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23 Dec 2024, 10:33
Kudos
Bookmarks
| Rate | Time | Distance | |
| Home 2 School | 15 | t1 | D |
| School 2 Home | x | t2 | D |
Avg speed = Total dist/total time = 2D/(t1+t2)
(I)
t1 = d/15
t2 = 60/100 * d/15 = d/25
total time = 8d/75
Avg speed = total dist/total time = 18.75mph
Sufficient
(II)
t1/t2 = 5/3
t2 = 3/5 t1
t1 = d/15
t2 = d/25
This is the same relation as given in statement 1
Avg speed = total dist/total time = 18.75mph
Sufficient
Option D
Bunuel
23 Dec 2024, 10:36
Kudos
Bookmarks
Avg speed = Total Distance/Total Time
Let
distance one way = d
time from home to school = t1
time from school to home = t2
(I)
t1 = d/15
t2 = 0.6*d/15 = d/25
we have time in terms of 'd' and distance in terms of 'd'
They will cancel out in the ratio
Therefor sufficient
(II)
t1/t2 = 5/3
t1 = d/15
t2 = d/25
we have time in terms of 'd' and distance in terms of 'd'
They will cancel out in the ratio
Therefore sufficient
Answer is option D
Let
distance one way = d
time from home to school = t1
time from school to home = t2
(I)
t1 = d/15
t2 = 0.6*d/15 = d/25
we have time in terms of 'd' and distance in terms of 'd'
They will cancel out in the ratio
Therefor sufficient
(II)
t1/t2 = 5/3
t1 = d/15
t2 = d/25
we have time in terms of 'd' and distance in terms of 'd'
They will cancel out in the ratio
Therefore sufficient
Answer is option D
Bunuel
23 Dec 2024, 10:37
Kudos
Bookmarks
Option D
Avg speed = Total distance / total time
statement 1 : 2d/(d/15+d/s2) from the given.
And t = d/15 , t2= 60/100 t => d/15 (60/100) = d/25 thus speed from school to home is 25m/h.
Substituting in 2d/(d/15+d/25) => 2/(1/15+1/25) = 18.75 m/h is the average speed. so sufficient.
Statement 2: d/15 = 5t and d/s2 = 3t ; t=d/75 (3) = 3d/75 = d/25 . Speed 2 is 25m/h
Substituting in the equation like before we get 18.75m/h as the average speed.
Avg speed = Total distance / total time
statement 1 : 2d/(d/15+d/s2) from the given.
And t = d/15 , t2= 60/100 t => d/15 (60/100) = d/25 thus speed from school to home is 25m/h.
Substituting in 2d/(d/15+d/25) => 2/(1/15+1/25) = 18.75 m/h is the average speed. so sufficient.
Statement 2: d/15 = 5t and d/s2 = 3t ; t=d/75 (3) = 3d/75 = d/25 . Speed 2 is 25m/h
Substituting in the equation like before we get 18.75m/h as the average speed.
