Question

A businessman knows that the price of a commodity will increase by Rs.5 per packet. He bought some packets of this commodity for Rs.4500. If he bought this packet at the new price, then he gets 10 packets less. What is the number of packets bought by him?
A. 100
B. 72
C. 84
D. 97

Answer 1

Hint: First analyze what is given in the question. Total money spent on the commodity is provided using which we can form an algebraic equation. This helps us in finding the price of one commodity. Using the conditions above that is the per packet increase in price and the number of commodities brought less on new price proceed further.

Complete step by step answer:
Let the number of packets purchased by the businessman be \[x\].
Price of \[x\] packets = Rs. 4500
So, the price of 1 packet \[ = {\text{Rs}}{\text{.}}\dfrac{{4500}}{x}\]
Price of a commodity increase per packet = Rs.5
So, the new price of 1 packet \[ = {\text{Rs}}{\text{.}}\dfrac{{4500}}{x} + 5\]
According to question, He gets 10 packets less
So, number of packets \[ = x - 10\]
Price of \[\left( {x - 10} \right)\] packets \[ = \left( {x - 10} \right)\left( {\dfrac{{4500}}{x} + 5} \right)\]
According to question
Price of \[\left( {x - 10} \right)\] packets = 4500
\[
   \Rightarrow \left( {x - 10} \right)\left( {\dfrac{{4500}}{x} + 5} \right) = 4500 \\
   \Rightarrow 4500 - \dfrac{{45000}}{x} + 5x - 50 = 4500 \\
\]
Multiplying by \[x\] on both sides, we have
\[
   \Rightarrow 4500x + 5{x^2} - 45000 - 50x = 4500x \\
   \Rightarrow 5{x^2} - 45000 - 50x = 0 \\
\]
Divide by 5 on both side
\[ \Rightarrow {x^2} - 10x - 9000 = 0\]
We know that, the roots of the quadratic equation \[a{x^2} + bx + c = 0\] is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
So, the roots of the formed equation are given by
\[
   \Rightarrow x = \dfrac{{ - \left( { - 10} \right) \pm \sqrt {{{\left( { - 10} \right)}^2} - 4 \times 1 \times \left( { - 9000} \right)} }}{{2 \times 1}} \\
   \Rightarrow x = \dfrac{{10 \pm \sqrt {100 + 36000} }}{2} \\
   \Rightarrow x = \dfrac{{10 \pm \sqrt {36100} }}{2} = \dfrac{{10 \pm 190}}{2} \\
   \Rightarrow x = \dfrac{{10 + 190}}{2},\dfrac{{10 - 190}}{2} \\
  \therefore x = 100, - 90 \\
\]
Since \[x\] can’t be negative, \[x = 100\].
So, Number of packets = 100

So, the correct answer is “Option A”.

Note: The other method of solving the equation is hit and trial method. The roots of the quadratic equation \[a{x^2} + bx + c = 0\] is given by \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]. Although we have obtained two values of \[x\], we have taken \[x = 100\] as the number of packets is always a positive number.