Question

A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is

• A. 90
• B. 100
• C. 80
• D. 70

Answer 1

The Correct Option is A

The correct option is (A): \(90\)

In this particular case, we know Ratio of speeds = \(\sqrt{\text{Inverse ratio of times taken}}\)

\(C_1:C_2\) = \(\sqrt{20:45}\) i.e., \(2:3\)

As the speed of Car \(C_1\) is \(60\) kmph, the speed of Car \(C_2\) is \(90\) kmph

Answer 2

Let the speed of Car 2 be \(x\) kmph and the time taken by the two cars to meet be \(t\) hours.
In t hours, Car 1 travels \(= (60 \times t) \)km 
while Car 2 travels \(= (x \times t) \)km
Given that the time taken by Car 1 to travel \(= (x \times t) \) km is \(45\) minutes.
\(\frac {x\times t}{60}= \frac {45}{60}\)

\(\frac {x\times t}{60}= \frac {3}{4}\)

\(t=\frac {180}{4x}\)        ………. (i)
Similarly, the time taken by Car 2 to travel \((60 \times t)\) km_is \(20\) minutes.
\(\frac {60\times t}{x}= \frac {20}{60}\)

\(\frac {60\times t}{x}= \frac {1}{3}\)

\(t=\frac {x}{180}\)        ……….. (ii)
From eq (i) and eq (ii),
\(\frac {180}{4x}=\frac {x}{180}\)
\(4x^2 = 180 \times 180\)
\(4x^2 = 32,400\)
\(x^2=\frac {32400}{4}\)
\(x^2=8100\)
\(x=90\) km/hr

So, the correct option is (A): \(90\) km/hr