Question

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

Answer 1

Hint: First we have to let the numbers of days and the total expense. Use the quadratic equation method to solve the question. Let the number of days of the tour is x days and he extends his tour by 3 days, so the new days of the tour are $x + 3$ and if we let the expense of his daily life is y then the daily expense becomes $y - 70$.

Complete step-by-step answer:
The total expense for a tour is = 4200
Let the number of days for tour = x days
Let the daily expense is = y Rs.
So we have the equation
$xy = 4200$ ……………………….. (1)
Now if we extend the days of the tour and cut the daily expense we have an equation
Days = x+3
Expense = y-70
Equation
$(x + 3)(y - 70) = 4200$
After simplifying the equation we have
$xy + 3y - 70x - 210 = 4200$
Put the value of $xy = 4200$
$(4200 + 3y - 70x - 210) = 4200$
Solving the numbers having the same variables
$3y - 70x = 210$
Now put the value of y with the help of equation 1
$3.\dfrac{{4200}}{x} - 70x = 210$
Now dividing the whole equation by 70 we get
$3.\dfrac{{60}}{x} - x = 3$
$180 - {x^2} = 3x$
${x^2} + 3x - 180 = 0$
Now factorize the above equation we get
${x^2} + 15x - 12x - 180 = 0$
Taking the commons and make pairs
$x(x + 15) - 12(x + 15) = 0$
$(x + 15)(x - 12) = 0$
Solve the equation by taking equal to 0
$x = 12, - 15$
X can not be negative so, the value of x is 12

Hence, the original duration of the tour is 12 days.

Note: Student can solve this question by an alternative method which is used to find the value of x
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ you can use this formula for finding x in the quadratic equation $a{x^2} + bx + c = 0$.