Question

A person on tour has Rs. 4200 for his expenses. On his tour for 3 days, he cut down his daily expenses by Rs. 70. Find the original duration of the tour.

Answer 1

Hint: To solve this question, we will start with assuming the original duration of tour be x. So, we will calculate his expenses for the original tour and then his expenses for his tour after adding three days onto it. Now, he cut his expenses by equating his expenses, we will get the value of x, and hence the original duration of the tour.

Complete step by step solution:
Given,
A person on tour has Rs. 4200 for his expenses his tour for 3 days,
We have been given the total money the person has for tour = 4200
Let the number of original tour days be ‘x’.
So, his expense per day
 \[ = \dfrac{{4200}}{x}\]
Now, it is given that the tour is extended to 3 days. That is \[x + 3\] days.
Then, his expense per day
 \[ = \dfrac{{4200}}{{x + 3}}\]
It is given in the question that he has cut down his daily expenses by Rs. 70.
Now, to find the value of x, on equating the above equations, we get
 \[ \Rightarrow \dfrac{{4200}}{x} - \dfrac{{4200}}{{x + 3}} = 70\]
Taking LCM and simplifying we have,
 \[\dfrac{{(x + 3)\left( {4200} \right) - 4200x}}{{x(x + 3)}} = 70\]
 \[\dfrac{{4200x + 12600 - 4200x}}{{x(x + 3)}} = 70\]
 \[\dfrac{{12600}}{{x(x + 3)}} = 70\]
 \[12600 = 70x(x + 3)\]
 \[12600 = 70{x^2} + 210x\]
 \[70{x^2} + 210x - 12600 = 0\]
 \[{x^2} + 3x - 180 = 0\]
 \[{x^2} + 15x - 12x - 180 = 0\]
 \[x(x + 15) - 12(x + 15) = 0\]
 \[(x + 15)(x - 12) = 0\]
Using zero product principle we have,
 \[ \Rightarrow (x + 15) = 0\] and \[(x - 12) = 0\]
 \[ \Rightarrow x = - 15\] and \[x = 12\] .
We know that the duration of the tour cannot be negative so we neglect \[x = - 15\] .
Hence the required answer is \[x = 12\] .
Hence the original duration of the tour is 12 days.
So, the correct answer is “12”.

Note: We convert the given word problem into an algebraic expression and then we solve for the variable. That is the unknown value. We solved the obtained quadratic equation by factorization, if factorization is not suitable we use a quadratic formula to solve this.