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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # A car is moving from A to B at a constant speed. After 80 km, due to a

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Math Expert V
Joined: 02 Sep 2009
Posts: 65290
A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 64% (03:21) correct 36% (03:16) wrong based on 53 sessions

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A car is moving from A to B at a constant speed. After 80 km, due to an engine malfunction, the speed of the car decreases to 4/5th of the original speed. Because of that, the car reaches B 1 hour and 24 minutes later than intended. Had the same malfunction happened after the car travelled 120 km, the car would have reached B 1 hour later than intended. What was the original speed of the car?

A. 15 Km/hr
B. 20 Km/hr
C. 25 Km/hr
D. 30 Km/hr
E. 40 Km/hr

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Math Expert V
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Re: A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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Bunuel wrote:
A car is moving from A to B at a constant speed. After 80 km, due to an engine malfunction, the speed of the car decreases to 4/5th of the original speed. Because of that, the car reaches B 1 hour and 24 minutes later than intended. Had the same malfunction happened after the bus travelled 120 km, the car would have reached B 1 hour later than intended. What was the original speed of the bus?

A. 15 Km/hr
B. 20 Km/hr
C. 25 Km/hr
D. 30 Km/hr
E. 40 Km/hr

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A quick and logical approach would be..

Quote:
After 80 km, due to an engine malfunction, the speed of the car decreases to 4/5th of the original speed. Because of that, the car reaches B 1 hour and 24 minutes later than intended. Had the same malfunction happened after the bus travelled 120 km, the car would have reached B 1 hour later than intended.

What is the difference in two scenarios...
So, if the vehicle travels 120-80 km by normal speed and not 4x/5, it should save 24 min..

Thus, $$\frac{40}{\frac{4x}{5}}-\frac{40}{x}=\frac{24}{60}........$$..

$$\frac{50}{x}-\frac{40}{x}=\frac{2}{5}........x=10*\frac{5}{2}=25$$
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Re: A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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[quote="Bunuel"]A car is moving from A to B at a constant speed. After 80 km, due to an engine malfunction, the speed of the car decreases to 4/5th of the original speed. Because of that, the car reaches B 1 hour and 24 minutes later than intended. Had the same malfunction happened after the bus travelled 120 km, the car would have reached B 1 hour later than intended. What was the original speed of the bus?

As speed reduced by 1/5; time should increase by 1/4 to cover the remaining distance

if t1 is the original time to cover the remaining distance in case 1 then, t1*1/4 = 1hr24min = 7/5 => t1=28/5 hrs

Similarly if t is the original time is t2 in case 2 then, t2 * 1/4 = 1 => t2 = 4 hrs;

Hence, the difference in distance in the two cases (=120-80) was to be covered in 28/5-4 hrs = 8/5 hrs

=> original speed = 40 / 8/5 = 25 km/hr
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Re: A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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can someone please explain this questio to me again? i am unable to understand
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Re: A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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hassan6262 wrote:
can someone please explain this questio to me again? i am unable to understand

Same here. I found hard to understand the relation of the car and the bus!
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Re: A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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ZNLove wrote:
hassan6262 wrote:
can someone please explain this questio to me again? i am unable to understand

Same here. I found hard to understand the relation of the car and the bus!

Hi ZNLove and hassan6262,
There was a small typo in the question, which has now been corrected. Hope the question will be easier to approach now.
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Re: A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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Bunuel wrote:
A car is moving from A to B at a constant speed. After 80 km, due to an engine malfunction, the speed of the car decreases to 4/5th of the original speed. Because of that, the car reaches B 1 hour and 24 minutes later than intended. Had the same malfunction happened after the car travelled 120 km, the car would have reached B 1 hour later than intended. What was the original speed of the car?

A. 15 Km/hr
B. 20 Km/hr
C. 25 Km/hr
D. 30 Km/hr
E. 40 Km/hr

Let r = the original speed of the car and t = the regular time the car reaches its destination. We can create the equations:

rt = 80 + 4r/5 * (t + 1 + 24/60 - 80/r)

and

rt = 120 + 4r/5 * (t + 1 - 120/r)

Subtracting the second equation from the first, we have:

0 = -40 + 4r/5 * (24/60 + 40/r)

40 = 4r/5 * (2/5 + 40/r)

40 = 8r/25 + 32

8 = 8r/25

r = 8 * 25/8 = 25

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Re: A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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Can someone please post an easy to understand solution to the question!
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Re: A car is moving from A to B at a constant speed. After 80 km, due to a  [#permalink]

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kushtayal wrote:
Can someone please post an easy to understand solution to the question!

This question is easy when you practice various types of them. Lets jump into it

There are 3 scenarios : 1) When the vehicle has no Engine Malfunctions, 2) When Engine Malfunction occurs after 80 Kms, 3) When the Engine Malfunctions after 120 Kms

Let us take Scenario 1. When the engine doesn't malfunction, we can assume that the car travels a distance 'D' at a rate 'R' at time 'T'. We only have variables to work with so lets just keep it that way. We get D = RT (from the formula)

From Scenario 2, note that the distance traveled by the car at the end of the trip is same as above 'D', but the rate and the time it takes to reach the destination is different. Lets say that Scenario 2 has two trips [Trip 1 + Trip 2] [NOTE: Distance of Trip 1 + Distance of Trip 2 = D (When No Engine Malfunctions Occur]

For Trip 1, the car travels 80 Km with no engine malfunctions. In this case we get D = RT --> 80 = RT1
For Trip 2, the car travels the remaining distance (D - 80) at 4/5 the original rate. So we get D-80 = 4/5(R)T2

Note that we only have a relation between T & T1 & T2 from the question as follows -> T1 + T2 = T + 1hr 24 mins;

Now we can use the above equations to perform the following (Remember the criteria -- Distance traveled Is the same SAME regardless of when the malfunction occurred)

So we have D = 80 (Trip 1 Distance) + D-80 (Trip 2 Distance)

RT = 80 + 4/5(R)T2
RT = 80 + 4/5(R)[T + 1hr 24 mins - T1]
RT = 80 + 4/5(R)[T + 1hr 24 mins - 80/R] (The 80/R is the time from Trip 1 --> Refer to equation Trip 1) ---> 1

The same procedure is followed for Trip 2, we get the equation

RT = 120 + 4/5(R)[T + 1 - 120/R] --> 2

Subtracting 2 from 1, we can isolate variable R and end up with 25 Km/hr

Hope its clear now
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B School Aspirant, 2021 Intake Re: A car is moving from A to B at a constant speed. After 80 km, due to a   [#permalink] 29 May 2020, 08:56

# A car is moving from A to B at a constant speed. After 80 km, due to a  