**Q1. Supposing that telegraph poles on a railway track are 50 metres apart. How many pole will be passed by a train in 2 hours. If the speed of the train is 45 km/hr**

**(a) 1801**

(b) 1800

(c) 1601

(d) 1600

**Sol.**

**Distance travelled by train in 2 hours = 45 × 2 = 90 kms = 90000 m**

**No. of pole passed = 90000/50+1**

**= 1800 + 1**

**= 1801 poles**

**Q2. A train 150 metre long, passes a pole in 15 seconds and another train of the same length travelling in the opposite direction in 12 seconds. The speed of the second train is**

**(a) 45 km/hr**

(b) 48 km/hr

(c) 52 km/hr

(d) 54 km/hr

**Sol.**

**When distance is same, speed is indirectly proportional to the time**

**Speed of the first train = 150/15 = 10 m/s**

**Speed of the second train = 10/12×15**

**= 150/12 m/s**

**=150/12×18/5**

**= 45 km/hr**

**Q3. A train 100 metres long moving at a speed of 50 kmph crossed a train 120 metres long coming from opposite direction in 6 seconds. The speed of the second train is**

**(a) 82 km/hr**

(b) 84 km/hr

(c) 86 km/hr

(d) Can’t be determined

**Sol.**

**Total speed = (100 + 120)/6**

**= 220/6 m/sec.**

**= 220/6×18/5 km/hr.**

**= 132 km/hr.**

**Speed of second train = 132 – 50 = 82 km/hr.**

**Q4. A boat covers 12 km. upstream and 18 km downstream in 3 hours while it covers 36 km upstream and 24 km downstream in 6 hours, what is the velocity of the stream?**

(a) 1.5 km/hr

(b) 1 km/hr

**(c) 2 km/hr**

(d) 2.5 km/hr

**Sol.**

**12/y+18/x = 3 …(i)**

**36/y+24/x=13/2 …(ii)**

**To equate (i) & (ii)**

**Multiply (i) by 3 first**

**36/y+54/x = 9 …(iii)**

**36/y+24/x=13/2 …(iv)**

**Subtracting (iv) from (iii)**

**30/x=9-13/2=5/2**

**⇒ x = 12km/hr.**

**Now, put value of x in equation (i), we get y = 8 km./hr.**

**∴ Velocity of stream(Sc)=(12 – 8)/2 = 2 km./hr.**

**Q5. In a flight of 600 kms, an aircraft was slowed down due to bad weather. Its average speed for the trip is reduced by 200 kms/hr. and the time of flight increased by 30 min. The duration of the flight is –**

**(a) 1 hr.**

(b) 2 hrs.

(c) 3 hrs.

(d) 4 hrs.

**Sol.**

**Let the duration of flight be t hours.**

**S = D/T**

**S1-S2 = 200 kms/hr.**

**600/t-600/(t+1/2) = 200**

**600/t-(2×600)/(2t+1)=200**

**(2t + 1) 600 – t × 1200 = 200t (2t + 1)**

**3(2t + 1) – 6t = t (2t + 1)**

**6t + 3 – 6t = 2t^2 + t**

**2t^2 + t – 3 = 0**

**2t^2 + 3t – 2t – 3 = 0**

**t (2t + 3) – 1(2t + 3)**

**(2t + 3) (t – 1) = 0**

**t = 1 hour.**

**Q6. Without any stoppage a person travels a certain distance at an average speed of 42 km/hr. and with stoppage he covers the same distance at an average speed of 28 km/hr. How many minutes per hour does he stop?**

**(a) 20 min.**

(b) 30 min.

(c) 21 min.

(d) 23 min.

**Sol.**

**Here x = 42 and y = 28**

**∴ Stoppage time/hr. = (x – y)/x**

**⇒(42 – 28)/42⇒1/3 hr. ⇒ 20 min.**

**Q7. The speed of a car increases by 2 km/hr after every hour. If the distance travelled in the first one hour was 35 km, the total distance travelled in 12 hours was-**

(a) 456 km

(b) 482 km

**(c) 552 km**

(d) 556 km

**Sol.**

**Distance travelled in first hour = 35 km**

**Distance travelled in second hour = 37 km**

**common difference = 2 km**

**So, distance travelled in 12 hour = 12/2 [2 × 35 + (12 – 1) × 2]**

**= 12/2 × (70 + 22)**

**= 12 × 46**

**= 552 km**

**Q8. Walking at 6/7th of his usual speed a man is 12 minutes late. The usual time taken by him to cover that distance is-**

(a) 1 hr.

**(b) 1 hr. 12 min.**

(c) 1 hr. 15 min.

(d) 1 hr. 20 min.

**Sol.**

**Reduced in speed = (1-6/7)/1=1/7**

**So, increase in time 1/(7 – 1)=1/6 time=12**

**total time = 6 × 12**

**= 1 hr. 12 min.**

**Q9. A starts from a place P to go to a place Q. At the same time B starts from Q for P. If after meeting each other A and B took 4 and 9 hours more respectively to reach their destinations, the ratio of their speeds is**

**(a) 3:2**

(b) 5:2

(c) 9:4

(d) 9:13

**Sol.**

**(Speed of A)/(Speed of B)=√(9/4 )**

**(Speed of A)/(Speed of B)=3/2**

**Ratio=3:2**

**Q10. I have to be at a certain place at a certain time and I find that I shall be 15 minutes too late, if I walks at 4 km and hour; and 10 minutes too soon, if I walks at 6 km an hour. How far have I to walk?**

(a) 3 km

**(b) 5 km**

(c) 6 km

(d) 8 km

**Sol.**

**The difference = 15 – (–10) = 25 m**

**Increase in speed = 4 km to 6 km**

**= 2/4=1/2**

**So,**

**1/(2 + 1) × time = 25 minute**

**Time = 75 minute**

**Total walk = 4 × 75/60**

**= 5 km.**