Question

A man sold a chair and a table together for Rs.1520 thereby making profit of 25% on the chair and 10% on table. By selling them together for Rs.1535 he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each
A. C.P. of chair=Rs.800, C.P. of table=Rs.500
B. C.P. of chair=Rs.600, C.P. of table=Rs.700
C. C.P. of chair=Rs.800, C.P. of table=Rs.700
D. C.P. of chair=Rs.600, C.P. of table=Rs.500

Answer 1

Hint: First assume 2 variables for the cost price of each. Now using the first condition find the relation between 2 variables. Similarly from other conditions you have 2 equations of 2 variables. So, apply a substitution method to solve and find the solutions. The value of the variables is the required result.
\[\text{Use the relation:}\dfrac{100+P}{100}\times CP=SP\]

Complete step-by-step answer:
Given condition in the question can be written as:
Man gets 25% profit on a chair and 10% on table if he sells for Rs.1520.
Second condition can be written in the form of:
Man gets 10% profit on a chair and 25% on table if he sells for Rs.1535.
\[\text{Total Selling price}=\,\text{Selling}\,\,\text{price}\,\,\text{of}\,\,\text{chair}+\text{Selling}\,\,\text{price}\,\,\text{of}\,\,\text{table}\]
Let us assume Cost price of the chair to be x.
Let us assume the Cost price of the table to be y.
We know that the relation between CP, SP, and P is given by:
\[\dfrac{100+P}{100}\times CP=SP\]
Case-I: 25% profit on chair and 10% on table
By applying relation on chair substituting all values we get it as:
\[\dfrac{100+25}{100}\times x=\text{SP}\,\,\text{of}\,\,\text{chair}\]
By applying relation on table substituting all values we get it as:
\[\dfrac{100+10}{100}\times y=\text{SP}\,\,\text{of}\,\,\text{table}\]
By adding the above two equation, we get it as:
\[\dfrac{125}{100}x+\dfrac{110}{100}y=\text{SP}\,\,\text{of}\,\,\text{chair+SP}\,\,\text{of}\,\,\text{table=Total}\,\,\text{SP}\]
By substituting total SP of case-I as 1520, we get:
\[\dfrac{5}{4}x+\dfrac{11}{10}y=1520\]
By taking L.C.M. we get it as follows:
\[25x+22y=30400\ldots \ldots \ldots \ldots \ldots ..\left( \text{1} \right)\]

Case-II: 10% profit on chair and 25% on table we get Rs.1535
From the case-I, we can write the equation as:
\[\dfrac{110}{100}x+\dfrac{125}{100}y=1535\]
By taking L.C.M. we can write the equation follows:
\[22x+25y=30700\ldots \ldots \ldots \ldots \ldots ..\left( \text{2} \right)\]
By subtracting 25y and dividing 22 on both sides, we get:
\[x=\dfrac{30700-25y}{22}\ldots \ldots \ldots \ldots \ldots ..\left( \text{3} \right)\]
Substitution method: The method of solving a system of equations. It works by solving one of the variables to get in terms of another variable, then plugging this back into another equation, and solving for the other variable. By this you can find both the variables. This method is generally used when there are 2 variables. For more variables it will be tough to solve.
Substituting this in equation (1), we get it as:
\[25\left( \dfrac{30700-25y}{22} \right)+22y=30400\]
By simplifying the above equation, we get the equation as:
\[767500-625y+484y=668800\]
By taking variables to one side, constants to other side, we get:
\[484y-625y=668800-767500\]
By simplifying the above equation, we get it in form of:
\[141y=98700\]
By dividing with 141 on both sides, we get it as:
\[\dfrac{141y}{141}=\dfrac{98700}{141}\]
By simplifying the equation, we get it as follows:
\[y=700\ldots \ldots \ldots \ldots \ldots ..\left( \text{4} \right)\]
By substituting equation (4) in equation (3), we get it as:
\[x=\dfrac{30700-25\left( 700 \right)}{22}\]
By simplifying term with minus sign, we get it as:
\[x=\dfrac{30700-17500}{22}\]
By simplifying the term in the numerator, we get it as:
\[x=\dfrac{13200}{22}\]
By calculating the above term, we get it as:
\[x=600\]
By values of x, y we can say that CP of the chair is 600 and CP of table is 700.
So, the correct answer is “Option B”.

Note: Be careful while calculating the value of SP as you must add both of them. In both equations only coefficients of x, y exchange because their percentage exchanges some students write the same in conclusion but remember they must exchange. As condition is like coefficient exchange you can also use the elimination method as \[110\times \left( 1 \right)-125\times \left( 2 \right)\] then also you will get the same results.