Time speed, work, Alligations and mixtures problems

[mattoo]

nair can walk a certain distance in 52 days when he rests 10 hrs a day. how long will he take for twice the distance, if he walks twice as fast and rests twice as long as each day?
a)104 days
b)26 days
c)78 days
d)182 days
e)136 days

[pscoe2]

Originally Posted by mattoo

nair can walk a certain distance in 52 days when he rests 10 hrs a day. how long will he take for twice the distance, if he walks twice as fast and rests twice as long as each day?
a)104 days
b)26 days
c)78 days
d)182 days
e)136 days

he walks 14 hrs a day...
for covering distance x he takes 14*52 hrs of walking
now for twice distance..hrs req=104*14
given he now walks for 4 hrs a day and at twice speed...
so new time in days=(104*14)/4*2=182 days

[srikar2097]

Originally Posted by mattoo

nair can walk a certain distance in 52 days when he rests 10 hrs a day. how long will he take for twice the distance, if he walks twice as fast and rests twice as long as each day?
a)104 days
b)26 days
c)78 days
d)182 days
e)136 days

Initially, Total time = 52*24=1248hrs
rests = 52*10=520hrs
Walk time = 728hrs
=>d=s*728 --->1

Later days be x.
=> Total time =24x hrs
rests = 20x hrs
walks = 4x hrs
=> 2d = 2s * 4x --->2
solve 1 & 2 to get x =182 days.

Hence new time taken is 182 days or (d)

[mattoo]

two trains move from station A and B towards each other at the 50 km/hr and 60 km/hr. at meeting point, the driver of second train felt that the train has covered 120 km more. what is the distance between A and B?
a) 1320
b)1200
c)1100
d)960 kms

[srikar2097]

Originally Posted by mattoo

two trains move from station A and B towards each other at the 50 km/hr and 60 km/hr. at meeting point, the driver of second train felt that the train has covered 120 km more. what is the distance between A and B?
a) 1320
b)1200
c)1100
d)960 kms

For train A; d=50*t
for train B; d+120=60*t
solve, we get d=600
Total distance = (d) + (d+120) = 1320km.

Hence(a)

[mattoo]

shan starts 3 minutes after bonny from same point,for a place at a distance of 4 1/2 miles. bonny on reaching her destination immediately returns and afer walking a mile meets shan. if shans speed be a mile in 18 minutes, what is the bonny's speed in miles per minute?
a) 1/6
b)1/24
c)2/15
d)1/4
e)1/12

[pscoe2]

Originally Posted by mattoo

shan starts 3 minutes after bonny from same point,for a place at a distance of 4 1/2 miles. bonny on reaching her destination immediately returns and afer walking a mile meets shan. if shans speed be a mile in 18 minutes, what is the bonny's speed in miles per minute?
a) 1/6
b)1/24
c)2/15
d)1/4
e)1/12

distance covered by shan till the time of meet=3.5 miles
distance covered by bonny=5.5 miles
speed of shan=(1/1mile/min
time taken by shan=3+63=66min
speed of bony=5.5/66=1/12 miles/min

[srikar2097]

Originally Posted by mattoo

shan starts 3 minutes after bonny from same point,for a place at a distance of 4 1/2 miles. bonny on reaching her destination immediately returns and afer walking a mile meets shan. if shans speed be a mile in 18 minutes, what is the bonny's speed in miles per minute?
a) 1/6
b)1/24
c)2/15
d)1/4
e)1/12

shan-
d=3.5 miles
speed 1mile ->18minutes
=>for 3.5 miles ->63 minutes

Bonny-
takes 3 min. more than shan, i.e. 66min.
d = 4.5 + 1 = 5.5 miles
=>time taken, t=5.5/66 or 1/12 miles/min.
Hence (e)

[mattoo]

a hare pursued by a grayhound is 50 of her own laps ahead of him. While the hare takes 4 leaps,the grahound takes 3 leaps . In 1 leap the hare goes 1 3/4 metres and grayhound 2 3/4 metres. In how many leaps will the grayhound overtake hare?
Answer is 210 leaps.
Solve it.

[pscoe2]

Originally Posted by mattoo

a hare pursued by a grayhound is 50 of her own laps ahead of him. While the hare takes 4 leaps,the grahound takes 3 leaps . In 1 leap the hare goes 1 3/4 metres and grayhound 2 3/4 metres. In how many leaps will the grayhound overtake hare?
Answer is 210 leaps.
Solve it.

let us say that the hare will be caught after x m
we know that the hare is 175/2 m ahead of hound...
also....the time in which hare moves 7m,hound moves 33/4 m i.e. 5/4 m more...
i.e. for every 7m the hare runs...the hound gains 5/4 m on him...
so, x=7*(175/2)/(5/4)
=490m
leaps taken by hound to catch the hare=(x+(175/2))/(11/4)=577.5*4/11=210...
i think u must have understood