A meter scale that is rigged to measure 90 cms . However in summers it expands by 20 % of its actual length. The trader sells goods at a markup of 10 % but he does not know that the wholesaler from where the trader buys goods uses a meter scale which measures 80 cms. What profit or loss percentage does the trader make ?

My approach

There are 2 transactions taking place here Trader-Wholesaler and Trader-Customer

So firstly I focused on Trader-Wholesaler, lets assume trader buys 100 cm cloth from him, and let the cost for 1 cm cloth is £1, so his cost price here becomes £100, but in turn receives only 80 m cloth.

Now Transaction 2 (Trader-Customer)

Lets say the customer orders the 100 cm cloth, but the trader has 80 cm cloth (which he thinks is of 100 cm length), so he measures that on his scale which would show a measurement of 80cm * (90/100) * (120/100) = 86.4 cm (I did 90/100 as the scale which trader owns shows 90 cm for every 100 cm, and 120/100 because of the climate effect if he measures 100 cm , it would be actually measuring to 120 cm.

Now he sells this 86.4 cm to the customer , by claiming to sell 100cm , therefore his cost price here is £86.4 and selling price would be £100 * (1.1) = £110 {due to the mark-up}

Therefore overall in 2 transactions his SP= 110, and CP=100(From 1st transaction) + 86.4 (From 2nd transaction) =186.4, therefore loss =186.4-100=86.4 , and loss percentage= 86.4/186.4 = 46.35% loss

I am getting completely different answer than my textbook, please help me identify my mistakes


1 Answer 1


Say, the trader thinks that he is trading $x$ m of cloth with the customer for $\$y.$

  • So, the trader thinks that the wholesale rate is $\displaystyle\frac{\$\frac{10y}{11}}{\frac{10x}9\textrm m}=\frac{\$9y}{11x\:\textrm m}.$

    So, the actual wholesale rate is $\displaystyle\frac{\$9y}{11x\:\textrm m}\div0.8=\frac{\$45y}{44x\:\textrm m}.$

  • So, the trader is actually handing $1.2x$ m of cloth to the customer.

Therefore, the trader is making a loss of $$(1.2x\:\textrm m)\frac{\$45y}{44x\:\textrm m}-\$y=$\frac{5y}{22},$$ i.e., a percentage loss of $$\frac{\frac{5y}{22}}{y+\frac{5y}{22}}=18.519\%.$$


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