Question

Two cities P and Q are 110 km apart. Person A started from city P at 7 a.m. at the speed of 20 km/hr and B started from city Q at 8 a.m. at the speed of 25 km/hr, they meet at
A. 9 a.m.
B. 10 a.m.
C. 12 a.m.
D. 11 a.m.

Answer 1

Hint: Let us denote the meeting time as x from 7:00 am. We know that $\text{Distance=Speed}\times \text{Time}$
Distance covered by A starting from P in x hours $=20x\text{ km/h}$ and the distance covered by B starting from Q in $\left( x-1 \right)$ hours $=25\left( x-1 \right)\text{ km/h}$ . We are also given that the total distance $=110\text{ km}$ $\Rightarrow 20x+25\left( x-1 \right)=110$ . The required time will be $7:00\text{ am}+x$ .

Complete step-by-step answer:
We have to find the time at which A and B meet. Let us denote the meeting time as x from 7:00 am.

We are given that the speed of A from P \[=20\text{ km/h}\]
We know that $\text{Distance=Speed}\times \text{Time}$
Hence, distance covered by A starting from P in x hours $=20x\text{ km/h}$
We are given that the speed of B from Q \[=25\text{ km/h}\]
Hence, distance covered by B starting from Q in $\left( x-1 \right)$ hours $=25\left( x-1 \right)\text{ km/h}$
We are also given that the total distance $=110\text{ km}$ .
$\Rightarrow 20x+25\left( x-1 \right)=110$
Let us solve this equation to find the value of x.
$\Rightarrow 20x+25x-25=110$
Let us collect constants in RHS and variables in the LHS.
$\begin{align}
  & \Rightarrow 20x+25x=110+25 \\
 & \Rightarrow 45x=135 \\
\end{align}$
Let us find the value of x by taking 45 to RHS.
$x=\dfrac{135}{45}=3$
We can now find the meeting time. We considered x from 7:00 am. Hence,
The required time is $7+3=10:00\text{ am}$

So, the correct answer is “Option B”.

Note: We took $\left( x-1 \right)$ since we considered x from 7:00 am and since B is starting at 8:00 am and that at 7:00 am are of 1 hour difference. If we have considered x from 8:00, this solution can be solved in the following way:
Distance covered by A starting from P in $\left( x+1 \right)$ hours $=20\left( x+1 \right)\text{ km/h}$
Distance covered by B starting from Q in $x$ hours $=25x\text{ km/h}$
Now, we can find the total distance as
$\begin{align}
  & 20\left( x+1 \right)+25x=110 \\
 & \Rightarrow 20x+20+25x=110 \\
\end{align}$
$\begin{align}
  & \Rightarrow 45x=110-20 \\
 & \Rightarrow 45x=90 \\
 & \Rightarrow x=\dfrac{90}{45}=2 \\
\end{align}$
Hence, the required time is $2+8:00\text{ am}=10:00\text{ am}$