Question

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) \[24\] days
(b) \[16\] days
(c) \[18\] days
(d) \[20\] days

Answer 1

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 \[x\] days he has Rs. \[360\] for his expenses.
So for 1 day he has Rs. \[\dfrac{360}{x}\] for his expenses\[..............(1)\]
Now if he extends his tour for 4 days then total days of the tour turns out to be \[x+4\] 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,
\[\dfrac{360}{x+4}+3=\dfrac{360}{x}\]
By bringing all the terms with variable on one side we get,
\[\Rightarrow \dfrac{360}{x}-\dfrac{360}{x+4}=3\]
Now we will take the L.C.M so we get,
\[\Rightarrow \dfrac{360(x+4)-360x}{x(x+4)}=3\]
Now we will do cross multiplication and we get,
\[\Rightarrow 360(x+4)-360x=3\,x(x+4)\]
\[\Rightarrow 360x+360\times 4-360x=3{{x}^{2}}+12x\]
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,
\[\begin{align}
  & \Rightarrow {{x}^{2}}-20x+24x-480=0 \\
 & \Rightarrow x(x-20)+24(x-20)=0 \\
 & \Rightarrow (x+24)\,(x-20)=0 \\
 & \Rightarrow x=-24,\,20 \\
\end{align}\]
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\[a{{x}^{2}}+bx+c=0\].