Time and Work
 A man, on tour travel first 160 km, at 64 km/hr. and the next 160km at 80km/hr. the average speed for the first 320 km. of the tour is ?

 35.55km/hr.
 71.11km/hr.
 36km/hr.
 72km/hr.

View Hint View Answer Discuss in Forum
Average speed = 2t_{1} x t_{2}/( t_{1} + t_{2})
Correct Option: B
Average speed = 2t_{1} x t_{2}/( t_{1} + t_{2})
= (2x64x80)/(64+80)km
= (2 x 64 x 80 ) / 144 km/hr.
=71.11km/hr.
 By walking at 3/4 of his usual speed, a man reaches his office 20 minute later than usual. this usual time is ?

 30 minute
 60 minute
 75 minute
 1 hour 30 minute

View Hint View Answer Discuss in Forum
At a speed of 3/4 of the usual speed, the time taken is 4/3 of the usual time
Correct Option: B
At a speed of 3/4 of the usual speed, the time taken is 4/3 of the usual time
= (4/3 of usual time ) – (usual time ) = 20min.
⇒ 4x/3 x = 20
⇒ x/3 =20
⇒ x=60 min
 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

 9 hours
 7 hours
 4.5 hours
 8 hours

View Hint View Answer Discuss in Forum
Suppose the man covers first distance in x hrs. and second distance in y hrs.
Correct Option: D
Suppose the man covers first distance in x hrs. and second distance in y hrs.
Then , 4x + 5y = 35 and 5x +4y= 37
Solving these equation , we get
X=5 and y=3
Total time taken = (5+3) hrs= 8hrs
 6 men or 10 women can reap a field in 15 days, then the number of days that 12 men and 5 women will take to reap the same field is :

 5
 6
 8
 12

View Hint View Answer Discuss in Forum
Work done by 6 men = work done by 10 women.
Work done by 1 man = work done by 10 / 6 = 5/3 women
∴ 12 men + 5 women = 12 x ( 5 / 3) + 5 = 25 women
∴ W_{1} x D _{1} = W_{2} xCorrect Option: B
Work done by 6 men = work done by 10 women.
Work done by 1 man = work done by 10 / 6 = 5/3 women
∴ 12 men + 5 women = 12 x ( 5 / 3) + 5 = 25 women
∴ W_{1} x D _{1} = W_{2} x D_{2} W = women, D=days
10 X 15 = 25 x D_{2}
D_{2} = 6
 6 children and 2 men complete a certain piece of work in 6 days. Each child takes twice the time taken by a man to finish the work. In how many days will 5 men finish the same work?

 6
 8
 9
 15

View Hint View Answer Discuss in Forum
6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
∴ 36+12X2 = 1 day
60 children can do the work in 1 dayCorrect Option: A
6C + 2M = 6days
36C + 12M = 1 days
Again 1M = 2C
∴ 36+12X2 = 1 day
60 children can do the work in 1 day
Now, 5 men = 10 children
∴ 10 children can do the work in 6 days.