**Q1. A train passes a man standing on a platform in 8 seconds and also crosses the platform which is 264 metres long in 20 seconds. The length of the train (in metres) is:**

(a) 188

**(b) 176**

(c) 175

(d) 96

**Let the length of train = L meter**

**And speed of train = S m/s**

**L = S × 8**

**S = L/8 m/s**

**S = (L + 264)/20**

**L/8=(L+264)/20**

**5L/2-L=264**

**L = 264 ×2/3**

**L = 176 meter**

**Q2. A 150 metre long train crosses a 500 metre long bridge in 30 seconds. What time will it take to cross a platform 370 metre long?**

(a) 36 seconds

(b) 30 seconds

**(c) 24 seconds**

(d) 18 seconds

**Train travels 150 + 500 meter in 30 second**

**Speed of train = 650/30**

**= 65/3 m/s**

**Now, time take to cross a platform 370 metre long**

**=(150+370)/(65/3)**

**=(520×3)/65**

**= 24 sec.**

**Q3. A passenger standing on a railway platform observes that a train going in one direction takes 4 seconds to pass him. Another train of same length going in opposite direction takes 5 seconds to pass him. The time taken (in seconds) by the two train to cross each other will be:**

(a) 35

(b) 36.5

**(c) 40/9**

(d) None of these

**Let the length of each train be x metre and the speeds of the trains are S1 and S2 respectively.**

**S1=x/4, S2=x/5**

**Now, Required time = 2x/(x/4 + x/5)**

**[∴ Trains are moving opposite] = 40/9 seconds**

**Q4. A passenger sitting in a train of length l m, which is running with speed of 60 km/h passing through tow bridges, notices that he crosses the first bridge and the second bridge in time intervals which are in the ratio of 7 : 4 respectively. If the length of first bridge is 280 m, then the length of second bridge is:**

(a) 490 m

(b) 220 m

**(c) 160 m**

(d) Can’t be determined

**Passenger sitting in a train pass through two bridges then the train will travel the only length of the bridge.**

**So, ratio of length of bridge = ratio of time to pass the bridge. (Distance α time)**

**= 7: 4**

**So, length of second train = (280 × 4)/7**

**= 160 metre.**

**Q5. Two guns are fired from the same place at an interval of 6 minutes. A person approaching the place observes that 5 minutes 52 seconds have elapsed between the hearing of the sound of the two guns. If the velocity of the sound is 330 m/sec, the man was approaching that place at what speed (in km/hr)?**

(a) 24

**(b) 27**

(c) 30

(d) 36

**Difference of time = 8 second**

**So, distance covered by 352 s (5 × 60 + 52) second by man**

**= distance covered by sound in 8 second**

**S × 352 = 330 × 8**

**Here S is the speed of man in m/s**

**S = 15/2 m/s**

**So, speed in km/hr**

**S = 15/2×18/5=27 km/hr**

**Q6. If the arithmetic mean of 3a and 4b is greater than 50, and a is twice of b, then the smallest possible integer value of a is**

(a) 18

(b) 19

(c) 20

**(d) 21**

**According to the question,**

**(3a+4b)/2 > 50**

**3a + 4b > 100**

**3a + 2a > 100 (given that a = 2b)**

**5a > 100**

**a > 20**

**Hence, smallest possible integer value of a = 21**

**Q7. A profit of 12% is made mobile phone is sold at Rs. P and there is 4% loss when the phone is sold at Rs. Q. Then Q : P is**

(a) 4 : 5

(b) 3 : 1

(c) 1 : 1

**(d) 6 : 7**

**Let the cost of mobile phone = 100**

**So, P = 112 Rs.**

**Q = 96 Rs.**

**Required ratio Q : P = 96 : 112 = 6 : 7**

**Q8. From 1980-1990, the population of a country was increased by 20%.**

**From 1990-2000, the populations of the country was increased by 20%.**

**From 2000-2010, the populations of the country was increased by 20%.**

**Then the overall increased population (in percentage) of the country from 1980-2010 was**

(a) 60%

(b) 62.8%

(c) 72.2%

**(d) 72.8%**

**Let the population of the country in the year 1980 = P**

**The overall increase in population of a country from 1980 to 2000 is = (P × 1.2 × 1.2 × 1.2 – P)/P**

**=(1.728P-P)/P**

**% increase in population = .728 ×100**

**= 72.8%**

**Q9. If the ratio of cost price to selling price is 10 : 11, then the rate of percent of profit is**

(a) 0.1%

(b) 1%

(c) 1.1%

**(d) 10%**

**Let the cost prices = 10 Rs.**

**Selling price will be = 11 Rs.**

**Profit % = (S.P – C.P)/(C.P)**

**=(11-10)/10=10%**

**Q10. A farmer travelled a distance of 61 km in 9 hours. He travelled party on foot at the rate 5km/hour and party on bicycle at the rate 9 km/hour. The distance travelled on foot is**

(a) 20 km

(b) 15 km

**(c) 16 km**

(d) 17 km

**Let distance travelled on foot = x km**

**According to the question,**

**x/4+ (61-x)/9=9**

**x/4-x/9+61/9=9**

**5x/36=9-61/9**

**x = 16**

**The distance, travelled on foot = 16 km.**