Question

Ashish goes to his friend's house which is $ 12km $ away from his house. He covers half of the distance at a speed of $ x{\text{ }}km{\text{ }}per{\text{ }}hour $ and the remaining at $ \left( {x + 2} \right)km{\text{ }}per{\text{ }}hour $ . If he takes $ 2{\text{ }}hrs{\text{ }}30{\text{ }}min $ to cover the whole distance, find the value of $ x $ .

Answer 1

Hint: As per the question, first we will convert the given time taken of the whole distance into its fractional form and then we will calculate the time for the first half by applying the time formula, i.e. $ Time = \dfrac{{Distance}}{{Speed}} $ , similarly we will calculate the time of the second half. And then we will add both the half of time taken which is equal to the whole time taken. That’s how we will solve for the $ x $ .

Complete Step By Step Answer:
Given-
Distance, $ s = 12km $
Time taken to cover the whole distance, $ t = 2hrs\,30\min = \dfrac{5}{2} $
Now, for the first half:-
 $ \because Speed = xkm.h{r^{ - 1}} $
So, time taken for first half, $ \therefore Time\,for\,1st\,half = \dfrac{{Dis\tan ce}}{{Speed}} = \dfrac{{\dfrac{{12}}{2}}}{x} = \dfrac{6}{x} $ ………….(i)
Again, in the second half:-
 $ \because Speed = x + 2km.h{r^{ - 1}} $
So, time taken for second half, $ \therefore Time\,for\,2nd\,half = \dfrac{{Dis\tan ce}}{{Speed}} = \dfrac{{\dfrac{{12}}{2}}}{{x + 2}} $ ……………(ii)
Now, we will add the time taken of both halves to make the whole time taken to cover the distance.
By adding eq(i) and eq(ii), we get:
 $
  \therefore \dfrac{6}{x} + \dfrac{6}{{x + 2}} = \dfrac{5}{2} \\
   \Rightarrow \dfrac{{6x + 12 + 6x}}{{{x^2} + 2x}} = \dfrac{5}{2} \\
  $
Now, we will do cross-multiplication to solve for $ x $ :-
 $
   \Rightarrow 2(12x + 12) = 5({x^2} + 2x) \\
   \Rightarrow 24x + 24 = 5{x^2} + 10x \\
   \Rightarrow 5{x^2} - 14x - 24 = 0 \\
   \Rightarrow (5x + 6)(x - 4) = 0 \\
  $
 $ \therefore x = - \dfrac{6}{5} $ (time can’t be negative, so $ - \dfrac{6}{5} $ can’t be the value of distance here).
And,
 $
  \therefore x - 4 = 0 \\
   \Rightarrow x = 4 \\
  $
Hence, the required speed of the first half $ = xkm.h{r^{ - 1}} = 4km.h{r^{ - 1}} $

Note:
We can also solve this question in a very short time by directly assuming both the half of times as $ {t_1}\,and\,{t_2} $ and then find both the half quickly by just applying the formula of time. And in the next line, we will add both the time which is equal to the whole time, i.e. $ {t_1} + {t_2} = t $ .