The price of a table is Rs.400 more than that of a chair. If 6 tables and 6 chairs together cost Rs.4800, by what percent is the price of the chair less than that of the table?
A.\[33\dfrac{1}{3}\% \]
B. \[50\% \]
C. \[66\dfrac{2}{3}\% \]
D. None of these
Answer
585.6k+ views
Hint: We assume the price of chair as a variable and use the relation given in the question to write the price of table in terms of price of chair. Form an equation for number of items multiplied with the initial amount of item and write sum from tables and chair and equate with given amount. Calculate the value of the variable. Use a method of percentage to calculate percentage of price of chair.
* Percentage of a particular item with respect to another item is given by dividing the value of that item by the value of another item and multiplying by 100.
Complete answer:
Let us assume price of chair as ‘x’
Then we are given that the price of a table is Rs.400 more than that of a chair
So, price of table \[ = x + 400\]
Now we are given that there are 6 tables and 6 chairs, they together have price Rs.4800
We form an equation where we add the product of price of a chair with 6 chairs and price of a table with 6 tables and equate that sum to Rs.4800
\[ \Rightarrow x \times 6 + (x + 400) \times 6 = 4800\]
Since we can write \[4800 = 6 \times 800\]
\[ \Rightarrow x \times 6 + (x + 400) \times 6 = 6 \times 800\]
Cancel same factors i.e. 6 from both sides of the equation
\[ \Rightarrow x + x + 400 = 800\]
Shift all constant values to RHS of the equation
\[ \Rightarrow 2x = 800 - 400\]
\[ \Rightarrow 2x = 400\]
Cancel same factors i.e. 2 from both sides of the equation
\[ \Rightarrow x = 200\]
So, the price of 1 chair is Rs.200
Also, we know price of a table \[ = 200 + 400\] i.e. 600
So, the price of 1 table is Rs.600
Now we have to calculate by what percentage is the price of the chair less than that of the table.
So, we have to calculate the percentage of price by which the chair is less than the table.
\[ \Rightarrow \]Percentage \[ = \left( {\dfrac{{200}}{{600}} \times 100} \right)\% \]
Cancel same factors from numerator and denominator
\[ \Rightarrow \]Percentage \[ = \left( {\dfrac{1}{3} \times 100} \right)\% \]
Convert division of 100 by 3 into mixed fraction where 33 is quotient, 3 is divisor and 1 is the remainder
\[ \Rightarrow \]Percentage \[ = 33\dfrac{1}{3}\% \]
\[\therefore \]Correct option is A.
Note:
Many students are likely to get confused when calculating the percentage as they tend to take the sum of prices of both table and chairs in the denominator, keep in mind we take that term in denominator to whose comparison we are making, in this case we want to see what percent is price of chair less than price of table so price of table will be in denominator.
* Percentage of a particular item with respect to another item is given by dividing the value of that item by the value of another item and multiplying by 100.
Complete answer:
Let us assume price of chair as ‘x’
Then we are given that the price of a table is Rs.400 more than that of a chair
So, price of table \[ = x + 400\]
Now we are given that there are 6 tables and 6 chairs, they together have price Rs.4800
We form an equation where we add the product of price of a chair with 6 chairs and price of a table with 6 tables and equate that sum to Rs.4800
\[ \Rightarrow x \times 6 + (x + 400) \times 6 = 4800\]
Since we can write \[4800 = 6 \times 800\]
\[ \Rightarrow x \times 6 + (x + 400) \times 6 = 6 \times 800\]
Cancel same factors i.e. 6 from both sides of the equation
\[ \Rightarrow x + x + 400 = 800\]
Shift all constant values to RHS of the equation
\[ \Rightarrow 2x = 800 - 400\]
\[ \Rightarrow 2x = 400\]
Cancel same factors i.e. 2 from both sides of the equation
\[ \Rightarrow x = 200\]
So, the price of 1 chair is Rs.200
Also, we know price of a table \[ = 200 + 400\] i.e. 600
So, the price of 1 table is Rs.600
Now we have to calculate by what percentage is the price of the chair less than that of the table.
So, we have to calculate the percentage of price by which the chair is less than the table.
\[ \Rightarrow \]Percentage \[ = \left( {\dfrac{{200}}{{600}} \times 100} \right)\% \]
Cancel same factors from numerator and denominator
\[ \Rightarrow \]Percentage \[ = \left( {\dfrac{1}{3} \times 100} \right)\% \]
Convert division of 100 by 3 into mixed fraction where 33 is quotient, 3 is divisor and 1 is the remainder
\[ \Rightarrow \]Percentage \[ = 33\dfrac{1}{3}\% \]
\[\therefore \]Correct option is A.
Note:
Many students are likely to get confused when calculating the percentage as they tend to take the sum of prices of both table and chairs in the denominator, keep in mind we take that term in denominator to whose comparison we are making, in this case we want to see what percent is price of chair less than price of table so price of table will be in denominator.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

