
An increase of 30% in the price of oranges enables a man to buy 6 kg less for Rs. 300. Find the increased price per kg.
A) 16
B) 15
C) 17
D) 18
Answer
585k+ views
Hint: Cent means hundred and percent simply means a part per hundred, it is a way of expressing amount per hundred. It is denoted by ‘%’
For example, 30 % of ‘x’ means \[\dfrac{{30}}{{100}}x\]
Percentage formula = \[\dfrac{{value}}{{totalvalue}} \times 100\]
Example, if a student gets 15 marks out of 20 then,
Percentage of the marks \[\dfrac{{15}}{{20}} \times 100 = 75\% \]
Complete step-by-step answer:
Given,
The increase in price of the oranges \[ = 30\% \]
Money in hand of the person \[ = Rs.300\]
Let us suppose the initial price of the oranges be X Rs/kg
Then amount of oranges the man can buy initially,
\[ = \dfrac{{Money{\text{ }}in{\text{ }}hand{\text{ }}of{\text{ }}the{\text{ }}person}}{{the{\text{ }}initial{\text{ }}price{\text{ }}of{\text{ }}the{\text{ }}oranges}} = \dfrac{{300}}{x}\] ----( i )
Now the percentage increase in the price of oranges \[ = 30\% \]
Now the increased price per kg yhat is the new price\[ = \dfrac{{30}}{{100}}x + x = 1.3x\]
Amount of oranges man can buy after the increase in price \[ = \dfrac{{Money{\text{ }}in{\text{ }}hand{\text{ }}of{\text{ }}the{\text{ }}person}}{{The{\text{ final }}price{\text{ }}of{\text{ }}the{\text{ }}oranges}} = \dfrac{{300}}{{1.3x}}\] -----( ii )
To find \[ \to \] the increased price per kg of oranges that is the new price = 1.3 x,
As per the question the man buys 6 kg less than initially,
So the difference between ( i ) and ( ii ) will be equal to 6 kg,
Hence, mathematically
\[\begin{gathered}
= \dfrac{{300}}{x} - \dfrac{{300}}{{1.3x}} = 6 \\
\Rightarrow \dfrac{{1.3 - 1}}{{1.3x}} = \dfrac{6}{{300}} \\
\Rightarrow \dfrac{{0.3}}{{1.3x}} = \dfrac{6}{{300}} \\
\therefore 1.3x = 15 \\
\end{gathered} \]
Hence the increased price of the oranges per kg is 15 Rupees, which is the required answer.
Thus the answer is option (B)
Note: x% of y = y% of x
Also in the above question, our assumption of x depends on how we are approaching the question. We could have also assumed the increased price to be x. So don’t confuse one question with multiple ways of solving it.
For example, 30 % of ‘x’ means \[\dfrac{{30}}{{100}}x\]
Percentage formula = \[\dfrac{{value}}{{totalvalue}} \times 100\]
Example, if a student gets 15 marks out of 20 then,
Percentage of the marks \[\dfrac{{15}}{{20}} \times 100 = 75\% \]
Complete step-by-step answer:
Given,
The increase in price of the oranges \[ = 30\% \]
Money in hand of the person \[ = Rs.300\]
Let us suppose the initial price of the oranges be X Rs/kg
Then amount of oranges the man can buy initially,
\[ = \dfrac{{Money{\text{ }}in{\text{ }}hand{\text{ }}of{\text{ }}the{\text{ }}person}}{{the{\text{ }}initial{\text{ }}price{\text{ }}of{\text{ }}the{\text{ }}oranges}} = \dfrac{{300}}{x}\] ----( i )
Now the percentage increase in the price of oranges \[ = 30\% \]
Now the increased price per kg yhat is the new price\[ = \dfrac{{30}}{{100}}x + x = 1.3x\]
Amount of oranges man can buy after the increase in price \[ = \dfrac{{Money{\text{ }}in{\text{ }}hand{\text{ }}of{\text{ }}the{\text{ }}person}}{{The{\text{ final }}price{\text{ }}of{\text{ }}the{\text{ }}oranges}} = \dfrac{{300}}{{1.3x}}\] -----( ii )
To find \[ \to \] the increased price per kg of oranges that is the new price = 1.3 x,
As per the question the man buys 6 kg less than initially,
So the difference between ( i ) and ( ii ) will be equal to 6 kg,
Hence, mathematically
\[\begin{gathered}
= \dfrac{{300}}{x} - \dfrac{{300}}{{1.3x}} = 6 \\
\Rightarrow \dfrac{{1.3 - 1}}{{1.3x}} = \dfrac{6}{{300}} \\
\Rightarrow \dfrac{{0.3}}{{1.3x}} = \dfrac{6}{{300}} \\
\therefore 1.3x = 15 \\
\end{gathered} \]
Hence the increased price of the oranges per kg is 15 Rupees, which is the required answer.
Thus the answer is option (B)
Note: x% of y = y% of x
Also in the above question, our assumption of x depends on how we are approaching the question. We could have also assumed the increased price to be x. So don’t confuse one question with multiple ways of solving it.
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