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A shopkeeper, while selling, uses a meter scale which measures 10% less cloth than it is actually suppose to measure. In winter, it further shrinks to by 10%. For purchasing, he uses a different metre scale which measures 20% more cloth than it is actually suppose to measure and there is no effect of winter on this metre scale. Find the approximate net profit or loss percentage made by the shopkeeper (in winter) if he sells cloth at the same rate at which it was purchased.
A. 48 B. 56 C. 60 D. 64 E. 72
adapted from CL
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A shopkeeper, while selling, uses a meter scale which measures 10% less cloth than it is actually suppose to measure. In winter, it further shrinks to by 10%. For purchasing, he uses a different metre scale which measures 20% more cloth than it is actually suppose to measure and there is no effect of winter on this metre scale. Find the approximate net profit or loss percentage made by the shopkeeper (in winter) if he sells cloth at the same rate at which it was purchased.
A. 48 B. 56 C. 60 D. 64 E. 72
adapted from CL
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1m = 100 cms
Let's assume that the purchase price of 1cm of cloth is $1.
The shopkeeper intends to purchase 100 cm of the cloth. The price that the shopkeeper will pay = 100 * $1 = $100
However, due to the faulty meter instead of 100 cm of cloth, the shopkeeper will have actually got 20% extra. Hence, the shopkeeper buys 120 cm of cloth for $100.
Purchasing price of the cloth / cm = \(\frac{100}{120} = \frac{5}{6}\)
While selling, the shopkeeper sells 10% less cloth on normal days, and in winter the meter scale further shrinks by 10%. Therefore, a customer who intends to purchase 100 cm of cloth on winter, will actually end up purchasing only 81 cm of cloth (\(100 * \frac{90}{100} * \frac{90}{100} = 81\)). However, the customer pays for 100 cm of the cloth at a purchase price of $1. In a nutshell, the shopkeeper sells 81 cm of cloth for $100.
Selling price of the cloth / cm = \(\frac{100}{81} \approx \frac{100}{80} \approx \frac{5}{4}\)
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