do these calculations without a calculator, you could use the geometric series expansions $\frac{1}{1 - h} = 1 + h + h^2 \cdots$ and $\frac{1}{1 + h} = 1 - h + h^2 \cdots$ and ignore the higher-order terms. This applies when $-1 < h < 1$ and ideally when $h$ is small. $h = 0.1$ works fine. – Toby Mak Dec 17 '20 at 08:24

1 Answers1

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As said before, there is no easier way to solve this. However, you can avoid any rounding errors by calculating at the very last step, instead of using approximations.

Calculate $\frac{1200}{1.1} + \frac{1200}{0.9}$ directly, as to get the original price, you undo the $1.1$ and $0.9$ multipliers on the tables.

Then the percentage change is:

$$\frac{\text{new} - \text{old}}{\text{old}} = \frac{2400-(1200/1.1 + 1200/0.9 )}{1200/1.1 + 1200/0.9} \times 100 \% \approx -1.01 \%$$

or approximately a $1 \%$ decrease.

Toby Mak
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