To arrive at its destination on time the bus should have maintained a speed of vv kilometers per hour throughout the journey. Instead, after going the first third of the distance at vv kilometers per hour, the bus increased its speed and went the rest of the distance at 1.2v1.2v kilometers per hour. As a result, the bus arrived at its destination xx minutes earlier than planned. What was the actual duration of the trip?

The answer to the question is

The actual duration of the trip is 8/9 times the normal duration or the bus arrived 1/9 times earlier than it would have at a constant speed of v kph

Step-by-step explanation:

Speed of bus in the first third distance = v kph

Speed of bus for the remaining two thirds = 1.2 v kph

To solve the question, we note that speed = distance/time

therefore time = distance/speed

Let the distance of the destination be D, then

Normal duration at speed v = D/v

Actual duration (D/3)÷v + (2D/3)÷(1.2v) ⇒ (1.2*D/3 + 2D/3)1.2v

= = (8/9)×D/v

Therefore the actual duration is 8/9 times the normal duration

11 feet 6 inches and 8 feet 6 inches

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent