GATE Chemical 2020 | GA | Question: 8

Question

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?

• A. $66.6$
• B. $75.2$
• C. $88.2$
• D. $116.5$

Answer 1

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 $(B).$

Comments

75/60 i am not geeting why div by 60 plz help anyone

---

Convert minutes to hours.

---

it is for minutes to hours conversion.

LHS is in Hours and RHS is in minutes.

so we'll divide RHS by 60 to convert it into hours

Answer 2

Total distance between P and Q = 233 Km.

Now distance = speed x time

At the point they meet the sum of the distance travelled by both will be 233.

233 = (speed of Q x time taken by Q) + (speed of P x time taken by P) ............................ (1)

 

Now let speed of Q be  q km/h

Speed of P = (q + 10) km/hr

time taken by Q to reach the point = 75 mins =1.25 hours

Time taken by P to reach = 1+ (time taken by Q) as it started 1 hour earlier = 2.25 hours

Substituting in  (1)

233 = (q x 1.25) + ((q+10) x 2.25)

solving q = 60.14 Km/hr

distance travelled by Q = Speed x time = 60.14 x 1.25 = 75.2 km 

(B) is the correct answer

Answer 3

Given:

Distance between Delhi and Agra = 233 km

Car P starts first from Delhi.

Car Q starts 1 hour later from Agra.

The cars meet 75 minutes = 1.25 hours after Q starts.

Speed of car P is 10 km/h more than speed of car Q.

 

Solution:

Time travelled by car Q = 1.25 hours

Time travelled by car P = 1 + 1.25 = 2.25 hours

Let the speed of car Q be x km/h.

Speed of car P = (x + 10) km/h

Distance travelled by car Q:
\[
1.25x
\]

Distance travelled by car P:
\[
2.25(x + 10)
\]

Since total distance is 233 km,
\[
1.25x + 2.25(x + 10) = 233
\]

\[
1.25x + 2.25x + 22.5 = 233
\]

\[
3.5x = 210.5
\]

\[
x = 60.14
\]

Distance travelled by car Q:
\[
1.25 \times 60.14 = 75.2 \text{ km}
\]

Answer: 75.2 km