Answered: A merchant sold two-thirds (2/3) of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

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Course: Financial Mathematics
A merchant sold two-thirds (2/3) of cement he had bought at a profit of 10%; half of remainder at cost; and, what was left, at such a PERCENTAGE LOSS that HE CAN ONLY MAKE ONLY COST OF ALL THIS BUSINESS. What was this PERCENTAGE LOSS? Teacher's solution: 40% loss. Detail your procedure to get 40%

Expert Solution

Step 1: Finding the percentage of loss

Given the merchant sold $\frac{2}{3}$ at a profit of $10 %$

Let the cost be $x$

$10 %$ of $\frac{2}{3}$ is

$= \frac{10}{100} \times \frac{2 x}{3} = \frac{x}{15}$

Therefore selling price of $\frac{2}{3}$ portion of cement is $= \frac{2 x}{3} + \frac{x}{15}$

Remaining portion of cement $= \frac{1}{3}$

Half of $\frac{1}{3}$ is sold at cost price .

Half of $\frac{1}{3}$ portion is $\frac{1}{6}$ portion

Therefore selling price of $\frac{1}{6}$ portion is $\frac{x}{6}$

Let remaining $\frac{1}{6}$ is sold at $r %$ discount

Therefore loss on this portion $\frac{1}{6}$ is $= \frac{x}{6} \times \frac{r}{100}$

Therefore selling price while making loss is $\frac{x}{6} (1 - \frac{r}{100})$

Now given that he make only the cost

Therefore,

$\Rightarrow \frac{2 x}{3} + \frac{x}{15} + \frac{x}{6} + \frac{x}{6} (1 - \frac{r}{100}) = x \Rightarrow \frac{2}{3} + \frac{1}{15} + \frac{1}{6} + \frac{1}{6} - \frac{r}{600} = 1 \Rightarrow 1 + \frac{1}{15} - 1 = \frac{r}{600} \Rightarrow \frac{r}{600} = \frac{1}{15} \Rightarrow r = 40 %$

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Follow-up Question

I'm confused way you answer exercise. Please could you be more cleared? May be if you transcript answer to a piece of paper? Number 23 must be 2/3 and that's confusing

Please rewrite your answer. Thanks a lot

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