156. P and Q are two cities. A boy travels on cycle from P to Q at a speed of 20 km per hr. and returns at the rate of 10 km per hr. Find his average speed for the whole journey.

1. 1. 13	2	kmph
      	3

   
   
   

   2. 12	1	kmph
      	3

   
   
   

   3. 13	1	kmph
      	3

   
   
   

   4. 12	2	kmph
      	3

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the speed between P and Q be x km.
   

   Then time taken to cover x km. P to Q =	x	hrs.
   	20

   
   Time taken to cover x km from Q to P at 10 km per hr. P to Q
   

   =	x	hrs.
   	10

   
   ∴  Total distance covered
   = x + x = 2x km.
   Time taken to cover 2x km.
   

   =	x	+	x	=	x + 2x	=	3x	hrs.
   	20		10		20		20

   
   

   ∴ Average Speed =	2x	=	2x × 20
   	3x/20		3x

   
   

   =	40	km per hr.
   	3

   
   

   = 13	1	km per hr.
   	3

   
   
   Second Method : Here, x = 20 km per hr.
   y = 10 km per hr
   ∴ Average Speed [∵  Distance is same]
   

   =	2xy	=	2 × 20 × 10
   	x + y		20 + 10

   
   

   =	400	=	40	= 13	1	km per hr.
   	30		3		3

   Correct Option: C

   Let the speed between P and Q be x km.
   

   Then time taken to cover x km. P to Q =	x	hrs.
   	20

   
   Time taken to cover x km from Q to P at 10 km per hr. P to Q
   

   =	x	hrs.
   	10

   
   ∴  Total distance covered
   = x + x = 2x km.
   Time taken to cover 2x km.
   

   =	x	+	x	=	x + 2x	=	3x	hrs.
   	20		10		20		20

   
   

   ∴ Average Speed =	2x	=	2x × 20
   	3x/20		3x

   
   

   =	40	km per hr.
   	3

   
   

   = 13	1	km per hr.
   	3

   
   
   Second Method : Here, x = 20 km per hr.
   y = 10 km per hr
   ∴ Average Speed [∵  Distance is same]
   

   =	2xy	=	2 × 20 × 10
   	x + y		20 + 10

   
   

   =	400	=	40	= 13	1	km per hr.
   	30		3		3

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157. A man walked a certain distance. One-third he walked at 5 km per hr. Another one-third he walked at 10 km per hr. and the rest at 15 km per hr. Find his average speed.

1. 1. 8	1	kmph
      	11

   
   
   

   2. 7	1	kmph
      	11

   
   
   

   3. 7	2	kmph
      	11

   
   
   

   4. 8	2	kmph
      	11

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here, the man covers equal distance at different speeds. Using the formula, the Average Speed is given by
   

   =	3
   	1	+	1	+	1
   	5		10		15

   
   

   =	3
   	6 + 3 + 2
   	30

   
   

   =	90	= 8	2	km per hour.
   	11		11

   Correct Option: D

   Here, the man covers equal distance at different speeds. Using the formula, the Average Speed is given by
   

   =	3
   	1	+	1	+	1
   	5		10		15

   
   

   =	3
   	6 + 3 + 2
   	30

   
   

   =	90	= 8	2	km per hour.
   	11		11

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158. An aeroplane travels a distance in the form of a square with the speed of 400 km per hr, 600 km per hr, 800 km per hr. and 1200 km per hr respectively. Find the average speed for the whole distance along the four sides of the square.

1. 1. 640 kmph

   
   
   

   2. 620 kmph

   
   
   

   3. 630 kmph

   
   
   

   4. 650 kmph

   
   
   

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1. View Hint View Answer Discuss in Forum

   As distance is covered along four sides (equal) of a square at different speeds, the average speed of the aeroplane
   

   =	4
   	1	+	1	+	1	+	1
   	400		600		800		1200

   
   [∵  All the sides of square are equal, so distance between them is same]
   

   =	4
   	30 + 20 + 15 + 10
   	12000

   
   

   =	48000	= 640 km per hr.
   	75

   Correct Option: A

   As distance is covered along four sides (equal) of a square at different speeds, the average speed of the aeroplane
   

   =	4
   	1	+	1	+	1	+	1
   	400		600		800		1200

   
   [∵  All the sides of square are equal, so distance between them is same]
   

   =	4
   	30 + 20 + 15 + 10
   	12000

   
   

   =	48000	= 640 km per hr.
   	75

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159. Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is

1. 1. 3 km

   
   
   

   2. 4 km

   
   
   

   3. 3.5 km

   
   
   

   4. 2 km

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the distance of the office be x km, then
   

   x	−	x	=	8
   5		6		60

   
   

   ⇒	6x − 5x	=	2
   	30		15

   
   ⇒  x = 2 × 2 = 4 km
   
   Second Methdod :
   Here, S₁ = 5, t₁ = 6
   S₂ = 6, t₂ = 2
   

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

   
   

   =	(6 × 5)(6 + 2)
   	6 − 5

   
   

   = 30 ×	8	= 4 km
   	60

   Correct Option: B

   Let the distance of the office be x km, then
   

   x	−	x	=	8
   5		6		60

   
   

   ⇒	6x − 5x	=	2
   	30		15

   
   ⇒  x = 2 × 2 = 4 km
   
   Second Methdod :
   Here, S₁ = 5, t₁ = 6
   S₂ = 6, t₂ = 2
   

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

   
   

   =	(6 × 5)(6 + 2)
   	6 − 5

   
   

   = 30 ×	8	= 4 km
   	60

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160. A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 minutes early. The distance to his school is

1. 1. 9 km

   
   
   

   2. 5 km

   
   
   

   3. 4 km

   
   
   

   4. 6 km

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the required distance be x km.
   According to the question,
   

   x	−	x	=	18
   4		5		60

   
   

   ⇒	5x − 4x	−	3
   	20		10

   
   

   ⇒  x =	3	× 20 = 6 km
   	10

   
   
   Second Methdod :
   Here, S₁ = 4, t₁ = 9
   S₂ = 5, t₂ = 9
   

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

   
   

   =	(4 × 5)(9 + 9)
   	5 − 4

   
   

   = 20 ×	18	= 6 km
   	60

   Correct Option: D

   Let the required distance be x km.
   According to the question,
   

   x	−	x	=	18
   4		5		60

   
   

   ⇒	5x − 4x	−	3
   	20		10

   
   

   ⇒  x =	3	× 20 = 6 km
   	10

   
   
   Second Methdod :
   Here, S₁ = 4, t₁ = 9
   S₂ = 5, t₂ = 9
   

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

   
   

   =	(4 × 5)(9 + 9)
   	5 − 4

   
   

   = 20 ×	18	= 6 km
   	60

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