
A shopkeeper sells 100kg of sugar partly at 10% profit and the remaining at 20% profit. If he gains 12% on the whole transaction how much sugar did he sell at 20% profit?
$
{\text{A}}{\text{. 25kg}} \\
{\text{B}}{\text{. 40kg}} \\
{\text{C}}{\text{. 20kg}} \\
{\text{D}}{\text{. 30kg}} \\
$
Answer
611.1k+ views
Hint – To find out the amount of sugar he sold at 20% profit, we assume the cost of one kg sugar as a variable x, then compute the price of kg, then compute the profit on it at 12%. Then we divide the sugar into parts, compute profits at 10% and 20% respectively and equate to 12% profit, solve for x to get the answer.
Complete step by step answer:
Let us assume the cost of 1kg of sugar = Rs x.
So the price of 100kg sugar = 100x.
Now the profit of 100kg sugar at 12% = ${\text{100x }} \times {\text{ }}\dfrac{{12}}{{100}}{\text{ = 12x}}$ --- (1)
Now let the sugar be divided into p kg and (100 – p) kg.
So the price of p kg of sugar = px.
Then the profit on p kg at 10% = ${\text{px }} \times {\text{ }}\dfrac{{10}}{{100}}{\text{ = }}\dfrac{{{\text{px}}}}{{100}}$ --- (2)
Now, the price of (100 – p) kg of sugar = (100 – p)x.
Then the profit on (100 – p) kg at 20% = $\left( {100 - {\text{p}}} \right){\text{x }} \times {\text{ }}\dfrac{{20}}{{100}}{\text{ = }}\dfrac{{\left( {200{\text{ - 2p}}} \right){\text{x}}}}{{10}}$ --- (3)
Now we add (2) and (3) to compute the total profit,
$
\Rightarrow \dfrac{{{\text{px}}}}{{100}} + \dfrac{{\left( {200 - 2{\text{p}}} \right){\text{x}}}}{{10}} \\
\Rightarrow \dfrac{{{\text{x}}\left( {200 - {\text{p}}} \right)}}{{10}} \\
$
Now we compare the total profit obtained with the profit earned from overall transaction, i.e. with equation (1)
$
\Rightarrow \dfrac{{{\text{x}}\left( {200 - {\text{p}}} \right)}}{{10}} = 12{\text{x}} \\
\Rightarrow {\text{200 - p = 120}} \\
\Rightarrow {\text{p = 80kg}} \\
$
i.e. (100-p) = 100-80 = 20kg.
Hence, the amount of sugar sold at 20% profit is 20kg.
Note – In order to solve this type of question the key is to consider an unknown variable and convert all the given sentences in the question into equations and eventually reduce them in terms of our variable. Using the conditions given we solve for the variable which gives us the answer.
Profit percent = price × percentage of profit.
Complete step by step answer:
Let us assume the cost of 1kg of sugar = Rs x.
So the price of 100kg sugar = 100x.
Now the profit of 100kg sugar at 12% = ${\text{100x }} \times {\text{ }}\dfrac{{12}}{{100}}{\text{ = 12x}}$ --- (1)
Now let the sugar be divided into p kg and (100 – p) kg.
So the price of p kg of sugar = px.
Then the profit on p kg at 10% = ${\text{px }} \times {\text{ }}\dfrac{{10}}{{100}}{\text{ = }}\dfrac{{{\text{px}}}}{{100}}$ --- (2)
Now, the price of (100 – p) kg of sugar = (100 – p)x.
Then the profit on (100 – p) kg at 20% = $\left( {100 - {\text{p}}} \right){\text{x }} \times {\text{ }}\dfrac{{20}}{{100}}{\text{ = }}\dfrac{{\left( {200{\text{ - 2p}}} \right){\text{x}}}}{{10}}$ --- (3)
Now we add (2) and (3) to compute the total profit,
$
\Rightarrow \dfrac{{{\text{px}}}}{{100}} + \dfrac{{\left( {200 - 2{\text{p}}} \right){\text{x}}}}{{10}} \\
\Rightarrow \dfrac{{{\text{x}}\left( {200 - {\text{p}}} \right)}}{{10}} \\
$
Now we compare the total profit obtained with the profit earned from overall transaction, i.e. with equation (1)
$
\Rightarrow \dfrac{{{\text{x}}\left( {200 - {\text{p}}} \right)}}{{10}} = 12{\text{x}} \\
\Rightarrow {\text{200 - p = 120}} \\
\Rightarrow {\text{p = 80kg}} \\
$
i.e. (100-p) = 100-80 = 20kg.
Hence, the amount of sugar sold at 20% profit is 20kg.
Note – In order to solve this type of question the key is to consider an unknown variable and convert all the given sentences in the question into equations and eventually reduce them in terms of our variable. Using the conditions given we solve for the variable which gives us the answer.
Profit percent = price × percentage of profit.
Recently Updated Pages
Questions & Answers - Ask your doubts

Master Class 9 Social Science: Engaging Questions & Answers for Success

Class 9 Question and Answer - Your Ultimate Solutions Guide

Master Class 8 Science: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 English: Engaging Questions & Answers for Success

Trending doubts
The average rainfall in India is A 105cm B 90cm C 120cm class 10 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

Who Won 36 Oscar Awards? Record Holder Revealed

Indias first jute mill was established in 1854 in A class 10 social science CBSE

Indias first jute mill was established in 1854 in A class 10 social science CBSE

