Chemical Engineering

+1 vote

The distance between Delhi and Agra is $233$ km. A car $P$ started travelling from Delhi to Agra and another car $Q$ started from Agra to Delhi along the same road $1$ hour after the car $P$ started. The two cars crossed each other $75$ minutes after the car $Q$ started. Both cars were travelling at constant speed. The speed of car $P$ was $10$ km/hr more than the speed of car $Q$. How many kilometers the car $Q$ had travelled when the cars crossed each other?

- $66.6$
- $75.2$
- $88.2$
- $116.5$

+1 vote

Best answer

It is given that the distance between Delhi and Agra is $233\;\text{km}.$

Let the speed of $Q$ be $x\;\text{km/hr}.$ Then the speed of $P$ will be $(x+10)\;\text{km/hr}.$

We know that, $\text{Speed (S)} = \dfrac{\text{Distance (D)}}{\text{Time (T)}}$

In $1$ hour, distance travelled by $P= (x+10) \times 1 = (x+10)\;\text{km}$

Now the remaining distance between $P \& Q= 233 – (x+10)\;\text{km}$

Both trains are going in the opposite direction and so the relative speed will be their sum of speeds.

$i.e., x+x+10 = (2x+10)\;\text{km/hr}$

The two cars crossed each other $75$ minutes after car $Q$ started.

So, $\dfrac{233-(x+10)}{2x+10} = \dfrac{75}{60}$

$\implies \dfrac{233-x-10}{2x+10} = \dfrac{5}{4}$

$\implies 892– 4x = 10x + 50$

$\implies 14x = 842$

$\implies x = 60.142\;\text{km/hr}$

Now the distance travelled by $Q=60.142 \times \dfrac{75}{60} = 60.142 \times \dfrac{5}{4} = 75.177 \approx 75.2\;\text{km}.$

$\therefore$ Car $Q$ had travelled $75.2\;\text{km}$, when the cars crossed each other.

So, the correct answer is $(B).$

Let the speed of $Q$ be $x\;\text{km/hr}.$ Then the speed of $P$ will be $(x+10)\;\text{km/hr}.$

We know that, $\text{Speed (S)} = \dfrac{\text{Distance (D)}}{\text{Time (T)}}$

In $1$ hour, distance travelled by $P= (x+10) \times 1 = (x+10)\;\text{km}$

Now the remaining distance between $P \& Q= 233 – (x+10)\;\text{km}$

Both trains are going in the opposite direction and so the relative speed will be their sum of speeds.

$i.e., x+x+10 = (2x+10)\;\text{km/hr}$

The two cars crossed each other $75$ minutes after car $Q$ started.

So, $\dfrac{233-(x+10)}{2x+10} = \dfrac{75}{60}$

$\implies \dfrac{233-x-10}{2x+10} = \dfrac{5}{4}$

$\implies 892– 4x = 10x + 50$

$\implies 14x = 842$

$\implies x = 60.142\;\text{km/hr}$

Now the distance travelled by $Q=60.142 \times \dfrac{75}{60} = 60.142 \times \dfrac{5}{4} = 75.177 \approx 75.2\;\text{km}.$

$\therefore$ Car $Q$ had travelled $75.2\;\text{km}$, when the cars crossed each other.

So, the correct answer is $(B).$

650 questions

12 answers

0 comments

3,092 users