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

We use unitary method to solve this question. Assume total number of initial days as a variable and calculate the expense for one day using a unitary method. Similarly, add 3 days to the number of initial days and calculate the expense for one day using a unitary method. Form an equation where expense for one day for initial days equates the expense for one day after adding three days but we subtract the given expense for 3 days from expense for one day with initial number of days. Solve the equation by cross multiplication for initial number of days.

• Unitary methods help us to calculate the value of a single unit when we are given the value of multiple units by dividing the value of multiple units by the number of units.

Step-By-Step answer:
Let us assume the initial number of days as ‘x’.
Then we can write expense for ‘x’ days $= 4200$
We use unitary method to calculate expense for one day
$\Rightarrow$Expense for 1 day $= \dfrac{{4200}}{x}$ … (1)
Now we know number of days is increased by 3, i.e. number of days $= x + 3$
Then we can write expense for $x + 3$ days $= 4200$
We use unitary method to calculate expense for one day
$\Rightarrow$Expense for 1 day $= \dfrac{{4200}}{{x + 3}}$ … (2)
Now we know he has to cut down his daily expenses by Rs.70 if he has to extend his trip for 3 days.
We cut the amount Rs.70 from expenses per day where the number of days is the initial number of days.
$\Rightarrow \dfrac{{4200}}{x} - 70 = \dfrac{{4200}}{{x + 3}}$
Take LCM on LHS of the equation
$\Rightarrow \dfrac{{4200 - 70x}}{x} = \dfrac{{4200}}{{x + 3}}$
$\Rightarrow \dfrac{{70(60 - x)}}{x} = \dfrac{{4200}}{{x + 3}}$
Cancel same terms from numerator of both sides of the equation
$\Rightarrow \dfrac{{60 - x}}{x} = \dfrac{{60}}{{x + 3}}$
Cross multiply both sides of the equation
$\Rightarrow (60 - x)(x + 3) = 60x$
Form quadratic equation in LHS by multiplying the brackets
$\Rightarrow 60x + 180 - {x^2} - 3x = 60x$
Cancel same terms from both sides of the equation
$\Rightarrow 180 - {x^2} - 3x = 0$
Multiply both sides by -1
$\Rightarrow {x^2} + 3x - 180 = 0$
Use factorization method to form factors of quadratic equation
$\Rightarrow {x^2} + 15x - 12x - 180 = 0$
Take x common from first two terms and -12 common from last two terms
$\Rightarrow x(x + 15) - 12(x + 15) = 0$
Combine the factors
$\Rightarrow (x - 12)(x + 15) = 0$
Equate both factors to 0
$\Rightarrow x = 12;x = - 15$
Since x is initial number of days, so it cannot be negative.
Initial number of days is 12

$\therefore$Original duration of tour is 12 days.

Note: Many students make the mistake of forming the equation in terms of number of days which is wrong, since we are given fixed expenses for fixed number of days we calculate expense for 1 day, we don’t calculate days for Rs.1 expense.