Q.1 Vikas has an appointment near the train station in the city at 10:30 am.To get to the city, Vikas needs to catch a bus & then a train. The bus leaves from Vikas’s stop every 10 min. from 9:00 am.The bus trip to the train station takes 25 min. From 9.30 am the train leaves for the city every 20 min. the train takes 15 min. to get to the city. If Vikas wants to arrive as close as possible to the appointment time , what is the latest bus he would catch ?
A) 9.10 am
B) 9.20 am
C) 9.30 am
D) 9.40 am
Ans. D
Given, The train leaves for the city every 20 minutes from 9:30 AM. So,
The timing of train = 9:30 AM, 9:50 AM, 10:10 AM, 10:30 AM and so on.
Train takes 15 minutes to get to the city but Vikas has an appointment at 10:30 AM. So, he will take train at 10:10 AM.
Now, the bus trip to the train station takes 25 minute. So,
10:10 AM – 25 min = 9:45 AM
Given, bus leaves from Vikas’s stop every 10 minutes from 9:00 AM. So,
The bus timing are = 9:00 AM, 9:10 AM, 9:20 AM, 9:30 AM, 9:40 AM, 9:50 AM and so on.
Thus, the latest bus which Vikas can catch is 9:40 AM.
Q.2 A bus left point for point y. Two hours later a car left x for y and arrived at y at the same time as the bus. If the car and the bus left simultaneously from the opposite ends x and y towards each other, they would meet 1.33 hours after the start. How much time for the bus to travel from x to y.
A) 3
B) 4
C) 5
D) 6
Ans. B
take d as distance btwn x-y,
car Speed=c
bus speed=b
by prob: d/c+b=80/60=4/3——(1)
Now, as both car and bus reach y at same time let time taken by car is t
Thus d/c=t and d/b=t+2
From (1)-> c+b/d=3/4 or c/d+b/d=3/4 Thus 1/t +1/t+2 =3/4
Forming eqn: 3t^2-2t-8=0
Solving: t=2 So bus time= t+2=4
Q.3 Rohan and sohan travel with uniform speed from P to Q and Q to P respectively, via the same route. Rohan starts at 5:00 a.m. while Sohan starts four hours later. They meet, on the way, at 1:10 p.m. At what time do they reach their destinations, if both of them reach at the same time?
A) 7
B) 8
C) 6:30
D) 7:30
Ans. A
Speed of Rohan=x
Speed of Sohan=y
Rohan begins from 5 am and continues for 4 hrs alone till 9 am. So, distance covered is 4x.At 9 am, Sohan begins with a speed of y.Now, after 25/6 hrs (4 hrs 10 mins), Rohan meets Sohan . So, both have covered 25/6x and 25/6y distances respectively.
Thus, total distance becomes 4x+(25/6)x+(25/6)y.
Now, it is given both reach end point at the same time. Thus, total time taken by Rohan will be 4 hours more than Sohan
This gives us the equation
{4x+(25/6)x+(25/6)y}/x = [{4x+(25/6)x+(25/6)y}/y] +4
Solving it, 5y=7x=>y/x= 7/5
Now we need to find out time taken by Rohan to travel (25/6)y distance as we know the time taken to traverse previous distances. So, it will be 25y/6x=(25/6)*(7/5)= 35/6= 5 hrs 50 mins
So, total time taken by Rohan= 4+ 4 hrs 10 mins + 5 hrs 50 mins= 14 hrs i.e. 7 pm
Q.4 Tripti leaves office at 6.00 p.m. and catches a 6.30 p.m. train that arrives in her town at 7.00 p.m. Her father leaves home to pick her up at 7.00 p.m. from the station as she gets off the train. Yesterday, Tripti left her office early and took a 6.00 p.m. train and arrived at 6.30 p.m. As her father was not there to pick her up, she started walking towards home. Her father left home at the usual time, saw his daughter walking, turned around, picked her up and drove home arriving there 10 min earlier than usual. For how long did Tripti walk before her father picked her up?
A) 10
B) 15
C) 20
D) 25
Ans. D
As her father reached home 10 minutes earlier than the usual time, he saw his daughter 5 minutes before the scheduled time and he turned back (saving 5 minutes each side).
So, he met her at 6.55 pm
i.e. she walked for 25 minutes from the station towards her home before her father picked her up
Q.5 A leaves Mumbai at 6am and reaches Bangalore at 10am. B leaves Bangalore at 8am and reaches Mumbai at 11:30am. At what time do they cross each other ?
A) 10am
B) 8:32 am
C) 8:56am
D) 9 :20 am
Ans. C
Let us assume that the distance between Mumbai and Banglore is 28kms (LCM of their time travelling so that we can make calculations easy)
Now :
speed of A is 28/4 = 7 kmph
speed of B is 28/3.5 = 8 kmph
Since A to hours earlier than B and the total distance travelled by both we be equal to the total distance between the cities.
Let us assume that they met after t hours from the time A started.
Hence,
7t+8(t-2) = 28
t= 44/15
hence t = 2 hrs and 56 mins which is 8:56 AM
