
A dealer earned a profit of 5% by selling radio for ₹ 714. Find the cost price of the radio.
Answer
141.3k+ views
Hint: As we know that the Cost Price (C.P) means the amount which is paid by the seller to acquire that product and Selling Price (S.P) is the money that is finally received by the seller after selling that same product to any customer.
If the S.P is more than the C.P then there will be profit and if C.P is more than the S.P then there will be a loss for the seller.
Hence, Profit % is the percent of the profit gained by the seller and loss % is the percent of loss suffered by the seller.
Profit and loss for any product are given by:
$Profit = S.P - C.P$
$Loss = C.P - S.P$
Now Profit and Loss percentage can be calculated by:
\[Profit\% = \dfrac{{Profit}}{{C.P}} \times 100\]
\[Loss\% = \dfrac{{Loss}}{{C.P}} \times 100\]
Complete step-by-step answer:
Given,
\[\begin{gathered}
Profit = 5\% \\
S.P{\text{ }} = \;Rs.714 \\
\\
\end{gathered} \]
Let us denote Profit by ‘P’
We know that
\[Profit\% = \dfrac{{Profit}}{{C.P}} \times 100\]
Then,
\[C.P = \dfrac{{Profit}}{{Profit\% }} \times 100\]
By putting the value of Profit % we get
\[C.P = \dfrac{P}{5} \times 100\]
i.e. \[C.P{\text{ }} = {\text{ }}20P\] eqn (i)
Also,
\[Profit{\text{ }} = {\text{ }}S.P{\text{ }} - {\text{ }}C.P\]
By using eqn(i) and putting value of S.P, we get
\[\begin{array}{*{20}{l}}
{21P{\text{ }} = {\text{ }}714} \\
{P = 34}
\end{array}\]
Using the value of P in equation (i), we get
\[C.P = 20 \times 34\]
Hence,
\[C.P = \;Rs.{\text{ }}680\]
Note: Cost Price can also be calculated by simply using the formula given below:
\[C.P = \dfrac{{100}}{{100 + P\% }} \times S.P\]
Never Confuse C.P with S.P the C.P means the cost price and S.P means selling price.
And there would be Profit only if \[S.P{\text{ }} > {\text{ }}C.P\]
If the S.P is more than the C.P then there will be profit and if C.P is more than the S.P then there will be a loss for the seller.
Hence, Profit % is the percent of the profit gained by the seller and loss % is the percent of loss suffered by the seller.
Profit and loss for any product are given by:
$Profit = S.P - C.P$
$Loss = C.P - S.P$
Now Profit and Loss percentage can be calculated by:
\[Profit\% = \dfrac{{Profit}}{{C.P}} \times 100\]
\[Loss\% = \dfrac{{Loss}}{{C.P}} \times 100\]
Complete step-by-step answer:
Given,
\[\begin{gathered}
Profit = 5\% \\
S.P{\text{ }} = \;Rs.714 \\
\\
\end{gathered} \]
Let us denote Profit by ‘P’
We know that
\[Profit\% = \dfrac{{Profit}}{{C.P}} \times 100\]
Then,
\[C.P = \dfrac{{Profit}}{{Profit\% }} \times 100\]
By putting the value of Profit % we get
\[C.P = \dfrac{P}{5} \times 100\]
i.e. \[C.P{\text{ }} = {\text{ }}20P\] eqn (i)
Also,
\[Profit{\text{ }} = {\text{ }}S.P{\text{ }} - {\text{ }}C.P\]
By using eqn(i) and putting value of S.P, we get
\[\begin{array}{*{20}{l}}
{21P{\text{ }} = {\text{ }}714} \\
{P = 34}
\end{array}\]
Using the value of P in equation (i), we get
\[C.P = 20 \times 34\]
Hence,
\[C.P = \;Rs.{\text{ }}680\]
Note: Cost Price can also be calculated by simply using the formula given below:
\[C.P = \dfrac{{100}}{{100 + P\% }} \times S.P\]
Never Confuse C.P with S.P the C.P means the cost price and S.P means selling price.
And there would be Profit only if \[S.P{\text{ }} > {\text{ }}C.P\]
Recently Updated Pages
Difference Between Mutually Exclusive and Independent Events

Difference Between Area and Volume

JEE Mains 2025 April 4 Shift 2 Chemistry Question Paper Analysis - PDF

JEE Mains 2025 April 4 Shift 2: Physics Question Paper PDF and Analysis

JEE Main 2025 April 4 Shift 2 Maths Paper Analysis & Solutions PDF

JEE Main 2025 April 4 Shift 2 Paper Analysis and Solutions FREE PDF

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main Syllabus 2025 (Updated)

JEE Main Marks Vs Percentile Vs Rank 2025: Calculate Percentile Using Marks

JEE Main Question Papers 2025

JEE Mains 2025 Cutoff: Expected and Category-Wise Qualifying Marks for NITs, IIITs, and GFTIs

Raoult's Law with Examples

Other Pages
NCERT Solutions for Class 9 Maths Chapter 11 Surface Area and Volume

NCERT Solutions for Class 9 Maths Chapter 11 Surface Areas And Volumes Ex 11.3

NCERT Solutions for Class 9 Maths Chapter 9 Circles

NCERT Solutions for Class 9 Maths Chapter 12 Statistics

NCERT Solutions for Class 9 Maths Chapter 10 Heron'S Formula

NCERT Solutions for Class 9 Maths In Hindi Chapter 1 Number System
