Question : A faulty weighing machine reads 1 kg when 900 gm is actually weighted on it. The shopkeeper marked the price of his goods by 10%, but unfortunately, he was caught by the metrology department and then they ordered him to repair his weighing machine and gave punishment to sell the goods at a 10% discount on the cost price for a month. If each customer is now paying INR 20 for 1 kg, then before the raid _____ amount they have paid for the same quantity. (Rounded off to two places of decimal)
Option 1: INR 27.16
Option 2: INR 27.04
Option 3: INR 28.05
Option 4: INR 28.15
Correct Answer: INR 27.16
Solution :
Given,
A faulty weighing machine reads 1 kg instead of 900 gm
Shopkeeper marked the price of goods by 10%
We know,
SP = $\frac{100 + P\%}{100}$ × CP
CP = $\frac{100}{100 - L\%}$ × SP
Profit% = $\frac{\text{Profit}}{CP} × 100$
Successive profit% = $x + y + \frac{xy}{100}$
Where, SP = Selling price, CP = Cost price, $x$ and $y$ = profit% respectively.
It is given that initially, the weighing machine weighs 900 gm instead of 1000 gm
⇒ It means the shopkeeper earns 100 gm profit on selling 900 gm goods
⇒ Profit% = $\frac{\text{Profit}}{CP} × 100$
⇒ Profit% = $\frac{100}{900} × 100$ = 11.11%
He earns this by alteration of weight along with marking up the price by 10%.
⇒ Overall profit = $11.11 + 10 + \frac{11.11 × 10}{100}$ = 22.22%
Now the shopkeeper sells the goods at Rs. 20 per kg after giving a 10% discount
⇒ Actual cost price of 1 kg sugar = $\frac{100}{100 - 10}$ × 20 = $\frac{100}{90} × 20$ = Rs. 22.22
So, the selling price before the raid = $\frac{100 + 22.22}{100} × 22.22$
= $\frac{122.22}{100} ×22.22$ = Rs. 27.157 ~ Rs. 27.16
Hence, the correct answer is Rs. 27.16.
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