**1. The sum of money that will give Rs 2 as simple interest per day at the rate of 8% per annum is –**

**A. Rs 9125**

B. Rs 7236

C. Rs 6382

D. Rs 9262

**Annual interest = Rs 365 × 2 = Rs 730**

**Rate = 8%**

**Sum = (100 * 730)/8 = Rs 9125**

**2. X can do a piece of work in 20 days and Y can do it in 10 days. They started together, but after 6 days X leaves off. Y will do the rest work in –**

**A. 1 day**

B. 2 days

C. 7/3 days

D. 3 days

**(X + Y)’s 6 days’ work = 6 * (1/20 + 1/10) = 9/10**

**Remaining work = 1 – 9/10 = 1/10**

**1/10 work is done 2y Y in 10 * 1/10 = 1 day**

**3. A radio is sold for Rs 990 at a profit of 10%. What would have been the gain or loss in percentage had it been sold for Rs 880?**

A. 19/3% gain

B. 17/3% loss

C. 17% gain

**D. 20/9% loss**

**Let the CP is Rs x.**

**Then, x * 110/100 = 990**

**x = 900**

**Now CP is Rs 900 and SP is 880.**

**% loss = 20/900 * 100 = 20/9%**

**4. The ratio between the rates of walking of P and Q is 2 : 3. If the time taken by Q to cover a certain distance is 36 minutes, the time taken by P to cover that much distance is –**

A. 24 min

**B. 54 min**

C. 48 min

D. 21.6 min

**Ratio of times taken = 12:13**

**12:13 = x : 36 or 12×36:13×x?18=13×x or x = 54 min.**

**5. A man, on tour, travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. the average speed for the first 320 km of the tour, is –**

A. 35.55 km/hr

**B. 71.11 km/hr**

C. 36 km/hr

D. 72 km/hr

**Average speed = (2×64×8064+80) km/hr = (2×64×80144)km/hr.**

**=71.11 km/hr.**

**6. A man travels 35 km partly at 4 km/hr and at 5 km/hr. If he covers former distance at 5 km/hr and later distance at 4 km/hr, he could cover 2 km more in the same time. The time taken to cover the whole distance at original rate is –**

A. 9 hours

B. 7 hours

C. 412 hours

**D. 8 hours**

**Suppose the man covers first distance in x hrs.and second distance in y hrs. Then,**

**4x+5y=35 and 5x+4y = 37**

**Solving the equations,**

**we get x = 5 and y = 3**

**Total time taken = (5+3)hrs = 8 hrs**

**7. A 270 meters long train running at the speed of 120 km/h crosses another train running in opposite direction at the speed of 80 km/h in 9 seconds. What is the length of the other train?**

**A. 230 m**

B. 245 m

C. 260 m

D. 275 m

**Relative speed = (120 + 80) km/hr**

**= 200 x 5/18**

**= 500/9 m/sec**

**Then, (x+270)/9 = 500/9**

**–> x + 270 = 500**

**–> x = 230.**

**8. Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is –**

**A. 10.8 sec**

B. 9.5 sec

C. 7.4 sec

D. 8.9 sec

**Relative speed = (60 + 40) km/hr = 100 x 5/18 = 250/9 m/ sec.**

**Distance covered in crossing each other = (140 + 160) m = 300 m.**

**Required time = 300 x 9/250 = 54/5 = 10.8 sec.**

**9. Two trains are running in opposite directions with the same speed. If the length of each train is 120 meters and they cross each other in 12 seconds, then the speed of each train (in km/hr) is –**

A. 18 km/hr

B. 26 km/hr

**C. 36 km/hr**

D. 42 km/hr

**Let the speed of each train be x m/sec.**

**Then, relative speed of the two trains = 2x m/sec.**

**So, 2x = (120 + 120)/12**

**–> 2x = 20**

**–> x = 10.**

**–> Speed of each train = 10 m/sec = 10 x 18/5 km/hr = 36 km/hr.**

**10. Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 meters, in what time (in seconds) will they cross each other travelling in opposite direction?**

A. 8 sec

**B. 12 sec**

C. 15 sec

D. 10 sec

**Speed of the first train = 120/10 m/sec = 12 m/sec.**

**Speed of the second train = 120/15 m/sec = 8 m/sec.**

**Relative speed = (12 + 8) = 20 m/sec.**

**Required time = (120 + 120)/20 sec = 12 sec.**