1. Mr A started half an hour later than usual for the marketplace. But by increasing his speed to $$\frac{3}{2}$$ times his usual speed he reached 10 minutes earlier than usual. What is his usual time for this journey?
a) 1 hours
b) 2 hours
c) 3 hours
d) 45 minutes
Explanation: Let the usual speed be S.
increased speed = $$\frac{{3{\text{S}}}}{2}$$
Let usual time taken be T
Thus, S × T = D, and also $$\frac{{3{\text{S}}}}{2}$$ × (T - 40) = D. (T - 40, Because he started 30 minutes later and reached 10 minutes earlier, means he saved 40 minutes)
$$\frac{{3{\text{S}}}}{2}$$ × (T - 40) = ST
$$\frac{{3 \times {\text{ST}}}}{2}$$ - 60S = ST
$$\frac{1}{2}$$ × ST = 60S
T = 120 minutes = 2 hours
2. Page and Plant are running on a track AB of length 10 metres. They start running simultaneously from the ends A and B respectively. The moment they reach either of the ends, they turn around and continue running. Page and Plant run with constant speeds of 2m/s and 5m/s respectively. How far from A (in metres) are they, when they meet for the 23rd time?
a) 0 meters
b) 10 meters
c) 40 meters
d) $$\frac{{60}}{7}$$ meters
Explanation: The ratio of speeds is 2 : 5. So when slower one completes 2 one way journeys (and reaches its starting point), faster one travels 5 one way journeys (and reaches the other end). So that means after the faster one has traveled 5 one way journeys, both of them have reached same end and in next 5 one-way journey of faster runner, both reach their starting position simultaneously. Now most important to observe is that FASTER one will always meet the SLOWER one EXACTLY ONCE in each of its one-way journey, except when both of them have started with the same starting point.
Once you reduce that for the 5th time, they'll meet at A, and the next 5 rounds will have 4 meetings.
Just add 5 + 4 + 5 + 4 + 5 = 23 or just go by the position of Plant after every 5 rounds : A, B, A, B, A. The cycle repeats.
So, total distance will be 0 meters.
3. Two cars start simultaneously from cities A and B, towards B and A respectively, on the same route. Once the two cars reach their destinations they turned around and move towards the other city without any loss of time. The two cars continue shuttling in this manner for exactly 20 hours. If the speed of the car starting from A is 60km/hr and the speed of the car starting from B is 40km/hr and the distance between the two cities is 120 km, find the number of times the two cars cross each other?
a) 8
b) 10
c) 12
d) 20
Explanation: Suppose both moves with the same speed 60 km/h then they will meet max 10 times. So answer will be less than 10. Thus answer will be 8.
4. Two persons A and B start simultaneously from P and Q respectively. A meets B at a distance of 60m from P. After A reaches Q and B reaches P, they turn around and start walking in opposite direction, now B meets A at a distance of 40m from Q. Find distance between P and Q?
a) 160 m
b) 100 m
c) 140 m
d) 105 m
Explanation: P_____60m___R__Xm____40m____Q
Total distance = 100 + X
Initially, A traveled 60m and B traveled (40 + X)m. Since time is same. So, Speed ∝ Distance.
$$\frac{{{{\text{S}}_{\text{a}}}}}{{{{\text{S}}_{\text{b}}}}} = \frac{{60}}{{40 + {\text{X}}}}$$
Second time,
So, now total distance traveled by A = 100 + X + 40
Total distance traveled by B = 100 + X + 60 + X
These will be in the ratio of Speed of A and B
$$\frac{{{{\text{S}}_{\text{a}}}}}{{{{\text{S}}_{\text{b}}}}} = \frac{{60}}{{40 + {\text{X}}}} = \frac{{100 + {\text{X}} + 40}}{{100 + {\text{X}} + 60 + {\text{X}}}}$$
X = 40
Total distance = 100 + X = 100 + 40 = 140 m
5. Rohan, Shikha's boyfriend, had to pick her from her home for a live concert on her 23rd birthday. The venue of the concert and Shikha's home were in opposite directions from Rohan's office. He got late because of some work at office and realised that if he goes to pick Shikha from her home, which was a 48-minute drive from his office, they would be late for the show by 16 minutes. He asked her to start from her home towards his office in an auto-rickshaw and himself started driving towards her home. Both of them started simultaneously, he picked her as soon as they met and they managed to reach the venue just in time for the concert. If Rohan drives at an average speed of 60 km/hr, find the speed (in km/hr) of the auto-rickshaw.
a) 16 km
b) 20 km
c) 12 km
d) 15 km
Explanation: Let the speed of the rickshaw = S km/h. Let after time T they meet.
So, 60t + ST = $$\frac{{4 \times 60}}{5}$$ $$\left[ {{\text{Since,}}\,\,48\,{\text{min}} = \frac{4}{5}\,\,{\text{hour}}} \right]$$
Rohan saves 16 min by making Shikha move as well. So, she saves 16 min by moving for distance X (Let).
2X = $$\frac{{60 \times 16}}{5}$$
X = 8 km
Speed of Auto rickshaw = 12 km. $$\left( {{\text{ST}} = 8\,{\text{km,}}\,\,{\text{T}} = \frac{2}{3}} \right)$$
6. A train 350 m long is running at the speed of 36 km/h. If it crosses a tunnel in 1 minute, the length of the tunnel in metres:
a) 200 meters
b) 250 meters
c) 300 meters
d) 150 meters
Explanation: Let the length of the tunnel be x m
Time = $$\frac{{{\text{Length of train}} + {\text{Length of tunnel}}}}{{{\text{Speed}}}}$$
60 = $$\frac{{350 + {\text{x}}}}{{10}}$$
$$ {{\text{Speed}} = \frac{{36 \times 5}}{{18}} = 10\,\,{\text{m/sec}}} $$
x = 250 meters
So, length of tunnel is 250 meters
7. If a 250 m long train crosses a platform of same length as that of the train in 25 seconds, then the speed of the train is:
a) 150 m/sec
b) 200 m/sec
c) 20 km/h
d) 72 km/h
Explanation: Speed of train
= $$\frac{{{\text{length of the train}} + {\text{length of platform}}}}{{{\text{Time}}}}$$
= $$\frac{{250 + 250}}{{25}}$$
= 20 m/sec
= 72 km/h
8. A postman goes with a speed of 36 km/h what is the speed of postman in m/s?
a) 4.5 m/s
b) 6 m/s
c) 5 m/s
d) 10 m/s
Explanation: Speed = 36 km/h = $$\frac{{36 \times 5}}{{18}}$$ = 10 m/s
9. Sabarmati express takes 18 seconds to pass completely through a station 162 m long and 15 seconds through another station 120 m long. The length of the train is :
a) 132 m
b) 100 m
c) 80 m
d) 90 m
Explanation: Let length of the train be x m
Speed of train,
$$\frac{{{\text{x}} + 162}}{{18}} = \frac{{{\text{x}} + 120}}{{15}}$$
x = 90 m
10. If two incorrect watches are set at 12:00 noon at correct time, when will both watches show the correct time for the first time given that the first watch gains 1 min in one hour and second watch looses 4 min in two hours.
a) 6 PM, 25 days later
b) 12 noon, 30 days later
c) 12 noon, 15 days later
d) 6 AM, 45 days later
Explanation: First watch:
It shows correct time when it creates difference of 12 hours.
So, to create difference of 12 hour, time required = $$\frac{{60 \times 12}}{{24}}$$ = 30 days
Second watch:
It shows correct time when it creates difference of 12 hours.
So, to create difference of 12 hour, time required = $$\frac{{30 \times 12}}{{24}}$$ = 15 days
LCM of 30 and 15 gives the time when they show correct time together.
Required time = 30 days, at same time.