166. A train covers a distance between station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then the same distance is covered in 48 minutes. The distance between station A and B is

1. 1. 60 km

   
   
   

   2. 64 km

   
   
   

   3. 80 km

   
   
   

   4. 55 km

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Let the distance between stations be x km, then speed of train
   

   =	2x					= (4x/3) kmph
   	45
   	60

   
   

   ∴	x					= (48/60)
   	4x	− 5
   	3

   
   

   ⇒	3x	=	4
   	4x − 15		5

   
   ⇒  16x – 60 = 15x
   ⇒  x = 60 km

   Correct Option: A

   Let the distance between stations be x km, then speed of train
   

   =	2x					= (4x/3) kmph
   	45
   	60

   
   

   ∴	x					= (48/60)
   	4x	− 5
   	3

   
   

   ⇒	3x	=	4
   	4x − 15		5

   
   ⇒  16x – 60 = 15x
   ⇒  x = 60 km

---

167. A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be :

1. 1. 10 minutes

   
   
   

   2. 13 minutes 20 sec

   
   
   

   3. 13 minutes

   
   
   

   4. 11 minutes 20 sec

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Speed of train =	Distance
   	Time

   
   

   =	10					kmph
   	12
   	60

   
   

   =	10 × 60	= 50 kmph
   	12

   
   New speed = 45 kmph
   

   ∴   Required time =	10	hour
   	45

   
   

   =	2	× 60 minutes
   	9

   
   

   =	40	minutes
   	3

   
   or 13 minutes 20 seconds

   Correct Option: B

   Speed of train =	Distance
   	Time

   
   

   =	10					kmph
   	12
   	60

   
   

   =	10 × 60	= 50 kmph
   	12

   
   New speed = 45 kmph
   

   ∴   Required time =	10	hour
   	45

   
   

   =	2	× 60 minutes
   	9

   
   

   =	40	minutes
   	3

   
   or 13 minutes 20 seconds

---

---

168. A student walks from his house at

     a speed of 2	1	km per hour and reaches his school
     	2

     6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house ?

1. 1. 5	km
      4

   
   
   

   2. 7	km
      4

   
   
   

   3. 9	km
      4

   
   
   

   4. 11	km
      4

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Let the required distance be x km.
   Difference of time
   

   = 6 + 6 = 12 minutes =	1	hr.
   	5

   
   According to the question,
   

   ⇒	x	−	x	=	1	⇒	2x	−	2x	=	1
   	5/2		7/2		5		5		7		5

   
   

   ⇒	14x − 10x	−	1
   	35		5

   
   

   ⇒	4x	=	1	⇒ x =	35	=	7	km.
   	35		5		20		4

   
   
   Second Methdod :
   

   Here, S1 = 2	1	, t1 = 6
   	2

   
   

   Here, S2 = 3	1	, t2 = 6
   	2

   
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	5	×	7	× (6 + 6)
   	2		2
   	5	−	7
   	2		2

   
   

   =	35	×	12	=	7	km
   	4		60		4

   Correct Option: B

   Let the required distance be x km.
   Difference of time
   

   = 6 + 6 = 12 minutes =	1	hr.
   	5

   
   According to the question,
   

   ⇒	x	−	x	=	1	⇒	2x	−	2x	=	1
   	5/2		7/2		5		5		7		5

   
   

   ⇒	14x − 10x	−	1
   	35		5

   
   

   ⇒	4x	=	1	⇒ x =	35	=	7	km.
   	35		5		20		4

   
   
   Second Methdod :
   

   Here, S1 = 2	1	, t1 = 6
   	2

   
   

   Here, S2 = 3	1	, t2 = 6
   	2

   
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	5	×	7	× (6 + 6)
   	2		2
   	5	−	7
   	2		2

   
   

   =	35	×	12	=	7	km
   	4		60		4

---

169. If a boy walks from his house to school at the rate of 4 km per hour, he reaches the school 10 minutes earlier than the scheduled time. However, if he walks at the rate of 3 km per hour, he reaches 10 minutes late. Find the distance of his school from his house.

1. 1. 5 km

   
   
   

   2. 4 km

   
   
   

   3. 6 km

   
   
   

   4. 4.5 km

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Let the distance of school be x km, then
   

   x	−	x	=	20
   3		4		60

   
   

   ⇒	x	=	1	⇒ x =	12	= 4 km
   	12		3		3

   
   
   Second Methdod :
   Here, S₁ = 3, t₁ = 10
   S₂ = 4, t₂ = 10
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	(3 × 4)(10 + 10)
   	4 − 3

   
   

   = 12 ×	20	= 4 km
   	60

   Correct Option: B

   Let the distance of school be x km, then
   

   x	−	x	=	20
   3		4		60

   
   

   ⇒	x	=	1	⇒ x =	12	= 4 km
   	12		3		3

   
   
   Second Methdod :
   Here, S₁ = 3, t₁ = 10
   S₂ = 4, t₂ = 10
   

   Distance =	(S1 ×S2)(t1 + t2)
   	S2 −S1

   
   

   =	(3 × 4)(10 + 10)
   	4 − 3

   
   

   = 12 ×	20	= 4 km
   	60

---

---

170. A train travelling at a speed of 55 km/hr travels from place X to place Y in 4 hours. If its speed is increased by 5 km/hr., then the time of journey is reduced by

1. 1. 25 minutes

   
   
   

   2. 35 minutes

   
   
   

   3. 20 minutes

   
   
   

   4. 30 minutes

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Distance between stations X and Y = Speed × Time
   = 55 × 4 = 220 km.
   New speed = 55 + 5 = 60 kmph
   

   ∴   Required time =	220
   	60

   
   

   =	11	hours
   	3

   
   = 3 hours 40 minutes.
   ∴   Required answer
   = 4 hours – 3 hours 40 minutes
   = 20 minutes

   Correct Option: C

   Distance between stations X and Y = Speed × Time
   = 55 × 4 = 220 km.
   New speed = 55 + 5 = 60 kmph
   

   ∴   Required time =	220
   	60

   
   

   =	11	hours
   	3

   
   = 3 hours 40 minutes.
   ∴   Required answer
   = 4 hours – 3 hours 40 minutes
   = 20 minutes

---