Question

A trader has 600 kgs of rice, a part of which he sells at 15% profit and the remaining quantity at 20% loss. On the whole. he incurs an overall loss of 6%. What is the quantity of rice he sold at 20% loss?

Answer 1

Hint: Assume the quantity of rice sold at 20% loss to be \[x\] kg because that is what the question is asking us to find and hence the remaining quantity of rice is \[600-x\] kg which is sold at 15% profit.

Complete step-by-step solution:
Before proceeding with the question, we should know the definition of cost price, selling price, profit and loss.
The price at which any article is purchased is its cost price (C.P).
The price at which any article is sold is its selling price (S.P).
Profit is the difference between S.P. and C.P., if S.P. is greater than C.P.
Loss is the difference between C.P and S.P., if C.P. is greater than S.P.
Now, let the quantity of rice sold at 20% loss be \[x\] kg and the quantity of rice sold at 15% profit be \[600-x\] kg. Let the price of 1 kg of rice be Rs. 100.
As the loss is 20% so (100-20) is 80 and profit is 15% so (100+15) is 115 and the overall loss is 6% so (100-6) is 94.
\[x\] kg sold at amount \[=x\times \dfrac{80}{100}........(1)\]
\[600-x\] kg sold at amount \[=(600-x)\times \dfrac{115}{100}........(2)\]
Now it is mentioned in the question that the overall loss for 600 kg of rice is 6%.
So the overall amount received by the trader after having a loss of 6% \[=600\times \dfrac{94}{100}........(3)\]
As it is given in the question that some quantity of rice the trader sells at 20% loss and some quantity at 15% profit so that is why adding equation (1) and equation (2) we get,
\[\Rightarrow x\times \dfrac{80}{100}+(600-x)\times \dfrac{115}{100}...........(4)\]
Equating equation (4) to equation (3) because overall is mentioned in the question so we get,
\[\Rightarrow x\times \dfrac{80}{100}+(600-x)\times \dfrac{115}{100}=600\times \dfrac{94}{100}...........(5)\]
As denominator is same we can directly add the numerators in the left hand side of equation (5),
\[\Rightarrow \dfrac{80x+(600\times 115)-115x}{100}=600\times \dfrac{94}{100}\]
Cancelling 100 from the denominator of both sides we get,
 \[\Rightarrow 80x+(600\times 115)-115x=600\times 94\]
Bringing all the terms with \[x\] on one side and all the numbers on another side we get,
\[\Rightarrow 115x-80x=(600\times 115)-(600\times 94)\]
After simplifying, we get,
\[\Rightarrow 35x=12600\]
\[\Rightarrow x=360\]
Hence 360 kg is the quantity of rice he sold at 20% loss.

Note: We could have taken the quantity of rice sold at 15% profit as \[x\] but then it would have consumed more time. In a hurry, we may make simple calculation mistakes so we need to take care of that by doing each and every step. Understanding overall meaning in the question is important as it means that after taking into account the 15% profit and 20% loss we have to equate its summation with the overall 6% loss.