SameerGupta2001
Using speed-distance-time Relationship
we could easily form 2 eqn
And with initial condition given
we could use substitute these vakues in the eqn we formed and solving them accordinngly
I guess answer is E
Posted from my mobile deviceYou might have solved it but if you are looking for inputs from people do post a well written solution. That would also help others who could not solve this problem. Here's the solution:
Quote:
A bus travels from A to B at a constant speed, at which it should reach B in x minutes. After 30 km, due to an engine malfunction, the speed of the bus decreases to 4/5th of the original speed. Because of that, the bus reaches B in x + 45 minutes, instead of intended x minutes. Had the same malfunction happened after the bus travelled 48 km, it would have reached B, in x + 36 minutes, instead of intended x minutes. What is the distance between A and B?
A. 20 km
B. 40 km
C. 60 km
D. 90 km
E. 120 km
Let the distance between A and B = D
Speed of bus be = S
So, \(\frac{D}{S} = x\)
First Condition,
\(\frac{30}{S} + \frac{5}{4S}*(D-30) = \frac{D}{S} + \frac{45}{60}\) Eqn. 1
Second Condition,
\(\frac{48}{S} + \frac{5}{4S}*(D-48) = \frac{D}{S} + \frac{36}{60}\) Eqn. 2
Eqn. 1 - Eqn. 2
\(\frac{30}{S} + \frac{5}{4S}*(D-30) - \frac{48}{S} - \frac{5}{4S}*(D-48) = \frac{D}{S} + \frac{45}{60} - \frac{D}{S} - \frac{36}{60}\)
\(-\frac{18}{S} + \frac{5}{4S}* 18 = \frac{9}{60}\)
S = 30kmph
NOW,
\(\frac{30}{30} + \frac{5}{4*30}*(D-30) = \frac{D}{30} + \frac{45}{60}\)
\(\frac{(D-30)}{24} - \frac{D}{30} = \frac{3}{4} - 1\)
\(\frac{D}{24} - \frac{D}{30} = \frac{5}{4} -\frac{1}{4}\)
D = 120km
Answer E.