Question
Download Solution PDFA dishonest shopkeeper pretends to sell his goods at cost price, but uses a false weights and gains \(11\frac{1}{9}\%\). For a weight of a kg, he uses:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Gain percentage = \(11\frac{1}{9}\%\) = \(\frac{100}{9}\%\)
Formula used:
Gain % = (Error / (True Value - Error)) × 100
Calculation:
\(\frac{100}{9}\) = (Error / (1000 - Error)) × 100
⇒ \(\frac{1}{9}\) = Error / (1000 - Error)
⇒ 1000 - Error = 9 × Error
⇒ 1000 = 10 × Error
⇒ Error = 100
True value - Error = 1000 - 100 = 900
∴ He uses 900 grams instead of 1000 grams.
Last updated on Jun 16, 2025
-> The RRB Technician 2025 Vacancies have been announced in the Employment Newspaper.
-> As per the Notice, around 6180 Vacancies will be announced for the Technician 2025 Recruitment.
-> The Online Application form for RRB Technician will be open from 28th June 2025 to 28th July 2025.
-> The Pay scale for RRB Technician posts ranges from Rs. 19900 - 29200.
-> Prepare for the exam with RRB Technician Previous Year Papers.