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for car 1 ; for the first 20 km it must have covered in 20/100 ; .2 hrs or say 12 mins
given the total time taken for both cars for 150 km distance = ( 50/100+ 50/50 + 50/25) ; 350/100 ; 3.5 hours
car 2 is lagging by 12 mins when car 1 reaches point B
so car 2 must be at a distance of 12*25/60 = 5 km
OPTION C


Bunuel
Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. What is the distance, in km, between car 2 and B when car 1 reaches B ?

A. 3
B. 4
C. 5
D. 6
E. 7


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Bunuel
Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. What is the distance, in km, between car 2 and B when car 1 reaches B ?

A. 3
B. 4
C. 5
D. 6
E. 7


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A----------50km @100kmph in 30mins---------------P-----------50km @50kmph in 1 hours------------Q-----------50km @25kmph in 2 hours----------B

Total time taken to reach B from A = 3 hours 30 mins

Time taken by car 1 to reach B = 18 mins + 1 hour + 2 hours = 3 hours 18 mins

Distance travelled by car 3 in last 12 mins = 12/60 * 25 = 5 km

IMO C
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When car 1 hits first milestone and slows to 50, car 2 is 20miles back
20=100t
t=1/5

car 1 goes 50*1/5 = 10 miles --> car 2 is now 10 miles back

When car 1 hits 2nd miltestone and slows to 25, car 2 is 10miles back
10=50t
t=1/5

25*1/5 = 5 --> car 2 is now only 5 miles back
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