An ambulance, an hour after starting, meets with an accident which....

And the simple method using Ratio Logic:


Ignoring the 30 minute lay-over,

driving at 3/4th Speed made him late by 3 hours

if he had driven for +90 km more at his Usual Speed before he had the accident, he would have been late by 2.5 hours


Based on his reduced Speed,

driving an Extra +90 km at reduced Speed ----------> Results in .5 hour of Late

thus, 3 hours of Late (= .5 * 6) ---------------> would require him to drive (+90) * (6) = 540 km

Distance from accident to Destination in Scenario 1 = 540 km


(2nd) Find the Usual Speed R

(Time at Reduced Speed) - (Time at Usual Speed) = 3 hours

(540 / (3R/4)) - (540 / R) = 3

(720/R) - (540/R) = 3

(180) / (R) = 3

R = (180) / (3) = 60 km per hour = Usual, pre-accident Speed

He Drives 1 hour at 60 km per hour before the accident -----> + 60 km

+ 540 km driven after the accident
___________________________

600 km Total ------> Converted to miles = 373 miles approximately

-B-

Let the normal speed of the ambulance be 's', the time it would have taken had it not met with the accident and been able to travel the entire distance at its normal speed be 't' and the distance it travels be 'd'. Then, t=(d/s).
Now, had the accident occurred 90kms further along the route the ambulance could have traveled 90kms more at its normal speed and 0.5 hrs could thus have been saved. So we can conclude that the difference between the times for the ambulance to cover 90kms at 's' km/hr and 90 kms at (0.75)s km/hr is 0.5 hrs. So, 90/(0.75)s - 90/s = 0.5. So, 's'=60 km/h.
Now, as per the first scenario, the ambulance has the accident 1 hr after starting, there is an half hour delay and it travels at three-fourth of its normal speed the rest of the trip and thus takes 3.5 hrs more to reach its destination. So before the accident it travels 60 kms in 1 hr and for the rest of the trip (d-60)kms it travels at 45 km/hr. We can thus formulate the following equation summing up the different time segments of the trip:
1hr + 0.5hr (delay) + {(d-60 kms)/45}hrs = (t + 3.5)hrs. But t=d/s=d/60
Therefore, (d-60)/45=t+2. D=600 kms= 372.8 miles. Ans B.

A man goes to the fair with his and dog. Unfortunately man misses his son which he realizes 20 minutes later. The son comes back towards his home at the speed of 20 m/min and man follows him at 40 m/min. The dog runs to son and comes back to the man to show him the direction of his son. He keeps moving to and fro at 60 m/min between son and father, till the man meets the son. What is the distance traveled by the dog in the direction of the son? ans 1000 m
Detail solution please

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(1st) take the 1/2 hour stop out of the Late Times because the Stop Time does NOT Contribute to the Distance and Find the Usual Pre-Accident Speed, which we will call R

Focusing just on the 90km Difference in each Scenario:

Scenario 1:

---------------{A} ----------------Rest @(3/4)R----------------> 3 hours Late


Scenario 2:

-----------------------------------{A - 90 km later} -------------Rest @ (3/4)R-----------> 2.5 hours Late


Logic:

in Scenario 2, he was driving an Extra 90 km @ Usual Speed R ------> and was only 2.5 hours late

in Scenario 1, during this SAME 90 km stretch, because the accident occurred earlier, he was driving @ (3/4)R ------> and was 3 hours late


Over this Constant Distance of 90 km, the Speed is Inversely Proportional to the Time taken

Speed driven over this 90 km in Scenario 2 is R --------> in Scenario 1 it is (3/4)R, thus Speed DECREASED by (1/4)

therefore, the Time taken to cover the 90 km from Scenario 2 --------> to Scenario 1 INCREASED by (1/3)


because the accident occurred 90km EARLIER in Scenario 1, the extra time he was Late by in Scenario 1 was:

(3 hours) - (2.5 hours) = 1/2 hour



in Summary, over this 90 km Distance:

When the Speed DECREASED by (1/4) ---------> the Time INCREASED by (1/3) and this Increase in Time corresponds to an Extra + (1/2) hour of Lateness

Let the Usual Time needed to cover 90km The Usual Speed of R = T

T + (1/3)T = T + 1/2 hour

T = (3/2) hour


thus, driving @ Usual, pre-accident Speed R, he covers 90 km in (3/2) hours

R = (90 km) / (3/2 hours) = 60 km/hr

which means

(3/4)R = 45 km/hr = Reduced Speed after the Accident



(2nd) Using the Regular Speed R = 60 and the Reduced Speed = 45 ------> find the Total Distance of the Trip

First, just focus on the portion AFTER the Accident took place in Scenario 1 above.

in Scenario 1, call the Distance from the Accident {A} to the Destination = D

had he NOT got into an accident, he would have covered this D driving @ 60 km/hr without being Late. Call the Usual "on-time" Time = T

because he actually DID get into this accident, he covered this D driving a Reduced Speed of 45 km/hr and was + 3 hours Late (Removing the 1/2 hour lay-over which does NOT Contribute to the Distance)


Time taken 45 km/hr = (Time @ 60 km/hr) + 3 more hours of "lateness"

Time = Distance/Speed

D/45 = D/60 + 3 ------------- where D is in km and the Time is in hours

(D/45) - (D/60) = 3

(60D - 45D) / (2,700) = 3

(15) / (2,700) * D = 3

D = (3 * 2,700) / (15)

D = 540 km = the Distance form the point of the Accident to the Final Destination


There was ALSO 1 hour that he drove before he had the accident
he drove for 1 hour at his Usual, pre-accident Speed = 60 km/hr


Total Distance = 600 km


Finally, need to convert to miles-------> 1 km = .6213 mile

we can approximate:

(600 km) * (.62) = approximately 372 miles (which is estimated a LITTLE Under the Actual Value)

-B-
373 Miles

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