Kevin drove from A to B at a constant speed of 60 mph. Once he reache

This is a straightforward problem if we carefully consider the concept of rate / time problems type. and answer is indeed , B


so, here we go , the driver went from point A to B in the CONSTANT speed of 60 mph , the simple concept here is , 60 mph / 60 min =1 it means to cover every mile

, the driver has to spend one minute , so keep it in the mind,

Next : the problem says that Exactly 4 hours before the end of the trip , drive was 15 miles away point B and approaching to the point B , it means , 4 hours was the time to cover

15 miles to cover 15 miles toward the point B and also cover the distance from reverse direction from point B to the point A . already we know that 15 miles with the constant speed of 60 mph

, takes 15 minutes to cover , so if we deduct 15 minutes form 4 hours , we can get the time required to get from point B to the point A WITH 80 mph.

so , we have : 4 hours - 15 minutes = 3 hours and 45 minutes

here we have time required to travel from point B to point A , and we are given the rate of travel 80 mph , so we can get the distance between point B to the point A ,


3.45 h = 3 hours and 45 minutes , 80 mph for 3 hours is : 80 *3 =240 miles and 45 minutes = 45/60 = 3/4 hours then , 80 * 3/4 = 60 miles


so, Total distance = 240+ 60 = 300 miles..... ANSWER : B

regards,

How did you solve this equation to get X = 300 ?

MAGOOSH OFFICIAL SOLUTION:

In the last 15 miles of his approach to B, Kevin was traveling at 60 mph, so he traveled that distance in ¼ hr, or 15 minutes. That means, when he arrived at B, 15 minutes had elapsed, and he took (4 hr) – (15 min) = 3.75 hr to drive the distance D at 80 mph. It will be easier to leave that time in the form (4 hr) – (15 min).

D = RT = (80 mph)[ (4 hr) – (15 min)] = 320 mi – 20 mi = 300 mi

Answer = (B)

Answer = B =300

Refer distance diagram below:



Say the distance between A & B = d

Given that

Time required for 15 miles (From A to B) + Break (x minutes) + Time required for d miles (from B to A) = 4 hours (Shown in double line in the diagram)

\(\frac{15}{60} + x + \frac{d}{80} = 4\)

x = Time consumed pausing/breaking at point B. This is already included in 4 hours. Equation setup showing prominence of x.

\(d = \frac{15}{4} * 80 = 300\)

let distance be d.

d/60 = t1 --- (1)
d/80 = t2 --- (2)
(d-15)/60 = t1+t2-4 --- (3)

adding 1 & 2 and substituting in 3

d = 300

How did we reach to this?

In the last 15 miles of his approach to B, Kevin was traveling at 60 mph, so he traveled that distance in ¼ hr, or 15 minutes.

ThxQ

"Once he reached B, he turned right around with pause". IS this statement right?

Should it not be "without" pause?

the question is worded incorrectly. it says that there will be a pause.
if we get 15 min left to travel to B, then the remaining distance will travel at distance of 80mph for 3h45m. if 300 is the distance, then he MUST have turned right away to A (thus no PAUSE). because of this, B is incorrect, and the only correct answer that can be is A.
are you sure it is not given that there is no pause?

let d=distance between A and B
round trip time=d/60+d/80=7d/240
7d/240-4=d/60-1/4
d=300 miles

A|---------------------------C|------15mi-----|B

It's given that exactly 4 hours before the end of the trip, kevin was 15 miles was away from B.
It means that From point C till end of the journey(return from B to A took 4 hours)

So total time T(C-B)+T(Return journey B-A)=4 hours
Since they are moving at constant speeds
So
15/60+T(return)=4 hrs
or T return=4-1/4=15/4 hrs.

Now,Return speed=80 mph
So Distance=speed*time=80*15/4=300 mi

I also think a mistake was made in the question.

Kevin drove from A to B at a constant speed of 60 mph. Once he reached B, he turned right around with NO pause, and returned to A at a constant speed of 80 mph. Exactly 4 hours before the end of his trip, he was still approaching B, only 15 miles away from it. What is the distance between A and B?

In that case:

\(\frac{15}{60}\) + \(\frac{d}{80}\) = 4
\(\frac{d}{80}\) = 4 - \(\frac{1}{4}\)
d = \(\frac{15 * 80}{4}\) = 300

Do you agree ?

hi bunuel,

What does "he turned around with pause" mean?...this has got me confused. As i can see you have solved by taking into consideration that Kevin turned around and started driving immediately without pause. Please clarify?

Refer attached diagram. Entire distance that Kevin has to cover is in blue line.

The diagram clarifies the exact location of "15 miles".

When Kevin was approaching B, his speed was 60 mph.

Time taken by Kevin in covering those 15 miles, before he touched B: [15 miles / 60mph] = (1/4) hours.

(Note: In order to cover the entire distance shown in the diagram, Kevin takes 4 hours)

His trip would end when he reaches back A.

So

(entire time taken from travelling back from B to A) + (1/4) hrs = 4 hours

Time taken in return journey from B to A = 4 - (1/4) = 3 hours + (3/4) hours
Speed: 80 Mph

80 * 3 + 80 * (3/4) = 240 +60 = 300 Km.

Answer: B.

In the question it is mentioned that Kevin turned around with pause.
I thought that some time T should also account for that pause in the total journey. So the answer that I came up with
1/4 + Pause(p) + D/80 = 4
D= 20[15-4p]
So max value of dist = 300 when p=0 but it is given that there was a pause, so the ans should be <300 which is only statement 1

average speed for kevin for the whole journey = 2*60*80/140=6*80/7 =480/7
in 4 hours he should cover (480/7)*4 miles ,which is ~275 miles
from answer choices -let the total distance be 300 miles ,
distance travelled in 4 hours = 300 -15 =275 miles
therefore, B

60 mph means 1 mile per minute.
If, 4 hours before the end of the trip, he's 15 miles from B, then he'll arrive at B in 15 minutes.
That means that, on his way back, he'll travel 3.75 hours at an average speed of 80 mph. D = AS * T -> D = 3.75 * 80 = 300 miles.

The total mileages he travel within 4hours before the end of the trip is 15 + AB.
We have : 60/15 + AB/80 = 4 => AB = 300 miles