Question

A car travels from P to Q at a constant speed. If the speed were increased by $10$ km/hr, it would have taken $1$ hour less to cover the distance. It would have taken further $45$ minutes lesser if the speed was further increased by $10$ km/hr. What is the distance between the two cities?

Answer 1

Hint: Let the time for traveling at a constant speed from P to Q be \[\dfrac{{\text{x}}}{{\text{y}}}\] where x be the distance and y be the speed. Now if the speed was increased by $10$ km/hr then the time will be $\dfrac{{\text{x}}}{{{\text{y + 10}}}}$ and since it would have taken $1$ hour less then \[\dfrac{{\text{x}}}{{\text{y}}} - \dfrac{{\text{x}}}{{{\text{y + 10}}}} = 1\]. Simplify it and form an equation. Form another equation by following the same process for when speed is increased further by $10$ km/hr. Solve the equations to find the value of x.

Complete step-by-step answer:
Given, a car travels from city P to Q at a constant speed. Let the distance be x m and speed be y km/hr. Then the time taken will be=$\dfrac{{{\text{distance}}}}{{{\text{speed}}}} = \dfrac{{\text{x}}}{{\text{y}}}$
Now if the speed was increased by $10$ km/hr, it. So the time taken will be $\dfrac{{\text{x}}}{{{\text{y + 10}}}}$.Since it would have taken $1$ hour lesser to cover the distance, then
$ \Rightarrow $ $\dfrac{{\text{x}}}{{\text{y}}} - 1 = \dfrac{{\text{x}}}{{{\text{y + 10}}}}$ $ \Rightarrow \dfrac{{\text{x}}}{{\text{y}}} - \dfrac{{\text{x}}}{{{\text{y + 10}}}} = 1$
On simplifying the equation we get,
$ \Rightarrow \dfrac{{{\text{xy + 10x - xy}}}}{{{\text{y(y + 10)}}}} = 1 \Rightarrow 10{\text{x = }}{{\text{y}}^2} + 10{\text{y}}$
$ \Rightarrow {{\text{y}}^2}{\text{ + 10y - 10x = 0}}$ ---- (i)
Now if the speed is further increased by $10$ km/hr then the total speed will be ${\text{y + 20}}$ km/hr. The time will be further decreased by $45$ minutes, so the lesser time will be $1{\text{hr45min}}$.
$1{\text{hr45min}}$$ = 1 + \dfrac{{45}}{{60}} = 1 + \dfrac{9}{{12}} = \dfrac{{21}}{{12}} = \dfrac{7}{4}{\text{hrs}}$
So then according to question,
$ \Rightarrow \dfrac{{\text{x}}}{{\text{y}}} - \dfrac{{\text{x}}}{{{\text{y + 20}}}} = \dfrac{7}{4} \Rightarrow \dfrac{{{\text{xy + 20x - xy}}}}{{{{\text{y}}^2}{\text{ + 20y}}}} = \dfrac{7}{4}$
On simplifying we get,
$ \Rightarrow 80{\text{x = 7}}{{\text{y}}^2}{\text{ + 140y}} \Rightarrow {\text{7}}{{\text{y}}^2}{\text{ + 140y - }}80{\text{x = 0}}$ --- (ii)
On putting the value of x from eq. (i) into (ii) we get,
$ \Rightarrow {\text{7}}{{\text{y}}^2}{\text{ + 140y - }}80\left( {\dfrac{{{{\text{y}}^2}{\text{ + 10y}}}}{{10}}} \right){\text{ = 0}}$
$ \Rightarrow {\text{7}}{{\text{y}}^2}{\text{ + 140y - }}8{{\text{y}}^2}{\text{ - 80y = 0}}$
$ \Rightarrow - {{\text{y}}^2}{\text{ + 60y - = 0}} \Rightarrow {{\text{y}}^2}{\text{ = 60y}}$
$ \Rightarrow $ ${\text{y = 60}}$
On putting the value of y in eq.(i) we get,
$ \Rightarrow {60^2}{\text{ + 10}} \times {\text{60 - 10x = 0}} \Rightarrow {\text{3600 + 600 - 10x = 0}}$
$ \Rightarrow 10{\text{x = 4200}} \Rightarrow {\text{x = }}\dfrac{{4200}}{{10}} = 420$ Km
Hence the distance between the 2 cities is $420$ km.

Note: The question can also be solved by this method-
We know that distance (d) = speed(s) × time (t) and we have to find the distance (d) in the given question. Since when speed (s) is increased by 10 the time (t) is decreased by 1 hour. Now in the second statement it says that when speed is increased by 20 the time is decreased by 45 minutes which can be written as 7/4hours, then we can write the coefficients of s and t as

We can cross multiply the coefficients of s and t to form the equation. So the equations will be $ - {\text{s + 10t = 10 and - }}\dfrac{7}{4}{\text{s + 20t = 35}}$. On solving these eq., we get ${\text{t = 7hrs and s = 60km/hr}}$.Now put the values in the formula of distance and you’ll get the answer.