Directions: Read the following information to answer the question.
A passenger train running from Sitapur to Gitapur meets with an accident 50 kms. From Sitapur, after which it travels at 3/5 times its original speed and arrives 3 hours late at Gitapur. If the accident had occurred 50 kms farther, it would have been only 2 hours late.
S & G denote Sitapur and Gitapur SC = 50 km & CD = 50 km
From C, the train goes at 3/5th of its original speed, hence it should take 5/
3 of its original time to make up for the lag or 2/3 rd of the original time more,
i.e , 3 hrs. late (given). Hence total time required to go from C to G = 4.5 hrs.
Similarly, had the accident occurred 50 kms. further on, the total time to go
from D to G (exactly as above)= 3 hrs.
Thus for covering a distance of 50 kms the train requires 4.5 – 3 = 1.5 hrs.
Speed = 50/1.5 = 33.33 km/hr
Distance CG = 4.5 × (33.33) = 150 km
Thus, total distance SG = 150 + 50 = 200 km.
Directions: Read the following information to answer the question.
A passenger train running from Sitapur to Gitapur meets with an accident 50 kms. From Sitapur, after which it travels at 3/5 times its original speed and arrives 3 hours late at Gitapur. If the accident had occurred 50 kms farther, it would have been only 2 hours late.
S & G denote Sitapur and Gitapur SC = 50 km & CD = 50 km
From C, the train goes at 3/5th of its original speed, hence it should take 5/
3 of its original time to make up for the lag or 2/3 rd of the original time more,
i.e , 3 hrs. late (given). Hence total time required to go from C to G = 4.5 hrs.
Similarly, had the accident occurred 50 kms. further on, the total time to go
from D to G (exactly as above)= 3 hrs.
Thus for covering a distance of 50 kms the train requires 4.5 – 3 = 1.5 hrs.
Speed = 50/1.5 = 33.33 km/hr
Distance CG = 4.5 × (33.33) = 150 km
Thus, total distance SG = 150 + 50 = 200 km.
Directions: Read the following information to answer the question.
Let A is the point at which the
man is standing and from
there he throws ball and
moves towards the wall. After
1 second he catches the ball
which bounced off the wall, so
the distance covered by ball
will be AP + PB = 222 ⇒ 2 + X
+ X = 222.
A 2m B X P
Hence X = 110. So PA = 110 + 2 = 112m
Directions: Read the following information to answer the question.
Let the speed of bird be X miles/hr then
7/(X – 0.125X) + 7/(X + 0.05X) = 22/60
X = 39.98 miles/hr ≈ 40 miles/hr.
7/40 + 7/40 = 14/40 OR 21 minutes.
Directions: Read the following information to answer the question.
Let the usual speed of the man be x km/hr.
So, walking at the usual speed, in 1.25 hrs. he can cover 5/4 x km i.e, walking
at increased speed, he would cover (5x/4 + 1) km/hr. Now, usual time – new
time = 2, 96/x – 96/(5x/4 + 1) = 2
Solving the above equation we get, x = 12.
i.e. his usual speed = 12km/hr and his
increased speed = 16km/hr So, his usual
time = 8 hours and his new time = 6
hours
So, ratio of times = 8 : 6 = 4 : 3.