A bus travels from A to B at a constant speed, at which it should reac

Question

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

Are You Up For the Challenge: 700 Level Questions

A. 20 km
B. 40 km
C. 60 km
D. 90 km
E. 120 km

Are You Up For the Challenge: 700 Level Questions

nick1816 · Accepted Answer

Distance between A and B= D

Original speed = v

Because bus travelled (48-30=) 18Km at the 4/5th of the original speed, it reached destination (x+45-x-36=) 9mins late

\frac{18}{(0.8v)} - \frac{18}{v} = \frac{9}{60}

\frac{18}{4v} = \frac{9}{60}

v= 30Km/h

Also, Because bus travelled D-30 Kms at the 4/5th of the original speed, it reached destination (x+45-x) 45 mins late.

\frac{D-30}{24} -\frac{D-30}{30} = \frac{45}{60}

\frac{D-30}{120} = \frac{3}{4}

D=120Kms

freedom128 · Answer

Distance between A and B= D km

At 80% speed,
(D-30) km causes additional 45 mins
(D-48) km causes additional 36 mins

(D-30)/(D-48) = 45/36

If D=.....
A. 20 km --> (D-30)<0 (NO!)
B. 40 km --> (D-48)<0 (NO!)
C. 60 km --> 30/12 = 45/18 (NO!)
D. 90 km --> 60/42 = 45/31. 5 (NO!)
E. 120 km --> 90/72 = 45/36 (Yes!)

FINAL ANSWER IS (E)

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SameerGupta2001 · Answer

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

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unraveled · Answer

You 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:
 
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.

monikakumar · Answer

d =original dist
x = original time
when it traveled 30 km, it took x+45min,
when it travelled 48 km, it took x+36 min,
so differnce of distance 18km, it has taken extra 9 minutes 
so speed = 150km/hr
the extra time in x+45 min is due to this reduction in speed, which is 120km/hr
so distance covered in 120km/hr speed is, d1=120*45/60
d1=90km
so original full distance is 90+30=120km
D

Shobhit7 · Answer

Considering both the case, we can deduce that-
For 18 extra KM (30 to 48) on 4/5th speed, increase in time = 9 min (45 min -36 min)
Therefore, for every 1 km on 4/5th speed, increase in time= 0.5 min

Now, in first case, extra time = 45 min, so distance traveled during this time= 45*0.5= 90KM
Total distance= 30+90= 120 KM

Similarly, in second case, extra time= 36 min, so distance traveled during this time= 36*0.5= 72KM
Total distance= 48+72= 120 KM

Ans E (120 KM)

Fdambro294 · Answer

If the accident happened 18 km later (48 - 30), then the Time saved would have been 9 minutes (X + 45 - X + 36)

The difference in speed over this 18 km is what caused the extra 9 minutes of lateness

From hypothetical scenario 2 to scenario 1:

18 more miles at reduced speed ————> results in extra +9 limits of lateness (x + 45 -x+ 36)

How many miles would have to be traveled at the reduced speed to be +45 minutes late in scenario 1?

+9 min. ————-> 18 km at reduced Speed

* 5...............................* 5
____________________

+45 min late would require ———> (18) (5) = 90 miles driven at reduced speed

First 30 miles driven at normal speed

+

Last 90 miles driven at (4/5)th speed
_______________________________
Total of 120 miles to be 45 minutes late

(E)

Posted from my mobile device

ramlala · Answer

2 conditions.
After 30kms...time x + 45 minutes
After 48kms...time x + 36 minutes

Difference 18 kms....time 9 minutes
So 45 minutes (9 * 5).....18 * 5 = 90

+30 = 90 + 30 =  120 kms  total distance.

Bambi2021 · Answer

48 - 30 km gives an extra 9 min, i.e. 18 km gives an extra 9 min. The delay results in an extra 0,5 min per 1 km.

After 30 km malfunction we have 45 min extra, which is 0,5 min per km left to drive. So the distance left is 45*2 km = 90 km.

30 + 90 = 120 km

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A bus travels from A to B at a constant speed, at which it should reac

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Prep Toolkit

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