1)A train covers a distance in 50 min ,if it runs at a speed of 48kmph on an average.The speed at which the train must run to reduce the time of journey to 40min will be.

**Answer and Solution :**

**Time=50/60 hr=5/6hr**

Speed=48mph

distance=S*T=48*5/6=40km

time=40/60hr=2/3hr

New speed = 40* 3/2 kmph= 60kmph

Speed=48mph

distance=S*T=48*5/6=40km

time=40/60hr=2/3hr

New speed = 40* 3/2 kmph= 60kmph

2)Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is?

**Answer and Solution :**

**Let total distance be S**

total time=1hr24min

A to T :: speed=4kmph

diistance=2/3S

T to S :: speed=5km

distance=1-2/3S=1/3S

21/15 hr=2/3 S/4 + 1/3s /5

84=14/3S*3

S=84*3/14*3

= 6km

total time=1hr24min

A to T :: speed=4kmph

diistance=2/3S

T to S :: speed=5km

distance=1-2/3S=1/3S

21/15 hr=2/3 S/4 + 1/3s /5

84=14/3S*3

S=84*3/14*3

= 6km

3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr. the usual time is.

**Answer and Solution :**

**Usual speed = S**

Usual time = T

Distance = D

New Speed is ¾ S

New time is 4/3 T

4/3 T – T = 5/2

T=15/2 = 7 ½

Usual time = T

Distance = D

New Speed is ¾ S

New time is 4/3 T

4/3 T – T = 5/2

T=15/2 = 7 ½

4)A man covers a distance on scooter .had he moved 3kmph faster he would have taken 40 min less. If he had moved 2kmph slower he would have taken 40min more.the distance is.

**Answer and Solution :**

**Let distance = x m**

Usual rate = y kmph

x/y – x/y+3 = 40/60 hr

2y(y+3) = 9x ————–1

x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2

divide 1 & 2 equations

by solving we get x = 40

Usual rate = y kmph

x/y – x/y+3 = 40/60 hr

2y(y+3) = 9x ————–1

x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2

divide 1 & 2 equations

by solving we get x = 40

5)Excluding stoppages,the speed of the bus is 54kmph and including stoppages,it is 45kmph.for how many min does the bus stop per hr.

**Answer and Solution :**

**Due to stoppages,it covers 9km less.**

time taken to cover 9 km is [9/54 *60] min = 10min

time taken to cover 9 km is [9/54 *60] min = 10min