**Q1. Two trains, 130 m and 110 m long, while going in the same direction, the faster train takes one minute to pass the other completely. If they are moving in opposite direction, they pass each other completely in 3 seconds. Find the speed of each train. **

(a) 42 m/s 38 m/s

(b) 38 m/s 36 m/s

(c) 36 m/s 42 m/s

(d) None of these

**Q2. A train leaves the station at 5 am at 60 km/h. Another train leaves the same station at 6.30 am at 75 km/h and travels in the direction of the first train. At what time and at what distance from the station will they meet?**

(a) 12.30 am 450 km

(b) 1.30 pm 375 km

(c) 11.30 am 425 km

(d) None of these

**Q3. A and B are two stations. A train goes from A to B at 64 km/h and returns to A at a slower speed. If its average speed for the whole journey is 56 km/h, at what speed did it return? **

(a) 48 km/h

(b) 49.77 km/h

(c) 30 km/h

(d) 47.46 km/h

**S3. Ans.(b)**

**Sol.** Let the required speed b x km/h

Then, (2 × 64 × x)/(64 + x)=56

⇒ 128x = 64 × 56 + 56x

∴x=(64 × 56)/72 = 49.77 km/h

**Q4. A 100 metres long train completely passes a man walking in the same direction at 6 km/h in 5 seconds and also a car travelling in the same direction in 6 seconds. At what speed was the car travelling? **

(a) 18 km/h

(b) 48 km/h

(c) 24 km/h

(d) 30 km/h

**S4. Ans.(a)**

**Sol.** Let, x km/h be the speed of the train

⇒ (x-6)×5/18×5=100

⇒ x = 78 km/h

Let, the speed of the car be y km/h

⇒ (78-y)×5/18×6=100 ⇒ y = 18.

**Q5. If a 110 metres long train passes a man walking in opposite direction at a speed of 6 km/h in 6 seconds, it will pass another man walking at the same speed in the same direction in time of: **

(a) 28/3 seconds

(b) 32/3 seconds

(c) 8 second

(d) 22/3 seconds

**S5. Ans.(d)**

**Sol. **Let, the speed of the train = x km/h

Relative speed = (x + 6) km/h

=(x+6)×5/18 m/s

∴(x+6)×5/18×6=110

∴ x = 60

∴ Speed of train = 60 km/h for the person,

Relative speed = (60 – 6)km/h

=54×5/18 m/s = 15 m/s

∴ Time taken the cross the speed person

=110/15=22/3 seconds

**Q6. If a distance of 50 m is covered in 1 minute, 90 m in 2 minutes and 130 m in 3 minutes find the distance covered in 15 minutes. **

(a) 610 m

(b) 750 m

(c) 1000 m

(d) 650 m

**S6. Ans.(a)**

**Sol.**Distance covered in 2nd minute = 90 – 50 = 40 metres

**Q7. If a train runs at 40 km/h, it reaches its destination late by 11 minutes, but if it runs at 50 km/h, it is late by 5 minutes only. Find, the correct time for the train to complete its journey. **

(a) 19 minutes

(b) 20 minutes

(c) 21 minutes

(d) 18 minutes

**S7. Ans.(a) **

**Sol.** 40 km/h =40/60 km/h minute =2/3 km/minute

50 km/h =50/60=5/6 km/minute

Let, distance be x km and the actual time be t minutes,

Then, according to the question,

x/(2/3)=t+11 ….(i)

x/(5/6)=t+5 …(ii)

By equations (i) and (ii), we get

3x/2-6x/5=6

⇒ (15x – 12x)/10=6

⇒ 3x = 60 km

⇒ x = 20 km

From equation (i),

3x/2=t+11

⇒ (3 × 20)/2=t+11

⇒ t = (30 – 11) = 19 minutes

**Q8. Two trains, A and B, start from stations X and Y towards Y and X respectively. After passing each other, they take 4 hours 48 minutes and 3 hours 20 minutes to reach Y and X respectively. If train A is moving at 45 km/h, then the speed of the train B is: **

(a) 60 km/h

(b) 64.8 km/h

(c) 54 km/h

(d) 37.5 km/h

**Q9. A train covers a distance between station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/h, then the same distance is covered in 48 minutes. The distance between stations A and B is: **

(a) 60 km

(b) 64 km

(c) 80 km

(d) 55 km

**Q10. Two places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at a constant speed, they meet in 5 hours. If the cars travel towards each other, they meet in 1 hour. What is the speed of the car running faster? **

(a) 60 km/h

(b) 50 km/h

(c) 40 km/h

(d) 32 km/h

S10. Ans.(a)

Sol. Let the speeds of the cars starting from A and B be x km/h and y km/h respectively.

Now, according to the question,

Case I: When both the cars are running in the same direction

Relative speed = (x – y) km/h

∴ 5x – 5y = 100 …(i)

Case II: When both the cars are running in the opposite directions

Relative speed = (x + y) km/h

∴ (x + y) × 1 = 100

⇒ x + y = 100 …(ii)

Now, solving equation (i) and equation (ii), we have

x = 60 and y = 40

Therefore, speed of the faster car is 60 km/h

**Q11. A bus started its journey from Ramgrah and reached Devgrah in 44 minutes at an average speed of 50 km/h. If the average speed of the bus is increased by 5 km/h, how much time will it take to cover the same distance?**

(a) 40 minutes

(b) 38 minutes

(c) 36 minutes

(d) 31 minutes

**S11. Ans.(a)**

**Sol.** Distance =44/60×50=x/60×55

∴ x = 40 minutes

**Q12. 0.05 is what % of 0.5?**

(a) 1%

(b) 0.1%

(c) 10%

(d) 100%

**S12. Ans.(c)**

**Sol.**

Let x% of 0.5 is 0.05

x×.5/100=.05

x=(.05 × 100)/.5=10%

**Q13. One liter of water evaporates from 6 liters of sea water containing 4% salt. Find the % of salt in the remaining solution?**

(a) 7.5%

(b) 4.5%

(c) 4.8%

(d) 8%

**S13. Ans.(c)**

**Sol. **Salt in sea water = 4% × 6 = .24

After evaporation of 1L of sea water remaining water is = 5 liters

Required % of salt in the remaining solution =.24/5=4.8%

**Q14. In an election with only two contestants the candidates who gets 30% of the votes polled is defeated by 15000 votes. What is the number of votes polled?**

(a) 112350

(b) 10580

(c) 26250

(d) 37500

**S14. Ans.(d)**

**Sol.** Let the total vote polled = 100x

According to question,

70x – 30x = 15000

x =15000/40

x = 375

So, total vote polled = 37500

**Q15. If in a fraction, numerator is increased by 200% and denominator by 500%, then the new fraction is what % of previous fraction?**

(a) 25%

(b) 45%

(c) 60%

(d) 50%

**S15. Ans.(d)**

**Sol.**

Let the fraction is x/y

New fraction =(x (100 + 200))/(y (100 + 500))

=3x/6y=x/2y

% of new fraction of pervious fraction

⇒(x/2y×100)/(x/y)=50%