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Question: A person on tour has Rs. 360 for his expenses. If he extends his tour for 4 days, he has to cut down...

A person on tour has Rs. 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by Rs. 3. Find the original number of days of the tour.
(a) 2424 days
(b) 1616 days
(c) 1818 days
(d) 2020 days

Explanation

Solution

Hint: By looking at the question we can see that the original duration of tour is not mentioned, so we will assume the original duration of tour to be x and then proceed with the other details given in the question. We will be using a unitary method to solve this question.

Complete step-by-step solution:
Let the original duration of the tour be x.
So by unitary method,
For xx days he has Rs. 360360 for his expenses.
So for 1 day he has Rs. 360x\dfrac{360}{x} for his expenses..............(1)..............(1)
Now if he extends his tour for 4 days then total days of the tour turns out to be x+4x+4 days.
So now his daily expense turns out to be Rs. \dfrac{360}{x+4}$$$$..............(2)
It is mentioned that if he extends his tour for 4 days he has to cut his daily expenses by Rs. 3. So according to this we will add Rs. 3 to equation (2) and then equate it to equation (1). We get,
360x+4+3=360x\dfrac{360}{x+4}+3=\dfrac{360}{x}
By bringing all the terms with variable on one side we get,
360x360x+4=3\Rightarrow \dfrac{360}{x}-\dfrac{360}{x+4}=3
Now we will take the L.C.M so we get,
360(x+4)360xx(x+4)=3\Rightarrow \dfrac{360(x+4)-360x}{x(x+4)}=3
Now we will do cross multiplication and we get,
360(x+4)360x=3x(x+4)\Rightarrow 360(x+4)-360x=3\,x(x+4)
360x+360×4360x=3x2+12x\Rightarrow 360x+360\times 4-360x=3{{x}^{2}}+12x
Cancelling similar terms we get a quadratic equation,
\Rightarrow 3{{x}^{2}}+12x-1440=0$$$$..............(3)
After taking out 3 which is a common factor to all the terms from equation (3) we get,
\Rightarrow {{x}^{2}}+4x-480=0$$$$..............(4)
Factoring the above equation (4) we get,

& \Rightarrow {{x}^{2}}-20x+24x-480=0 \\\ & \Rightarrow x(x-20)+24(x-20)=0 \\\ & \Rightarrow (x+24)\,(x-20)=0 \\\ & \Rightarrow x=-24,\,20 \\\ \end{aligned}$$ Answer is 20 because -24 is not possible as days cannot be negative. So option (d) is the answer. Note: Reading the question 4 to 5 times is important here. Using a unitary method is the key. In a hurry we can make mistakes in taking the L.C.M. We can also factorize by using direct formula $$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ for quadratic equations of the form$$a{{x}^{2}}+bx+c=0$$.